Haritha

Oscillation And Waves

Simple Harmonic Motion Preliminary Topics

Periodic Motion Definition: Any motion, that repeats itself at regular intervals of time, is known as periodic motion.

The motion of planets around the sun, the hands of a clock, and the blades of a revolving electric fan are some examples of periodic motion. A characteristic of this motion is that each moves along a definite circular or elliptical path repeatedly in a regular time interval.

• The motion along an elliptical path is called elliptical or orbital periodic motion, while the other motions are called rotational periodic motion.
• On the other hand, if the bob of a simple pendulum is pulled aside slightly and then released, it swings to the other side passing through its equilibrium position.
• On its way back, the bob again passes through its equilibrium position and returns to its point of release.
• Then it goes on repeating this to and fro motion. Thus the bob covers a definite path repeatedly. If the angular displacement of oscillation of the pendulum is less than 4°, the amplitude or range of oscillation is small compared to the radius of curvature of the oscillatory path and so the path of the bob may be taken as a straight line.

Hence, this motion is called linear periodic motion. The motion of an elastic spring and that of a piston in the cylinder of an automobile engine are examples of linear periodic motion.

Oscillation Or Vibration: If a body undergoing periodic motion has an equilibrium position somewhere inside its path, it experiences no net external force at that point. Hence, if it is left there at rest, it remains there forever.

Now, if the body is given a small displacement from the equilibrium position, a restoring force comes into play which tries to bring the body back to its equilibrium point, giving rise to oscillations or vibrations.

Oscillation Or Vibration Definition: If a particle that executes periodic motion moves to and fro along the same path, the motion is called oscillation or vibration.

Simple Pendulum: A pendulum is suspended from a rigid support. Let P be the point of suspension and O be the mean or equilibrium position of the body. If the bob is pulled to position B and then released, it oscillates along the path BOC.

If the angular displacement, i.e., ∠OPB or∠OPC is less than 4°, then the path BOC or COB may be taken as a straight line. The pendulum is then called a simple pendulum and its oscillation is a linear periodic motion.

Elastic Spring: The upper end of an elastic spring is attached with a rigid support, Let P be the point of suspension. If a heavy body is suspended from its lower end, the spring stretches and the body hangs at rest, at some position of equilibrium O.

If the body is slightly pulled to B and then released, the spring executes an up-and-down oscillatory motion along the straight path BOC and COB. The oscillation of the elastic spring is therefore a linear periodic motion.

Stretched String: A string XY is attached to two rigid supports at points X and Y, It remains at a position of equilibrium along the straight line XY, If the string is pulled slightly upwards or downwards and then released, it vibrates about its position of equilibrium.

• If we consider a point O on the string, it is seen that the point vibrates along the straight path BOC, This vibration is also a linear periodic motion. Although any kind of oscillation or vibration is a periodic motion, the converse is not true. All periodic motions are not oscillations or vibrations.
• For example, the earth completes one revolution around the sun in 1 year, but it is not a to-and-fro motion about any mean position. Hence the motion is periodic but not oscillatory.

Some Quantities Related To Oscillation: It is evident from the different examples of oscillation above that the motion of the particle is restricted to a line segment, say BC.

Complete Oscillation: If a vibrating particle starts its motion from any point on its path towards a certain direction, returns to the same point, and then follows the same path in the same direction then it is said to have executed one complete oscillation or one complete vibration.

• If the particle starts its motion from B and after tracing the paths BOC and COB returns to B, then the particle executes one complete oscillation. It is seen that for a complete oscillation, the particle moves along the entire straight path twice.
• So, if the particle starts its motion from D and after tracing the paths DOC, COB, and BD, finally returns to D, then also it can be said that the particle has executed one complete oscillation. A complete oscillation is also known as a period.

Time Period: Time period of oscillation of a vibrating particle is defined as the time taken by it to execute one complete oscillation. Its dimension is T and its unit in all systems is second (s).

Frequency: The frequency of oscillation of a vibrating particle is defined as the number of complete oscillations executed by it in 1 second.

In time T the particle executes one complete oscillation, Thus the number of complete oscillations executed by the particle In ls is \(\frac{1}{T}\). Hence, the frequency n of the particle is n = \(\frac{1}{T}\).

The dimension of frequency is T^-1 and its unit in all systems is second^-1 or per second. This unit of frequency Is called hertz (Hz). So, 1 Hz = ls^-1.

Amplitude: The amplitude of oscillation of a vibrating particle is defined as Its maximum displacement from Its equilibrium position.

• Amplitude, A = OB or OC. The dimension of amplitude is 1, and Its units In the CGS system and SI are centimeters (cm) and meters (m) respectively. It is to be noted that in the above discussions, sometimes we have used the term ‘oscillation1 and sometimes ‘vibration’. In fact, oscillation and vibration are synonymous.
• Usually, when the time period of the particle is large, i.e., frequency is low, the motion of the particle is called oscillation.

Amplitude Example: Oscillation of a simple pendulum or an elastic spring. On the other hand, when the time period of the particle is small, i.e., frequency is high, the motion of the particle is called vibration. Examples of vibration are the vibration of a stretched string or that of a tuning fork.

Periodic Functions: A function f(t) is periodic if the function repeats itself after a regular interval of the independent variable t.

The simplest examples of a periodic motion can be represented by any of the following functions.

f(t) = \(A \cos \frac{2 \pi}{T} t\)….(1)

and \(g(t)=A \sin \frac{2 \pi}{T} t\)….(2)

Here T is the time period of the periodic motion.

To check the periodicity of these functions t is to be replaced by (t+ T), In the above equations simultaneously.

Hence equation (1) gives us,

f(t+T) = \(A \cos \left[\frac{2 \pi}{T}(t+T)\right]=A \cos \left[\frac{2 \pi}{T} t+2 \pi\right]\)

= \(A \cos \frac{2 \pi}{T} t[because \cos (\theta+2 \pi)=\cos \theta]\)

= f(t)

And equation (2) gives us,

g(t+T) = \(A \sin \left[\frac{2 \pi}{T}(t+T)\right]=A \sin \left(\frac{2 \pi}{T} t+2 \pi\right)\)

= \(A \sin \frac{2 \pi}{T} t[because \sin (\theta+2 \pi)=\sin \theta]\)

= g(t)

So, for a periodic function with period T, f(t+T) = f(t) or, g(t+T) = g(t)

The result will be the same if we consider a linear combination of sine and cosine functions of period T,

i.e., f(t) = \(A \sin \frac{2 \pi}{T} t+B \cos \frac{2 \pi}{T} t\)

Another example of a periodic function is, f(t)= sinωt + cos2ωt + cos4ωt But, f(t) = e^-ωt is not a periodic function, because it decreases monotonically with the increase in time and tends to zero as t → ∞

Displacement As A Function Of Time: Displacement can be represented by a mathematical function of time. In the case of periodic motion, this function is periodic in nature. One of the simple periodic functions is, f(t) = Acoscot

When the argument cot, is increased by an integral multiple of 2π radians, the value of the function remains the same.

Properties Of Simple Harmonic Motion Or SHM: Simple harmonic motion is the simplest form of oscillation. From the properties of simple harmonic motion, we can analyze any complex oscillation or vibration.

Any type of oscillatory motion can be considered to be the result of two or more simple harmonic motions acting on a particle. Thus, it is of great importance to discuss SHM in detail.

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Restoring Force: When a vibrating particle is at its position of equilibrium, the resultant force acting on it is zero. For example, when a simple pendulum is at its equilibrium position the downward force due to the weight of the bob is balanced by the upward tension of the string.

• So, the resultant force acting on the bob is zero. If the bob is displaced slightly from its equilibrium position and released, then a resultant force acts on the bob which tries to bring it to its equilibrium position.
• This force is called the restoring force. Since force is a vector quantity, this restoring force has a magnitude and a direction.
• If the magnitude and the direction of the restoring force satisfy the following two conditions, the motion of the particle is termed simple harmonic motion (SHM),

1. The restoring force Is always directed towards the position of equilibrium of the particle.
2. The magnitude of the restoring force is proportional to the displacement of the particle from its position of equilibrium.

WBBSE Class 11 Simple Harmonic Motion Notes

Equation Of Simple Harmonic Motion: Suppose, a particle is executing linear periodic motion along x-axis, and the point O, which is the origin (x = 0), is the position of equilibrium of the particle. Let D be any point on the path of the particle with position coordinate x.

According to condition (2), if F is the restoring force acting on the particle at D, then F ∝ x

Again from condition (1), as the restoring force F is directed towards the equilibrium position, it is taken as negative since the displacement OD = x is taken as positive.

So, F = -kx …..(1)

k is called the force constant and it is positive. Therefore, the magnitude of the restoring force acting on the particle when it is at a position of unit displacement is called the force constant. The units of k are dyn · cm^-1 (CGS) and N · m^-1 (SI).

If m is the mass of the particle and a is its acceleration then F = ma. So, from equation (1) we get, ma = -kx

or, a = \(-\frac{k}{m} x=-\omega^2 x\)…(2)

Here, \(\omega=+\sqrt{\frac{k}{m}}=\text { constant }\)…… (3)

Any one of the equations (1) or (2) is called the equation of simple harmonic motion. As the forms of these equations are identical, it can be said that the properties of acceleration of the particle and those of the restoring force are identical. Simple harmonic motion can be defined with reference to the properties of acceleration.

Simple Harmonic Motion Definition: The motion of a particle is said to be simple harmonic if its acceleration

1. Is proportional to its displacement from the position of equilibrium and
2. Is always directed towards that position.

It is to be noted that the acceleration of a particle executing simple harmonic motion is expressed as a = -ω²x. Conversely, if the acceleration of a particle obeys the equation a = -ω²x, then we can say that the motion of the particle is simple harmonic.

 

Oscillation And Waves

Simple Harmonic Motion Energy Of Simple Harmonic Motion

Understanding Simple Harmonic Motion (SHM)

Let m be the mass of a particle executing simple harmonic motion, A be the amplitude and T be the time period of the motion. The particle possesses kinetic energy due to its velocity all along its path of motion except at the extremities.

Again restoring force acts on the particle all along its path of motion except at the position of equilibrium. So, the work that is to be done to move the particle against the restoring force remains stored in the particle as potential energy.

Kinetic Energy: At a displacement x from the position of equilibrium, the velocity of the particle, v = \(\omega \sqrt{A^2-x^2}\)

So, the kinetic energy of the particle of mass m at that instant, K = \(\frac{1}{2} m v^2=\frac{1}{2} m \omega^2\left(A^2-x^2\right)\)….(1)

1. When the particle is just at the position of equilibrium, the kinetic energy at that instant, K = \(\frac{1}{2}\)mω²A² this is the maximum value of the kinetic energy.
2. As the particle reaches any end of its path, X = ± A, the kinetic energy at that instant, K = 0 this is the minimum value of the kinetic energy.

So, the kinetic energy of a particle executing SHM is maximum at the position of equilibrium and zero at the two ends, i.e., at the two extremities of its path of motion.

Potential Energy: When the particle is at a distance x from its position of equilibrium, the restoring force acting on it is, F = mω²x.

Again, when the particle is just at the position of equilibrium, x = 0, the restoring force F = 0, i.e., no force acts on the particle.

So, within the displacement from 0 to x, the average force acting on the particle =  \(\frac{0+m \omega^2 x}{2}=\frac{1}{2} m \omega^2 x\)

Now potential energy of the particle, U = work done to move the particle through the distance x against this average force = average force x displacement acting on the particle = \(\frac{1}{2} m \omega^2 x \cdot x=\frac{1}{2} m \omega^2 x^2\)…(2)

1. When the particle is just at the position of equilibrium, x = 0, the potential energy at that instant, U = 0  this is the minimum value of the potential energy.
2. As the particle reaches any end of its path, x = ± A, the potential energy at that instant, U = \(\frac{1}{2}\) mω²A² this is the maximum value of the potential energy.

Calculation Of Potential Energy With The Help Of Calculus: When the displacement of the particle is x from the position of equilibrium, the restoring force acting on it is, mω²x. Let dx be a further infinitesimal displacement of the particle such that the above force acting on the particle remains constant throughout this displacement.

So, work done in moving the dx = mω²x · dx

Hence, the total work done in moving it from 0 to x, i.e., the potential energy of the particle is,

U  = \(\int_0^x m \omega^2 x d x=\frac{1}{2} m \omega^2 x^2\)

Total Mechanical Energy: Total energy of a particle executing simple harmonic motion,
E = K+ U = \(\frac{1}{2} m \omega^2\left(A^2-x^2\right)+\frac{1}{2} m \omega^2 x^2=\frac{1}{2} m \omega^2 A^2 .\)

• Since m and co are constants, if the amplitude A remains unchanged, then E = constant, i.e., the total energy of the particle does not depend on its displacement. As the particle moves away from the position of equilibrium, kinetic energy gradually gets transformed into potential energy.
• While returning from the extreme position towards the position of equilibrium, its potential energy gradually gets converted into kinetic energy. This conversion is in accordance with the law of conservation of energy.
• Due to any external cause (for example, air resistance), if the amplitude of the motion decreases, the total energy will also decrease. In that case, the energy of the particle is transferred to the surroundings (for example, different air particles).
• The conversion of kinetic energy to potential energy and vice versa with the change in displacement of the particle executing SHM are shown in the following table. Using this table, the energy-displacement curve is plotted.

From the table, we see that when the displacement of the particle is ± \(\frac{A}{\sqrt{2}}\)

then kinetic energy = potential energy = \(\frac{1}{4} m \omega^2 A^2=\frac{E}{2}\)

Oscillation And Waves

Simple Harmonic Motion Energy Of Simple Harmonic Motion Numerical Examples

Equations of Simple Harmonic Motion

Example 1. When a particle executing SHM is at a distance of 0.02 m from its mean position, then its kinetic energy is thrice its potential energy. Calculate the amplitude of motion of the particle.
Solution:

According to the question, K = 3U

i.e., \(\frac{1}{2} m \omega^2\left(A^2-x^2\right)=3 \times \frac{1}{2} m \omega^2 x^2\) or, \(A^2-x^2=3 x^2 or, A^2=4 x^2\) or, \(A=| \pm 2 x|=0.04\)

[since x = 0.02m]

∴Amplitude = 0.04 m.

Example 2. When a particle executing SHM is at a distance of 0.02 m from its position of equilibrium, then its kinetic energy is twice its potential energy. Calculate the distance from the position of equilibrium where its potential energy is twice its kinetic energy.
Solution:

In the first case, when x = 0.02 m, K = 2 U

i.e., \(\frac{1}{2} m \omega^2\left(A^2-x^2\right)=2 \times \frac{1}{2} m \omega^2 x^2\)

or, \(A^2-x^2=2 x^2 or, A^2=3 x^2\)

or, \(A^2=3 \times(0.02)^2=0.0012\)

In the second case, 2 K=U

i.e., \(2 \times \frac{1}{2} m \omega^2\left(A^2-x^2\right)=\frac{1}{2} m \omega^2 x^2\)

or, \(2\left(A^2-x^2\right)=x^2 or, 3 x^2=2 A^2\)

or, \(x^2=\frac{2}{3} A^2=\frac{2}{3} \times 0.0012=.0008\)

or, x = \(\sqrt{0.0008}=0.0282 \mathrm{~m} \approx 0.03 \mathrm{~m}\)

Example 3. A particle of mass 0.2 kg is executing SHM along the x-axis with a frequency of \(\frac{25}{\pi}\)Hz. If its kinetic energy is 0.5 J at x = 0.04 m, then find its amplitude of vibration.
Solution:

Here, n = \(\frac{25}{\pi} \mathrm{Hz}\); so, \(\omega=2 \pi \cdot \frac{25}{\pi}=50 \mathrm{rad} \cdot \mathrm{s}^{-1}\)

We know, K = \(\frac{1}{2} m \omega^2\left(A^2-x^2\right) or, \frac{2 K}{m \omega^2}=A^2-x^2\)

or, \(A^2=\frac{2 K}{m \omega^2}+x^2=\frac{2 \times 0.5}{0.2 \times(50)^2}+(0.04)^2=0.0036 or, A=0.06 \mathrm{~m}\).

Example 4. The total energy of a particle executing SHM is 3 J. A maximum force of 1.5 N acts on it the Time period and epoch of the SHM are 2 s and 30° respectively. Establish the equation of this SHM and also find the mass of the particle.
Solution:

Total energy = \(\frac{1}{2} m \omega^2 A^2=3\)…(1)

Maximum force = mass x maximum acceleration

= m x ω²A = 1.5

Dividing (1) by (2) we get, \(\frac{1}{2}\) A = 2 or, A = 4 m

Time period, T = 2s

∴ \(\omega=\frac{2 \pi}{T}=\frac{2 \pi}{2}=\pi \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Epoch, \(\alpha=30^{\circ}=\frac{\pi}{6}\)

So, the equation of SHM is x = Asin(ωt+α)

or, x = \(4 \sin \left(\pi t+\frac{\pi}{6}\right) \mathrm{m}\)

Again from equation (2), m = \(\frac{1.5}{\omega^2 A}=\frac{1.5}{\pi^2 \cdot 4}=0.038 \mathrm{~kg}\)

Example 5. The equation of motion of a particle executing SHM is x = Asin(ωt+ θ). Calculate the velocity and acceleration of the particle. If m is the mass of the particle, then what is the maximum value of its kinetic energy?
Solution:

Given equation ofSHM, x = Asin(ωt+ θ)….(1)

So, velocity v = \(\frac{d x}{d t}=\omega A \cos (\omega t+\theta)\)

From equation (1) we get, \(\sin (\omega t+\theta)=\frac{x}{A}\)

∴ \(\cos (\omega t+\theta)=\sqrt{1-\sin ^2(\omega t+\theta)}=\sqrt{1-\frac{x^2}{A^2}}=\sqrt{\frac{A^2-x^2}{A^2}}\)

∴ v = \(\omega A \sqrt{\frac{A^2-x^2}{A^2}}=\omega \sqrt{A^2-x^2}\)

Again, the acceleration of the particle,

a = \(\frac{d \nu}{d t} =\frac{d}{d t}\{\omega A \cos (\omega t+\theta)\}=-\omega^2 A \sin (\omega t+\theta)\)

= \(-\omega^2 x\)  (x = \(A \sin (\omega t+\theta)\))

As the maximum value of \(\cos (\omega t+\theta)\) is 1 ,

the maximum value of velocity, \(v_{\max }=\omega \mathrm{A}\)

∴ Maximum kinetic energy of the particle

= \(\frac{1}{2} m v_{\max }^2=\frac{1}{2} m \omega^2 A^2 .\)

Example 6. The amplitude of a particle of mass 0.1 kg executing SHM is 0.1m. At the mean position, its kinetic energy is 8×10^-3 J. Find the time period of its vibration.
Solution:

At the mean position, the potential energy of the particle =0

Hence, kinetic energy of the particle at the mean position = total energy = \(\frac{1}{2}\)A

∴ \(\frac{1}{2} m \omega^2 A^2=E \text { or, } \omega^2=\frac{2 E}{m A^2} \text { or, } \omega=\sqrt{\frac{2 E}{m A^2}}\)

i.e., time period,

T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m A^2}{2 E}}\)

= \(2 \times 3.14 \times \sqrt{\frac{0.1 \times(0.1)^2}{2 \times 8 \times 10^{-3}}}\)

= \(2 \times 3.14 \times \sqrt{\frac{1}{16}}=\frac{2 \times 3.14}{4}=1.57 \mathrm{~s} .\)

Example 7. An object of mass 10 kg executes SHM. Its time period and amplitude are 2 s and 10 m respectively. Find Its kinetic energy when it is at a distance of

1. 2 m and
2. 5 m respectively from its position of equilibrium.

Justify the two different results for (1) and (2).

Solution:

T = \(2 \mathrm{~s}\), i.e., \(\omega=\frac{2 \pi}{T}=\frac{2 \pi}{2}=\pi \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Kinetic energy, K = \(\frac{1}{2} m \omega^2\left(A^2-x^2\right)\)

1. When x = 2m, K = \(\frac{1}{2} \times 10 \times(\pi)^2 \times\left\{(10)^2-(2)^2\right\}\) = \(5 \times(3.14)^2 \times 96=4732.6 \mathrm{~J} .\)
2. When x = 5m, K = \(\frac{1}{2} \times 10 \times(\pi)^2 \times\left\{(10)^2-(5)^2\right\}\) = \(5 \times(3.14)^2 \times 75=3697.35 \mathrm{~J}\)

The velocity of a particle executing SHM becomes maximum at its mean position. As the displacement increases, i.e., as it moves away from the equilibrium position, its velocity decreases and so the kinetic energy also decreases.

Example 8. A particle of mass m executes SHM with amplitude and frequency n. What is the average kinetic energy of the particle during its motion from the position of equilibrium to the end?
Solution:

KE in the mean position = \(\frac{1}{2} m a^2 w^2\)

= \(\frac{1}{2} m a^2(2 \pi n)^2\)

= \(2 \pi^2 m a^2 n^2\)

KE at the end = 0

∴ Average kinetic energy = \(\frac{0+2 \pi^2 m a^2 n^2}{2}=\pi^2 m a^2 n^2\)

Oscillation And Waves

Simple Harmonic Motion Simple Pendulum

A simple pendulum is nothing but a small, heavy body suspended from a rigid support with the help of a long string. The heavy body remains in its lowest position A when the string is vertical.

• This position OA is called the position of equilibrium of the pendulum. The heavy body is called the pendulum bob. The support (O) from which the bob is suspended is called the point of suspension.
• The center of gravity of the suspended bob Is the point of oscillation, When the bob Is displaced from the equilibrium position by a little distance and then released, the pendulum oscillates about its equilibrium position on either side.

Applications of Simple Harmonic Motion in Real Life

For convenience in the mathematical treatment of the properties of a pendulum, an Ideal simple pendulum Is considered. A simple pendulum will be ideal If

1. The string is weightless,
2. The string is inextensible,
3. No frictional resistance acts on the bob during its oscillation and
4. The bob is a point mass. Conforming to all the conditions stated above is not practically possible; we never get an ideal simple pendulum. Hence, for laboratory use, a small, heavy metal ball is tied to one end of a long, light string and the system is suspended from a rigid support.

Simple Pendulum Definition: If a small, heavy body, suspended from a rigid support by a long, weightless, and inextensible string, can be set into oscillation, then the arrangement is called a simple pendulum.

1. When the pendulum bob is slightly displaced from position A to B and then released, it starts oscillating along arc BAC and CAB. This means that the position of the pendulum periodically changes from OB to OC.
2. This to-and-fro oscillatory motion is periodic. But practical experience shows that due to air resistance and friction at the suspension point, this oscillation slowly subsides and ultimately the pendulum comes to rest along OA, its equilibrium position.

A Few Definitions Related To Simple Pendulum

Plane Of Oscillation: In the given diagram, the straight lines OA, OB, and OC lie on the same vertical plane. The pendulum does not leave that plane during oscillation. This plane is called the plane of oscillation.

Effective Length: The distance of the center of gravity of the bob from the point of suspension is called the effective length of the pendulum. In the case of a spherical bob of radius r, the center of gravity lies at the center of the sphere, and if the length of the string is l, then the effective length L = l+r.

Amplitude And Angular Amplitude: The maximum displacement of The pendulum hob on either side of its equilibrium position is called amplitude. In the given diagram, AB or AC is the amplitude of the pendulum.

The angle subtended at the point of suspension by the equilibrium position and the maximum displaced position of the pendulum bob is the angular amplitude of the pendulum. The angular amplitude ∠AOB = ∠AOC = θ.

It is to be noted that, the angular amplitude should be less than 4° so that the arc CAB is almost a straight line.

Complete Oscillation And Period Of Oscillation Or Time Period: Starting from an endpoint, when the pendulum bob again and returns to the same point, the pendulum completes one complete oscillation.

• The bob starting from point B reaches C and returns to B. This completes one complete oscillation. During one complete oscillation, the pendulum bob covers twice the total path of its movement. Hence, let us assume that the bob starts from point A towards point B.
• After reaching point B, the bob starts moving in the opposite direction, crosses A, and reaches point C, then from C the bob reaches A. A complete oscillation is executed in this manner also. After executing one complete oscillation the pendulum returns to its initial phase.
• The time taken by a pendulum to complete one oscillation is called the period of oscillation or the time period. In other words, the time period is the minimum time taken by the pendulum to return to its starting phase.

The movement from B to C is a half oscillation of the pendulum, and the time required for it is called the half-time period or half the period of oscillation.

Simple Pendulum Frequency: The number of complete oscillations executed in one second by a pendulum is its frequency. If the time period of a pendulum is T, then as per definition, the number of complete oscillations in time T = 1. Hence, in unit time, the number of complete oscillations = \(\frac{1}{t}\).

Now, from definition, the frequency n = \(\frac{1}{t}\).

The unit of frequency is s^-1 or hertz or Hz; the dimension is T^-1.

Motion Of A Simple Pendulum: A simple pendulum of effective length*¹ L is oscillating with m angular amplitude not exceeding 4°. The bob of the pendulum oscillates from B to C on either side of its position of equilibrium, O.

Let at any instant of motion, the bob of mass m be at P and its displacement from the position of equilibrium OP = x. If the angular displacement is θ, then θ = \(\frac{x}{L}\) rad, provided θ is small and sinθ ≈ θ ≈ tanθ.

At P, the weight mg of the bob acts vertically downwards. The component mg sinθ tries to bring the bob to the position of equilibrium. As this force acts in a direction opposite to that of displacement, it is the restoring force, expressed as

F = -mg sinθ

= -mgθ[since # is less than 4°]

= -mg\(\frac{X}{L}\)

Now, the acceleration of the bob,

a = \(\frac{F}{m}=\frac{-g}{L} x=-\omega^2 x \quad\left[\text { where, } \omega=\sqrt{\frac{g}{L}}\right]\)

As the motion of the bob obeys the equation, a = -ω²x, it can be said that the motion of a simple pendulum with an angular amplitude less than 4° is simple harmonic.

Time Period: Time period of the pendulum,

T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\frac{g}{L}}}\)

= \(2 \pi \sqrt{\frac{L}{g}}\)…(1)

Mechanical Energy Of The Pendulum: The kinetic energy of the pendulum,

K = \(\frac{1}{2} m \omega^2\left(A^2-x^2\right)=\frac{1}{2} m \frac{g}{L}\left(A^2-x^2\right)\)

The potential energy of the pendulum,

U = \(\frac{1}{2} m \omega^2 x^2=\frac{1}{2} m \frac{g}{L} x^2\)

∴ The total mechanical energy of a simple pendulum,

E = K+ U = \(\frac{1}{2} m \omega^2 A^2=\frac{1}{2} m \frac{g A^2}{L}=\frac{m g A^2}{2 L}\)

Thus, if the angular amplitude is less than 4°, the total mechanical energy of a simple pendulum is

1. Directly proportional to the mass of the bob,
2. Inversely proportional to the effective length of the pendulum and
3. Directly proportional to the square of the amplitude of the SHM executed by the pendulum.

Tension On The String: When the pendulum oscillates about its point of suspension, a centripetal force is required for the circular motion of the bob. At the instant when the bob passes through the position of equilibrium O, its velocity becomes maximum. So, the upward centripetal force, Fc, becomes maximum at O. The resultant of downward weight mg and tension F’ on the string becomes equal to this force F.

Therefore, F_c = F’ – mg or, F’ = F_c + mg

The kinetic energy of the pendulum at P,

K = \(\frac{1}{2} m v^2=\frac{1}{2} m \omega^2\left(A^2-x^2\right)=\frac{1}{2} m \frac{g}{L}\left(A^2-x^2\right)\)

So, \(m v^2=\frac{m g\left(A^2-x^2\right)}{L}\)

Therefore, the centripetal force at P,

∴ \(F_c=\frac{m v^2}{L}=\frac{m g\left(A^2-x^2\right)}{L^2}\)

∴ \(F^{\prime}=m g+F_c=m g+m g \frac{\left(A^2-x^2\right)}{L^2}\)

= \(m g\left(1+\frac{A^2-x^2}{L^2}\right)\)

When the bob passes through O (where x = 0 ), the tension F1 on the string becomes the maximum

∴ \(F_{\max }^{\prime}=m g\left(1+\frac{A^2}{L^2}\right)\)

Seconds Pendulum Pendulum Clock Definition: A simple pendulum of a time period of 2 seconds, or a half-time period of 1 second, is called a seconds pendulum.

A seconds pendulum has a time period, T = 2s

It is known, T = \(2 \pi \sqrt{\frac{L}{g}} or, 2=2 \pi \sqrt{\frac{L}{g}}\)

or, \(1=\pi \sqrt{\frac{L}{g}} or, L=\frac{g}{\pi^2}=\frac{981}{\pi^2}=99.40 \mathrm{~cm}=0.9940 \mathrm{~m}\).

Pendulum Dock Running Fast Or Slow: A pendulum clock marks time by means of its time period. From the equation T = \(2 \pi \sqrt{\frac{L}{g}},\), it can be shown that if

1. The effective length l of a pendulum changes, or
2. The value of the acceleration due to gravity changes, the time period of a second pendulum does not remain 2 s, but increases or decreases accordingly.

An increase in the time period means that the pendulum oscillates slowly, i.e., the clock goes slow. On the other hand, a decrease in the time period makes the clock run fast.

• For example, when a pendulum clock is taken from the poles to the equator, or from the sea level to the top of a mountain, or from the earth’s surface to deep inside a mine, the acceleration due to gravity g decreases; hence, the time period increases. Thus, in each case, the clock goes slow.
• A pendulum clock may run slow or fast depending on the temperature. At a higher temperature, due to the expansion of the metallic suspender and the bob, the effective length increases, and the time period also increases.

On the other hand, at low temperatures, the time period decreases. Therefore, a pendulum clock runs slow in summer and fast in winter. To get the correct time from the same pendulum at different temperatures, compensated pendulums are used.

Some Uses Of A Simple Pendulum

Finding The Value Of g: From equation (1) we get, g = \(4 \pi^2 \frac{L}{T^2}\)…(1)

Using this equation, the value of acceleration due to gravity (g) of a place can be determined. For different lengths of strings, the time period (T) and the effective length (L) of a pendulum can be evaluated by experiment.

Then \(\frac{L}{T^2}\) value in each case is determined and the average value of \(\frac{L}{T^2}\) is calculated. Substituting this average value in equation (1), the value of g can be obtained.

Determining The Height Of A Place: Let R = radius of the earth, h = height of a place (say, top of a hill), g = value of acceleration due to gravity on earth’s surface, g’ = value of acceleration due to gravity at height h, T =time period of a simple pendulum al a fixed place on the earth surface and T’ = time period of the same pendulum at height h.

From equation (1) Newtonian Gravitation and Planetary Motion,

⇒ \(\sqrt{\frac{g}{g^{\prime}}}=\frac{R+h}{R}=1+\frac{h}{R}\)…(2)

Also \(T=2 \pi \sqrt{\frac{L}{g}} and T^{\prime}=2 \pi \sqrt{\frac{L}{g^{\prime}}}\)

∴ \(\frac{T}{T^{\prime}}=\sqrt{\frac{g}{g}}\)

or,  \(\sqrt{\frac{g}{g^{\prime}}}=\frac{T^{\prime}}{T}\)

From (2) and (3),

1 + \(\frac{h}{R}=\frac{T^{\prime}}{T} \text { or, } \frac{h}{R}=\frac{T^{\prime}}{T}-1\)

or, h = \(\frac{R}{T}\left(T^{\prime}-T\right)\)

Knowing the value of R and determining the values of T and T’ using a stopwatch, the value of h can be determined from equation (4).

Finding The Depth Of A Mine: Let h = depth of the mine, g = value of acceleration due to gravity on earth’s surface, g’ = value of acceleration due to gravity at the depth h below earth’s surface, T = time period of a simple pendulum on the earth surface, T’ = time period of the same pendulum at depth h.

In this case, \(T=2 \pi \sqrt{\frac{L}{g}}\) and \(T^{\prime}=2 \pi \sqrt{\frac{L}{g^{\prime}}}\)

or, \(\frac{T}{T^{\prime}}=\sqrt{\frac{g^{\prime}}{g}} or, g^{\prime}=g \frac{T^2}{T^{\prime 2}}\)…(5)

From equation (4) Newtonian Gravitation and Planetary Motion, we get

⇒ \(g^{\prime}=g\left(1-\frac{h}{R}\right)\)…(6)

From equations (5) and (6), \(\frac{T^2}{T^{\prime 2}}=1-\frac{h}{R}\)

or, \(h=R\left(1-\frac{T^2}{T^{\prime 2}}\right)\)

Knowing the value of R and determining the values of T and T’ using a stopwatch, the value of h can be determined from equation (7).

Oscillation And Waves

Simple Harmonic Motion Simple Pendulum Numerical Examples

Short Answer Questions on Simple Harmonic Motion

Example 1. A simple pendulum executes 40 complete oscillations in a minute. What is the effective length of the pendulum? g = 980 cm · s^-2.
Solution:

Time period (T) = \(\frac{60}{40}=\frac{3}{2} \mathrm{~s}\)

Now, T = \(2 \pi \sqrt{\frac{L}{g}} or, T^2=4 \pi^2 \frac{L}{g}\)

or, \(L=\frac{g T^2}{4 \pi^2}=\frac{980 \times\left(\frac{3}{2}\right)^2}{4 \times \pi^2}=55.9 \mathrm{~cm}\).

Example 2. What will be the percentage increase in the time period of a simple pendulum when its length is increased by 21%?
Solution:

Given, the increase in length of the pendulum = 0.21 L, where L is the initial length

Hence, increased length L’ = L + 0.21L = 1.211

Let, the time period change to T’ from T due to the change in length.

As T ∝ √L,

⇒ \(\frac{T}{\sqrt{L}}=\text { constant }\)

Thus, \(\frac{T}{\sqrt{L}}=\frac{T^{\prime}}{\sqrt{L}}\)

or, \(T^{\prime}=\sqrt{\frac{L^{\prime}}{L}} T=\sqrt{\frac{1.21 L}{L}} T=1.1 T\)

∴ Increase in time period = \(T^{\prime}-T=1.1 T-T\)

= 0.1 T = 10% of T

Hence, the time period increases by 10%.

Example 3. Two simple pendulums of lengths 100 cm and 101 cm are set Into oscillation at the same time. After what time does one pendulum gain one complete oscillation over the other?
Solution:

The length of the First pendulum is comparatively less, and hence, its time period is also less; thus the first pendulum oscillates faster.

By the time the second pendulum executes n oscillations, suppose the first one completes (n + 1) oscillations.

Hence, if T₁ and T₂ are time periods of the first and the second pendulums, \((n+1) T_1=n T_2 \text { or, } \frac{T_2}{T_1}=\frac{n+1}{n}=1+\frac{1}{n}\)…(1)

Again \(\frac{T_2}{T_1}=\sqrt{\frac{L_2}{L_1}}=\sqrt{\frac{101}{100}}=\left(1+\frac{1}{100}\right)^{1 / 2}\)

= \(1+\frac{1}{2} \times \frac{1}{100}\)=\(1+\frac{1}{200}\)…(2)

From equations (1) and (2) \(\frac{1}{n}=\frac{1}{200} \text { or, } n=200 \text {, i.e., } n+1=201\)

Thus, the required time = time of 201 complete oscillations of the first pendulum = \(201 \times T_1\)

= \(201 \times 2 \pi \sqrt{\frac{L_1}{g}}\)

= \(201 \times 2 \times \pi \sqrt{\frac{100}{980}} \approx 403 \mathrm{~s}=6 \mathrm{~min} 43 \mathrm{~s} .\)

Example 4. Find the length of a simple pendulum on the surface of the moon that has a time period same as that of a simple pendulum on the earth’s surface. The mass of Earth is 80 times that of the moon and the radius of Earth is 4 times that of moon.
Solution:

If masses of the earth and the moon are and respectively, \(\frac{M_1}{M_2}=80\)

Again, their radii are R₁ and R₂ (say).

So, \(\frac{R_1}{R_2}=4\)

Acceleration due to gravity on earth, \(g_1=\frac{G M_1}{R_1^2}\); acceleration due to gravity on moon, \(g_2=\frac{G M_2}{R_2^2}\)

∴ \(\frac{g_1}{g_2}=\frac{M_1}{M_2} \times \frac{R_2^2}{R_1^2}=\frac{M_1}{M_2} \times\left(\frac{R_2}{R_1}\right)^2=80 \times\left(\frac{1}{4}\right)^2=5\)

Also, in the case of a simple pendulum, time period on earth’s surface T = \(2 \pi \sqrt{\frac{L_1}{g_1}}\)

and time period on the moon’s surface T = \(2 \pi \sqrt{\frac{L_2}{g_2}}\)

∴ \(\sqrt{\frac{L_1}{g_1}}=\sqrt{\frac{L_2}{g_2}} \text { or, } \frac{L_1}{L_2}=\frac{g_1}{g_2}=5\)

∴ \(L_2=\frac{L_1}{5}\)

Hence, the length of the pendulum on the moon’s surface should be 1/5 th of its length on the earth’s surface.

Example 5. Two pendulums of time periods 1.8 s and 2 s are set Into oscillation at the same time. After how many seconds will the faster-moving pendulum execute one complete oscillation more than the other? How many oscillations will the faster-moving pendulum execute during this time?
Solution:

Suppose the faster pendulum executes one more oscillation than the other after t s.

Number of complete oscillations of the first pendulum in t s = \(\frac{t}{1.8}\)

Number of oscillations of the second pendulum in t s = \(\frac{t}{2}\)

According to the given condition, \(\frac{t}{1.8}\) – \(\frac{t}{2}\) = 1

or, 0.2t= 2×1.8 or, t = 18 s

Number of complete oscillations executed by the faster pen¬dulum (time period 1.8 s) = \(\frac{1.8}{1.8}\) = 10.

Example 6. A pendulum clock runs 20s slow per day. What should be the change in length of the clock so that it records the correct time? Take the pendulum as a simple pendulum.
Solution:

Half time period of a simple pendulum t = \(\frac{T}{2}=\pi \sqrt{\frac{L}{g}}\)

For a perfect seconds pendulum t = 1s

∴ 1 = \(\pi \sqrt{\frac{L}{g}} \text { or, } L=\frac{g}{\pi^2}\)….(1)

Also, 1d = 24 x 60 x 60 = 86400 s.

Number of half oscillations executed in a day by a pendulum that runs 20 s slow per day = 86400 – 20 = 86380.

Hence, the time period of that pendulum \(t_1=\frac{86400}{86380} \mathrm{~s}\)

If the length of this simple pendulum is L₁, then \(t_1=\pi \sqrt{\frac{L_1}{g}} \text { or, } L_1=\frac{g t_1^2}{\pi^2}\)…(2)

Subtracting equation (1) from equation (2), \(L_1-L=\frac{g}{\pi^2}\left(t_1^2-1\right)\) = \(\frac{980}{\pi^2}\left[\left(\frac{86400}{86380}\right)^2-1\right]\)

Hence, to get the correct time, the length of the pendulum is to be decreased by 0.46 mm.

Alternative Method: Half-time period of a perfect seconds pendulum = 1 s.

Number of Half oscillations of a clock that runs slow by t₀ s in a day = 86400 – t₀.

Hence, half time period = \(\frac{86400}{86400-t_0} \mathrm{~s} \)

∴ Increase in the value of half time period, dt = \(\frac{86400}{86400-t_0}-1=\frac{t_0}{86400-t_0} \mathrm{~s}\)

If the value of t₀ is negligibly smaller than 86400, dt \(=\frac{t_0}{86400} \mathrm{~s}\)

Now, t = \(pi \sqrt{\frac{L}{g}} or, \log t=\log \pi+\frac{1}{2} \log L-\frac{1}{2} \log g\)

Differentiating, \(\frac{d t}{t}=\frac{1}{2} \frac{d L}{L}-\frac{1}{2} \frac{d g}{g}\)…(1)

Here, dL = increase in length and dg = increase in acceleration due to gravity.

Putting values of t and dt in (1), \(\frac{t_0}{86400}=\frac{1}{2}\left(\frac{d L}{L}-\frac{d g}{g}\right)\)

or, \(t_0=43200\left(\frac{d L}{L}-\frac{d g}{g}\right)\)…(2)

This equation can be used as a rule for a seconds pendulum. For a second pendulum L = \(\frac{g}{\pi^2}\) and on the surface of the earth g = 980 cm · s^-2. If the clock runs fast, the value of t₀ is negative.

In the given problem, t₀ = 20 s;

If there is no change in the value of g, dg = 0

∴ \(t_0=43200 \times \frac{a L}{L}\)

or, \(d L=\frac{20 \times L}{43200}=\frac{20}{43200} \times \frac{980}{\pi^2}=0.046 \mathrm{~cm}=0.46 \mathrm{~mm}\)

Hence, the length of the defective clock has increased by 0.46 mm. Thus to get the correct time, its length needs to be decreased by 0.46 mm.

Example 7. A pendulum of length 60 cm is suspended inside an airplane. The aeroplane is flying up with an acceleration of 4 m · s^-2 making an angle of 30° with the horizontal. Find the time period of oscillation of the pendulum.
Solution:

Horizontal component of acceleration a of the plane = acos30° and vertical component = asin30°

Hence, the downward acceleration experienced by the pendulum bob = g- (-a sin30°) = g+ a sin30°

So, the acceleration of the pendulum bob

g’ = \(\sqrt{\left(g+a \sin 30^{\circ}\right)^2+\left(a \cos 30^{\circ}\right)^2} \)

= \(\sqrt{\left(9.8+4 \times \frac{1}{2}\right)^2+\left(4 \times \frac{\sqrt{3}}{2}\right)^2}=\sqrt{(11.8)^2+12}\)

= \(\sqrt{151.24}=12.3 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

∴ The time period of the pendulum

T = \(2 \pi \sqrt{\frac{L}{g^{\prime}}}=2 \times \pi \sqrt{\frac{0.60}{12.3}}=1.38 \mathrm{~s} .\)

Example 8. The effective length of a simple pendulum is 1 m and the mass of its bob is 5 g. If the amplitude of motion of the pendulum is 4 cm, what is the maximum tension on the string to which the bob is attached?
Solution:

The velocity of the bob is maximum at its position of equilibrium.

The maximum velocity \(v_{\max }=\omega A=\frac{2 \pi A}{T} .\)

At this position, centripetal force is also maximum, whose value is \(F_c=\frac{m \nu_{\max }^2}{L}=\frac{m \omega^2 A^2}{L}\)

[L = effective length of the pendulum]

The resultant of the weight mg of the bob and tension F on the string becomes equal to this force F_c .

∴ F = \(m g+F_c=m g+\frac{m \omega^2 A^2}{L}=m\left(g+\frac{\omega^2 A^2}{L}\right)\)

Again, \(T=2 \pi \sqrt{\frac{L}{g}} ; so, \omega=\frac{2 \pi}{T}=\sqrt{\frac{g}{L}} or, \omega^2=\frac{g}{L}\)

∴ F = \(m\left(g+\frac{g A^2}{L^2}\right)=m g\left(1+\frac{A^2}{L^2}\right)\)

= \(5 \times 980\left(1+\frac{4^2}{100^2}\right)[because L=1 \mathrm{~m}=100 \mathrm{~cm}]\)

= \(4908 \mathrm{dyn}=0.04908 \mathrm{~N} \approx 0.05 \mathrm{~N}\)

Example 9. Prove that the change in the time period t of a simple pendulum due to a change AT of temperature is, \(\Delta t=\frac{1}{2} \alpha t \Delta T\), where α = coefficient of linear expansion.
Solution:

If L is the effective length of a simple pendulum, then its time period is, t = \(2 \pi \sqrt{\frac{L}{g}}\)

For a change ΔT of temperature, the length becomes, L’ = L(1 +αΔT)

Therefore the time period,

t’ = \(2 \pi \sqrt{\frac{L}{g}}=2 \pi \sqrt{\frac{L}{g}(1+\alpha \Delta T)}\)

= \(2 \pi \sqrt{\frac{L}{g}}(1+\alpha \Delta T)^{1 / 2}=t\left(1+\frac{1}{2} \alpha \Delta T\right)=t+\frac{1}{2} \alpha t \Delta T\)

[neglecting the terms containing a², a³, etc. since a is very small]

The change in time period, Δt = t’ -t = \(\frac{1}{2}\)αtΔT (Proved).

Alternative Method: t = \(2 \pi \sqrt{\frac{L}{g}}\)

log t = \(\log 2 \pi+\frac{1}{2} \log L-\frac{1}{2} \log g\)

Differentiating with respect to L, \(\frac{1}{t} d t=\frac{1}{2 L} d L\)

∴ \(\frac{d t}{t}=\frac{1}{2} \frac{d L}{L}\)…(1)

Now \(L_t=L(1+\alpha \Delta T)\)

or, \(L_t-L=L \alpha \Delta T or, d L=L \alpha \Delta T\)

or, \(\frac{d L}{L}=\alpha \Delta T\)…(2)

From equations (1) and (2) we get, \(\frac{d t}{t}=\frac{1}{2} \alpha \Delta T \text { or, } d t=\frac{1}{2} \alpha \Delta T \cdot t \text { (Proved). }\)

Example 10. The bob of a simple pendulum is made of brass and its time period is T. It is completely immersed in a liquid and is allowed to oscillate. If the density of the liquid is 1/8 th of the density of brass, what will be the time period of oscillation of the pendulum now?
Solution:

Initial time period of the simple pendulum, T = \(2 \pi \sqrt{\frac{L}{g}}\)

[L = effective length of the pendulum]

If m is the mass, V is the volume of the bob and d is the density of brass, then the apparent weight of the hob inside the liquid,

W₁ = W- buoyancy (weight of the displaced liquid)

= \(V d g-V \frac{d}{8} g=\frac{7}{8} V d g\)

If g₁ is the effective acceleration due to gravity in the immersed condition, then

⇒ \(W_1=m g_1=\frac{7}{8} V d g \text { or, } g_1=\frac{7}{8} \frac{m g}{m}=\frac{7}{8} g\)

∴ Final time period, \(T_1=2 \pi \sqrt{\frac{L}{g_1}}=2 \pi \sqrt{\frac{8 L}{7 g}}=\sqrt{\frac{8}{7}} T\).

Example 11. A brass sphere is hung from one end of a massless and inextensible thread. When the sphere is set into oscillation, it oscillates with a time period of T. If now the sphere is dipped completely into a non-viscous liquid, then what will be the time period of its oscillation? (The density of the liquid is 1/10th of that of brass)
Solution:

Let the volume of the sphere be V, the density of brass be ρ, the density of the liquid be ρ’.

∴ Apparent weight of the sphere when immersed in the liquid = real weight – weight of displaced liquid = Vρg-Vρ’g =Vg(ρ-ρ’)

∴ Apparent acceleration due to gravity of the sphere immersed in the liquid,

g’ = \(\frac{\text { apparent weight }}{\text { mass }}=\frac{V g\left(\rho-\rho^{\prime}\right)}{V \rho}=g\left(1-\frac{\rho^{\prime}}{\rho}\right)\)

According to the problem, \(\frac{\rho^{\prime}}{\rho}=\frac{1}{10}\);

hence \(g^{\prime}=g\left(1-\frac{1}{10}\right)=\frac{9}{10} g \text {. }\)

In the case of a simple pendulum, \(T\propto \frac{1}{\sqrt{g}}\); so, if the rime period of oscillation of the sphere, when unmersed in die liquid, is T’, then \(\frac{T^{\prime}}{T}=\sqrt{\frac{g}{g^{\prime}}}=\sqrt{\frac{10}{9}} \text { or, } T^{\prime}=\frac{\sqrt{10}}{3} T \text {. }\)

Oscillation And Waves

Simple Harmonic Motion A Few Examples Of SHM

Oscillation Of A Mass Attached To A Vertical Elastic Spring: Let a body of mass m be attached to the bottom of a vertical elastic spring of negligible mass suspended from a rigid support, As a result of this, let the increase in length of the spring be l.

So the force constant of the spring.

k = force required for a unit increase in length = \(\frac{mg}{l}\)

Now the mass is pulled downwards through a distance x from its position of equilibrium O. If the extension of the spring does not exceed its elastic limit, then a reaction force, – kx, equal and opposite to the applied force is developed in the spring.

This force acts as the restoring force. If a is the acceleration of the suspended body, then restoring force -kx = ma

or, a = \(\frac{-k}{m} x=-\omega^2 x \quad\left[\text { where } \omega=\sqrt{\frac{k}{m}}\right]\)

As the motion of the body of mass m obeys the equation a = -ω²x, it can be said that the motion of the body attached to the spring is simple harmonic.

In this case, the time period of oscillation,

T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{k}} .\)

Now, k = \(\frac{m g}{l}\); therefore T = \(2 \pi \sqrt{\frac{m}{\frac{m g}{l}}}=2 \pi \sqrt{\frac{l}{g}}\)

Here, the initial increase in length of the spring due to suspension of the body of mass m is l. So by measuring this increase in length with a meter scale and the time period T with a stopwatch, acceleration due to gravity g can be calculated from the above relation.

Oscillation Of A Mass Attached To A Horizontal Elastic Spring: Let one end of an elastic spring of negligible mass be attached to a vertical support and its other end to a body of mass m. The body lies on a smooth horizontal plane. At this moment, no force acts on the body due to the spring as it is not stretched. So the body is at rest.

If the body is now moved towards the right, the spring will be elongated and a restoring force F will act on the body towards the left, trying to bring the mass to its equilibrium position.

If the force constant of the spring is k and the body is moved through a distance x towards the right, then F = – kx.

∴ Acceleration of the body, a = \(\frac{F}{m}=\frac{-k x}{m}=-\omega^2 x\left[\text { where } \omega=\sqrt{\frac{k}{m}}\right]\)

As the motion of the body obeys the equation, a = -ω²x, it can be said that the motion of the body attached to the spring is simple harmonic.

Time period of oscillation, T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{k}}\)

It is to be noted that the time periods of vertical oscillation and horizontal oscillation of a spring are equal.

Oscillation Of A Liquid In A U-Tube: Consider that a U-tube of uniform cross-section a contains a liquid of density ρ. Let the length of the liquid column in each limb at equilibrium be L. Therefore the total length of the liquid column is 2L, if the horizontal separation between the two limbs is negligibly small.

Then the mass of the liquid column, m = 2Lαρ

If the liquid in one limb is depressed by x then the liquid level in the other limb will be raised by x. Hence the difference in the height of the liquid levels in the two limbs will be 2x.

Weight of this liquid head = 2xαρg.

This weight provides a restoring force trying to bring the liquid to its initial equilibrium. This force acts opposite to the direction of displacement x in the two limbs.

Thus, restoring force = -2xαρg.

Energy in Simple Harmonic Motion

Due to this force, if a is the acceleration of the liquid level, then ma = -2xαρg

or, 2Lαρa = -2xαρg

or, a = \(\frac{-g \cdot x}{L}=-\omega^2 x\left[\text { where } \omega=\sqrt{\frac{g}{L}}\right]\)

As the motion of the liquid level obeys the equation a = -ω²x, it follows a simple harmonic motion. So, if the liquid in one limb of a U-tube is depressed and then released, the up and down motion of the liquid column would be simple harmonic.

Time period of this motion, T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{L}{g}}\)

Oscillation Of A Piston In A Gas Cylinder: Suppose, some amount of gas is enclosed in a cylinder fitted with a frictionless piston. Let the piston be initially at C, the position of equilibrium, the pressure of the gas enclosed be P and the length of the gas column be L.

The piston is now pushed down slightly to B very slowly and then released. The compressed gas will then expand and cause the piston to oscillate up and down.

When the piston is moved through a distance x from C to B, suppose the pressure of the enclosed gas increases from P to P + p and the volume decreases to (V- v) from V. If this change takes place isothermally, then according to Boyle’s law, PV = (P + p) (V- v)

or, PV = PV-Pv+ pV-pv

or, Pv = pV [neglecting pν as it is very small]

or, p = \(\frac{P v}{V}=\frac{P \alpha x}{\alpha L}=\frac{P x}{L}\)

[α = cross-sectional area of the piston]

An additional force acts on the piston for this excess pressure p and tries to bring the piston to its initial position of equilibrium.

So, the restoring force = \(-p \alpha=\frac{-P \alpha x}{L}\)

∴ Acceleration of the piston,

a = \(\frac{\text { restoring force }}{\text { mass of the piston }}=-\frac{P \alpha x}{L M}\)

[M = mass of the piston]

or, a = \(-\omega^2 x, \text { where } \omega=\sqrt{\frac{P \alpha}{L M}}\)

The motion of the piston obeys the equation a = -ω2x.

So this motion is simple harmonic. –

Time period of this motion, T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{L M}{P \alpha}}=2 \pi \sqrt{\frac{L \alpha M}{P \alpha^2}}=2 \pi \sqrt{\frac{V M}{P \alpha^2}} .\)

 

Simple Harmonic Motion Synopsis

Periodic Motion: Any motion, which repeats itself at regular intervals of time is called periodic motion.

Oscillation Or Vibration: If a particle executing periodic motion moves to and fro along the same path, the motion is called oscillation or vibration.

Complete Oscillation: If an oscillating particle starting from any point on its path towards a certain direction returns to the same point and then moves in the same direction, it is said to have executed a complete oscillation.

Time Period: The time period of oscillation of a vibrating particle (T) is defined as the time taken by it to execute one complete oscillation.

Frequency: The frequency of oscillation of a vibrating particle (n) is defined as the number of complete oscillations executed by it in 1 second.

Amplitude: The magnitude of the maximum displacement of a vibrating particle on either side of its position of equilibrium is called the amplitude (A) of vibration.

Phase: The state of the motion at any instant of a particle executing simple harmonic motion is called its phase.

Initial Phase Or Epoch: Epoch or phase constant is the phase of the particle executing simple harmonic motion at the initial instant, i.e., at t = 0.

Simple Harmonic Motion: If the acceleration of a vibrating particle is

1. Directly proportional to the displacement of the particle from the position of equilibrium and
2. Is always directed towards the equilibrium position, then the motion is called a simple harmonic motion.

• All simple harmonic motions are periodic motions but all periodic motions are not simple harmonic.
• If it is possible to oscillate a small but heavy body suspended from a rigid support by means of a long, weightless, and inextensible string, then that system is called a simple pendulum.
• The motion of a simple pendulum is simple harmonic in nature if the angular amplitude of oscillation of the pendulum is less than 4°.
• A simple pendulum that has a time period of 2s or a half time period of Is is called a seconds pendulum.

Simple Harmonic Motion Useful Relations For Solving Numerical Problems

For a SHM, F ∝-x or, F = -kx; [where, F = restoring force, k = force constant, x = displacement of the particle from its equilibrium position,]

a = -ω²x and ω = 2πn [where, a = acceleration of the particle, n = frequency, ω = angular frequency]

Differential Equation Of Simple Harmonic Motion: \(\frac{d^2 x}{d t^2}=-\omega^2\)

The General Equation For Displacement In A Simple Harmonic Motion: x = A sin(ωt+ α)

(where, A = amplitude, α = initial phase]

The velocity of a particle executing simple harmonic motion, v = \(\pm \omega \sqrt{A^2-x^2}\) maximum velocity v_max = ± Aω [where x = A]; minimum velocity v_min = 0 (where x = ±A)

Acceleration, a = -ω²x, maximum acceleration, a_max = ω²A [where x = ±A]; minimum acceleration, a_min = 0 (where x = 0)

Kinetic energy, K = \(\frac{1}{2} m v^2=\frac{1}{2} m \omega^2\left(A^2-x^2\right)\)

Potential energy, U = \(\frac{1}{2} m \omega^2 x^2\)

Total energy, E = \(K+U=\frac{1}{2} m \omega^2 A^2=\text { constant }\)

Time period, T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{x}{a}}=2 \pi \sqrt{\frac{\text { displacement }}{\text { acceleration }}}\)

Frequency, n = \(\frac{1}{T}\)

Time period of a simple pendulum oscillating at an angular amplitude less than 4° is T = \(2 \pi \sqrt{\frac{L}{g}}\)

[where, L = effective length of the pendulum = length from the point of suspension of the pendulum to the center of gravity of the bob.]

In case of oscillation of a mass attached to a vertical elastic spring, time period, T = \(2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{l}{g}}\)

[where, k = spring constant,  l = initial elongation of the spring due to the attachment of the mass m]

In case of oscillation of a mass attached to a horizontal elastic spring, time period, T = \(2 \pi \sqrt{\frac{m}{k}}\)

In case of oscillation of a liquid in a U-tube, time period, T = \(2 \pi \sqrt{\frac{L}{g}}\)

[Where L = length of the liquid column in each limb at equilibrium].

Simple Harmonic Motion Very Short Answer Type Questions

Question 1. If the time period of an SHM is 2 s, then what will be its frequency?
Answer: \(\frac{1}{2}\)Hz

Question 2. If the frequency of an SHM is 200 Hz, then what will be its time period?
Answer: \(\frac{1}{200}s\)

Question 3. What is the unit of force constant of SHM in SI?
Answer: N • m^-1

Question 4. The motion of the earth around the sun is a ______ motion.
Answer: Periodic

Question 5. What is the maximum displacement of a vibrating particle from the equilibrium position called?
Answer: Amplitude

Question 6. What is the phase difference between the displacement and the velocity of a particle executing SHM?
Answer: 90°

Question 7. What is the phase difference between the displacement and the acceleration of a particle executing SHM?
Answer: 180°

Question 8. If the time period is T then what will be the time taken by a particle executing SHM to traverse from the position of equilibrium to an extremity?
Answer: \(\frac{T}{4}\)

Question 9. What will be the change in the time period of a simple pendulum if the metallic bob of the pendulum is replaced by a wooden bob of the same radius? provided both bobs are of uniform density
Answer: Time Period Remains The same

Question 10. In which direction is the acceleration of a particle executing SHM directed?
Answer: Equilibrium

Question 11. At which points of its path, the velocity of a particle executing SHM becomes zero?
Answer: Extreme

Question 12. At which position, the velocity of a particle executing SHM becomes maximum?
Answer: Equilibrium

Question 13. At which position, the acceleration of a particle executing SHM become zero?
Answer: Equilibrium

Question 14. At which points of its path, the acceleration of a particle executing SHM becomes maximum?
Answer: Extreme

Question 15. A simple pendulum is oscillating in a vertical plane with a small amplitude. State whether the total energy at any point in its motion will be equal to that at an extreme point.
Answer: Yes

Question 16. At which position, the kinetic energy of a particle executing SHM becomes maximum?
Answer: Equilibrium

Question 17. At which points of its path, the potential energy of a particle executing SHM becomes maximum?
Answer: Extreme

Question 18. Total mechanical energy of a simple pendulum is directly proportional to the mass of the pendulum. Is the statement true or false?
Answer: True

Question 19. How is the total mechanical energy of a simple pendulum related to the length of the pendulum?
Answer: Inversely proportional

Question 20. Total mechanical energy of a simple pendulum is directly proportional to the amplitude of the pendulum. Is the statement true or false?
Answer: False

Question 21. The bob of a simple pendulum is made of iron. A powerful magnetic pole is placed below the bob in its equilibrium position. How will the time period of the pendulum change?
Answer: Decrease

Question 22. A body attached to a spring is executing SHM. If the force constant of the spring is increased then what will be the change in the frequency of oscillation?
Answer: Frequency will increase

Question 23. If a straight tunnel is bored from the north pole to the south pole of the earth and if a body is dropped into that tunnel then what time will the body take to move from one end to the other end of the tunnel?
Answer: 42 min

Question 24. What is the type of motion of a body along the tunnel passing through the centre of the earth?
Answer: Simple harmonic

Simple Harmonic Motion Assertion Reason Type Question And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: The total energy of a particle performing simple harmonic motion could be negative.

Statement 2: The potential energy of a system could be negative.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 2.

Statement 1: The spring constant of a spring is k. When it is divided into n equal parts, then the spring constant of each piece is k/n.

Statement 2: The spring constant is independent of the material used for the spring.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 3.

Statement 1: A particle performs a simple harmonic motion with amplitude A and angular frequency ω. To change the angular frequency of the simple harmonic motion to 3ω and amplitude to A/2, we have to supply an extra energy of (5/4) mω²A², where m is the mass of the particle executing simple harmonic motion.

Statement 2: The angular frequency of simple harmonic motion is independent of the amplitude of oscillation.

Answer: 4. Statement 1 is false, statement 2 is true.

Simple Harmonic Motion Match Column 1 With Column 2

Question 1. A particle of mass 2 kg is moving on a straight line under the action of force F = (8 – 2x)N. The particle is released from rest at x = 6m. For the subsequent motion match the following (all the values in Column 2 are in their SI units.)

Answer: 1. C, 2. D, 3. D, 4. B

Question 2. Two particles ‘ P ’ and ‘ Q ’ start SHM at t = 0. Their positions as a function of time are given by \(x_p=A \sin \omega t; \quad x_Q=A \sin \left(\omega t+\frac{\pi}{3}\right)\)

Answer: 1. B, 2. A, 3. D, 4. C

Simple Harmonic Motion Comprehension Type Question And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A block of mass m is connected to a spring of spring constant k as shown. The block is found at its equilibrium position at t = 14 and it has a velocity of 0.25 m · s^-1 at t = 2s. The time period of oscillation is 6s.

1. The amplitude of oscillation is

1. \(\frac{3}{2 \pi} \mathrm{m}\)
2. \(3 \mathrm{~m}\)
3. \(\frac{1}{\pi} \mathrm{m}\)
4. \(1.5 \mathrm{~m}\)

Answer: 1. \(\frac{3}{2 \pi} \mathrm{m}\)

2. Determine the velocity of a particle at t = 5s.

1. -0.4 m · s^-1
2. 0.5 m · s^-1
3. -0.25 m · s^-1
4. None of these.

Answer: 3. -0.25 m · s^-1

Question 2. Two identical blocks A and B, each of mass m = 3 kg, are connected with the help of an ideal spring and placed on a smooth horizontal surface as shown. Another identical block C moving with velocity v₀ = 0.6 m · s^-1 collides with A and sticks to it. As a result, the motion of the system takes place in some way.

1. After the collision of C and A, the combined body and block B would

1. Oscillate about the center of mass of the system and the center of mass is at rest
2. Oscillate about the center of mass of the system and the center of mass is moving
3. Oscillate but about different locations other than the center of mass
4. Not oscillate

Answer: 2. Oscillate about the center of mass of the system and the center of mass is moving

2. Oscillation energy of the system, i.e., part of the energy which is oscillating (changing) between potential and kinetic forms, is

1. 0.27 J
2. 0.09 J
3. 0.18 J
4. 0.45 J

Answer: 2. 0.09 J

3. The maximum compression of the spring is

1. 3√30 mm
2. 3√20 mm
3. 3√10 mm
4. 3√50 mm

Answer: 3. 3√10 mm

Question 3. A particle suspended from a vertical spring oscillates 10 times per second. At the highest point of oscillation, the spring becomes unstretched. Take g = π² m • s^-2

1. The maximum speed of the particle is

1. 5π cm · s^-1
2. 4π cm · s^-1
3. 3π cm · s^-1
4. 2π cm · s^-1

Answer: 1. 5π cm · s^-1

2. The speed of the particle when the spring is stretched by 0.2 cm is

1. 15.4 cm · s^-1
2. 12.8 cm · s^-1
3. 10.8 cm · s^-1
4. 11.4 cm · s^-1

Answer: 1. 15.4 cm · s^-1

Question 4. Two identical balls A and B each of mass 0.1 kg are attached to two identical massless springs. The spring-mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle as shown. The pipe is in the horizontal plane. The centers of the balls can move in a circle of radius 0.06m. Each spring has a natural length of 0.06πm and a spring constant of 0.1 N · m^-1. Initially, both the balls are displaced by an angle θ = \(\frac{\pi}{6}\) radian with respect to the diameter PQ of the circle.

1. The frequency of oscillation of ball B is

1. π HZ
2. π^-1 HZ
3. π² HZ
4. π^-2 Hz

Answer: 1. π HZ

2. Speed of ball A when A and B are at the two ends of the diameter PQ

1. 0.05 m · s^-1
2. 0.071 m · s^-1
3. 0.0628 m · s^-1
4. 0.083 m · s^-1

Answer: 3. 0.0628 m · s^-1

3. The total energy of the system is

1. 4 x 10^-4 J
2. 5 x 10^-3 J
3. 4 x 10^-3J
4. 5 x 10^-4 J

Answer: 1. 4 x 10^-4 J

Question 5. A man has an antique pendulum clock of 1832 which bears the signature of the purchaser. He does not want to replace it in the fond memory of his great-grandparents. It ticks off one second in each side-to-side swing. It keeps the correct time at 20 °C. The pendulum shaft is made of steel and its mass can be ignored as compared to the mass of the bob. The linear expansion coefficient of steel is 1.2x 10^{-5 °}C^-1

1. What is the fractional change in length if the shaft is cooled to 10 °C?

1. 0.01 %
2. 1.2 x 10^-1 %
3. 1.2 x 10^-3 %
4. 1.2 x 10^-4 %

Answer: 1. 1.2 x 10^-2 %

2. How many seconds will the clock gain or lose in a day at 10 ºC?

1. Gains 5.2 s
2. Loses 5.2 s
3. Gains 10.4 s
4. Loses 10.4 s

Answer: 1. Gains 5.2 s

3. How closely must the temperature be controlled so that it does not gain or lose more than a second in a day?

1. ± 0.2 °C
2. ± 0.1 °C
3. ± 1 °C
4. ± 2 °C

Answer: 4. ± 2 °C

4. The pendulum mentioned in the paragraph is called _____ and its time period is_______

1. Seconds pendulum, 1s
2. Seconds pendulum, 2s
3. 2 Second pendulum, 2s
4. None

Answer: 2. Seconds pendulum, 2s

Simple Harmonic Motion Integer Type Question And Answers

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9.

Question 1. If the displacement (x) and velocity (v) of a particle executing SHM are related through the expression 4v² = 25-x², what should be the value of (T/π)? [T is the time period (in second) of the SHM.]
Answer: 4

Question 2. Starting from the origin, a body oscillates simply harmonically with a period of 2s. After a certain time (t) its kinetic energy will be 75% of the total energy. What should be the value of 1/t (in s^-1)?
Answer: 6

Question 3. A particle executing SHM can be expressed by the equation x = 3cosωt+ 4sinωt. Find the amplitude of the resultant SHM.
Answer: 5

Question 4. Two pendulums of lengths 100 cm and 225 cm start oscillating in phase simultaneously. After how many oscillations will they again be in phase together?
Answer: 2

Question 5. At a certain temperature, the pendulum of a clock keeps the correct time. The coefficient of linear expansion for the pendulum material = 1.85 x 10^-5 K^-1. How much will the clock gain or lose in 24 h if the ambient temperature is 10 °C higher?
Answer: 8

 

 Viscosity And Surface Tension

The flow of fluids is of two types

1. Laminar or streamline flow and
2. Turbulent flow.

Laminar or Streamline Flow Definition: A smooth, uninterrupted flow in ordered layers, without any energy transfer between the layers, is called a laminar streamline or steady flow.

Mathematically speaking, for a one-dimensional fluid flow along the x- direction, the fluid velocity \(\vec{v}\) is a function of position x and time t, i.e., \(\vec{v}\) = f(x, t).

For a laminar or streamline motion, the fluid satisfies the condition that \(\vec{v}\) is a function of x only, and not of r. It means that, at any particular point along the fluid flow, the magnitude and the direction of the fluid velocity do not change with time, although the velocity may be different at different points. In short, \(\vec{v}\) = f(x), but \(\vec{v}\) ≠ f(t)

In Fig a laminar flow for a liquid is shown. At points A, B, C, and D, let the flow velocities at any instant be ν_A, ν_B, ν_c, and ν_D respectively. Also at any subsequent time, a liquid particle that reaches the point A will have the velocity ν_A similarly, at point B, the velocity will be ν_B at point C the velocity will be ν_C and at point D. the velocity will be ν_D. It means that each particle of the liquid follows the velocity of its preceding particle and moves along the same path.

Streamline: In the case of streamline motion, the paths along which the particles of the fluid move are called streamlines. A tangent drawn at any point on this path indicates the direction of motion of the fluid at that point.

Properties of streamlining:

1. Two streamlines never intersect each other. Otherwise, at the point of intersection of two streamlines, two tangents can be drawn and hence two directions of motion of the particle are possible. Hut, in streamlined motion, any particle can move in one direction only and hence two streamlines can never intersect.

2. In the flow tube, where the streamlines are crowded together, the velocity of flow is higher. Where they are spaced, the velocity of flow is lower.

A special case of streamlined flow is a steady flow, for which the fluid velocity is a constant at all points along this flow at all times. So this velocity in neither a function of time, not of position. Example: a sufficiently slow liquid flow along a narrow uniform horizontal tube.

Turbulent Flow: In general, the motion of a fluid is streamlined, if its velocity does not exceed a definite limiting value. The limiting value of velocity is called the critical velocity.

If the velocity of a fluid exceeds the critical velocity, then the flow becomes turbulent, and in some regions, eddies and vortices are formed. This kind of flow is called a turbulent flow.

Turbulent Flow Definition: If the velocity of a fluid along its flow continuously and randomly changes in magnitude and direction, then it is called a turbulent or disorderly flow.

The path of fluid particles in turbulent flow is shown in Fig At every point along the flow, both the magnitude and the direction of fluid velocity change with time.

Turbulent Flow Experiment: Reynolds demonstrated the difference between streamline and turbulent flow by a simple experiment. a discharge pipe Q is attached horizontally to a vertical water-filled cylinder P. The flow rate may be varied with a valve at the end of the pipe.

Understanding Viscosity in Fluids

• To make the flow behavior visible, we use the KMnO₄ solution, which is injected centrally into the horizontal pipe through a very narrow tube (diameter < 1 mm), positioned to avoid additional turbulence.
• If the velocity of the transparent liquid is low, the colored liquid is observed to travel continuously in the form of a thread, indicating a streamlined flow.
• As the flow of the transparent liquid is increased gradually, the colored thread gets disrupted and later on the colored liquid begins to move randomly, or forms eddies and vortices and mixes with the transparent liquid.
• The velocity of the transparent liquid at which this disturbance starts is called the critical velocity. It depends on the nature of the liquid, the cross-section of the tube, etc.

Viscosity And Surface Tension Viscosity: When a liquid flows slowly over a fixed horizontal surface, i.e., when the flow is laminar, the layer of the liquid in contact with the fixed surface remains at rest due to adhesion.

• The layer just above it moves slowly over the lower one, the third layer moves faster over the second one, and so on. The velocities of the layers of liquid increase with the increase in distance from the horizontal rigid surface.
• For two consecutive horizontal layers inside the liquid, the upper layer moves with a velocity greater than that of the lower one.
• The upper layer tends to accelerate the lower layer, while the lower layer tends to retard the upper one. In this way, the two adjacent layers tend to decrease their relative velocity—as if a tangential force acts on the upper layer and tries to oppose its motion.
• This tangential force is called viscous force. Therefore, to maintain a constant relative motion between the layers, an external force must act. If no external force acts, then the relative motion between the layers will cease and the flow of the liquid will stop.

WBBSE Class 11 Viscosity and Surface Tension Notes

Viscosity Definition: The property by which a liquid opposes the relative motion between its adjacent layers is called the viscosity of the liquid.

Comparison of viscosity with friction: Viscosity is a general property of a fluid. The frictional force acting between two solid surfaces resembles in many ways the viscosity of a liquid.

• Hence, viscosity is called the internal friction of a liquid. Like friction, the viscous force is absent if a liquid is at rest.
• The difference between the frictional force in solids and viscosity in liquids is that the viscous force depends on the area of the liquid surface while the frictional force does not.

Viscosity and mobility of different liquids: Viscosities of different liquids are different. If alcohol and oil are poured separately into two identical vessels and stirred, then oil will come to rest earlier. This shows that the viscosity of oil is greater.

Velocity profile: The surface formed by joining the end points of the velocity vectors of different layers of any section of a flowing liquid is called its velocity profile. The velocity profile for flow above a horizontal surface is shown in Fig.

Velocity profile of a non-viscous liquid: An ideal liquid is non-viscous. For such a liquid, there is no resistance due to viscosity. The velocities of the different layers are the same.

Every particle in a given cross-section of the liquid moves forward with the same velocity. On joining the ends of these velocity vectors, we get a plane surface. Therefore, we can say that the velocity profile of a non-viscous liquid is linear (on 2D graph).

Velocity profile of a viscous liquid: When a viscous liquid flows through a horizontal tube, the layer of liquid in contact with the wall of the tube remains stationary due to adhesion. So the velocity of that layer is zero.

• The layer of the liquid which flows along the axis of the tube has the maximum velocity. As we progress from the center towards the walls, the velocity decreases.
• Therefore, on joining the ends of the velocity vectors, we get a parabolic surface. The velocity profile of a viscous liquid is a parabola (on the 2D graph).

Coefficient of Viscosity: Let PQ be a solid horizontal surface. A liquid is in streamlined motion over the surface PQ. Two liquid surfaces CD and MN are at distances x and (x+dx) respectively from the fixed solid surface. The velocity of layer CD is ν and that of layer MN is ν+ dν.

Due to the viscosity of the liquid, an opposing force acts between these two layers and tries to slow down the relative motion of the layers. If this opposing viscous force is F, then for streamline motion of the liquid, Newton proved that

1. F ∝ A; A = area of cross-section of the liquid surface, and
2. \(F \propto \frac{d v}{d x}; \frac{d v}{d x}\) = velocity gradient = rate of change of velocity with distance perpendicular to the direction of flow.

∴ \(F \propto A \frac{d v}{d x} \text { or, } F=-\eta A \frac{d v}{d x}\) ….(1)

Here, η is a constant known as the coefficient of viscosity. Its value depends on the nature of the liquid.

Equation (1) is known as Newton’s formula for the streamlined flow of a viscous liquid. Liquids that obey this law are called Newtonian liquids and liquids that do not obey this law are called non-Newtonian liquids.

From equation (1), we get, \(\eta=\frac{F}{A \frac{d v}{d x}}\)

If A = 1 and \(\frac{d v}{d x}=1\), then η = F; from this, we can define the coefficient of viscosity.

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Coefficient of Viscosity Definition: The coefficient of viscosity of a liquid is defined as the required tangential force acting per unit area to maintain unit relative velocity between two liquid layers unit distance apart.

Units of coefficient of viscosity: \(\eta=\frac{F}{A \frac{d v}{d x}}=\frac{F d x}{A d \nu}\)

So, unit of \(\eta=\frac{\mathrm{N} \cdot \mathrm{m}}{\mathrm{m}^2 \cdot\left(\mathrm{m} \cdot \mathrm{s}^{-1}\right)}=\mathrm{N} \cdot \mathrm{s} \cdot \mathrm{m}^{-2}=\mathrm{Pa} \cdot \mathrm{s}\)

Unit:

• dyn · s · cm^-2 CGS System or g · cm^-1 · s^-1
• N · s · m^-2 or Pa · s or kg · m^-1 · s^-1 SI

Relation: \(1 \mathrm{~kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-1}=\frac{1 \mathrm{~kg}}{1 \mathrm{~m} \times 1 \mathrm{~s}}=\frac{1000 \mathrm{~g}}{100 \mathrm{~cm} \times 1 \mathrm{~s}}\)

= 10 g · cm^-1 · s^-1

Poise and decompose: The coefficient of viscosity of a liquid is 1 poise when a tangential force of 1 Dyn is required to maintain a relative velocity of 1 cm · s^-1 between two parallel layers of the liquid 1 cm apart where each layer has an area of 1 cm².

So, 1 poise is the CGS unit of the coefficient of viscosity η.

1 poise = 1 dyn • s • cm^-2 = 1 g • cm^-1 • s^-1.

As, 1 kg · m^-1 · s^-1 = 10g · cm^-1 • s^-1 = 10 poise,

the SI unit of η is called 1 decapoise = 10 poise.

The coefficient of viscosity of a liquid is 1 decompose when a tangential force of 1 newton is required to maintain a relative velocity of 1 m · s^-1 between two parallel layers separated by distance of 1 m, where each layer has an area of 1 m².

Dimension of coefficient of viscosity: \([\eta]=\frac{[\mathrm{F}]}{[\mathrm{A}]\left[\frac{d \nu}{d x}\right]}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2 \cdot \frac{\mathrm{LT}^{-1}}{\mathrm{~L}}}=\mathrm{ML}^{-1} \mathrm{~T}^{-1}\)

Effect of pressure and temperature on the coefficient of viscosity:

Effect of pressure: Usually, viscosity increases with pressure. In less viscous liquids, the viscosity increases at a low rate with pressure.

• However for highly viscous liquids, an increase in pressure results in a rapid rise in its viscosity. However, water behaves differently and, with an increase in pressure, its viscosity decreases.
• From the kinetic theory of gases, it is known that a change in pressure does not affect the viscosity of a gas. But for a large increase (or decrease) in pressure, viscosity is affected.

Effect of temperature: Usually, the coefficient of viscosity of liquids decreases with a temperature rise. The relation between temperature and coefficient of viscosity is rather complicated. One commonly used equation relating these two is

⇒ \(\eta_t=\frac{A}{(1+B t)^C}\)

where, η_t = coefficient of viscosity of a liquid at t°C and A, B, and C are constants for a particular fluid.

For gases, the coefficient of viscosity increases with an increase in temperature.

Critical Velocity and Reynolds Number

Critical velocity: On gradually increasing the velocity of a fluid, the streamline flow does not become turbulent abruptly. Rather this change occurs gradually.

With the help of experimental demonstration and also by dimensional analysis, it can be proved that the critical velocity (ν_c) of a fluid is

1. Inversely proportional to the density (ρ) of the fluid,
2. Directly proportional to the coefficient of viscosity (η) of the fluid, and
3. Inversely proportional to the characteristic length (l) of the channel. So,

⇒ \(v_c \propto \frac{\eta}{\rho l} \text { or, } v_c=N_c \cdot \frac{\eta}{\rho l}\) …..(1)

In the case of a tube, the characteristic length is the diameter of the tube while, for a canal, the characteristic length is its breadth.

If, for a liquid, ρ and η are known and its critical velocity ν_c can be determined experimentally during its flow through a tube of diameter l, then from equation (1), the value of the constant N_c for that liquid can be determined. This value is nearly 2300.

For any velocity ν of the fluid flow, equation (1) can also be written in an equivalent form as

⇒ \(v=N \cdot \frac{\eta}{\rho l} \text { or, } N=\frac{\rho l v}{\eta}\)…….(1)

N is called the Reynolds number.

Special cases:

1. If ν<ν_c, i.e., the velocity of fluid flow is less than the critical velocity, then comparing equations (1) and (2), we can say that N<N_c. It means that the value of the Reynolds number is less than 2300. So, if the value of the Reynolds number is less than 2300, then the flow will be streamlined.

2. On the other hand, if ν>ν_c, i.e., the velocity of the fluid is greater than the critical velocity, then N >N_c, and hence the value of the Reynolds number will be greater than 2300. If the Reynolds number is greater than 2300, then the flow will be turbulent.

Dimension of Reynolds number: From equation (2) we get, the dimension of N

= \(\frac{\text { dimension of } \rho \times \text { dimension of } l \times \text { dimension of } \nu}{\text { dimension of } \eta}\)

= \(\frac{M L^{-3} \cdot L \cdot L T^{-1}}{M L^{-1} T^{-1}}=1\)

So, N is a dimensionless quantity; it is a pure number.

Reynolds number: A dimensionless number N= \(\frac{\rho l v}{\eta}\) can be formed by combining the characteristic length (l) of a fluid channel and the velocity (v), density (ρ) and coefficient of viscosity (η) of the fluid the magnitude of N determines whether the fluid flow is streamlined or turbulent. This number N is called the Reynolds number.

• In the above discussion, 2300 is an approximate value of Nc. Usually, for N < 2000, the fluid flow is streamlined, and for N> 3000 the fluid flow is turbulent. If N lies between 2000 and 3000, the streamlined flow of fluid gradually changes into a turbulent flow.
• In the above discussion, 2300 is an approximate value of N_c. Usually, for N < 2000, the fluid flow is streamlined, and for N > 3000 the fluid flow is turbulent. If N lies between 2000 and 3000, the streamlined flow of fluid gradually changes into a turbulent flow.
• As N is a pure number, its value does not depend on the system of units chosen. For a particular flow, the value of N remains the same.

If the radius of a tube of flow is considered, instead of its diameter, then the effective value is N_c ≈ 1150.

Viscosity And Surface Tension Numerical Example

Example: A plate of area 100 cm2 is floating on an oil of depth 2 mm. The coefficient of viscosity of oil is 15.5 poise. What horizontal force is required to move the plate horizontally with a velocity of 3 cm · s^-1?
Solution:

The viscous force, F = \(\eta A \frac{d v}{d x}\)

Here, A = 100 cm², η = 15.5 poise,

dν = 3 cm · s^-1 and dx = 2 mm = 0.2 cm.

∴ F = 15.5 x 100 x 3/0.2 = 23250 dyn

So the required horizontal force is 23250 dyn.

Terminal Velocity of a Body in a Viscous Medium and Stokes’ Law: When a body falls through a viscous medium (liquid or gas), it drags a layer of the fluid adjacent to it due to adhesion. But fluid layers at a large distance from the body are at rest.

• As a result, there is relative motion between different layers of the fluid at different distances from the body. However the viscosity of the fluid opposes this relative motion.
• The opposing force due to viscosity increases with an increase in the velocity of the body due to the gravitational acceleration g. If the body is small in size, then after some time the opposing upward force (i.e., viscous force and buoyant force) becomes equal to the downward force (weight of the body).
• Then the effective force acting on the body becomes zero and the body begins to fall through the medium with a uniform velocity, called the terminal velocity. A graph representing the change in velocity of a falling object with time is shown in Fig.

Stokes’ law: Stokes proved that, if a small sphere of radius r is falling with a terminal velocity ν through a medium of coefficient of viscosity η, then the opposing force acting on the sphere due to viscosity is

F = 6 πηrν ………..(1)

Equation (1) expresses Stokes.

1. To establish Stokes’ law, the following assumptions are
   made.
2. The fluid medium must be infinite and homogeneous. E3D The sphere must be rigid with a smooth surface.
3. The sphere must not slip when falling through the medium.
4. The fluid motion adjacent to the falling sphere must be streamlined.
5. The sphere must be small in size, but it must be greater than the intermolecular distance of the medium.

The equation for terminal velocity: if the density of the material of the sphere is ρ, then the weight of the sphere = \(\frac{4}{3} \pi r^3 \rho g.\)

If the density of the fluid medium is σ, then the upward buoyant force acting on the sphere = \(\frac{4}{3} \pi r^3 \sigma g\)

∴ The resultant downward force acting on the sphere =

= \(\frac{4}{3} \pi r^3 \rho g-\frac{4}{3} \pi r^3 \sigma g=\frac{4}{3} \pi r^3(\rho-\sigma) g\)….(2)

If the sphere attains terminal velocity, then

⇒ \(6 \pi \eta r \nu=\frac{4}{3} \pi r^3(\rho-\sigma) g \text { or, } \nu=\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\) ….(3)

So, from equation (3), we see that the terminal velocity obeys the following rules.

1. Terminal velocity is directly proportional to the square of the radius of the sphere.
2. It is directly proportional to the difference of densities of the material of the sphere and that of the medium.
3. It is inversely proportional to the coefficient of viscosity of the medium.

If the density of the body is less than the density of the medium, i.e., ρ < σ, then it is clear that the terminal velocity becomes negative. Hence, the velocity of the body will be in the upward direction. For this reason, air or other gas bubbles move upwards through water.

Applications of Stokes’ law:

1. Falling of rain drops through air: Water vapour condenses on the particles suspended in air far above the ground to form tiny water droplets. The average radius of these tiny water droplets is 0.001 cm (approx.)

• Assuming the coefficient of viscosity of air as 1.8 x 10^-4  poise (approx.) the terminal velocity of these droplets is calculated as 1.2 cm · s^-1 (approx.) which is negligible. So, these water droplets float in the sky. Collectively these droplets form clouds.
• But as they coalesce to form larger drops, their terminal velocities increase. For example, the terminal velocity of a water droplet of radius 0.01cm becomes 120 cm · s^-1 (approx.). As a result, they cannot float any longer and so they come down as rain.

2. Coming down with the help of a parachute: When a soldier jumps from a flying airplane, he falls with acceleration due to gravity but due to viscous drag in air, the acceleration goes on decreasing till he acquires terminal velocity.

The soldier then descends with constant velocity and opens his parachute close to the ground at a pre-calculated moment, so that he may land safely near his destination.

Terminal Velocity Numerical Examples

Example 1. An oil drop of density 950 kg · m^-3 and radius 10^-6 m is falling through air. The density of air is 1.3 kg · m^-3 and its coefficient of viscosity is 181 x 10^-7 SI unit. Determine the terminal velocity of the oil drop, [g = 9.8 m · s^-2]
Solution:

Terminal velocity, \(\nu=\frac{2}{9} \cdot \frac{r^2(\rho-\sigma) g}{\eta}\)

[Here, ρ = 950 kg · m^-3 ; r = 10^-6 m; σ = 1.3 kg · m^-3; η = 181 x 10^-7 SI]

= \(\frac{2}{9} \cdot \frac{\left(10^{-6}\right)^2(950-1.3) \times 9.8}{181 \times 10^{-7}}\)

= \(1.14 \times 10^{-4} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Example 2. An air bubble of radius 1 cm is rising from the bottom of a long liquid column. If its terminal velocity is 0.21 cm · s^-1, calculate the coefficient of viscosity of the liquid. Given that the density of the liquid is 1.47 g · cm^-3. Ignore the density of air.
Solution:

Coefficient of viscosity of the liquid, \(\eta=\frac{2}{9} \cdot \frac{r^2(\rho-\sigma) g}{v}\)

[Here, r = 1cm; v = -0.21 cm · s^-1; ρ = 0; cσ = 1.41 g · cm^-3]

= \(\frac{2}{9} \times \frac{(1)^2(0-1.47) \times 980}{-0.21}=1524.4 \text { poise. }\)

 

Viscosity And Surface Tension Surface Tension Of Liquids

All liquids possess a special property—a tendency to minimize its surface area. This tendency of a liquid surface to contract its area is called surface tension. From our practical experience, we know that water droplets, or a small amount of mercury always take the shape of a sphere.

• In the absence of external forces, all liquids always take a spherical shape. For a given volume, the surface area of a sphere is the least and hence a liquid drop has a natural tendency to take the shape of a sphere.
• If a clean dry needle is placed horizontally on the surface of water, then it is observed that the needle floats on water. The water surface under the needle is slightly depressed.
• Insects like spiders and mosquitoes can walk on the surface of water. Where their legs touch, the water surface becomes slightly depressed.
• From such observations, it seems that the surface of water behaves like a stretched rubber membrane.

Experimental demonstration: A wire loop is dipped into a soap solution. A thin soap film will be formed in the loop when it is taken out of the solution. This film acts as the free surface of the liquid.

A wet cotton thread loop, after being dipped in soap solution, is put on the film. No change is seen in the shape of the loop. The soap film inside the loop is now punctured with a fine needle and it is observed that the cotton thread pulls itself into a circle. What is the reason behind this observation?

• Initially, there was soap film inside and outside the cotton loop. Every point on the loop experienced equal and opposite forces tangential to the surface of the film. These two forces balanced each other. As a result, no resultant force acted on the loop.
• After the film inside the loop was removed, the inward force vanished and only the film outside the loop exerted a force on the thread. We know that among all plane surfaces having the same boundary length, the area of a circle is the greatest.
• Hence, a circle formed by the loop occupies the maximum area. So, the area of the film in between the loop and the thread reduces to a minimum. From this, it can be inferred clearly that the film tends to minimize its area.
• It can be concluded that tension always acts on the free surface of a liquid and that the free surface behaves as a stretched-thin membrane. Due to this tension, the free surface of any liquid tends to contract to occupy the minimum area. This tension is known as surface tension.

Surface Tension Explained

Surface tension: Surface tension is the property of the free surface of a liquid due to which the liquid behaves as a stretched-thin membrane and has a tendency to contract to minimize the surface area.

• As a reason behind the origin of surface tension, it can be said that the molecules of a liquid attract each other by cohesive force [the force of attraction which acts between the molecules of the same material is called cohesive force).
• The equal cohesive force acts on the molecule inside the liquid from all directions.
• Consequently, the resultant cohesive force on the molecule is zero. However, no cohesive forces act on the molecules of the free surface of the liquid in the outward direction. So, the cohesive force inside the liquid is not balanced.
• As a result, a resultant cohesive force acts on each molecule of the free surface in the downward direction. Thus, the free surface of a liquid tends to have the least surface area.
• Let us imagine a straight line on the free surface of a liquid. Due to the tendency of the liquid surface to contract, the molecules on the opposite sides of the line try to move away from each other. This can be seen in the following experiment.

If a matchstick is placed on a water surface, it remains at rest. But when a drop of alcohol is put on water on one side of the stick, the stick moves in the opposite direction.

1. The water surface exerts an equal pressure on both sides of the stick. This force is normal to the stick.

2. When a drop of alcohol is put on water, the force on that side is weaker and the stick is pulled away by the stronger tangential force towards the opposite side.

Hence, surface tension can also be defined as follows:

Surface tension Definition: The tangential force per unit length on a liquid surface, that acts along the normal on either side of an imaginary line on that surface, is called the surface tension of the liquid.

Units:

• dyn · cm^-1 CGS System
• N · m SI

Relation: \(1 \mathrm{~N} \cdot \mathrm{m}^{-1}=\frac{1 \mathrm{~N}}{1 \mathrm{~m}}=\frac{10^5 \mathrm{dyn}}{10^2 \mathrm{~cm}}=10^3 \mathrm{dyn} \cdot \mathrm{cm}^{-1}\)

Dimensional: \([\text { Surface tension }]=\frac{[\text { force }]}{\text { [length }]}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}}=\mathrm{MT}^{-2}\)

Surface Energy: We know that on the free surface of a liquid, the surface tension always tries to minimize the surface area. So, to increase the area of the surface of the liquid, an external force is needed.

• The external force does work to increase the area of the surface of the liquid, and the work done remains stored inside the surface of the liquid as potential energy.
• The surface energy of a liquid is measured by the work done to increase the area of the surface of a liquid by unity.

Units:

• erg · cm^-2 CGS System
• J · m^-2 SI

Relation: \(1 \mathrm{~J} \cdot \mathrm{m}^{-2}=\frac{1 \mathrm{~J}}{1 \mathrm{~m}^2}=\frac{10^7 \mathrm{erg}}{10^4 \mathrm{~cm}^2}=10^3 \mathrm{erg} \cdot \mathrm{cm}^{-2}\)

Dimension: \([\text { Surface energy }]=\frac{[\text { work }]}{[\text { area }]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^2}=\mathrm{MT}^{-2}\)

Relation between surface tension and surface energy: A rectangular wire frame PQRS is taken. A wire AB can move along PQ and SR.

When the frame is dipped into a soap solution and taken out, a thin film is formed within the frame.

As a result, the surface tension acts normally on the wire AB and tangentially to the surface of the film. This force tries to contract the film surface and pulls the wire AB towards QR. To keep the wire AB in its position, an equal but opposite force needs to be applied on it.

Let the length of the wire AB be l; the surface tension of the liquid be T.

∴ The net force acting on the wire AB in the direction QR = 2lT. [The film has two surfaces and surface tension acts on each surface, therefore the net force has a factor of 2]

∴ To keep the wire AB still, the force required to be applied in the opposite direction, F = 2lT

Now, the amount of work done in displacing the wire AB through a short distance δx against the surface tension (assuming the force F to be a constant throughout the displacement) so that it comes to the new position A’B’ is

Fδx = 2lTδx

Due to this, the total increase in the area of the film surface = 2 lδx.

This work remains stored as potential energy on the film surface.

∴ Work done for unit increase in area against the surface tension = \(\frac{2 l T \delta x}{2 l \delta x}=\) = T

So, the potential energy stored per unit area or the surface energy is numerically equal to the surface tension of the liquid. Note that the temperature is assumed to be constant.

Alternative definition of surface tension: Keeping the temperature constant, the amount of work done in increasing the area of a liquid surface by unity is called the surface tension of that liquid at that temperature.

Units:

• erg · cm^-2 CGS System
• J · m^-2 SI Units

Dimension: According to the alternative definition,

⇒ \([\text { surface tension }]=\frac{[\text { work }]}{[\text { area }]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^2}=\mathrm{MT}^{-2}\)

Total surface energy: in the discussion, it was assumed that, during increase in the area of a liquid surface under the influence of an applied force, the temperature remains constant. But, in practice, some molecules from inside the liquid rise to its surface during the expansion of the surface.

• A resultant attraction exerted by the other molecules inside the liquid acts on these moving molecules. Therefore, these molecules on reaching the surface lose their linear kinetic energy and the average linear kinetic energy of the total liquid decreases.
• Since the temperature is directly proportional to the average kinetic energy, the temperature of the surface of the liquid decreases with an increase in its area. To keep the temperature constant, the liquid surface absorbs heat from its surroundings.

To increase the area of a liquid surface keeping the temperature constant, energy may be supplied in two ways

1. Mechanical energy to increase the surface area and
2. Heat energy to keep the temperature constant. The total of these two energies should actually be the surface energy.

So, the increase in potential energy per unit surface area or stored surface energy (E) = mechanical energy or work done (T) + heat (h) required for unit area

i.e., E = T+h……(1)

From thermodynamics, it can be proved that, h = \(-\theta \frac{d T}{d \theta}\)

[θ = temperature in absolute scale and dT/dθ = rate of increase in surface tension due to increase in temperature]

∴ E = \(T-\theta \frac{d T}{d \theta}\)…..(2)

Now, with the increase in temperature, the surface tension decreases and hence \(-\theta \frac{d T}{d \theta}\) is a negative quantity.

So, h is a positive quantity.

∴ E = \(T+\theta \frac{d T}{d \theta}\)…..(3)

when only the magnitudes are considered.

Again, at the absolute zero temperature, i.e., when θ = 0, E = T.

So, at any temperature except absolute zero, the total surface energy of a liquid is always greater than the surface tension of that liquid.

Factors Affecting Surface Tension of a Liquid: Surface tension of a liquid depends on the following factors.

1. Temperature of the liquid: With an increase in the temperature, the surface tension of almost all liquids decreases. For a small change in temperature, the relation between surface tension and temperature is

T’ = \(T\left[1-\alpha\left(t^{\prime}-t\right)\right]\)

Here, T and T’ are the surface tensions at temperatures t and t’ respectively]

For a given liquid, α is a constant quantity. It is called the temperature coefficient of surface tension.

It is experimentally verified that at a specific temperature of every liquid, the surface tension of the liquid disappears. This temperature is called the critical temperature of that liquid.

2. Pollution: If impurities are present on a liquid surface, then the surface tension of that liquid usually decreases. For example, when an oil or a fat-like substance is poured over water, it forms a thin film over the surface of water. This decreases the original surface tension of water.

3. Presence of dissolved substances: if a liquid contains dissolved inorganic substances, then the surface tension of that liquid increases. Again, if the liquid contains dissolved organic matter, then its surface tension decreases. For example, the surface tension of pure water is 0.072 N · m^-1.

If common salt (inorganic substance) is dissolved in water, then its surface tension becomes 0.083 N · m^-1 (approx.), but the surface tension of soap-water (organic substance) is approximately 0.030 N · m^-1.

4. Medium above the liquid surface: The surface tension of a liquid depends on the nature of the medium above the free surface of that liquid. For example, the surface tension of water is about 72 dyn · cm^-1 in the presence of dry air above the surface of water, but is about 70 dyn · cm^-1 when there is moist air above the surface of water at the same temperature.

5. Presence of electric charge: The surface tension of a liquid decreases due to the presence of electric charge on the surface of the liquid.

Some Phenomena in Connection with Surface Tension

1. Camphor darts to and fro when put on the Surface Of water: Camphor is soluble in water. When put on water, the portion that comes into contact with the water begins to dissolve. The part which gets dissolved in water contaminates the water and the surface tension of that part decreases. Due to this difference in surface tension, a net unbalanced force acts on the piece of camphor, and consequently, the piece of camphor darts to and fro on the surface of water.

2. Hair of a paint brush cling together when the brush is brought out of water: The hair of a brush lie apart while immersed in water because, inside water, the surface tension is absent. But when the brush is brought out of the water, a thin film of water clings to the hair, and the surface tension tries to contract the area of the film hence the hair clings together.

3. Turbulent sea calms down If oil is poured on the water: The surface tension of pure water is more than that of oily water. When oil is poured over sea water, the oil spreads in the direction of motion of the waves leaving uncovered sea water at the rear. Hence, the surface tension of the water ahead of the waves is lower than that of the water behind the waves. The water at the rear pulls the water at the front and, as a result, high waves become lower.

4. When oil is poured on water, it spreads readily Over the entire surface: Since the surface tension of pure water is greater than the surface tension of oil, a tensile force acts on the surface of oil. Due to this tensile force, oil spreads readily over the entire surface of water.

5. When chalk dust is sprinkled on water and a few drops of alcohol is added, then the dust particles rapidly spread on the water surface: Alcohol decreases the surface tension. Due to unequal surface tension on different parts of the water surface, the chalk particles spread rapidly on the surface of water.

6. Water cannot seep in through the cloth of raincoats, umbrellas and tents: The minute pores in the cloth of raincoats etc. trap air molecules. However, these pores are too small to let rain droplets enter, because the droplets retain their spherical shape due to surface tension, and the diameters of the spheres are greater than that of the pores. So, the rainwater falling on the cloth simply flows off.

7. A needle Coin float on water surface: A needle floats due to the surface tension of water. The surface of water where the needle is placed experiences a slight depression due to the surface tension of water. So, the water exerts an upward force on the needle which balances its weight (acting downwards). Therefore, a needle can float if it is placed carefully on a calm water surface.

Viscosity And Surface Tension Surface Tension Of Liquids Numerical Examples

Example 1. The surface tension of water at 20 °C is 72 dyn · cm^-1 and for water dT/dθ = -0.146 dyn · cm^-1 · K^-1. What is the total surface energy of water?
Solution:

We know that the total surface energy of water,

E = T – θ dT/dθ = 72 – 293 x (-0.146) [20°C = 293 K]

= 72 + 42.778 = 114.778 erg · cm^-2.

Example 2. A drop of water of radius 1 mm is to be divided into 10⁶ point drops of equal size. How much mechanical work should be done? The surface tension of water = 72 dyn · cm^-1.
Solution:

Let the radius of each point drop be r

∴ \(\frac{4}{3} \pi r^3 \times 10^6=\frac{4}{3} \pi\left(\frac{1}{10}\right)^3 \quad \text { or, } r=0.001 \mathrm{~cm}\)

The surface area of the original drop = \(4 \pi\left(\frac{1}{10}\right)^2 \mathrm{~cm}^2\)

and the total surface area of 10⁶ point drops = 10⁶ x 4π(0.001)² = 4π cm².

∴ Increase in surface area = \(4 \pi-4 \pi\left(\frac{1}{10}\right)^2=4 \pi \times 0.99 \mathrm{~cm}^2\)

= \(4 \pi-4 \pi\left(\frac{1}{10}\right)^2=4 \pi \times 0.99 \mathrm{~cm}^2\)

∴ Mechanical work done = increase in surface area x surface tension

= 72 x 4π x 0.99 = 895.73 erg

Mathematical Problems for Viscosity and Surface Tension

Example 3. 1000 water droplets having a radius of 0.01 cm each coalesce to form a single big drop. What will be the decrease in energy? The surface tension of water = 72 dyn · cm^-1
Solution:

Let the radius of the single big drop be R.

∴ \(\frac{4}{3} \pi R^3=\frac{4}{3} \pi(0.01)^3 \times 1000 \quad \text { or, } R=0.1 \mathrm{~cm}\)

Surface area of the big drop = 4π (0.1)² cm²

Total surface area of 1000 droplets

= 4π (0.01)² x 1000 cm²

∴ Decrease in area

= 4π (0.01)² x 1000-4π(0.1)²

= 4π(0.1 -0.01) = 47 x 0.09 cm³

∴ Decrease in energy = 4π x 0.09 x 72 = 81.43 erg.

Example 4. A rectangular glass slab measures 0.1 m x 0.0154 m x 0.002 m and its weight in air is 80.36 x 10^-3 N. The slab is immersed half in water keeping its length and thickness horizontal. What will be the apparent weight of the slab? The surface tension of water is 72 x 10^-3 N · m^-1.
Solution:

While it is immersed, the following forces act on the glass slab

1. Weight of the slab acting downwards,
2. Upward buoyant force due to the weight of displaced water and
3. Downward force due to surface tension.

Now, weight of the slab = 80.36 x 10^-3 N

Buoyant force = weight of displaced water

= \(0.1 \times \frac{0.0154}{2} \times 0.002 \times 1000 \times 9.8\)

= 15.092 x 10^-3 N

Force due to surface tension

= 2 x (0.1 + 0.002) x 72 x 10^-3 N

= 14.688 x 10^-3 N

∴ The apparent weight of the slab

= 80.36 x 10^-3 – 15.092 x 10^-3 + 14.688 x 10^-3

= 79.956 x 10^-3 N.

Example 5. The radius of a soap bubble is increased from 1 cm to 3 cm. What amount of work is done for this? The surface tension of soap-water is 26 dyn • cm^-1•
Solution:

Work done = increase in area x surface tension =

47{(3)²-(1)²} x 26 x 2 [the soap-bubble has two surfaces]

= 5227.6 erg = 5.2276 x 10^-4 J.

Example 6. Determine the surface energy of a liquid film formed on a ring of area 0.15 m². The surface tension of the liquid = 5 N · m^-1.
Solution:

Surface energy, E = 2 x surface tension x area =2 x 5 x 0.15= 1.5 J

Short Answer Questions on Viscosity and Surface Tension

Example 7. Determine the surface energy of a soap-water film formed on a frame of area 10^-3 m². Surface tension of soap-water = 70 x 10^-3 N · m^-1.
Solution:

Surface energy = 2 x surface tension x area

= 2 x 70 x 10^-3 x 10^-3 = 14 x 10^-5 J.

Example 8. What will be the work done to form a soap bubble of radius 5 cm? The surface tension of soap-water = 70 dyn · cm^-1
Solution:

Work done =2 x 4πr² x T =8πr² T

[r = radius of the soap-bubble and T = surface tension of soap-water]

= 8 x 22/7 x (5)² x 70 = 44000 erg

= 0.0044 J.

Example 9.  If a large number of water droplets of diameter 2rcm each coalesce to form a large water drop of diameter 2jRcm, then prove that the rise in tem-perature of water is \(\frac{3 T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)\). Here, T is the surface tension of water and J is the mechanical equivalent of heat.
Solution:

If the number of small water droplets is n, then the dissipation of surface energy, W = (4πr²n-4πR²)T.

We know that W = JH and H = heat absorbed = msθ

[where m = mass of the larger water drop, s = specific heat of water, 6θ = temperature increase]

H = \(\frac{W}{J}=\frac{T}{J} \times 4 \pi\left(n r^2-R^2\right)\)

or, \(\frac{4}{3} \pi R^3 \cdot 1 \cdot \theta=\frac{T}{J} \times 4 \pi\left(n r^2-R^2\right)\)

⇒ \({\left[because m=\frac{4}{3} \pi R^3 \cdot 1=\frac{4}{3} \pi R^3 \text { and } s=1 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}-1\right]}\)

or, \(\theta= \frac{3 T}{J} \frac{\left(n r^2-R^2\right)}{R^3}\)

⇒ \({\left[\text { here, } n \cdot \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3 \text { or, } R^3=n r^3 \text { or, } n=\frac{R^3}{r^3}\right]}\)

= \(\frac{3 T}{J}\left[\frac{R^3}{r^3} \cdot \frac{r^2}{R^3}-\frac{1}{R}\right]\)

= \(\frac{3 T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)\)

Example 10. Water is filled upto a height h in a beaker of radius R as shown. The density of water is p, the surface tension of water is T and the atmospheric pressure is P₀. Consider a vertical section ABCD of the water column through a diameter of the beaker. What is the force on water on one side of this section by water on the other side of this section?

Solution:

The force acted on the liquid of left side by the liquid of right side is equal to the resultant of the following two forces.

1. Force due to surface tension =2RT (right side).

2. Impulse due to the exerted pressure by the liquid of height h

= \(\left(P_0+\frac{\rho g h}{2}\right) \times 2 R h\) [where \(\frac{\rho g h}{2}\)= average pressure on the plane A B C D]

= \(2 P_0 R h+\dot{R} \rho g h^2\) (along left side)

∴ The resultant force on the plane ABCD = \(2 P_0 R h+R \rho g h^2-2 R T\)

Example 11. When water in a beaker is gradually heated, a bubble formed at the lower surface of the beaker starts to rise up from the bottom of the beaker. The radius of the spherical bubble is R and the radius of the circular region of the bubble touched with the lower surface of the container is r(r<<R). Show that, the value of r will be \(R^2 \sqrt{\frac{2 \rho_u g}{3 T}}\) just before the detached from the lower surface of the container [where, ρ_w = density of water, T = surface tension of water]. Consider, that though the density and surface tension are unchanged with the increase in temperature, the density of air changes significantly.

Solution:

At that instant when the bubble is just detached from the lower surface of the beaker, the buoyancy force = the force due to surface tension.

Now, the net force due to surface tension = (2πr)T sinθ (directed downward)

and the buoyancy force = \(\frac{4}{3} \pi R^3 \rho_w g\) (directed upward)

In this case, \(\frac{4}{3} \pi R^3 \rho_w g=(T)(2 \pi r) \sin \theta\)

[as θ is very small, sin/θ ≈ tanθ = r/R]

or, \(\frac{4}{3} \pi R^3 \rho_w g=(T)(2 \pi r)\left(\frac{r}{R}\right)\)

or, \(\frac{4}{3} \pi R^3 \rho_w g=T \cdot 2 \pi \frac{r^2}{R}\)

or, \(r^2=\frac{2\left(R^4 \rho_w g\right)}{3 T}\)

∴ r = \(R^2 \sqrt{\frac{2 \rho_w g}{3 T}}\)

Pressure Difference Between The Two Sides Of A Curved Liquid Surface

1. Suppose the free surface of a liquid is a plane. A molecule lying on its surface is attracted by other surface molecules equally in all directions. So the resultant tangential force on the molecule is zero.

2. If the free surface of the liquid is concave, then every molecule on the surface experiences an upward resultant force due to attraction by other surface molecules.

3. If the liquid surface is convex, then the resultant force on a molecule on the surface due to attraction by other surface molecules will be directed downwards.

• Obviously, there must be a difference of pressure between the two sides of a curved surface for equilibrium of it. This difference of pressure i.e., the excess pressure force will balance the resultant force due to surface tension. The pressure on the concave side must be greater than the pressure on the convex side.
• We shall now calculate the excess pressure on the concave side of spherical surfaces in case of a liquid drop, an air bubble in a liquid, and a soap bubble.

Excess pressure inside a liquid drop: Let us consider a liquid drop of radius R of a liquid of surface tension T. Every molecule on its surface experiences a resultant pull normally inwards due to surface tension.

So the internal pressure of the drop becomes greater than the pressure outside it. The internal excess pressure of the drop produces a force acting outwards which balances the force due to surface tension and maintains the equilibrium of the drop.

Suppose, external pressure on the drop =P, internal pressure of the drop = (P+ p).

So, the excess pressure inside the drop = p.

Suppose this internal excess pressure acting normally outwards increases the radius of the drop from R to R + ΔR i.e., it increases the surface area of the drop. Here ΔR is taken to be so small that the pressure inside the drop may be taken as unchanged.

Work done by the excess pressure,

W = Excess pressure x area x displacement

= p · 4πR² · ΔR ………(1)

Increase of surface area of the liquid drop,

⇒ \(\Delta A =4 \pi(R+\Delta R)^2-4 \pi R^2\)

= \(4 \pi\left\{R^2+2 R \cdot \Delta R+(\Delta R)^2-R^2\right\}\)

= \(8 \pi R \cdot \Delta R\) ; [neglecting the term \((\Delta R)^2\) which is very small]

∴ Increase in surface energy,

E = increase in surface area x surface tension

= 8πR · ΔR · T …….(2)

This increase in surface energy of the liquid drop takes place at the cost of work done by the excess pressure i.e., E = W.

So, from equation (1) and (2) we have, p · 4πR² · ΔR = 8π R ⋅ ΔR · T

or, p = 2T/R …..(3)

Excess pressure inside an air bubble in a liquid: Let us consider an air bubble of radius R formed in a liquid of surface tension T. Like a liquid drop the air bubble has also one surface in contact with the liquid.

So proceeding similarly as in the case of a liquid drop we can prove that the excess pressure inside the air bubble in a liquid is given by p = 2T/R.

Excess pressure inside a soap bubble: Let us consider a thin soap bubble of radius R formed from a soap solution of surface tension T.

Suppose, the pressure outside the bubble = P, internal pressure =(P+p)

So, the excess pressure inside the bubble = p. Suppose, the radius of the bubble increases from R to R + ΔR due to this internal excess pressure acting normally outwards, i.e., the surface area of the bubble increases.

Here ΔR is taken to be so small that the pressure inside the bubble may be taken as unchanged.

Work done by the excess pressure,

W = excess pressure x area x displacement = p · 4πR² · ΔR …..(1)

The soap bubble has two liquid surfaces in contact with air, one inside the bubble and the other outside the bubble.

So, increase of surface area of the soap bubble,

⇒ \(\Delta A =2\left[4 \pi(R+\Delta R)^2-4 \pi R^2\right]\)

= \(8 \pi\left[R^2+2 R \cdot \Delta R+(\Delta R)^2-R^2\right]\)

= \(16 \pi R \cdot \Delta R\) ; [neglecting the term] \((\Delta R)^2\) which is very small]

∴ Increase in surface energy,

E = increase in surface area x surface tension

= 16 πR · ΔR · T……….(2)

This increase in surface energy of the soap bubble takes place at the cost of work done by the excess pressure i.e., E = W

So, from equations (1) and (2) we have,

⇒ \(p \cdot 4 \pi R^2 \cdot \Delta R=16 \pi R \cdot \Delta R \cdot T\)

p = \(\frac{4 T}{R}\)

Viscosity And Surface Tension Bubble  Numerical Examples

Example 1. Find the excess pressure inside a rainwater drop of diameter 0.02cm. The surface tension of water = 0.072 N · m^-1.
Solution:

Water drop has only one curved surface.

So, excess pressure of a water drop, p =2T/r where, T = surface tension of water

r = radius of a water drop = 0.002/2 = 0.01 cm = 0.01 x 10m

∴ p = \(=\frac{2 \times 0.072}{0.01 \times 10^{-2}}=1440 \mathrm{~N} \cdot \mathrm{m}^{-2}\)

Example 2. Surface tension of soap solution = 27 dyn • cm^-1. Calculate the excess pressure (in N • m^-2) inside a soap bubble of radius 3 cm.
Solution:

The excess pressure inside a soap bubble,

p = \(\frac{4 T}{r}=\frac{4 \times 27}{3}=36 \mathrm{dyn} \cdot \mathrm{cm}^{-2}=3.6 \mathrm{~N} \cdot \mathrm{m}^{-2} \text {. }\)

Example 3. Find the pressure inside an air bubble of radius 0.1mm just inside the surface of water. Surface tension of water = 72 dyn · cm^-1
Solution:

Excess pressure inside an air bubble

= \(\frac{2 T}{r}=\frac{2 \times 72}{0.01}\)

[T = 72 dyn · cm^-1 and r = 0.1 mm = 0.01 cm] = 14400 dyn · cm^-2.

Atmospheric pressure = 76 x 13.6 x 980 dyn · cm^-2

∴ Total pressure inside an air bubble = (76 x 13.6 x 980 + 14400)

= 1.0274 x 106 dyn · cm^-2.

Real-Life Examples of Surface Tension Effects

Example 4. The excess pressure inside a soap bubble of radius 8mm raises the height of an oil column by 2mm. Find the surface tension of the soap solution. Density of the oil = 0.8 g · cm^-3.
Solution:

Excess pressure in a soap bubble (p) = 4πr.

Again, p = hρg, h = height of the oil column.

Now, \(\frac{4 T}{r}=h \rho g \text { or, } T=\frac{h \rho g \times r}{4}=\frac{0.2 \times 0.8 \times 980 \times 0.8}{4}\)

[h = 2mm = 0.2cm; ρ = 0.8 g • cm^-3, g = 980cm · s^-2 and r = 8mm = 0.8cm],

∴ T = 31.36 dyn · cm^-1

Example 5. In an isothermal process, two soap bubbles of radii a and b combine and form a bubble of radius c. If the external pressure is p, then prove that the surface tension of the soap solution is \(T=\frac{p\left(c^3-a^3-b^3\right)}{4\left(a^2+b^2-c^2\right)}\)
Solution:

We know, the excess pressure inside the soap bub¬ble = internal pressure – external pressure.

∴ For the bubble of radius a, excess pressure, \(\frac{4 T}{a}=p_a-p\)

∴ \(p_a=\left(p+\frac{4 T}{a}\right)\)

Similarly, for the bubble of radius b, \(p_b=\left(p+\frac{4 T}{b}\right)\)

For the bubble of radius c, \(p_c=\left(p+\frac{4 T}{c}\right)\)

Boyle’s law is applicable in isothermal process.

According to this law, \(p_a V_a+p_b V_b=p_c V_c\)

or, \(\left(p+\frac{4 T}{a}\right) \times \frac{4}{3} \pi a^3+\left(p+\frac{4 T}{b}\right) \times \frac{4}{3} \pi b^3\)

= \(\left(p+\frac{4 T}{c}\right) \times \frac{4}{3} \pi c^3\)

or, \(\left(p+\frac{4 T}{a}\right) a^3+\left(p+\frac{4 T}{b}\right) b^3=\left(p+\frac{4 T}{c}\right) c^3\)

or, \(4 T\left(a^2+b^2-c^2\right)=p\left(c^3-a^3-b^3\right)\)

∴ T = \(\frac{p\left(c^3-a^3-b^3\right)}{4\left(a^2+b^2-c^2\right)} .\)

Example 6. Two soap bubbles of radii 0.04 m and 0.03 m are combined in such a way that a common surface is formed between the two bubbles. What is the radius of curvature of the common surface?
Solution:

Let, the radii of the two soap bubbles are r₁ and r₂, and the internal pressures are p₁ and p₂ respectively.

The radius of the common surface = r, atmospheric pressure = p₀

For the first bubble, \(p_1-p_0=\frac{4 T}{r_1}\)

and for the second bubble, \(p_2-p_0=\frac{4 T}{r_2}\)

where, T = surface tension of soap solution.

Subtracting (1) from (2) we get,

⇒ \(p_2-p_1 = 4 T\left(\frac{1}{r_2}-\frac{1}{r_1}\right)=4 T\left(\frac{1}{0.03}-\frac{1}{0.04}\right)\)

= \(\frac{4 \times 100}{12} \times T\) ……(3)

But, for the common surface, \(p_2-p_1=\frac{4 T}{r}\)……..(4)

Comparing (3) and (4) we get, \(\frac{4 T}{r}=\frac{4 \times 100}{12} \times T\)

∴ r = 0.12 m.

 

Viscosity And Surface Tension Conclusion

Streamline flow: A smooth, uninterrupted flow of fluid in ordered layers, without any energy transfer between the layers, is called a laminar or streamline flow.

Streamline: In a smooth flow, the path along which any fluid particle moves is called a streamline.

Turbulent flow: If the velocity of a fluid along its flow continuously and randomly changes in magnitude and direction then it is called a turbulent or disorderly flow.

Viscosity: The property by virtue of which a liquid resists the relative motion between its adjacent layers is called viscosity of that liquid.

Velocity profile: The surface obtained by joining the terminal points of the velocity vectors of different layers of a flowing liquid at any section of it is called its velocity profile.

• In case of flow of a non-viscous liquid along a tube, the velocity profile becomes flat
• In case of flow of a viscous liquid along a tube, the velocity profile becomes parabolic.

Velocity gradient: In a horizontal streamline flow, the rate of change of velocity with distance \(\left(\frac{d u}{d x}\right)\) in a direction perpendicular to the flow of the liquid is called the velocity gradient. Dimension of velocity gradient \(\left(\frac{d u}{d x}\right)=\frac{\mathrm{LT}^{-1}}{\mathrm{~L}}=\mathrm{T}^{-1} .\)

Coefficient of viscosity: The tangential viscous force acting per unit area between two parallel liquid layers having unit velocity gradient between them is called the coefficient of viscosity of that liquid.

Units of coefficient of viscosity:

CGS system: dyn • s • cm^-2 or g • cm^-1 • s^-1 or poise

SI: N • s • m^-2 or Pa • s or kg • m^-1 • s^-1 or decapoise

Dimension of coefficient of viscosity: ML^-1T^-1.

• Usually the viscosity of a liquid increases with the increase in pressure and decreases with the increase in temperature.
• Pressure has almost no effect on the viscosity of a gas. With the increase in temperature, the viscosity of a gas increases.

Critical velocity: When the velocity of a fluid does not exceed a certain limiting value, the flow of the fluid remains streamline, but when the velocity exceeds that particular limiting value, the flow becomes turbulent. The limiting value of that velocity is called critical velocity.

Reynolds number N is a dimensionless quantity. For N< 2000, the fluid motion is streamlined. If N> 3000, then the fluid motion becomes turbulent. As the value of N gradually changes from 2000 to 3000, the pure streamline flow gradually changes into a fully turbulent flow.

Terminal velocity: When a body falls through a viscous medium under the influence of gravity, the viscosity of the medium offers resistance against its motion.

If the body is small, then after a certain time the magnitude of this upward viscous force become equal to the net force creating the motion. Then the body attains a uniform velocity through the medium. This uniform velocity of the body is called the terminal velocity.

Equation of continuity: In the case of streamline flow of a fluid (liquid or gas), the mass of fluid flowing per second through any cross-section of the tube of flow always remains constant. This is called the equation of continuity.

Bernoulli’s theorem: For the streamline flow of an ideal liquid, the sum of the potential energy, the kinetic energy, and the energy due to pressure per unit volume of the liquid remains constant at every point on the streamline. It leads to the relation

⇒ \(\frac{v^2}{2 g}+h+\frac{p}{\rho g}=\text { constant }\)

or, velocity head + elevation head + pressure head = constant.

Torricelli’s theorem: The velocity of efflux of a liquid, confined in a container through an orifice at some depth of the container is equal to the velocity acquired by a body falling freely from rest under gravity from the free surface of the liquid to the orifice.

Surface tension: Surface tension is a property of the free surface of a liquid due to which it behaves as a stretched thin membrane and has a tendency to con¬tract so as to minimise the surface area.

Units of surface tension:

• CGS system: dyn · cm^-1
• SI: N • m^-1

Dimension of surface tension: MT^-1.

Surface energy: The potential energy per unit area of the surface film is called surface energy.

or, surface energy = \(\frac{\text { work done in increasing the surface area }}{\text { increase in surface area }} \text {. }\)

• The surface energy per unit area is numerically equal to the surface tension of a liquid (if temperature remains constant)
• At any temperature except absolute zero, the total surface energy of a liquid is always greater than the surface tension.
• The surface tension of all liquids decreases with the rise in temperature. At the critical temperature of a liquid, the surface tension vanishes.
• If a liquid surface is contaminated with impurities, then the surface tension of that liquid usually decreases.
• If an inorganic substance is dissolved in a liquid, then the surface tension increases, but if an organic substance is dissolved, then the surface tension decreases.
• The presence of electric charges on the surface of a liquid causes a decrease in the surface tension.

Due to surface tension, capillary action is observed in liquids.

• When a liquid is in contact with a solid, the angle between the solid surface and the tangent to the free surface of the liquid at the point of contact, measured from inside the liquid is called the angle of contact for that specific pair of solid and liquid.
• If the angle of contact is less than 90°, then the liquid is said to wet the solid, and it rises in a capillary tube. But if the angle of contact is more than 90°, then the liquid does not wet the solid, and falls in a capillary tube.

Jurin’s law: The rise or fall of a liquid in a capillary tube is inversely proportional to the radius of the tube.

Viscosity And Surface Tension Useful Relations For Solving Numerical Problems

For two adjacent layers of a flowing liquid, if the opposing force acting is F, the area of the liquid surface is A and the velocity gradient is \(\frac{d v}{d x}\), then the coefficient of viscosity, \(\eta=\frac{F}{A \frac{d y}{d x}}\)

If Reynolds number is N, velocity of fluid flow is v, characteristic length of the fluid is l, the coefficient of viscosity is η and density of the fluid is ρ, then N = \(\frac{e l v}{\eta}\).

If a small sphere of radius r falls through a medium having a coefficient of viscosity η with a terminal velocity v, then the opposing force acting on the sphere due to viscosity is F = 6πηrv and the terminal velocity of the sphere is,

v = \(\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\)

where ρ and σ are the densities of the material of the sphere and the material of the medium respectively.

If the cross-sectional area at any place of a tube of flow is a and the velocity of the fluid at that place is v, then the equation of continuity is expressed as, vα = constant.

Bernoulli’s theorem: \(\frac{v^2}{2 g}+h+\frac{p}{\rho g}\) = constant,

where \(\frac{v^2}{2 g}\) is the velocity head, h is the elevation head, and \(\frac{p}{\rho g}\) is the pressure head.

Torricelli’s theorem: The velocity of efflux of a liquid through an orifice situated at a depth h of its container is, v = √2gh

Surface tension = \(\frac{\text { tangential force }}{\text { length }}\)

Work done to increase the area of a liquid surface by unity at constant temperature = surface energy stored per unit area = surface tension.

If the radius of a capillary tube is r, the density’ of liquid is ρ the angle of contact of the liquid with respect to the material of the tube is θ and the surface tension of the liquid is T, then the rise of the liquid in that capillary tube is

h = \(\frac{2 T \cos \theta}{r \rho g}\)

The excess pressure inside a spherical drop or bubble \(p=\frac{2 T}{r}\)

where T = surface tension and r = radius of curvature

The excess pressure inside a spherical soap bubble \(p=\frac{4 T}{r}\)

where T = surface tension and r = radius of curvature.

Viscosity And Surface Tension Assertion Reason Type Question And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true statement n is a correct explanation for statement 1.
2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, statement 2 is true.

Question 1.

Statement 1: The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down.

Statement 2: In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant.

Answer: 1. Statement 1 is true, statement 2 is true statement n is a correct explanation for statement 1.

Question 2.

Statement 1: The viscosity of liquid increases with rise in temperature.

Statement 2: Viscosity of a liquid is the property of the liquid by virtue of which it opposes the relative motion amongst its different layers.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 3.

Statement 1: All the rain drops hit the surface of the earth with the same constant velocity.

Statement 2: An object falling through a viscous medium eventually attains a terminal velocity.

Answer: 1. Statement 1 is true, statement 2 is true statement n is a correct explanation for statement 1

Question 4.

Statement 1: Air flows from a small bubble to a large bubble when they are connected to each other by a capillary tube.

Statement 2: The excess pressure because of surface tension inside a spherical bubble decreases as its radius increases.

Answer: 1. Statement 1 is true, statement 2 is true statement n is a correct explanation for statement 1

Question 5.

Statement 1: When height of the tube is less than the rise in liquid in a capillary tube, the liquid does not overflow.

Statement 2: Product of radius of meniscus and height of liquid In the capillary tube always remains constant.

Answer: 1. Statement 1 is true, statement 2 is true statement n is a correct explanation for statement 1

Question 6.

Statement 1: It is easier to spray water in which some soap is dissolved.

Statement 2: Soap is easier to spread.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 7.

Statement 1: A needle placed carefully on the surface of water may float, whereas a ball of the same material will always sink.

Statement 2: The buoyancy of an object depends both on the material and shape of the object.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 8.

Statement 1: A large force is required to draw apart normally two glass plates enclosing a thin water film.

Statement 2: Water works as glue and sticks two glass plates.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 9.

Statement 1: Tiny drops of liquid resist deforming forces better than bigger drop.

Statement 2: Excess pressure inside a drop is directly proportional to surface tension.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 10.

Statement 1: The uplift of the wing of an aircraft moving horizontally is caused by a pressure difference between the upper and lower faces of the wing.

Statement 2: The velocity of air moving along the upper surface is higher than that along the lower surface.

Answer: 1. Statement 1 is true, statement 2 is true statement n is a correct explanation for statement 1

 

Expansion Of Gases Introduction

WBBSE Class 11 Expansion of Gases Notes

Gases expand on heating and contract on cooling, like solids and liquids. Like liquids, gases too do not have any fixed shape and therefore the linear or the surface expansion of a gas is irrelevant.

Only the change in volume with the change in temperature is of importance. Other characteristic features of gaseous expansions, compared to those of solids and liquids, are discussed below.

1. The coefficient of volume expansion for a gas has a much higher value than that for solids and liquids. For a certain rise in temperature, the expansion of the container is negligible compared to the expansion of the gas in it. Hence, the apparent expansion of gas is practically the same as real expansion and is usually not reckoned separately unless a very high accuracy is required.
2. Unlike solids and liquids, the volume of a fixed mass of a gas is considerably affected due to any change in pressure. So, the effects of both temperature and pressure have to be studied in connection with the expansion or contraction of a gas. While studying the effect of one, the other is usually kept constant.
3. The rate of expansion or contraction, due to the change in either pressure or temperature, does not differ for different gases. Unlike solids and liquids, the coefficient of expansion is the same for all gases.

The state of a fixed mass of gas is therefore determined by the parameters

1. Volume,
2. Pressure and
3. Temperature.

The rules, that govern the change of one parameter with the change of another keeping the third one constant, are called gas laws.

1. It should be mentioned here that the pressure of a gas means the pressure exerted by the gas. In equilibrium (i.e., when each parameter of the gas is a constant) the pressure exerted by the gas and the pressure applied on the gas are equal.
2. The gas which follows Boyle’s law and Charles’ law at any temperature and pressure is called an ideal gas. In reality, an ideal gas does not exist. However, the ideal gas concept provides a very useful tool for the analysis of real gases.

 Expansion Of Gases Charles Law

The relationship between the volume and temperature of a fixed mass of gas at constant pressure was investigated experimentally by the French scientist Charles in 1787.

• He concluded that at constant pressure a fixed volume of all gases would expand by the same amount for an equal rise in temperature. Gay Lussac arrived at almost the same result in 1802.
• He found that the coefficient of volume expansion of all gases has the same value if pressure is kept constant. In 1842 Regnault showed that the experimental value of this coefficient is 1/273 per degree Celsius.
• Charles’ law is enunciated by combining the experimental results of Gay Lussac and Regnault.

Statement: When pressure is kept constant, the volume of a fixed mass of gas increases or decreases by 1/273 th part of its volume at 0°C, for each degree Celsius rise or fall in temperature.

Mathematical expression: Let V₀ be the volume of a mass m of a gas at 0°C. As per Charles’ law, the volume at 1°C will be

⇒ \(V_1=V_0+\frac{V_0}{273}=V_0\left(1+\frac{1}{273}\right)\)

The volume of the gas at \(2^{\circ} \mathrm{C}\),

⇒ \(V_2=V_0+\frac{2 V_0}{273}=V_0\left(1+\frac{2}{273}\right)\)

∴ The volume of the gas at \(t^{\circ} \mathrm{C}\),

⇒ \(V_t=V_0+\frac{V_0 t}{273}=V_0\left(1+\frac{t}{273}\right)\)….(1)

Similarly, if temperature is decreased by t°C, i.e., at a temperature -t°C, the volume of the gas becomes \(V_{-t}=V_0\left(1-\frac{t}{273}\right)\)

Therefore, at constant pressure volume of a fixed mass of gas changes linearly with its temperature. So, at fixed pressure, a graph plotted between the volume of a gas of a fixed mass and its temperature gives a straight line (AB).

Expansion Of Gases Pressure Law

The law that relates the change in pressure of a fixed mass of a gas at a fixed volume, with change in temperature is called pressure law or Regnault’s law.

Statement: When volume is kept constant, the pressure of a fixed mass of gas increases or decreases by1/273th part of its pressure at 0°C, for each degree centigrade rise or fall in temperature.

Mathematical expression: Let p₀ be the pressure of a fixed mass of a gas at 0°C. The pressure is p_t when the temperature is raised to t °C.

Therefore the pressure of the gas at 1°C, \(p_1=p_0+\frac{p_0}{273}=p_0,\left(1+\frac{1}{273}\right)\)

and the pressure of the gas at t°C, \(p_t=p_0+\frac{p_0 t}{273}=p_0\left(1+\frac{t}{273}\right)\)

Similarly, the pressure of the gas at -t°C, \(p_{-t}=p_0-\frac{p_0 t}{273}=p_0\left(1-\frac{t}{273}\right)\)

Hence the change in pressure of a fixed mass of gas is linearly related to the change in temperature at constant volume.

So, at constant volume, a graph plotted between the pressure of a gas of fixed mass and its temperature gives a straight line.

The increase in either the volume or the pressure of a gas with a rise in temperature is loosely termed as thermal expansion, although an increase in pressure is not actually an expansion.

Expansion Of Gases – Alternative Forms Of Charles Law And Pressure Law

Charles’ law: Suppose a fixed mass of a gas. at constant pressure, has volume V₀ at 0°C, at V₁ at t₁°C and V₂ at t₂°C. From Charles’ law.

⇒ \(V_1=V_0\left(1+\frac{t_1}{273}\right)=V_0\left(\frac{273+t_1}{273}\right)=\frac{V_0}{273} \cdot T_1\)

[where T₁ = t₁ + 273]

Obviously, the temperature t₁ in the Celsius scale is the same as the temperature T₁ K in the Kelvin scale.

Similarly, \(V_2=\frac{V_0}{273} \cdot T_2\)

where \(T_2=t_2+273\)

Here, \(t_2{ }^{\circ} \mathrm{C}=T_2 \mathrm{~K}\).

∴ \(\frac{V_1}{V_2}=\frac{T_1}{T_2}=\) constant

or, \(V \propto T\) at constant pressure.

Hence, Charles’ law can also be stated as the volume of a fixed mass of gas at constant pressure is directly proportional to its temperature in an absolute scale.

Pressure law: Suppose a fixed mass of a gas at constant volume has pressure p₀ at 0°C, p₁ at t₁ °C. and p₂ at t₂ °C. Hence, from pressure law.

⇒ \(p_1=p_0\left(1+\frac{t_1}{273}\right)=\frac{p_0}{273}\left(273+t_1\right)=\frac{p_0}{273} \times T_1\)

where \(T_1=t_1+273\)

It is dear that the temperature t°C is the same as the temperature T₁K.

Similarly, \(p_2=\frac{p_0}{273} \times T_2\) [where T₂ = t₂ + 273]

Here, t₂ °C = T₂ K.

∴ \(\frac{p_1}{p_2}=\frac{T_1}{T_2} \quad \text { or, } p \propto T\), at constant volume.

Hence, pressure law can be expressed as. the pressure of a fixed mass of a gas at constant volume is directly proportional to its temperature in an absolute scale.

Expansion Of Gases- Combination Of Boyles Law And Charles

Key Concepts in Gas Expansion for Class 11

Ideal or Perfect gas: The gases that obey Bodes and Charles’ law at any temperature are called ideal gases.

The equation obtained by combining Boyles’s law and Charles’ law is called the equation of the state of an ideal gas.

From Bovle’s law, \(V \propto \frac{1}{p}\) when T is constant.

From Charles’ law, V∝T when p is constant.

∴ \(V \propto \frac{T}{p}\) when both p and T vary

or, pV= kT ….(1)

where k is a constant whose value depends on the units of p, V, T, and the mass of the gas. If a gas of fixed mass occupies a volume V₁ at pressure p₁ and temperature T₁, and after a general change, a volume V₂, at pressure p₂, and temperature T₂, then

⇒ \(\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}\)….(2)

It is known that the physical property of an ideal gas of fixed mass depends on its pressure, volume, and temperature.

Hence, the equation pV = kT is called the equation of state of an ideal gas.

Note that no real gas totally follows this equation of state.

The deviation, though small at ordinary temperature and pressure, becomes significant at high pressure and at low- temperature.

Expansion Of Gases Absolute Scale Of Temperature

From Charles’ law, If a fixed mass of gas at constant pressure occupies a volume V₀ at 0°C, then at t°C its volume will be \(V_t=V_0\left(1+\frac{t}{273}\right)\)

Now if the temperature is reduced to t = -273°C

⇒ \(V_t=V_0\left(1-\frac{273}{273}\right)=0\), i.e., the volume of the gas vanishes at -273°C temperature.

Similarly from pressure law, if a fixed mass of a gas at constant volume has a pressure p at 0°C, then at t°C its pressure will be \(p_t=p_0\left(1+\frac{t}{273}\right) \text {. }\)

Now if the temperature is reduced to t = -273 °C, \(p_t=p_0\left(1-\frac{273}{273}\right)=0\), i.e., the pressure of the gas vanishes at -273°C temperature.

So, at -273 °C, the volume and pressure of a fixed mass of gas have worked out to be zero.

If the temperature could be lowered below -273 °C, the pressure and volume would have been negative, which is meaningless. Hence, the lowest conceivable temperature in this universe is -273 °C.

Therefore -273 °C is called the absolute zero of temperature, or simply, absolute zero. More precisely, the value of absolute zero is -273.15°C.

Definition: The temperature at which the volume and the pressure of a gas reduce to zero is called the absolute zero of temperature. This temperature is the lowest temperature in reality.

Taking absolute zero i.e., -273°C as zero, a new scale of temperature was developed by Lord Kelvin, and it is called the absolute scale of temperature or Kelvin scale after the inventor.

In this scale, the temperature reading is denoted by A (absolute) or K (Kelvin). Each degree in this scale is taken equal to a degree in the Celsius scale. Hence, the scale of temperature where -273°C is taken as zero (0) and each degree is equal to a degree in Celsius scale, is called the absolute scale or Kelvin scale.

The freezing point of water (0°C) in an absolute scale is 273 K and the boiling point of water (100°C) in this scale is 373 K.

If any temperature is represented by t°C in the Celsius scale and by TK on the Kelvin scale, then, T = t + 273.

In the Kelvin scale, a temperature may be zero or positive. A negative Kelvin temperature does not exist.

0° K In Fahrenheit scale: in the relation \(\frac{C}{5}=\frac{F-32}{9}\), putting C = -273°C (0K) we have, \(\frac{-273}{5}=\frac{F-32}{9}\) or, F = – 459.4°F.

Incidentally, gas laws are valid as long as the matter remains in a gaseous state. All gases get liquified much before they attain the absolute zero temperature. No gas can be cooled to the temperature of absolute zero, and as such, zero volume or zero pressure of a gas is never realized in practice.

Absolute scale of temperature—why it is so named: The selection of 0 degrees in Celsius and Fahrenheit scale has no scientific reason behind it. The freezing point of water is taken arbitrarily as 0°C and the boiling point as 100°C in the Celsius scale.

• But the selection of 0 degrees in absolute scale is scientific as it denotes the temperature at which the volume and pressure of any ideal gas would reduce to zero. This is the lowest conceivable temperature in this universe.
• In addition, the scale does not depend on the nature of the gas. Hence, it is justified to call it the absolute scale of temperature.

Expansion Of Gases Universal gas Constant

The equation of state for an ideal gas is pV = kT. The value of the constant k depends on the mass of the gas used. Keeping pressure (p) and temperature (T) constant, if the mass of a gas is doubled, the volume is also doubled.

• So k is doubled. We can conclude that k is directly proportional to the mass of the gas.
• When one gram-molecule or one mole of an ideal gas is taken, the constant k is written as R. As per Avogadro’s law, at the same temperature and pressure, volume of 1 gram-molecule of any gas is the same.
• Therefore, the value of the constant R, for all 1-gram-molecule ideal gases, is the same. Hence, R is called the universal gas constant or molar gas constant.

Hence, for 1 gram-molecule of any ideal gas,

pV = RT ….(1) f the volume of n gram-molecule of gas is V, then the volume of 1 gram-molecule of gas, = V/n

Then from equation (1) we get, \(p \cdot \frac{V}{n}=R T \quad \text { or, } p V=n R T\)….(2)

Obviously, for mg of a gas of molecular weight M the number of moles, n = m/n

∴ pV = \(\frac{m}{M} R T\)….(3)

Comparing equation(3) with the gas equation pV = kT, k = \(\frac{m}{M} R T\)

Therefore for 1 gram of gas, k = \(\frac{R}{M}\) = r (say), r is also a constant and is called the specific gas constant.

As the molecular weights of different gases are different, the magnitude of this constant is also different for different gases.

Thermal Expansion of Gases Explained

Magnitude of universal or molar gas constant: From equation (1) above, the relation R = \(\frac{p V}{T}\) is valid for any ideal gas at any temperature and pressure.

Hence, R is also equal to \(\frac{p_0 V_0}{T_0}\) where p₀ = 76 cm Hg
(standard pressure) = 76 x 13.596 x 980.665 dyn · cm^-2, T₀ (standard temperature) = 0 + 273 = 273 K and V₀ = the volume at STP of 1 mol of a gas = 22.4 litre, as per Avogadro’s law.

∴ R = \(\left(76 \times 13.596 \times 980.665 \mathrm{dyn} \cdot \mathrm{cm}^{-2}\right) \frac{\times 22.4 \times 1000 \mathrm{~cm}^3 \cdot \mathrm{mol}^{-1}}{273 \mathrm{~K}}\)

= 8.314 x 10⁷ dyn · cm · mol^-1 · K^-1

= 8.314 x 10⁷ erg · mol”1 · K^-1

= 8.314 J · mol^-1 · K^-1

Magnitude of specific gas constant: Value of r for agasis r= R/M.

1. For hydrogen, molecular weight = 2,

r = \(frac{R}{2}=\frac{8.314 \times 10^7}{2}\)

= \(4.16 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

2. For oxygen, molecular weight = 32

r = \(\frac{R}{32}=\frac{8.314 \times 10^7}{32}\)

= \(0.26 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

3. For nitrogen, molecular weight = 28

r = \(\frac{R}{28}=\frac{8.314 \times 10^7}{28}\)

= \(0.297 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

4. For carbon dioxide, molecular weight = 44

r = \(\frac{R}{44}=\frac{8.314 \times 10^7}{44}\)

= \(0.189 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Universal gas Constant Numerical Examples

Gas Laws and Expansion: Class 11 Notes

Example 1. The mass of 1 litre of hydrogen at STP is 0.0896 g. Calculate the value of R from this data.
Solution:

Given

The mass of 1 litre of hydrogen at STP is 0.0896 g.

Volume of 0.0896 g of hydrogen at STP = 1000 cm³.

Hence, a volume of 2 g or 1 mol of hydrogen at STP = \(\frac{1000 \times 2}{0.0896}=22321.4 \mathrm{~cm}^3.\)

Standard pressure = 76x 13.6×981 dyn · cm^-2; standard temperature = 0°C = 273 K

∴ R = \(\frac{p V}{T}=\frac{76 \times 13.6 \times 981 \times 22321.4}{273} \mathrm{erg} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

= \(8.29 \times 10^7 \mathrm{erg} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

Example 2. The mass of 3.76 litre of oxygen at 2 standard atmo¬sphere pressure and 20°C is 10 g. Find the value of R
Solution:

Given, p = 2 standard atmosphere = 2 x 76 x 13.6 x 981 dyn · cm^-2,

The mass of 3.76 litre of oxygen at 2 standard atmo¬sphere pressure and 20°C is 10 g.

V = 3.76 x 10³ cm³, T = 273 + 20 = 293 K and n = 10/32 mol, [where 32 is the molecular weight of oxygen]

∴ pV=nRT

R = \(\frac{p V}{n T}=\frac{2 \times 76 \times 13.6 \times 981 \times 3.76 \times 10^3 \times 32}{10 \times 293}\)

= \(8.33 \times 10^7 \mathrm{erg} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

Example 3. The density of air at STP = 1.293 g · L^-1 and that of mercury = 13.6 g ·  cm^-3. Find the value of the gas constant for 1 g of air.
Solution:

Given

The density of air at STP = 1.293 g · L^-1 and that of mercury = 13.6 g ·  cm^-3.

If k is the constant for 1 g of air, then k = pV/T

According to the problem, 1.293 g of air occupies a volume of 1000 cm³

∴ 1g of air occupies a volume of \(\frac{1000}{1.293} \mathrm{~cm}^3\)

Here, p = \(76 \times 13.6 \times 981 \mathrm{dyn} \cdot \mathrm{cm}^{-1}, T=273 \mathrm{~K}\)

∴ k = \(76 \times 13.6 \times 981 \times \frac{1000}{1.293} \times \frac{1}{273}\)

= \(0.287 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

Example 4. The masses, volumes, and pressures of two samples of oxygen and hydrogen gases are equal. Find the ratio of their absolute temperatures.
Solution:

Given

The masses, volumes, and pressures of two samples of oxygen and hydrogen gases are equal.

Let the volume of each gas = V, the pressure of each gas = p, and the absolute temperatures of the gases be T₁ and T₂ respectively.

If the mass of each sample is m g, then the number of moles of the gases are respectively, \(n_1=\frac{m}{32} \quad \text { and } n_2=\frac{m}{2} \text {. }\)

As per the equation pV = nRT, for oxygen gas, \(p V=\frac{m}{32} R T_1\) and for hydrogen gas, \(p V=\frac{m}{3} R T_2\)

∴ \(\frac{m}{32} R T_1=\frac{m}{2} R T_2 \quad \text { or, } \quad \frac{T_1}{T_2}=\frac{32}{2}=16: 1\)

 Expansion Of Gases Conclusion

Boyle’s law: At constant temperature, the volume (V) of a fixed mass of gas is inversely proportional to the pressure (p) of the gas.

i.e \(V \propto \frac{1}{p}\) or, pV = constant.

Charles’ law: At constant pressure, the volume of a fixed mass of gas increases or decreases by 1/273 of its volume at 0°C per degree Celsius rise or fall in temperature.

Pressure law: At constant volume, the pressure of a fixed mass of gas increases or decreases by 1/273 of its pressure at 0°C per degree Celsius rise or fall in temperature.

Volume coefficient: The volume coefficient of a fixed mass of a gas at a constant pressure is the increment of its volume per unit volume when its temperature is raised by 1°C from 0°C.

Pressure coefficient: The pressure coefficient of a fixed mass of a gas at a constant volume is the increment of its pressure per unit pressure when its temperature is raised by 1°C from 0°C.

• The volume coefficient of all gases is equal.
• The pressure coefficients of all gases are equal.
• In the case of any ideal gas, the volume coefficient and the pressure coefficient are equal, i.e., \(\gamma_p=\frac{1}{273}{ }^{\circ} \mathrm{C}^{-1}=\gamma_v.\)
• While determining the volume and pressure coefficients of a gas, we always refer to the initial volume or pressure at 0°C.
• The gases which obey Boyle’s and Charles’ laws at all temperatures are known as ideal gases. But in actual practice no real gas is ideal.

Absolute zero temperature: The temperature at which the volume and the pressure of an ideal gas theoretically become zero is called the absolute zero temperature. This is the lowest possible temperature in reality.

The scale of temperature in which the temperature -273°C is taken as zero and each degree interval is equal to a degree interval in the Celsius scale is called the absolute scale of temperature or Kelvin scale.

Absolute zero temperature = 0K = -273°C = 459.4°F.

Universal or molar gas constant.

R =8.314 x 10⁷ erg · mol^-1  K^-1

= 8.314 J ⋅ mol^-1 K^-1

At constant pressure, the density of a gas is inversely proportional to its absolute temperature.

At constant temperature, the density of a gas is directly proportional to its pressure.

Expansion Of Gases Useful Relations For Solving Numerical Problems

If the volume of a definite mass of a gas is V and its pressure is p, then according to Boyle’s law pV = constant

Let at constant pressure, the volume of some definite mass of a gas at 0°C be V₀. According to Charles’ law the final volume of the gas due to riC rise or fall in temperature,

⇒ \(V_t=V_0\left(1 \pm \frac{t}{273}\right)\)

Let at constant volume, the pressure of some definite mass of a gas at 0°C be p₀. According to pressure law. the final pressure of the gas due to t°C rise or fall in temperature,

⇒ \(p_t=p_0\left(1 \pm \frac{t}{273}\right)\)

Volume coefficient of a gas at constant pressure, \(\gamma_p=\frac{V_t-V_0}{V_0 \cdot t}\)

Pressure coefficient of a gas at constant volume, \(\gamma_v=\frac{p_t-p_0}{p_0 \cdot t}\)

⇒ \(\gamma_p=\gamma_v=\frac{1}{273}{ }^{\circ} \mathrm{C}^{-1}\)

T = t + 273 lf any temperature is represented by t and T in Celsius and Kelvin scales respectively]

For n mol of an ideal gas, pV = nRT = \(\frac{m}{M} R T\) [where m and M are the mass and molecular weight of the gas respectively]

If ρ is the density of an ideal gas, then \(\frac{p}{\rho T}=\frac{R}{M}\) [symbols have usual meanings]

Expansion Of Gases Very Short Answer Type Questions

Question 1. At what Celsius temperature, does the volume of a gas become zero according to Charles’ law?
Answer: -273°C

Question 2. How does the volume of a definite mass of gas change with pressure at constant temperature?
Answer: Inversely proportional

Question 3. How does the volume of a definite mass of gas change with its absolute temperature at constant pressure?
Answer: Directly proportional

Question 4. At constant volume, the pressure of a definite mass of gas is directly proportional to its absolute temperature. Is the statement true or false?
Answer: True

Question 5. What is the value of the volume coefficient of a gas?
Answer: 1/273 °C^-1

Question 6. What is the value of the pressure coefficient of a gas?
Answer: 1/273 °C^-1

Expansion Of Gases Assertion Reason Type Question And Answers

Direction: These questions have statement 1 and statement 2 Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: Equal masses of helium and oxygen gases are given equal quantities of heat. There will be a greater rise in the temperature of helium compared to that of oxygen.

Statement 2: The molecular weight of oxygen is more than the molecular weight of helium.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 2.

Statement 1: In the upper part of the atmosphere, the temperature of air is of the order of 1000 K, even then it is quite cold there.

Statement 2: Molecular density at high altitudes is low.

Answer:  1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: Shows the V – T graphs of a certain mass of an ideal gas at two pressures p₁ and p₂. It follows from the graph that p₁ is greater than p₂.

Statement 2: The slope of V- T graph for an ideal gas is directly proportional to pressure.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 4.

Statement 1: V-T graph in a process is a rectangular hyperbola. Then p-T graph in the same process will be a parabola.

Statement 2: If the V-T graph is a rectangular hyperbola, with an increase in T, the volume will decrease and hence, pressure will increase.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 5.

Statement 1: The size of a hydrogen balloon increases as it rises in the air.

Statement 2: The material of the balloon can be easily stretched.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Expansion Of Gases Match Column 1 With Column 2

Question 1. The V-T graph of two gases A and B is shown.

Answer: 1. D, 2. D, 3. D, 4. D

Expansion Of Gases Comprehension Type Questions And Answers

Short Answer Questions on Gas Expansion

Read the following passages carefully and answer the questions at the end of them.

Question 1. An air bubble starts rising from the bottom of a lake. Its diameter is 3.6 mm at the bottom and 4 mm at the surface. The depth of the lake is 250 cm and the temperature at the surface is 40°C. The atmospheric pressure is 76 cm of Hg and g = 980 cm · s^-2.

1. What is the pressure at the bottom of the lake?

1. 1279325 dyn • cm^-2
2. 1359943 dyn • cm^-2
3. 1257928 dyn • cm^-2
4. 1378174 dyn • cm^-2

Answer:  1. 1279325 dyn • cm^-2

2. What is the temperature at the bottom of the lake?

1. 9.77°C
2. 10.37°C
3. 11.31°C
4. 11.67°C

Answer: 2. 10.37°C

Question 2. A fixed thermally conducting cylinder has a radius of R and a height of L₀. The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L, from the top surface as shown. The atmospheric pressure is p₀.

1. The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be

1. \(\frac{p_0}{2}+\frac{M g}{\pi R^2}\)
2. \(\frac{p_0}{2}\)
3. \(p_0\)
4. \(\frac{p_0}{2}-\frac{M g}{\pi R^2}\)

Answer: 1. \(\frac{p_0}{2}+\frac{M g}{\pi R^2}\)

2. When the piston is at a distance 2L from the top the hole at the top is sealed. The piston is then released, at a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is

1. \(\left(\frac{2 p_0 \pi R^2}{\pi R^2 p_0+M g}\right)(2 L)\)
2. \(\left(\frac{p_0 \pi R^2-M g}{\pi R^2 p_0}\right)(2 L)\)
3. \(\left(\frac{p_0 \pi R^2+M g}{\pi R^2 p_0}\right)(2 L)\)
4. \(\left(\frac{p_0 \pi R^2}{\pi R^2 p_0-M g}\right)(2 L)\)

Answer: 4. \(\left(\frac{p_0 \pi R^2}{\pi R^2 p_0-M g}\right)(2 L)\)

3. The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown.

The density of the water is ρ. In equilibrium, the height H of the water column in the cylinder satisfies

1. \(\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)+L_0 p_0=0\)
2. \(\rho g\left(L_0-H\right)^2-p_0\left(L_0-H\right)-L_0 p_0=0\)
3. \(\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-L_0 p_0=0\)
4. \(\rho g\left(L_0-H\right)^2-p_0\left(L_0-H\right)+L_0 p_0=0\)

Answer: 3. \(\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-L_0 p_0=0\)

Expansion Of Gases Integer Type Questions And Answers

Real-Life Applications of Gas Expansion

In this type, the answer to each of the questions Is a single-digit integer ranging from 0 to 9.

Question 1. An ideal gas is heated from 27 °C to 627°C at constant pressure. If the initial volume was 3 m³, then what would be the final volume (in m³) of gas?
Answer: 9

Question 2. An air bubble of diameter 1 cm is formed at a depth of 238 ft in a lake. What will be the diameter (in cm) of the bubble when it reaches the free surface? Given that the temperature from top to bottom in the lake is the same and the height of a water barometer is 34 ft.
Answer: 2

Question 3. The volume of some air saturated with water vapor is 80 cm³ at a pressure of 74 cm Hg. Keeping the temperature fixed if pressure is made 146 cmHg then the volume becomes halved. What will be the pressure (in cm Hg) of water vapor then?
Answer: 2

Question 4. Find the percentage increase in the pressure when air enclosed at 30 °C is raised to 57 °C at a constant volume.
Answer: 9

Expansion Of Gases Short Answer Type Questions And Answers

Question 1. The temperature of an open room of volume 30 m³ increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1 x 10⁵ Pa. If n_i and n_f are the number of molecules in the room before and after heating, then n_f – n_i will be

1. 1.61 x 10²³
2. 1.38 x 10²³
3. 2.5 x 10²⁵
4. -2.5 x 10²⁵

Answer:

Given

The temperature of an open room of volume 30 m³ increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1 x 10⁵ Pa. If n_i and n_f are the number of molecules in the room before and after heating,

If n₁ and n₂ are the number of moles at the initial and final temperature inside the room respectively, then

⇒ \(n_1 =\frac{p_1 V_1}{R T_1}=\frac{10^5 \times 30}{8.31 \times(273+17)}\)

= \(1.245 \times 10^3\)

⇒ \(n_2 =\frac{p_2 V_2}{R T_2}=\frac{p_1 V_1}{R T_2}=\frac{10^5 \times 30}{8.31 \times(273+27)}\)

= \(1.203 \times 10^3\)

Therefore \(n_f-n_1=\left(n_2-n_1\right) \times 6.023 \times 10^{23}\)

= \(-2.5 \times 10^{25}\)

The option 4 is correct.

Question 2. Explain why air pressure in a car increases during driving.
Answer:

Air pressure in a car increases during driving because

The temperature of air inside the increases due to motion. We know from Charles’ law, p ∝ T, therefore air pressure inside the the increases during driving.

Calorimetry Introduction

An important characteristic of heat must be noted from the very beginning. Unlike the quantities volume, pressure, temperature, etc., heat is not related to any equilibrium state of a body. Heat manifests itself only when a body undergoes a transition from one state to another.

Statements like ‘heat of a body’, ‘initial heat’, and ‘final heat’ have no physical meaning. Physically significant are the quantities like ‘heat gained’, or ‘heat lost’ by a body during any process. In this sense, heat is regarded as an energy in transit.

The rise or fall in temperature of a body implies the addition or extraction of heat from it. If two objects are in contact with each other then there will be heat flow between them till both attain an equilibrium temperature.

In such situations a need arises to know the quantity of heat transferred from one body to another, or, from an object to its surroundings. The study of problems that involve heat exchange between different bodies is called calorimetry.

Calorimetry Factors Affecting Heat Transfer

We know that it takes longer to boil some water than to warm it. Also, boiling water takes longer to cool down than lukewarm water does. So for rise and fall of different amount of temperature, different amounts of heat are absorbed or released.

• Again, increasing the amount of water to be boiled, increases the time taken to boil it too. From this we can conclude that the amounts cf heat absorbed (or released) by the bodies of same material but different mass for a fixed rise (or fall) in temperature depend on their masses.
• When a piece of iron and some water of the same mass are heated, the piece of iron heats up faster than the water.
• Clearly, the material of an object plays a role in determining the amount of heat absorbed (or released) by the object for a certain rise (or fall) in temperature.
• From these observations, it can be inferred that, the amount of heat absorbed or released by a body depends on the change of temperature of the body, mass of the body, and material of the body, provided that the material does not undergo any phase change.

Temperature: Amount of heat gained or lost by a body of fixed mass is directly proportional to the rise or fall in its temperature.

Mass: Amount of heat gained or lost by a body, for a fixed rise or fall in temperature, is directly proportional to its mass.

Material: Amount of heat gained or lost by a body depends on the material of the body.

When a body of mass m receives heat H, the rise in temperature is t,
H ∝ mt or, H = smt……(1)

where s is specific heat whose value depends on the material of the body.

Therefore, H (the heat absorbed or released by a body) = m (mass of the body) x s (specific heat of the material of the body) x t (rise or fall in temperature)
It is to be noted that t in equation (1) is the change in temperature of the body.

So only the heat gained or lost can be measured by this equation.

Calorimetry Units Of Heat

Some useful useful units for the measurement of heat are defined as follows:

Calorie: It is the unit of heat in CGS system. The amount of heat required to raise the temperature of 1 g of water from 14.5 °C to 15.5 °C is called 1 calorie (1 cal).

• Practically, the heat required to raise the temperature of 1 g of pure water from 0°C to PC is not the same as the heat required to raise Its temperature, for example, from 45T to 46°C though the rise in temperature is the same.
• Hence, quantity of heat differs at the different ranges in the scale of temperature.
• So, most acceptable definition of calorie is, 1 cal = 1/100 X heat required to raise the temperature of 1 g of pure water from 0°C to 100°C. It is often known as mean calorie.
• Its value is equal to the amount of heat required to raise the temperature of 1 g of pure water from 14.5°C to 15.5°C. So it is sometimes called 15°C calorie.

Kilocalorie or kilogram calorie: The quantity of heat required to raise the temperature of 1 kg of pure water from 14.5°C to 15.5°C, is called 1 kilocalorie.

As 1 kg = 1000 g, 1 kilocalorie (kcal) = 1000 calorie (cal).

Joule: This is the unit of heat in SI. 1 cal = 4.2 J.

Calorimetry Notes for Class 11 WBCHSE

Calorimetry Specific Heat Or Specific Heat Capacity

It has been stated that heat gained (or lost) by a body depends on its mass, the rise (or fall) in temperature and the material of the body.

• Let us take two bodies of same mass but made of different materials. If we try to change the temperature of both the bodies by the same amount, we will find that the heat required is different for them.
• Therefore it is obvious, the quantity of heat required also depends on a special property. This property is known as specific heat or specific heat capacity of the body. It depends on the material of the body.
• It is obvious that different materials possess different specific heat capacities.

Specific Heat Or Specific Heat Capacity Definition: Specific heat capacity is the quantity of heat required to raise the temperature of unit mass of a substance by unit amount.

Definition of specific heat capacity in CGS system: It is the quantity of heat required to raise the temperature of 1 g of a substance through 1 °C. In this system the unit of specific heat is cal · g^-1 · °C^-1.

For example, ‘the specific heat capacity of copper is 0.093 cal · g^-1 · °C^-1 means that to raise the temperature of 1 g of copper by 1 °C, the heat required is 0.093 cal.

Definition of specific heat capacity in SI: The quantity of heat required to raise the temperature of 1 kg of a substance through 1 K is known as specific heat capacity of the substance.

The unit of specific heat in this system is J · kg^-1 · K^-1

In SI, the specific heat capacity of water is 4186 J · kg^-1 · K^-1.

Relation between CGS and SI unit of specific heat capacity: \(1 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}=\frac{4.186 \mathrm{~J}}{10^{-3} \mathrm{~kg} \times 1 \mathrm{~K}}=4186 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\)

Specific heat capacities of a few solids and liquids

 

Calorimetry Fundamentals Principal Of Calorimetry

When two bodies at different temperatures are brought in contact, heat is transferred from the body at higher temperature to that at lower temperature. So, the former starts to cool down whereas the latter starts to warm up.

The flow of heat continues till both reach the same temperature. The state where both the bodies are at the same temperature is called thermal equilibrium.

Now we know that heat is a form of energy. According to the law of conservation of energy, energy can neither be created nor be destroyed.

So during such heat transfers, if heat does not transform into another form of energy (or, if no other form of energy transforms into heat energy) then, heat released by a body = heat absorbed by another body

This is the fundamental principle of calorimetry. This is also true for any number of bodies brought in contact with one another. In that case, we can write down the principle of calorimetry as: heat released by the hot bodies = heat absorbed by the cold bodies.

Calorimetry Calorimeter

A calorimeter is a cylindrical metallic container mostly used in calorimetric experiments which is generally made of copper. The container is partly filled with a proper liquid called a calorimetric substance.

• A stirrer (S) made of copper is used to stir the liquid and a thermometer (T) is used to measure the temperature of the liquid. Both are immersed in the liquid.
• In a calorimetric experiment when an object is dipped into the liquid of the calorimeter there must be a difference in temperature between the object and the liquid. After some time the object, the calorimeter, and the liquid inside the calorimeter reach thermal equilibrium.

Calorimetry Experiments for Class 11

During this interval of time if

1. No heat exchange occurs between the calorimeter and the surrounding,
2. No exothermic or endothermic reaction (physical or chemical) happens inside the calorimeter and
3. The object does not dissolve in the liquid, then according to the fundamental principle of calorimetry, heat released by the hot objects = heat absorbed by the cold objects.

There will be error in the experimental result if the above three conditions are not satisfied. Heat exchange may take place between the calorimeter and the environment by means of

1. Conduction,
2. Convection and
3. Radiation.

A calorimeter is so constructed that heat exchange by any of the three processes can be minimised.

Calorimetric Substance

Water Advantages:

1. For the same rise or fall in temperature, heat gained or lost by water is more than that gained or lost by the same mass of other solids or liquids of lower specific heat.
2. Water is commonly used as a calorimetric substance because water is easily available.

Water Disadvantages:

1. Again due to high specific heat of water, for absorption of same amount of heat, rise in the temperature of water is less than that of other liquids of lower specific heat. Now if the increase of temperature is low then there is high chance of error in reading.
2. The boiling point of water is 100°C. If an object of temperature more than 100°C is mixed with water, a sudden vaporisation occurs. Due to these reasons water is not useful as a calorimetric substance.

Aniline: Recendy aniline has been recognized as the best calorimetric substance. Aniline is present in pure form. Its specific heat is low (0.62 cal · g^-1 • °C^-1) and boiling point is high (183.9°C).

 

Calorimetry Conclusion

If the state of a body remains unchanged, heat gained or lost by the body depends on

1. Rise or fall in temperature of the body,
2. Mass of the body and
3. Nature of the material of the body.

Heat absorbed or released by a body only for a change in its temperature can be measured. Total heat contained in a body at a particular temperature cannot be measured.

• The amount of heat required by lg of pure water for the rise of its temperature from 14.5°C tol5.5°C is called 1 cal.
• The amount of heat required by lg of pure water for the rise in its temperature from 0° to 100°, divided by 100, gives the value of mean calorie.
• The SI unit of heat is joule. 1 cal = 4.2 J.
• Specific heat is the quantity of heat required to raise the temperature of unit mass of a substance through 1 degree.

According to this definition, units of specific heat: cal · g^-1 · °C^-1 (in CGS system) and J · kg^-1 · K^-1 (in SI).

Thermal capacity of a body is defined as the quantity of heat required to raise its temperature by unity.

Unit of thermal capacity: cal • °C^-1 (in CGS system) and J · K^-1 (in SI).

Water equivalent of a given body is the mass of water for which the rise in temperature is the same as that for the body, when the amount of heat supplied is the same for both.

Unit of water equivalent: g (in CGS system) and kg (in SI).

During heat transfer, if heat energy is not converted into any other form of energy (or no other form of energy is converted into heat energy), then
heat lost by one body = heat gained by another body.

This is the basic principle of calorimetry.

Calorimetry Useful Relations For Solving Numerical Problems

If no change of state occurs, then the heat gained or lost by a body, H = mst.

[where m = mass of the body, s = specific heat, t = change in temperature]

Thermal capacity of a body = ms = H/t

Thermal capacity per unit mass of a body = specific heat of the material of the body

Water equivalent of a body, W = \(\frac{m s}{s_w}\)

According to the basic principle of calorimetry, for more than two bodies at different temperatures in contact, heat lost by hot bodies = heat gained by cold bodies.

In CGS system, specific heat of water = 1 cal · g^-1 · °C^-1, and in SI, the specific heat of water = 4200 J · kg^-1 · ^-1.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
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Calorimetry Very Short Answer Type Questions

Examples of Calorimetry Problems

Question 1.Which substance, used in our everyday life, has the highest specific heat?
Answer: Water

Question 2. For a body of mass m and specific heat capacity s, how much heat is required to increase its temperature by t?
Answer: mst

Question 3. There are two spheres of the same radius. One is solid but the other is hollow. If both the spheres are heated to the same temperature and then allowed to cool down, which sphere will cool faster?
Answer: Hollow

Question 4. What is the CGS unit of thermal capacity?
Answer:

The CGS unit of thermal capacity

“cal • °C^-1”

Question 5. What is the CGS unit of water equivalent?
Answer:

The CGS unit of water equivalent

“g”

Question 6. Which substance is treated as the best calorimetric substance nowadays?
Answer: Aniline

Calorimetry Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, statement 2 is true.

Question 1.

Statement 1: A gas have a unique value of specific heat.

Statement 2: Specific heat is defined as the amount of heat required to raise the temperature of unit mass of the substance through unit degree.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 2.

Statement 1: When a body of mass M loses heat, the time rate of fall of temperature for given amount of loss of heat is inversely proportional to mass.

Statement 2: ΔQ = MsΔT where, ΔQ = amount of heat, s = specific heat, andΔT = decrease in temperature.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: Specific heat capacity is the cause of formation of land and sea breeze.

Statement 2: The specific heat of water is more than land.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 4.

Statement 1: Specific heat of a body is always greater than its thermal capacity.

Statement 2: Thermal capacity is the heat required for raising temperature of a body by unity.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 5.

Statement 1: Two bodies at different temperatures, if brought in thermal contact do not necessarily settle to the mean temperature.

Statement 2: The two bodies may have different thermal capacities.

Answer: 1. Statement 1: Two bodies at different temperatures, if brought in thermal contact do not necessarily settle to the mean temperature.

 Calorimetry Match Column 1 And Column 2

Question 1. Three liquids A, B, and C having the same specific heat and masses m, 2 m, and 3 m have temperatures 20°C, 40°C, and 60°C respectively. Temperature of the mixture when

Answer: 1. E, 2. A, 3. D, 4. B

Calorimetry Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A vacuum flask contains 0.4 kg liquid paraffin whose temperature can be increased at the rate of 1°C per minute using an immersion heater of power 12.3 W. When a heater of 19.2 W is used to heat 0.5 kg of liquid paraffin, in the same flask, the rate of rise of temperature is 1.2°C per minute.

1. What is the specific heat of the paraffin?

1. 0.529cal · g^-1 · °C^-1
2. 0.429 cal · g^-1 · °C^-1
3. 0.472cal · g^-1 · °C^-1
4. 0.62.3 cal · g^-1 °C^-1

Answer: 1. 0.529 cal · g^-1 · °C^-1

2. What is the thermal capacity of the material of the flask?

1. 17.67cal · °C^-1
2. 16.85 cal · °C^-1
3. 17.01cal · °C^-1
4. 16.63 cal · °C^-1

Answer: 3. 17.01 cal · °C^-1

Question 2. When a block of metal of specific heat 0.1 caI · g^-1 · °C^-1 and weighing 110 g is heated to 100°C and then quickly transferred to a calorimeter containing 200 g of a liquid at 10°C, the resulting temperature is 1°C. On repeating the experiment with 400 g of the same liquid in the same calorimeter at same initial temperature, the resulting temperature is 14.5°C.

1. Find the specific heat of the liquid.

1. 0.42 cal · g^-1 · °C^-1
2. 0.52 cal · g^-1 · °C^-1
3. 0.48 cal · g-1 · °C^-1
4. 0.62 cal · g^-1 · °C^-1

Answer: 3. 0.48 cal · g^-1 · °C^-1

2. Find the water equivalent of calorimeter.

1. 15.5 g
2. 16.6 g
3. 17.3 g
4. 18.2 g

Answer: 2. 16.6 g

Calorimetry Integer Type Questions And Answers

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9.

Question 1. 50 g of copper is heated to increase its temperature by 10°C. If the same quantity is given to 10 g of water, what will be the rise in its temperature (in °C)?
Answer: 5

Question 2. The water of volume 2 L in a container is heated with a coil of 1 kW at 27°C. The lid of the container is open and energy dissipates at rate of 160 J · s^-1. In now much time (in minutes) temperature will rise from 27 °C to 75°C? [Given specific heat of water is 4.2 kJ • kg·^-1 ]
Answer: 8

Question 3. If 0.5 kg of coal on burning raises the temperature of 50 L of water from 20 °C to 90 °C, then the heat of combustion of coal is n x 10⁶ cal · kg^-1. Find the value of n.
Answer: 7

Question 4. When x gram of steam is mixed with y grams of ice at 0°C, we obtain (x + y) grams of water at 100°C. Find the value of y/x
Answer: 3

Calorimetry Short Answer Type Questions

Question 1. A 10 Watt electric heater is used to heat a container filled with 0.5 kg of water. It is found that the tempera¬ture of water and the container rises by 3°K in 15 minutes. The container is then emptied, dried, and filled with 2 kg of oil. The same heater now raises the tem¬perature of the container-oil system by 2°K in 20 minutes. Assuming that there is no heat loss in the process and the specific heat of water as 4200 J • kg^-1 • K^-1, the specific heat of oil in the same unit is equal to

1. 1.50 x 10³
2. 2.55 x 10³
3. 3.00 x 10³
4. 5.10 x 10³

Answer:

Given

A 10 Watt electric heater is used to heat a container filled with 0.5 kg of water. It is found that the temperature of water and the container rises by 3°K in 15 minutes. The container is then emptied, dried, and filled with 2 kg of oil. The same heater now raises the tem¬perature of the container-oil system by 2°K in 20 minutes. Assuming that there is no heat loss in the process and the specific heat of water as 4200 J • kg^-1 • K^-1,

In first case the temperature of water and the container is increased by 3K in 15 min or (15 x 60)s by heating.

So, 10 x (15 x 60) = (0.5 x 4200 x 3) + (Wx 3)

[W = water equivalent]

or, \(W=\frac{9000-6300}{3}=900 \mathrm{~kg}\)

In second case, the heater increases the temperature of the oil and the container by 2K in 20 min or (20 x 60)s.

So, 10 x (20×60) = (2 x s x 2) + (W x 2)

[s = specific heat of oil]

or, s = \(\frac{12000-2 W}{4}=\frac{12000-1800}{4}=\frac{10200}{4}\)

= 2550 = 2.55 x 10³ J • kg^-1 · K^-1

The option 2 is correct.

Thermal Properties of Matter and Calorimetry

Question 2. Three bodies of the same material and having masses m, m, and 3m are at temperatures 40°C, 50°C, and 60°C respectively. If the bodies are brought in the thermal contact, the final temperature will be

1. 45°C
2. 54°C
3. 52°C
4. 48°C

Answer:

Given

Three bodies of the same material and having masses m, m, and 3m are at temperatures 40°C, 50°C, and 60°C respectively. If the bodies are brought in the thermal contact,

If the final temperature is t°C,

ms(t-40) + ms(t-50) + 3ms(t-60) = 0

or, (t-40) + (t-50) + 3(t-60) = 0

or, 5t-270 = 0

∴ t = 54°C

The option 2 is correct.

Question 3. A copper ball of mass 100 g is at a temperature T. It is dropped in a copper calorimeter of mass 100 g, filled with 170 g of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by: (given- room temperature = 30°C, specific heat of copper= 0.1 cal/g °C)

1. 800°C
2. 885°C
3. 1250°C
4. 825°C

Answer:

Given

A copper ball of mass 100 g is at a temperature T. It is dropped in a copper calorimeter of mass 100 g, filled with 170 g of water at room temperature. Subsequently, the temperature of the system is found to be 75°C.

From the fundamental principle of calorimetry, heat released by copper ball = heat absorbed by calorimeter and water

or, 100 x0.1(t-75) = 100 x 0.1 x (75-30) + 170 x 1 x (75-30)

or, t = 885°C

The option 2 is correct.

Question 4. Explain why the coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
Answer:

This is due to the fact that heat absorbed by a substance is directly proportional to the specific heat of the substance.

Question 5. Explain why two bodies at different temperatures T₁ and T₂ if brought in thermal contact do not necessarily settle to the mean temperature (T₁ + T₂)/2.
Answer:

If two bodies, at different temperatures, come in thermal contact, heat flows from the body at higher temperature to the body at lower temperature till the temperature becomes same. The final temperature can be mean temperature, i.e., (T₁ + T₂)/2, only when both the bodies have equal thermal capacities.

 

Thermodynamics

• Elasticity Questions and Answers
• Hydrostatics Question And Answers
• Change Of State Of Matter Long Answer Questions
• Circular Motion Long Answer Questions
• Doppler Effect In Sound and Light Question and Answers
• Expansion Of Solid And Liquids Long Answer Questions
• For Calorimetry Questions and Answers
• Friction Long Answer Type Questions
• Kinetic Theory Of Gases Question and Answers
• Measurement And Dimension Of Physical Quantity Long Answer Questions
• Newton Law Of Motion Long Answer Questions
• Newtonian Gravitation And Planetary Motion Long Answer Questions
• One-Dimensional Motion Long Answer Questions
• Simple Harmonic Motion Long Answer Type Questions
• Statics Long Answer Questions
•  Superposition Of Waves Question and Answers
• Thermometry Question and Answers
• Transmission Of Heat Long Answer Questions
• Vector Question and Answers
• Viscosity And Surface Tension Long Answer Questions
• Wave Motion Question and Answers

First And Second Law Of Thermodynamics Introduction

Position, displacement, velocity, acceleration, etc., are external properties of a body. From them, we get no information about the nature of the body. They are discussed in the subject of mechanics.

• On the other hand, properties like mass (M), volume (V), pressure (p), temperature (T), density (ρ), etc., are internal properties of a body. These are bulk (macroscopic) properties and can be measured by simple experiments.
• Any change in these properties generally involves the transfer of energy in the form of heat or work. All of these are studied in a branch of Physics, known as thermodynamics. The above-mentioned properties and similar others are called thermodynamic properties.
• The subject is based only on experi¬mental data and every thermodynamic formula comes from the analysis of this data. For example, suppose a fixed mass of an ideal gas is studied at constant temperature.
• Any change of pressure produces a change in volume and different sets of p and V are obtained. An analysis of these measured values shows that the product pV = constant. This is the well-known Boyle’s law. As it is experiment-based, it is a thermodynamic law.

Thermodynamic Systems: A body or, a part of a body, or a combination of a number of bodies that is being studied is usually called a thermodynamic system. In general, such a system interacts with its surroundings or, environment and exchanges energy and mass.

The environment or surroundings include all external objects that have some influence on the system. For example, if a room in a house is considered as a system and its thermal properties are to be studied, the sun at a distance of 1.5 x 10⁸ km is an important component of the environment, but another house only 100 m away need not be included as a component. Thermodynamic systems are of three types.

Isolated system: This system does not exchange any energy or mass with the surroundings.

Closed system: Energy is exchanged with the surroundings, but the mass remains fixed.

Open system: Both energy and mass are exchanged. We shall not discuss open systems in this chapter.

Properties like a number of molecules, molecular velocity, etc., are internal microscopic properties of a body. They are not directly measurable and experimental data is absent. Hence, the study of this subject is based on theoretical assumptions and is known as kinetic theory.

Thermodynamics – First And Second Law Of Thermodynamics Nature Of Heat

Heat is a form of energy which can be identified only when it is in transit. It can be converted into other forms of energy and vice versa. To measure the amount of heat energy, different units are used:

CGS System: calorie: (cal)

SI: Joule (J)

• In SI, the unit of energy is directly used to measure some quantity of heat. So, Joule (J) is the unit of heat in SI, as it is the SI unit of energy. Similarly, other energy units like erg may also be used.
• Immediately it becomes clear that the energy units must have definite relations with the conventional units of heat. This topic is discussed in the next section.
• In thermodynamics, the definition of temperature comes earlier from the zeroth law of thermodynamics. The definition of heat is based on the concept of temperature.

Nature of heat Definition: Heat is a form of energy which is transmitted from one place to another due to their temperature differences only.

It is interesting to note the difference between this and the calorimetric definition of heat.

Thermodynamics – First And Second Law Of Thermodynamics Mechanical Equivalent Of Heat Joules Law

The widely used unit of heat is calorie (or cal, in CGS system). We know that heat is a form of energy; so energy units like erg (CGS system) and joule (J in SI) should somehow be related to the units of heat.

• James Prescott Joule was the first experimentalist who accurately measured the mechanical energy equivalent to some amount of heat energy i.e., the mechanical equivalent of heat.
• This measurement determines the amount of mechanical energy expressed in joule (or erg) which is equivalent to heat energy of 1 cal.
• Mechanical energy may be used to do work and work produces heat. On the other hand, heat may be converted to work or mechanical energy. This indicates that there is a natural relationship between work and heat. This relation is known as Joule’s law.

Joule’s law: If some amount of work is entirely converted to heat, then the work done and the heat produced are proportional to each other.

If W=workdoneand H=heat produced,then from Joule’slaw,

W ∝ H or, W = JH…..(1)

Here, J is a constant. This constant is called the mechanical equivalent of heat or Joule’s equivalent.

Definition of J and Hs magnitudes in different systems: In equation (1), if H = 1, then W = J. So mechanical equivalent of heat is defined as the amount of work done to produce a unit amount of heat.

Clearly, the unit and the magnitude of J depends on the unit of heat and work in different systems of units.

CGS system: The unit of H is calorie and the unit of W is erg.

As J = \({W}{H}\) the unit of J is erg · cal^-1 and the measured value of J = 4.2 x 10⁷ erg · cal^-1.

As 1 J = 10⁷ erg, the value of J = 4.2 J · cal-1.

The units like erg · cal^-1 and J · cal^-1 indicate that the mechanical equivalent of heat J actually denotes the conversion factor between two units. For example, J = 4.2 J · cal^-1.

This means that 1 cal = 4.2 J, i.e., each calorie should be multiplied by 4.2 to express the energy in joule.

In SI: Here W and H are expressed in the same unit, which is joule (J). So, no conversion factor is necessary. The mechanical equivalent of heat is J = 1. This means that the concept of J is unnecessary in SI.

Relation of calories with erg and joule: 1 cal = 4.2 x 10⁷ erg = 4.2 J.

Thermodynamics – First And Second Law Of Thermodynamics Mechanical Equivalent Of Heat Numerical Examples

Second Law of Thermodynamics Explained

Example 1. 800 g lead balls are kept in a 1 m long, vertical tube which is a bad conductor and the tube is then closed at both ends. The tube is suddenly inverted so that the balls fall from one end to the other. The I temperature of the lead balls increases by 3.89°C after 50 such inversions. Find out the mechanical equivalent of heat assuming that the lead balls have retained the entire amount of heat produced.
Solution:

Given

800 g lead balls are kept in a 1 m long, vertical tube which is a bad conductor and the tube is then closed at both ends. The tube is suddenly inverted so that the balls fall from one end to the other. The I temperature of the lead balls increases by 3.89°C after 50 such inversions.

Work done, W= loss of potential energy of the balls = n x mgh

= 50 x 800 x 980 x 100 erg (1 m = 100 cm)

Heat produced, H = mst = 800 x 0.03 x 3.89 cal

∴ J = \(\frac{W}{H}=\frac{50 \times 800 \times 980 \times 100}{800 \times 0.03 \times 3.89}=4.2 \times 10^7 \mathrm{erg} \cdot \mathrm{cal}^{-1}\).

Example 2. Water drops from a height of 50 m in a waterfall. Find out the rise in temperature of water, if 75% of its energy is converted into heat and absorbed by water. (\(J=4.2 \times 10^7 \mathrm{erg} \cdot \mathrm{cal}^{-1} ; \mathrm{g}=9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\))
Solution:

Given

Water drops from a height of 50 m in a waterfall.

Here, W = mgh x 75/100 and H = mst

Now, W = JH

or, \(m g h \times \frac{75}{100}=J m s t\)

or, \(t=\frac{m g h \times \frac{75}{100}}{J m s}\)

[s = \(J \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\) for water, g = \(9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\)=980 \(\mathrm{~cm} \cdot \mathrm{s}^{-2}, h=50 \mathrm{~m}=50 \times 100 \mathrm{~cm}\))

= \(\frac{g h}{J s} \times \frac{75}{100}=\frac{980 \times 50 \times 100}{4.2 \times 10^7 \times 1} \times \frac{75}{100}=0.0875^{\circ} \mathrm{C}\)

Real-Life Applications of the Second Law of Thermodynamics

Example 3. The velocity of a 42 kg celestial body reduces from 20 km · min^-1 to 5 km · min^-1 due to its passage through the earth’s atmosphere. Find out the heat produced in calories. (J = 4.2 x 10⁷ erg · cal^-1)
Solution:

Given

The velocity of a 42 kg celestial body reduces from 20 km · min^-1 to 5 km · min^-1 due to its passage through the earth’s atmosphere.

The initial velocity of the celestial body,

u = \(20 \mathrm{~km} \cdot \mathrm{min}^{-1}=\frac{20 \times 1000}{60} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Final velocity of the celestial body,

v = \(5 \mathrm{~km} \cdot \min ^{-1}=\frac{5 \times 1000}{60} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Work done, W = change in kinetic energy

= \(\frac{1}{2} m u^2-\frac{1}{2} m v^2=\frac{1}{2} m\left(u^2-v^2\right)\)

= \(\frac{1}{2} \times 42 \times\left(\frac{1000}{60}\right)^2\left(20^2-5^2\right)\)

= \(\frac{42 \times 10^4 \times 375}{2 \times 36} \text { joule }\)

∴ Heat produced,

H = \(\frac{W}{J}=\frac{42 \times 10^4 \times 375}{2 \times 36 \times 4.2} \)

(J = \(4.2 \times 10^7 \mathrm{erg} \cdot \mathrm{cal}^{-1}=4.2 \mathrm{~J} \cdot \mathrm{cal}^{-1}\))

= \(5.2 \times 10^5 \mathrm{cal}\).

Example 4. Find out the amount of work done to convert 100 g ice at 0°C to water at 100°C. (Latent heat of fusion of ice = 80 cal · g^-1; the mechanical equivalent of heat = 4.2 J · cal^-1).
Solution:

Total heat supplied,

H = 100 x 80 + 100 x 1 x (100-0)

= 8000 + 10000 = 18000 cal

∴ Work done, W = JH = 4.2 x 18000 = 75600 J.

Example 5. What will be the temperature difference between the top and the bottom of a 400 m high waterfall, assuming that 80% of the heat produced is retained by the water?
Solution:

Second Law of Thermodynamics and Entropy

Work done, W = loss in potential energy

= mg(h- 0) = mgh

∴ Heat produced, H = \(\frac{W}{J}=\frac{m g h}{J}\)

∴ Amount of heat retained by water = \(\frac{m g h}{J} \times \frac{80}{100}\)

If the increase in temperature of water is t, required heat = mst

According to the question,

mst = \(\frac{m g h}{J} \times \frac{80}{100}\)

or, t = \(\frac{g h}{J s} \times \frac{80}{100}=\frac{9.8 \times 400}{4.2 \times 1000} \times \frac{80}{100}\)

(specific heat of water in SI s = \(1000 \mathrm{cal} \cdot \mathrm{kg}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\))

or, t = 0.747 °C

Example 6. Find out the minimum height from which a piece of ice at 0°C should be dropped so that it melts completely due to its impact with the ground. Assume that half of the energy loss due to the fall is responsible for the fusion of ice. (Latent heat of fusion of ice = 80 cal · g^-1, g = 980 cm · s^-2, J = 4.2 J · cal^-1)
Solution:

Let the piece of ice of mass m be allowed to fall from height h.

Now, energy loss due to the fall = loss in potential energy = mgh = work done, W.

Half of it, i.e., W/2 amount of energy is converted into heat and it is responsible for the fusion of ice.

∴ Heat produced, \(H=\frac{W / 2}{J}=\frac{m g h}{2 J}\)

As mg of ice melts into water, required latent heat = 80 m cal

So, \(\frac{m g h}{2 J}=80 \mathrm{~m}\)

or, \(h=\frac{80 \times 2 J}{g}=\frac{80 \times 2 \times 4.2 \times 10^7}{980}\)

(J = \(4.2 \mathrm{~J} \cdot \mathrm{cal}^{-1}=4.2 \times 10^7 \mathrm{erg} \cdot \mathrm{cal}^{-1}\))

= \(6.857 \times 10^6 \mathrm{~cm}=68.57 \mathrm{~km}\)

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Example 7. A piece of ice at 0°C is dropped to the ground from some height. The piece of ice melts completely due to its impact on the ground. Find the height from which the piece was dropped considering that 60% of its energy is converted into heat. (J = 4.2 J · cal^-1)
Solution:

Given

A piece of ice at 0°C is dropped to the ground from some height. The piece of ice melts completely due to its impact on the ground.

The potential energy of the piece of ice of mass m at height h = mgh, kinetic energy = 0

So, total mechanical energy = mgh

This energy is conserved till it touches the ground. Due to the impact with the ground, 60% of this energy,

i.e., mgh x 60/100 or 0.6 mgh is converted into heat energy.

∴ Heat produced = \(\frac{0.6 \mathrm{mgh}}{J}\)

Again, heat required to melt m g of ice = mL

[L = latent heat of fusion of ice]

∴ \(\frac{0.6 m g h}{J}=m L\)

or, h = \(\frac{J L}{0.6 g}\left[L=80 \mathrm{cal} \cdot \mathrm{g}^{-1}=80 \times 1000 \mathrm{cal} \cdot \mathrm{kg}^{-1}\right]\)

= \(\frac{4.2 \times(80 \times 1000)}{0.6 \times 9.8}=5.71 \times 10^4 \mathrm{~m}=57.1 \mathrm{~km}\)

Example 8. A stirrer rotates in 1 l of water against a damping force of 0.1 N at 360 rpm in a circle of radius 5 cm. Calculate the rise in water temperature in 1 h, neglecting heat loss due to radiation. (J = 4.2 J · cal^-1)
Solution:

Given

A stirrer rotates in 1 l of water against a damping force of 0.1 N at 360 rpm in a circle of radius 5 cm.

Mass of 1 l of water, m = 1 kg.

If t is the rise in temperature of water in 1 h, then heat produced, H = mst.

Now, circumference of the circle =2πr; number of rotations in 1 h = 360 x 60

∴ Displacement of the stirrer, d = 360 x 60 x 2πr

Work done, W = force x displacement = Fd

∴ W = JH or, Fd = Jmst

or, t = \(\frac{F d}{J m s}=\frac{0.1 \times(360 \times 60 \times 2 \times \pi \times 0.05)}{4.2 \times 1 \times 1000}\)

(F = \(0.1 \mathrm{~N} ; r=5 \mathrm{~cm}=0.05 \mathrm{~m}\))

s = \(1000 \mathrm{cal} \cdot \mathrm{kg}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

= 0.162°C.

Understanding Heat Transfer in Thermodynamics

Example 9. 10 l of water is dropped from a height of 250 m. How much heat (in calories) will be generated when the water reaches the bottom? Assuming that the entire heat will be retained by the mass of water, what will be the rise in temperature of the water? (Given J = 4.18 J · cal^-1)
Solution:

Given

10 l of water is dropped from a height of 250 m.

Mass of 10 l of water, m = 10 kg.

Kinetic energy on impact with the ground

= initial potential energy = mgh = work done (W)

So, heat generated,

H = \(\frac{W}{J}=\frac{m g h}{J} \mathrm{cal}\)

The specific heat of water,

s = \(1000 \mathrm{cal} \cdot \mathrm{kg}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

If t is the rise in temperature, then

mst = H = \(\frac{m g h}{J}\)

or, t = \(\frac{g h}{J s}=\frac{9.8 \times 250}{4.18 \times 1000}=0.586^{\circ} \mathrm{C}\)

Example 10. What will be the time required to heat a 151 bucket full of water from 20°C to 40°C using a 1500W immersion heater?
Solution:

Mass of 15 1 of water, m = 15 kg specific heat of water, s = 1000 cal · kg^-1 · °C^-1

Now, W = JH or, Pt = Jmsθ

where P = power of the heater = 1500W; t = time required; θ = rise in temperature = 40 – 20 = 20°C.

So, t = \(\frac{J m s \theta}{P}=\frac{4.2 \times 15 \times 1000 \times 20}{1500}=840 \mathrm{~s}=14 \mathrm{~min} \text {. }\)

Example 11. The temperature of a piece of lead is 27°C. Find out the minimum velocity of its impact with a wall so that it melts completely. Suppose 58% of the heat generated is dissipated. Given, J = 4.2 J · cal^-1; melting point, specific heat, and latent heat of fusion of lead are 327°C, 0.03 cal · g^-1 · °C^-1 and 5 cal · g^-1, respectively.
Solution:

Given

The temperature of a piece of lead is 27°C.

⇒ \(H \times \frac{100-58}{100}=m s t+m l\)

or, \(H=(m s t+m l) \times \frac{100}{42}\)

Again, W = \(\frac{1}{2} m v^2\)

Now, W=J H

or, \(\frac{1}{2} m v^2=J(m s t+m l) \times \frac{100}{42}\)

or, \(v^2=2 J(s t+l) \times \frac{100}{42}\)

= \(2 \times\left(4.2 \times 10^7\right) \times[0.03 \times(327-27)+5] \times \frac{100}{42}\)

= \(28 \times 10^8\)

So, \(v=5.29 \times 10^4 \mathrm{~cm} \cdot \mathrm{s}^{-1}=529 \mathrm{~m} \cdot \mathrm{s}^{-1}\).

Example 12. A body of mass 2 kg is pulled with a velocity of 2 m · s^-1 on a horizontal surface. What will be the heat produced in 5s, if the coefficient of friction between the body and the surface is 0.2? Given, J = 4.21 • cal^-1; g = 9.8 m• s^-2.
Solution:

Given

A body of mass 2 kg is pulled with a velocity of 2 m · s^-1 on a horizontal surface.

Force of friction on the body, F = μmg

So the work done, W = Fs = μ mg · vt

∴ Heat produced, H = \(\frac{W}{J}=\frac{\mu \mathrm{mg \nu t}}{J}\)

= \(\frac{0.2 \times 2 \times 9.8 \times 2 \times 5}{4.2}\)

= 9.33 cal

Example 13. Two pieces of ice, of equal mass moving towards each other with the same velocity, collide head-on and are vaporized due to this collision. Find out the minimum initial velocity of the pieces of ice. Given, that the initial temperature, specific heat, and latent heat of fusion of ice are -12°C, 0.5 cal · g^-1 · °C^-1, and 80 cal · g^-1 respectively latent heat of steam is 540 cal · g^-1.
Solution:

Given

Two pieces of ice, of equal mass moving towards each other with the same velocity, collide head-on and are vaporized due to this collision.

The initial temperature, specific heat, and latent heat of fusion of ice are -12°C, 0.5 cal · g^-1 · °C^-1, and 80 cal · g^-1 respectively latent heat of steam is 540 cal · g^-1

Let m = mass of each piece of ice; v = minimum initial velocity.

Here, work done due to collision, W

= loss in kinetic energy = 2 x 1/2 mv² = mv²

Heat produced, H = 2[m x 0.5 x 12 + m x 80 + m x 1 x 100 + m x 540]

= 1452 m cal

∴ W = JH or, mv² = J x 1452m

or, v² = (4.2 x 10⁷)x 1452 = 6.1 x 10¹⁰

So, v = 2.47 x 10⁵ cm · s^-1 = 2.47 km · s^-1.

Example 14. The top of a waterfall is at a temperature 0.49°C below that of the bottom. The work done by water due to its fall is converted entirely into heat. Find the height of the waterfall. Given, g = 980 cm · s^-2, J = 4.2 J · cal^-1.
Solution:

Given

The top of a waterfall is at a temperature 0.49°C below that of the bottom. The work done by water due to its fall is converted entirely into heat.

Let m = mass of water falling from height h.

Work done, W = kinetic energy of impact

= initial potential energy = mgh.

Now, W = JH or, mgh = Jmst

or, h = \(\frac{J s t}{g}\)

= \(\frac{\left(4.2 \times 10^7\right) \times 1 \times 0.49}{980}\)

= 21000 cm =210 m.

Thermodynamics – First And Second Law Of Thermodynamics Intensive And Extensive Variables

Examples of the Second Law in Nature

Let us consider a wooden table and a wooden chair. The mass (m) and volume (V) of them are different because such properties depend on the whole body. These are the extrinsic thermodynamic properties.

On the other hand, the density (ρ) is the same for both the bodies. This property does not depend on the whole body, but only on the material (in this case, wood) of the body. This is an intrinsic thermodynamic property.

Generally, thermodynamic variables are of two types

1. Intensive variable and
2. Extensive variable.

The variables that do not depend on the amount of matter or mass in a thermodynamic system are called intensive variables. Pressure, temperature, density, surface tension, etc., are intensive variables.

• The variables that are proportional to the amount of matter or mass in a thermodynamic system are called extensive variables. Volume, magnetic moment, internal energy, entropy, etc., are extensive variables.
• As an illustration we may consider some amount of gas of mass m, volume V, pressure p, temperature T, and density ρ, enclosed in a container.

Now we divide the gas into 4 equal parts. Clearly, mass, volume, pressure, temperature, and density of each part arc \(\frac{m}{4}, \frac{V}{4}, p, T \text { and } \rho\), respectively.

• It means that m and V depend on the whole amount of the gas; they are extensive variables. But ρ, p, T do not depend on the amount; they are intensive variables.
• We consider a relation like mass = density x volume or, m = ρV. On the left-hand side, mass m changes proportionally with the amount of matter in a system. On the right-hand side, volume V changes similarly.
• Thus, the two sides remain equal only if ρ remain the same. So, whereas m and V are extensive, ρ is intensive. In general, in a product of two or more thermodynamic properties, only one is an extensive variable and the others are intensive variables.

So we can have a product like ρV which is extensive (V is extensive, but ρ is intensive). But a product like mV is not allowed, as both m and V are extensive.

Intensive and extensive variable in thermodynamics systems

State Function and Path Function: If a system is in equilibrium, the variables like its volume (V), pressure (p), temperature (T), etc. have definite values. So these values refer to the state of the system and do not depend on the path followed to reach that state. So variables like volume, pressure, temperature, etc. are called state functions or properties of the system.

• On the other hand, the two functions—work (W) and heat (Q) are not related to any state of the system, rather they are relevant to any process of the system. A statement—W_A is the amount of work done in the state A—has no meaning.
• Rather the statement ‘WAB is the amount of work done in die process A → B’ is meaningful. There can be various paths from the state A to the state B. The values of W_AB or Q_AB in each path are different, i.e., these values depend on the paths between the states A and B. So W and Q are called path functions.
• They are not state functions. It is to be noted that if V_A and V_B are the volumes of the system in the states A and B respectively, then the change in volume in the process A → B = V_B– V_A.
• Whatever may be the path taken from A to B the change in volume, V_B-V_A, remains the same. It does not depend on the path. So any change of a state function is independent of the path.

 Thermodynamics – First And Second Law Of Thermodynamics Thermodynamic Equilibrium And Process

Thermodynamic equilibrium: The macroscopic properties that are used to describe a system may change spontaneously or due to an external influence. During such a change, the system and its surroundings interact with each other.

1. The absence of unbalanced force or torque in the interior of a system or between a system and its surroundings implies mechanical equilibrium has been established.
2. For a system in mechanical equilibrium, when there is no spontaneous change of internal structure (by means of a chemical reaction) or transfer of matter from one part of the system to another (by means of diffusion) chemical equilibrium is said to be established.
3. For a system in mechanical and chemical equilibrium, thermal equilibrium is set to be attained if no exchange of heat occurs between the system and its surroundings. Hence it is obvious that in thermal equilibrium, the temperature remains the same throughout the system and is identical with that of the surroundings.

When all the three types of equilibrium stated above an attained by a system, it is said to be in a thermodynami equilibrium in an equilibrium state, or simply, in state.

• A closed thermodynamic system, i.e., a system having a fixed mass, can be described completely by three of its properties—volume (V), pressure (p), and temperature (T). All other properties of the system depend on these properties and therefore are functions of, V, p, and T.
• In thermodynamic equilibrium none of the tires properties of a system— volume, pressure, or temperature—changes with time. In our study, we shall deal with equilibrium states, and V, p, and T will not be treated as functions of time. So, the quantity time will never appear in our formulation of thermodynamic relations.

Moreover, volume, pressure, and temperature are related among themselves by an equation of state. Thus, if two of them are known, the third can be calculated using that equation of state. Hence, there are only two independent properties for a closed thermodynamic system. For example, let a fixed mass of 1 mol of a gas be taken at STP. Then,

p = 1 atm = 76 cm of mercury

= 76 x 13.6 x 980 dyn · cm^-2

T = 0 °C = 273K

If die gas is assumed to be ideal, then from the equation of state pV = RT, we have

V = \(\frac{R T}{p}=\frac{8.31 \times 10^7 \mathrm{erg} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 273 \mathrm{~K}}{76 \times 13.6 \times 980 \mathrm{dyn} \cdot \mathrm{cm}^{-2}}\)

= \(22400 \mathrm{~cm}^3 \cdot \mathrm{mol}^{-1}=22.4 \mathrm{l} \cdot \mathrm{mol}^{-1}\)

Here, p and T played the role of independent properties and V could be calculated from them.

Thermodynamic process: Let a system be initially at an equilibrium state A. Now, if the system exchanges energy, in the form of work or heat, with its surroundings, then the values of V, p, and T would change, in general. So, the system would deviate from the state A.

• But eventually, when the energy exchange stops, the system would attain a new equilibrium state, say B. The transition of a system from an initial state A to a final state B is known as a thermodynamic process and is usually denoted as A → B.
• It should be noted that the transition between two states may occur along different paths 1, 2, …. Each of these paths corresponds to a separate process.

A process, in general, involves simultaneous changes of all the three properties—volume (V), pressure (p), and temperature (T). However, for simplicity of analysis, some special processes are often considered:

1. Isochoric process: The volume of a system remains constant in this process. But, both pressure and temperature undergo some changes.
2. Isobaric process: Here, p remains constant and Vand T change their values.
3. Isothermal process: T is a constant, whereas there are changes in V and p.
4. Adiabatic process: The heat exchange between the system and its surroundings remains zero there are changes in all of V, p, and T.

Thermodynamics – First And Second Law Of Thermodynamics External Work

Hydrostatic system: A system that obeys Pascal’s law is called a hydrostatic system. Important characteristics of such a system are:

1. The pressure is uniform throughout the system and acts normally outwards at every point on the walls.
2. Any additional pressure applied at any point gets transmitted throughout the system in such a way that the pressure everywhere continues to remain uniform.

These conditions are obeyed by

1. Small amounts of liquids for which the effect of gravity may be neglected and
2. Gaseous systems.

However, all liquids are nearly incompressible, i.e., any change in pressure does not produce an appreciable change in volume. Due to this reason, for the thermodynamic analysis of a hydrostatic system, it is convenient to take a closed gaseous system.

Hydrostatic work: An amount of gas is enclosed by an airtight cylinder-piston arrangement. where the piston can move without friction along the inner walls of the cylinder.

These conditions are obeyed by

1. Small amounts of liquids for which the effect of gravity may be neglected and
2. Gaseous systems.

However, all liquids are nearly incompressible, i.e., any change in pressure does not produce an appreciable change in volume. Due to this reason, for the thermodynamic analysis of a hydrostatic system, it is convenient to take a closed gaseous system.

Hydrostatic work: An amount of gas is enclosed by an airtight cylinder-piston arrangement. where the piston can move without friction along the inner walls of the cylinder.

Let p = pressure of the enclosed gas,

a = area of cross-section of the piston.

So, the force acting on the piston = pa.

• Now, suppose the piston goes through a very small displacement outwards from position A to position B; displacement AB = dx.
• This displacement occurs when a slight difference develops between the pressures of this system and its surroundings. A small outwards pull on the piston, a gain of a small amount of heat from the surroundings, etc., may initiate such a displacement.
• During this displacement, however small, the pressure p of the system may change. But at even instant, the pressure must be uniform throughout the system. A sufficiently slow motion of the piston is required to satisfy this hydrostatic condition.

Due to the small displacement dx, the gas expands from volume V to V+dV. Clearly, dV = adx. So, the small work done in this infinitesimal process is,

dW = force x displacement

= (pα)(dx) = p(αdx) = pdV

or concisely, dW = pdV …(1)

Work done in a finite process, for which the volume changes from V₁ to V₂, is

W = \(\int d W=\int_{V_1}^{V_2} p d V\)…(2)

The relation (1) can be interpreted as:

1. When there is no change in volume, V = constant and dV = 0. So, dW = 0 no work is done in such a process.

2. When the volume increases, i.e., for an expansion, dV is positive, i.e., dV> 0. So, dW>0; positive amount of work is done in this process. The system releases some energy to its surroundings it is termed as work done by the system.

3. When the volume decreases, i.e., for a contraction occurring due to an inward motion of the piston, dV is negative, i.e., dV<0.

So, dW< 0; a negative work is done in this process. The system receives some energy from its surroundings it is termed as work done on the system.

The integral in relation (2) can be evaluated for special processes:

1. Isochoric process: V = constant and dV = 0.

So, W = 0 …(3)

2. Isobaric process:

p = constant.

So, W = \(\int_{V_1}^{V_2} p d V=p \int_{V_1}^{V_2} d V=p\left(V_2-V_1\right)\)….(4)

But, when the pressure p of a system changes in a process, the integral cannot be evaluated unless p can be expressed as a function of volume V.

It is only possible if the equation of state of the system is known. Later in this chapter, the expressions for isothermal work and adiabatic work for an ideal gas will be evaluated, using the ideal gas equation of state, pV = R.

There are systems in thermodynamics that are not hydrostatic. For them, the work equations (1) and (2) would be different. However, in this chapter, we shall restrict ourselves mainly to hydrostatic systems.

 

Unit 8 Thermodynamics Chapter 1 First And Second Law Of Thermodynamics Reverse And Irreversible Process

Thermodynamics and Energy Efficiency

Let a thermodynamic system be in an initial state A. Owing to heat exchange and external work, the system attains its final state B. That is, the process is A → B. Now, we shall state the conditions under which the process may be called ‘reversible’. A Process A → B is reversible, if

1. The process B → A occurs in nature and
2. After the processes A →B →A, there is no net change in the surroundings.

The conditions 1 and 2 are called the conditions of reversibility. Process A → B is an irreversible process, if these conditions are not satisfied.

• The concepts of reversibility and irreversibility are direct consequences of the second law of thermodynamics. The words ‘self-acting machine’ in Clausius or Kelvin-Planck statements are closely related to the words ‘no net change in the surroundings’ in the reversibility conditions.
• For example, we consider the process of heat flow from higher to lower temperatures. The reverse process does occur in nature — a domestic refrigerator actually transfers heat from its cool container to comparatively hotter surroundings.
• But it is not self-acting, because some work in the form of electrical energy must be supplied from the surroundings. As a result, the environment suffers a net change.

So we may say that heat transfer from higher to lower temperature is an irreversible process. This is because heat, on its own, cannot flow from lower to higher temperatures (Clausius statement).

• We now consider a process A →B on a p V diagram. Let the process occur along a particular path ACB.
• The opposite process B → A can take different paths from B to A. However, for reversibility, only the path BCA is important.
• Then for each elementary step (say, xy) the heat exchange and the work done in the forward process (x → y) are exactly equal and opposite to those in the reverse process (y → x) – This is essentially the condition for reversibility, equivalent to the conditions discussed earlier.

A process A→ B is reversible, if

1. The process B → A occurs in nature and
2. The heat exchange and the work done for each step in the forward process are exactly equal and opposite to those for the reverse process.

Conditions of reversibility:

1. A process is reversible if there is no dissipation of energy during this process. The origin of dissipative energy are friction, surface tension, resistance, etc. So, a process that occurs against friction, surface tension, etc., cannot be reversible.

2. A process is reversible if it occurs infinitesimally slowly. Every real process in nature is irreversible. A reversible process is only an ideal process, never occurring in nature. Still, the concept is very useful to formulate important thermodynamic relations.

Conditions of reversibility Example:

1. Let a gas be enclosed inside a cylinder-piston arrangement. Now the gas is allowed to get compressed very slowly through an isothermal process by applying force from its surroundings. So work is done on the gas by the piston.

• Now after this process, if the gas by itself expands very slowly pushing the piston up, it returns to the initial state. So work is done on the piston by the gas.
• This work done is nearly equal to the previous one. So apparently, the process is reversible. But actually, in each piston movement, some heat is generated due to friction.
• This heat is absorbed by the surroundings which cannot be restored. So strictly speaking, the process is irreversible, but becomes nearly reversible only when the piston movements are very slow and heat generated due to friction becomes almost negligible.

2. Let us take a container having 10 g of ice floating on 100 g of water at 0°C. If 80 cal of heat is supplied from the surroundings, 1 g of ice melts into 1 g of water.

• Now, if 80 cal of heat is taken away, we again get 10 g of ice on 100 g of water. Here, the surroundings also come to its initial state. So the process of fusion of ice is apparently reversible.
• However, some heat exchange with the surroundings can never be avoided. That heat cannot be recovered in any manner.
• So the surroundings suffer a permanent change. For this reason, processes like fusion and vaporisation are only approximately reversible.

3. Free fall of a body due to gravity from a height h to the ground is an irreversible process because the body on its own cannot move up to the height of h.

4. When containers with two different gases are connected, the gases mix with each other. The gases cannot separate themselves on their own and so the process is irreversible.

 

Unit 8 Thermodynamics Chapter 1 First And Second Law Of Thermodynamics Very Short Answer Type Questions

Question 1. Which branch of physics deals with the study of the relationship between heat and various forms of energy?
Answer: Thermodynamics

Question 2. How is the work done related to the heat produced when work is completely converted into heat?
Answer: Mutually proportional

Question 3. If some amount of work is completely converted into heat, what is the name of the ratio of work done and amount of heat?
Answer: Mechanical equivalent of heat

Question 4. What is the value of the mechanical equivalent of heat in a CGS system?
Answer: 4.2 x 10⁷ erg · cal^-1

Question 5. What is the value of the mechanical equivalent of heat in SI?
Answer: 1

Question 6. What would be the amount of heat generated if 4.2 x 10⁷ erg of work is completely converted into heat?
Answer: 1 cal

Question 7. Give an example of intensive variable.
Answer: Temperature

Question 8. Give an example of an extensive variable.
Answer: Volume

Question 9. Which one of the four quantities does not indicate the thermodynamic state of a substance—volume, temperature, pressure, and work?
Answer: Work

Question 10. What is the change in temperature of water when it falls from the top to the bottom in a waterfall?
Answer: Increases

Question 11. Is pressure intensive or extensive variable?
Answer: Intensive

Question 12. Is magnetic moment intensive or extensive variable?
Answer: Extensive

Question 13. What would be the amount of heat required to increase the temperature of 1 g of water by 1°C?
Answer: 4.2

Question 14. What is the change in internal energy of a substance when it is heated?
Answer: Increases

Question 15. On which factor does the internal energy of a certain amount of an ideal gas depend?
Answer: On its temperature

Question 16. Will there be any change in the internal energy of a certain amount of gas, if its pressure or volume changes at constant temperature?
Answer: No

Question 17. In case of expansion of a gas, would the work done by the gas be positive or negative?
Answer: Positive

Question 18. What is the SI unit of molar-specific heat?
Answer: J · mol^-1 · K^-1

Question 19. The first law of thermodynamics is the mathematical form of a universal law. Name it.
Answer: Law of conservation of energy

Question 20. How many types of specific heat of a gas are used in practice?
Answer: Two

Question 21. How many types of molar-specific heat of a gas are used in practice?
Answer: Two

Question 22. Which is greater — specific heat at constant volume, c_v, or specific heat at constant pressure, c_p?
Answer: c_p

Question 23. How is the heat supplied to a substance in an isothermal process used?
Answer: External work

Question 24. In which process does the heat accepted or rejected become zero?
Answer: Adiabatic

Question 25. Is the reversible process slow or fast?
Answer: Slow

Question 26. Which process does not suffer dissipative forces reversible or irreversible?
Answer: Reversible process

Question 27. Is the isothermal process slow of fast?
Answer: Very slow process

Question 28. Is the adiabatic process slow or fast?
Answer: Very fast process

Question 29. Is rusting of iron a reversible process?
Answer: No

Question 30. Hot milk is poured into a cup of tea a. d is mixed with a spoon. Is this an example of a reversible process?
Answer: No

Question 31. The sum of kinetic and potential energies of the molecules of a substance is equal to its ______
Answer: Internal energy

Question 32. In a process, if dU, dW, and dQ are changes in internal energy, work done, and heat accepted respectively for a system, what is the relation between dU, dQ, and dW?
Answer: dQ = dU+dW

Question 33. p and V are the pressure and volume respectively of a gas of a particular mass. If volume changes to V+ dV, what is the work done in the process?
Answer: pdV

Question 34. The molecular weight of a gas is M. If the specific heat and molar specific heat at constant volume of the gas is c_v and C_v respectively. Write down the relation between c_v > C_v and M.
Answer: C_v = Mc_v

Question 35. A bicycle pump becomes hot when air is pumped into the tube. Why?
Answer: Due to adiabatic compression

Question 36. Air coming out from a burst bicycle or motorcar tube appears to be cold. Why?
Answer: Due to adiabatic expansion

Question 37. If γ of a gas is equal to 1.66 then what is the number of atoms in a molecule of the gas?
Answer: 1

Question 38. Adiabatic curves are comparatively _______ than isothermal curves.
Answer: Steeper

Question 39. If the pressure and the temperature of a gas changes at constant volume, what is the work done by the gas?
Answer: Zero

Question 40. What is the change in internal energy in an isothermal process?
Answer: Zero

Question 41. In which expansion the internal energy of a gas be decrease?
Answer: Adiabatic

Question 42. If M is the molecular weight of a gas, what is the difference between the two specific heats of 1 g of an ideal gas?
Answer: \(\frac{R}{M}\)

Question 43. In case of 1 mol of an ideal gas, write down the value of C_v -C_p
Answer: -R

Question 44. An ideal gas rejects 10 cal of heat at constant volume. Find the work done
Answer: Zero

Question 45. What is the relation between p and V in an adiabatic process?
Answer: pV^γ = constant

Question 46. A process against frictional force cannot be ______
Answer: Reversible

Question 47. Two balls of the same mass, one of iron and the other of copper, are dropped from the same height. Which one would become hotter?
Answer: Copper

Question 48. In an isothermal process, the gas containers should be made of highly _____ materials.
Answer: Conducting

Question 49. In an adiabatic process, the gas containers should be made of highly _______ materials.
Answer: Non-conducting

Question 50. Is Joule’s heating process reversible or irreversible?
Answer: Irreversible

Question 51. Write down the coefficient of performance of an ideal refrigerator.
Answer: Infinite

Question 52. What is the value of the efficiency of an ideal heat engine?
Answer: 1

Question 53. Heat engine is a mechanical device that converts heat into _____
Answer: Work

Question 54. When heat is gained or lost by heat reservoir, its temperature ______
Answer: Remains constant

Unit 8 Thermodynamics Chapter 1 First And Second Law Of Thermodynamics Assertion Reason Type Question And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: In an isothermal process the whole heat energy supplied to the body is converted into internal energy.

Statement 2: According to the first law of thermodynamics ΔQ = ΔU+pΔV.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 2.

Statement 1: The specific heat of a gas in an adiabatic process is zero but it is infinite in an isothermal process.

Statement 2: Molar specific heat of a gas is directly proportional to heat exchanged with the system and inversely proportional to change in temperature.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: In an adiabatic process, the change in internal energy of a gas is equal to work done on or by the gas in the process.

Statement 2: The temperature of the gas remains constant in an adiabatic process.

Answer: 3. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 4.

Statement 1: The internal energy of an ideal gas does not depend on the volume of the gas.

Statement 2: This is because internal energy of an ideal gas depends on the temperature of the gas.

Answer: Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.

Unit 8 Thermodynamics Chapter 1 First And Second Law Of Thermodynamics Match Column 1 With Column 2

Question 1. For 1 mol of a monatomic gas match the following:

Answer: 1. D, 2. A, D, 3. D, 4. B, C

Question 2. Heat given to it pi recess is taken to be positive. Then mutch two options in column l with the corresponding options in column 2 for the given cycle.

Answer: 1. E, 2. E, 3. A, 4. B

Question 3. Column 1 contains a list of processes Involving the expansion of an ideal gas. Match this with column 2 describing the thermodynamic change during this process.

Answer: 1. B, 2. A, C, 3. A, D, 4. B, D

Question 4. Volume versus pressure curves are given for four processes as shown. Match the entries of column 1 with the entries of column 2.

Answer: 1. A, C, D, 2. C, D 3. C, 4. B

Question 5. A sample of gas goes from state A to state B in four different manners, as shown by the graphs. Let W be the work done by the gas and ΔU be the change in internal energy along the path AB. Match the graphs with the statements provided correctly.

Answer: 1. D, 2. B, 3. C, 4. B

Question 6. One mole of a monatomic ideal gas is taken along two cyclic processes E → F → G → E and E → F → H → E.

The processes involved are purely isochoric, isobaric, isothermal, or adiabatic. Match tire column 1 with the magnitudes of the work done in column 2.

Answer: 1. D, 2. C, 3. B, 4. A

Question 7. One mole of a monoatomic ideal gas is taken through a cycle ABCDA as shown in the p-V diagram. Column 2 gives the characteristics involved in the cycle. Match them with each of the processes given in column 1.

Answer: 1. A, C, E 2. A, C, 3. B, D 4. C, E

Unit 8 Thermodynamics Chapter 1 First And Second Law Of Thermodynamics Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A Fixed mass of gas is taken through a process A → B → C→ A. Here A→B is isobaric, B→ C is adiabatic and C→A is isothermal. (Take γ = 1.5)

1. Find pressure at C.

1. \(\frac{10^5}{64} \mathrm{~N} \cdot \mathrm{m}^{-2}\)
2. \(\frac{10^5}{32} \mathrm{~N} \cdot \mathrm{m}^{-2}\)
3. \(\frac{10^5}{12} \mathrm{~N} \cdot \mathrm{m}^{-2}\)
4. \(\frac{10^5}{6} \mathrm{~N} \cdot \mathrm{m}^{-2}\)

Answer: 1. \(\frac{10^5}{64} \mathrm{~N} \cdot \mathrm{m}^{-2}\)

2. Find volume at C.

1. 32 m³
2. 100m³
3. 64m³
4. 25m³

Answer: 3. 64m³

3. Find work done in the process.

1. 4.9 x 10⁵ J
2. 3.2 x 10⁵ J
3. 1.2 x 10⁵ J
4. 7.2 x 10⁵ J

Answer: 1. 4.9 x 10⁵ J

Question 2. One mole of an ideal gas has an internal energy given by U = U₀ + 2pV, where p is the pressure and V is the volume of the gas. U₀ is a constant. This gas undergoes the quasistatic cyclic process ABCD as shown in the U-V diagram.

1. The molar heat capacity of the gas at constant pressure is

1. 2R
2. 3R
3. 5/2R
4. 4R

Answer: 2. 3R

2. The work done by the ideal gas in the process AB is

1. Zero
2. \(\frac{U_1-U_0}{2}\)
3. \(\frac{U_0-U_1}{2}\)
4. \(\frac{U_1-U_0}{2} \log _e 2\)

Answer: 4. \(\frac{U_1-U_0}{2} \log _e 2\)

3. Assuming that the gas consists of a mixture of two gases, the gas is

1. Monatomic
2. Diatomic
3. A mixture of monatomic and diatomic gases
4. A mixture of diatomic and triatomic gases

Answer: 3. A mixture of diatomic and triatomic gases

Question 3. A cylinder containing an ideal gas and closed by a movable piston is submerged in an ice-water mixture. The piston is quickly pushed down from position X to position Y (process AB). The piston is held at position Y until the gas is again at 0°C (process BC). Then the piston is slowly raised back to position X (process CA).

1. Which of the following p-V diagrams will correctly represent the processes AB, BC, and CA and the cycle ABCA?

Answer: 4.

2. If 100 g of ice is melted during the cycle ABCA, how much work is done on the gas?

1. 8kcal
2. 5kcal
3. 2.1 kJ
4. 4.2 kJ

Answer: 1. 8kcal

3. If p is the atmospheric pressure acting on the piston and change in the volume is (V₁ – V₂). the work done (in N · m^-2) during the cycle is

1. \(\frac{p V}{2} \mathrm{~J}\)
2. \(\frac{2 p V}{3} J\)
3. \(p V J\)
4. None of these

Answer: 4. None of these

Question 4. 1 mol of an ideal monatomic gas undergoes thermodynamic cycle 1-2-3-1 as shown. The initial temperature of the gas is T₀= 300 K.

• Process 1 → 2 : p = aV
• Process 2 → 3 : pV = constant
• Process 3 → 1 : p = constant (Take In|3| = 1.09)

1. Find the net work done in the cycle.

1. 3.27 RT₀
2. 6.83 RT₀
3. 4.53 RT₀
4. 5.81 RT₀

Answer: 4. 5.81 RT₀

2. Determine the heat capacity of each process.

1. 20.75 J · mol^-1 -K^-1
2. 10.23 J · mol^-1 · K^-1
3. 22.37 J · mol^-1 · K^-1
4. 15.96 J · mol^-1 · K^-1

Answer: 1. 20.75 J · mol^-1 K^-1

Question 5. A container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are \(C_V=\frac{3}{2} R, C_p=\frac{5}{2} R\), and those for an ideal diatomic gas are \(C_V=\frac{5}{2} R, C_p=\frac{7}{2} R\)

1. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be

1. 550 K
2. 525 K
3. 513 K
4. 490 K

Answer: 4. 490 K

2. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be

1. 250R
2. 200R
3. 100R
4. -100R

Answer: 4. -100R

Unit 8 Thermodynamics Chapter 1 First And Second Law Of Thermodynamics Integer Type Questions And Answers

In this type, the answer to each of the questions Is a single-digit integer ranging from 0 to 9.

Question 1. A diatomic ideal gas is compressed adiabatically to 1/32 of its initial volume. If the initial temperature of the gas is T₁ (in kelvin) and the final temperature is aT₁ find the value of a.

Answer: 4

Question 2. 1 mol of an ideal gas p(atm) undergoes a cyclic change A ABCDA as shown. What is the net work done (in J) in the process? (Take. 1 atm = 10⁵Pa)
Answer: 4

Question 3. During adiabatic expansion of 10 mol of a gas, internal energy decreases by 700 J.Work clone during (the process is x x 10² J. What Is the value of x?
Answer: 7

Question 4. Two Carnot engines A and U operate respectively between 500 K and 400 K and 400 K and 300 K. What is the difference in their efficiencies (in percentage)?
Answer: 5

Question 5. Calculate the pressure required to compress a gas adiabatically at atmospheric pressure to one-third of its volume. Given γ = 1.47.
Answer: 5

Question 6. A thermodynamic system is taken from an initial state I with internal energy U_i = 100 J to the final state f along two different paths iaf and ibf, as shown.

The work done by the system along the paths af, ib, and bf is \(W_{af}=200 \mathrm{~J}, \quad W_{i b}=50 \mathrm{~J} \quad \text { and } \quad W_{b f}=100 \mathrm{~J}\) respectively. The heat supplied to the system along the path iaf, ib, and bf is Q_iaf, Q_ib, and Q_bf respectively. If the internal energy of the system in the state b is U_b = 200 J and Q_iaf = 500 J then find out the ratio of Q_bf/Q_ib.

Answer: 2

 

 

 

Statics – Rigid Bodies

Rigid Bodies: Statics is a branch of mechanics where the equilibrium of bodies under the action of a number of forces and the conditions for equilibrium are studied.

The concept of rigid bodies has to be introduced in the discussions of rotation, in particular. The particle description of material bodies is not sufficient in that context.

Rigid Bodies Definition: A body that does not change its shape or size under the action of forces is called a rigid body.

No substance is perfectly rigid. However, most material bodies may be assumed to be rigid when the forces acting on them do not exceed a certain limit.

Read and Learn More: Class 11 Physics Notes

Statics – Resultant Of Parallel Forces

Forces acting in the same direction are called parallel forces. \(\vec{F}_1, \vec{F}_2 \text { and } \vec{F}_3\) are examples of like parallel forces. When two forces act in opposite directions, they constitute a pair of, unlike parallel forces. \(\vec{F}_4\) and \(\vec{F}_5\) are two unlike parallel forces.

The straight line along which a force acts is called the line of action of that force. For example, AB is the line of action of forces \(\vec{F}_1\) and \(\vec{F}_2\), CD is the line of action of force \(\vec{F}_3\).

Resultant Of Two Like Parallel Forces: Let \(\vec{F}_1\) and \(\vec{F}_2\) be a pair of like parallel forces acting at points A and B. Their resultant has to be calculated. O is a point on the plane of the forces.

Line OX is drawn perpendicular to the lines of action of \(\vec{F}_1\) and \(\vec{F}_2\). OX is taken as the x-axis. OY is the y-axis on the same plane.

Since \(\vec{F}_1\) and \(\vec{F}_2\) do not have any components along the x-axis, the resultant \(\vec{R}\)(say) will also have no component along the x-axis and its line of action will be perpendicular to the line OX, i.e., parallel to the lines of action of \(\vec{F}_1\) and \(\vec{F}_2\). Hence, component of \(\vec{R}\) along y-axis is, R = F₁ + F₂….(1)

Hence, the magnitude of the resultant is the sum of the magnitudes of the forces and its direction will be the same as the direction of the parallel forces.

To find the line of action of R, let us assume that \(\vec{R}\) acts at point C. \(\vec{R}\) is the resultant of \(\vec{F}_1\) and \(\vec{F}_2\). It implies that the effect of forces \(\vec{F}_1\) and \(\vec{F}_2\) together, should be the same as that of R acting alone.

Hence, the algebraic sum of moments due to \(\vec{F}_1\) and \(\vec{F}_2\) about the point O should be equal to the moment of \(\vec{R}\) about O,

or, F₁ x OA + F₂ x OB = R x OC

or, F₁(OC-AC) + F₂(OC+CB) = (F₁ + F₂) x OC

or, F₁ x AC = F₂ x BC

or, \(\frac{F_1}{F_2}=\frac{B C}{A C}\)….(2)

Hence, the line of action of the resultant divides the distance AB into two parts in the inverse ratio of the magnitudes of the forces.

Suppose the line A’B’ is obtained by joining any two points on the lines of action of F₁ and F₂. Obviously, the line of action of the resultant divides it into two parts in the inverse ratio of the magnitude of the forces \(\vec{F}_1\) and \(\vec{F}_2\).

Position of the point C can also be determined in terms of points A and B. If OA = x₁, OB = x₂ and OC = x, then, taking moments about the point O of the forces \(\vec{F}_1\), \(\vec{F}_2\) and resultant \(\vec{R}\) we get,

∴ \(F_1 x_1+F_2 x_2=R \times x=\left(F_1+F_2\right) x\)

∴ x = \(\frac{F_1 x_1+F_2 x_2}{F_1+F_2}\)…(1)

WBBSE Class 11 Statics Notes

Resultant of Two Unlike Parallel Forces: Let \(\vec{F}_1\) and \(\vec{F}_2\) be two unlike parallel forces acting at points A and B as shown. Let \(F_1\) > \(F_2\). Suppose \(\vec{R}\) is the resultant of these two forces. Then, R = \(F_1\) – \(F_2\)

The resultant acts along the direction of the greater force.

Taking moments about point O and from the definition of resultant we get, F₁ x OA – F₂ x OB = R x OC

or, F₁(OC+ CA) – F₂(OC+ CB) = (F₁ – F₂) x OC

or, \(\frac{F_1}{F_2}=\frac{B C}{A C}\)

and x = \(\frac{F_1 x_1-F_2 x_2}{F_1-F_2}\)…(2)

Hence, the line of action of the resultant divides externally the distance of separation of the two forces in the inverse ratio of their magnitudes.

Resultant Of Three Or More Parallel Forces: When a body is acted upon by the parallel forces \(\vec{F}_1\), \(\vec{F}_2\), \(\vec{F}_3\),…… simultaneously, the resultant force \(\vec{R}\) can be obtained by framing an equation similar to equation (1).

R = \(F_1+F_2+F_3+\cdots=\sum_i F_i\)….(5)

Any one of the forces may be taken as positive. Correspondingly, forces in the same direction as this force are positive and those in the opposite direction are negative. If the forces are coplanar. the distance of the line of action of the resultant from any point O on that plane is,

x = \(\frac{F_1 x_1+F_2 x_2+F_3 x_3 \cdots}{R} \text { or, } x=\frac{\sum_i F_i x_i}{\sum_i F_i}\)…(6)

where x₁, x₂, x₃,…are the perpendicular distances of the point O from the lines of action of the forces, \(\vec{F}_1\), \(\vec{F}_2\), \(\vec{F}_3\)… etc.

The moment of the resultant about the point O can be positive or negative. The direction of rotation caused by the resultant about the point O is determined by this positive or negative sign.

Understanding Equilibrium in Statics

Statics- Centre Of Gravity

It is known that an extended object is an aggregation of a large number of particles. If the masses of these constituents are m₁, m₂, m₃, …., the earth pulls these particles with respective forces m₁g, m₂g, m₃g,…. towards its centre.

• The resultant of these parallel forces acting on the particles is the weight W of the object. This weight W, acts vertically downward through a definite point.
• This point is called the centre of gravity (G) of the body. Like the centre of mass, the centre of gravity may be located inside or outside the body.
• The respective distances between the particles remain fixed. Hence, the resultant of the weights also remains the same both in magnitude and in direction and acts through the same point.
• Thus, it can be stated that the position of the centre of gravity of a body does not depend on the orientation of the body.

Centre Of Gravity Definition: The force of gravity that is exerted on an extended object always acts through a unique point. This point; is the centre of gravity of the object.

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To Find The Position Of The Centre Of Gravity: The position of the centre of gravity (G) can be determined using the rules for finding the resultant of parallel forces. The magnitude of the resultant is equal to the weight of the body:

W = m₁g + m₂g + m₃g + …. = Mg

[Here, M=m₁ + m₂ + m₃ + = total mass of the object]

Let the coordinates of \(m_1, \quad m_2, \ldots\) be \(\left(x_1, y_1, z_1\right)\), \(\left(x_2, y_2, z_2\right)\), …respectively and the point of action of the resultant W, i.e., the coordinates of the centre of gravity be (x, y,z).

∴ \(\left.\begin{array}{l}
\begin{array}{rl}
therefore \quad x & =\frac{m_1 g x_1+m_2 g x_2+\cdots}{m_1 g+m_2 g+\cdots} \\
& =\frac{m_1 x_1+m_2 x_2+\cdots}{m_1+m_2+\cdots} \\
& =\frac{\sum_i m_i x_i}{\sum_i m_i}
\end{array} \\
\text { Similarly, } y=\frac{\sum_i m_i y_i}{\sum_i m_i} \text { and } z=\frac{\sum_i m_i z_i}{\sum_i m_i}
\end{array}\right\}\)

Some Important Information About The Centre Of Gravity: To establish equation (1), the magnitude of g is assumed to be the same for all particles constituting the body. This assumption is acceptable for small objects only, where the variation of g is negligible. But if the body is very large, the magnitude of g varies from point to point and equation (1) is not applicable.

Equation (1) of this section and equation (2) are identical. Hence, if the magnitudes of acceleration due to gravity at all points within a body are the same, the centre of mass and the centre of gravity become identical.

Uniqueness Of Centre Of Gravity: A body cannot have more than one centre of gravity. This is called the uniqueness of the centre of gravity. Let us assume a body has two centres of gravity G and G’. From the definition of centre of gravity, irrespective of the orientation of the body, its weight should act through G and G’.

But the weight of a body always acts vertically downwards. Therefore, GG’ must always be a vertical line. But for aflPoftlntations of the body, GG’ cannot be vertical. Hence a body having two centres of gravity is not a possibility.

Force Diagrams in Statics

Statics – Difference Between Centre Of Mass And Centre Of Gravity

The Centre of mass of a body is such a point that a force applied at this point produces linear motion only and does not produce any rotational motion.

1. The Centre of gravity is the point through which the weight of a body acts vertically downwards.
2. As the mass of a body remains constant anywhere in the universe, a body always has a definite centre of mass. But at places where the weight of the body becomes zero, there is no centre of gravity, for example, a body inside an artificial satellite rotating around the earth or a free-falling body is weightless. So in these cases, the body has no centre of gravity.
3. If the magnitude of acceleration due to gravity remains the same at all points of a body, the centre of mass and the centre of gravity will be identical.
   • But if the magnitude of gravitational acceleration is different at different points of a body, then the centre of mass will not coincide with the centre of gravity, For example, for very large mountains like the Himalayas acceleration due to gravity is greater at the bottom of the mountain, that’s why even for an object of uniform density earth’s attraction is greater at the bottom.
   • Again position of the centre of mass depends only on the density and not on the weight of the body. That is why for massive and tall objects like these centre of gravity remains below the centre of mass.

Statics – Centre Of Mass And Centre Of Gravity Numerical Examples

Short Answer Questions on Static Equilibrium

Example 1. A rod of weight W is kept horizontally on two knife edges with a distance d between them. The Centre of gravity of the rod is at a distance x from A. Find the normal reactions at points A and B.

Solution:

Let the centre of gravity of the rod be C. Normal reactions at A and B are R₁ and R₂. Weight W through C acting downwards, and R₁ and R₂ are the three coplanar forces that keep the rod in equilibrium.

Hence, W- R₁ – R₂ = 0 or, W = R₁ + R₂

Taking the moment of the forces about C, \(-R_1 \cdot x+R_2(d-x)=0 \text { or, } R_1 x=R_2(d-x)\)

or, \(x\left(R_1+R_2\right)=R_2 d \text { or, } W x=R_2 d\)

∴ \(R_2=\frac{x}{d} W\)

and \(R_1=W-R_2=W-\frac{W x}{d}=W\left(1-\frac{x}{d}\right)\)

Example 2. Show that the centre of gravity of three equal weights suspended from three vertices of a triangle coincides with the centre of mass of the triangle.
Solution:

Let the median of the triangle ABC be AD. Identical weights W are hung from each of the vertices A, B and C. Resultant of the weights suspended from B and C = 2W and it acts from the point D.

The resultant of the forces 2 W at D and W at A is 3 W. Suppose this resultant acts at G.

Hence, W x AG = 2 W x DG

or, AG = 2 DG

Thus, G is the point of intersection of the medians of the triangle, which is the centre of mass.

Calculating Moments and Torques in Statics

Example 3. A person of mass 80 kg is standing on the top of an 18 kg ladder of length 6 m. The upper end of the ladder rests on a smooth vertical wall and the lower end is on the ground 3 m away from the vertical wall. What should be the minimum coefficient of friction between the floor and the ladder so that the system remains in equilibrium?
Solution:

Given that, for the ladder weight W acts through the centre of gravity, C (mid-point of AB) of the ladder.

Weight W’ of the man acts at B downwards.

Normal reaction R’ by the wall acts at B.

Normal reaction R by the ground acts at A.

Limiting friction that acts along the floor at A, f = μR where, n is the coefficient of friction, required for equilibrium.

At equilibrium, the sum of all horizontal forces and the sum of all vertical forces will be zero separately.

⇒ \(\mu R-R^{\prime}=0 \text { or, } \mu R=R^{\prime}\)…(1)

⇒ \(R-W-W^{\prime}=0 \text { or, } R=W+W^{\prime}\)…(2)

Again the sum of the moments of all forces taken about A will be zero.

∴ \(R^{\prime} \times B D-W \times A E-W^{\prime} \times A D=0\)

or, \(\mu\left(W+W^{\prime}\right) \times B D=W \times A E+W^{\prime} \times A D\)

∴ \(\mu=\frac{W \times A E+W^{\prime} \times A D}{\left(W+W^{\prime}\right) \times B D}\)

Here, AB = \(6 \mathrm{~m}, A D=3 \mathrm{~m}\).

∴ AE = \(\frac{A D}{2}=\frac{3}{2} \mathrm{~m} ; B D=\sqrt{A B^2-A D^2}=5.2 \mathrm{~m}\)

W = \(18 \mathrm{~kg} \times g ; W^{\prime}=80 \mathrm{~kg} \times \mathrm{g}\)

∴ \(\mu=\frac{18 \times \frac{3}{2}+80 \times 3}{(18+80) \times 5.2}=0.52\).

Example 4. A uniform cylinder of diameter 8 cm is kept on a rough inclined plane, whose angle of inclination with the ground is 30°. What should be the maximum height of the cylinder so that it does not topple?
Solution:

The cylinder ABCD is kept on the plane of inclination 30°. Suppose the cylinder is on the verge of toppling over when its height is 2h. At this stage, the line of action of gravity must pass vertically through the endpoint A at the base of the cylinder.

From \(tan 30^{\circ}=\frac{A E}{E G}\)

or, \(\frac{1}{\sqrt{3}}=\frac{r}{h} or, h=\sqrt{3} r=4 \sqrt{3} \mathrm{~cm}\)

Hence, maximum permissible height = 2h = x 4√3 = 8√3 cm.

Example 5. A hollow cylinder of height 100 cm and diameter 8 cm has its top end open. It contains water up to a height of 50 cm. The mass per square centimetre of the hollow cylinder is 9 g. Find the height of the centre of gravity of the water cylinder system from the base of the cylinder.
Solution:

The Centre of gravity of the curved surface of the empty cylinder is 50 cm above its base at P.

The weight of the base of the empty cylinder

= π(4)² x 9 x g dyn [g = acceleration due to gravity]

Weight of curved surface of the cylinder = 2π x 4 x 100 x 9 x g dyn. Its centre of gravity is at G₁, where PG₁ = 50 cm.

Weight of water in the cylinder =π(4)² x 50 x 1 x g dyn and the centre of gravity of this water is at G₂ which is 25 cm above the base of the cylinder.

Suppose the centre of mass of the cylinder filled with water is at G, which is h cm above its base. The centre of gravity coincides with this centre of mass.

Taking moments of all the weights about P, \(2 \pi \times 4 \times 100 \times 9 \times G_1 P+\pi \times 16 \times 50 \times 1 \times G_2 P\)

= \((\pi \times 16 \times 9+2 \pi \times 4 \times 100 \times 9+\pi \times 16 \times 50 \times 1) \times G P\)

or, \(2 \pi \times 4 \times 100 \times 9 \times 50+\pi \times 16 \times 50 \times 1 \times 25\)

= \((\pi \times 16 \times 9+2 \pi \times 4 \times 100 \times 9 +\pi \times 16 \times 50 \times 1) \times h\)

or, h = 46.7 cm.

Example 6. A uniform narrow rod of mass M and length 2 L is kept vertically, along the y-axis, on a smooth horizontal plane. The lower end of the rod coincides with the origin (0,0). Due to a slight disturbance at time t = 0, the rod slides along the positive x-axis and begins to fall. Determine

1. The shift in the centre of gravity during the fall,
2. The equation of the locus of a point at a distance r from the lower end of the rod also mentions the shape of the locus.

Solution:

Initially (0, L) was the coordinates of the centre of gravity of the rod. Let B be the point at a distance r, from point O (0, 0), on the rod. So, (0, r) denotes the point B.

At t = 0, the rod is disturbed and it falls. During this fall there is no other force acting on it except the downward gravitational force.

1. We know that the centre of gravity is influenced only by an external force which is gravity in this case. So the centre of gravity will be shifted vertically downwards from (0, L) to (0, 0).
2. Let (x, y) be the position of B’ at any moment dining the fall of the rod, where B’ is the point at a distance r, from the lower end O’ of the rod.

From, \(\frac{O G^{\prime}}{O^{\prime} G^{\prime}}=\frac{C B^{\prime}}{O^{\prime} B^{\prime}}\)

or, \(\frac{a}{L}=\frac{y}{r}\)

Again, \(\frac{O^{\prime} O}{O C}=\frac{O^{\prime} G^{\prime}}{G^{\prime} B^{\prime}}\)

or, \(\frac{O C}{G^{\prime} B^{\prime}}=\frac{O^{\prime} O}{O^{\prime} G^{\prime}} or, \frac{x}{r-L}=\frac{b}{L}\)

∴ \(\frac{x^2}{(r-L)^2}=\frac{b^2}{L^2}=\frac{L^2-a^2}{L^2}=1-\frac{a^2}{L^2}=1-\frac{y^2}{r^2}\)

or, \(\frac{x^2}{(r-L)^2}+\frac{y^2}{r^2}=1\)

This is the equation of the locus of a point (x, y) and it is elliptical with its centre at the origin.

Example 7. A circular metal plate of uniform thickness has a radius of 10 cm. A hole of radius 4 cm is punched on the plate a little away from its centre. The centres of the plate and of the hole are 5 cm apart. Find the centre of gravity of the plate with the hole.
Solution:

The Centre of gravity of the plate before the hole was punched was at its centre G. The centre of gravity of the circular part before the punching was at its centre G₁. Suppose the centre of gravity of the plate with the hole is at G₂, on the line GG₁.

Let W₁ = weight of the plate with a hole, W₂ = weight of the disc that is cut off

∴ \(W_1 \times G G_2=W_2 \times G G_1 \text { or, } G G_2=\frac{W_2}{W_1} \times G G_1\)

But \(G G_1=5 \mathrm{~cm}, W_1=\pi\left(10^2-4^2\right) \times \rho\)(ρ= weight per unit area) and \(W_2=\pi \times 4^2 \times \rho\)

∴ \(G G_2=\frac{4^2}{10^2-4^2} \times 5=\frac{20}{21} \mathrm{~cm}\)

Hence, the centre of gravity of the system will be \(\frac{20}{21}\) cm to the left of the centre of the plate.

Real-Life Examples of Statics in Daily Life

Example 8. A mass of 10 kg is suspended using two strings. One string makes an angle of 60° with the vertical. What should be the angle made by the other string so that the tension in the string will be minimum? Find the tension in each wire.
Solution:

As the mass is in equilibrium, from the force diagram we get \(T_1 \cos \theta_1+T_2 \cos \theta_2=m g\)

or, \(T_1 \cos \theta_1=m g-T_2 \cos \theta_2\)….(1)

Also \(T_1 \sin \theta_1=T_2 \sin \theta_2\)…(2)

From (2) and (1) we get, \(\frac{\sin \theta_1}{\cos \theta_1}=\frac{T_2 \sin \theta_2}{m g-T_2 \cos \theta_2}\)

or, \( m g \sin \theta_1=T_2\left(\sin \theta_1 \cos \theta_2+\cos \theta_1 \sin \theta_2\right)\)

∴ \(T_2=\frac{m g \sin \theta_1}{\sin \left(\theta_1+\theta_2\right)}\)….(3)

As m, g and θ are constants, the minimum value of T₂ corresponds to the maximum value of sin(θ₁ + θ₂) i.e., sin(θ₁ + θ₂) = 1 = sin90°

∴ θ₁+ θ₂ = 90° or, θ₂ = 90° – θ₁ = 90° -60° = 30°

From equation (3) we get, \(T_2=\frac{10 \times g \sin 60^{\circ}}{1}=84.87 \mathrm{~N}\)

From equation (2) we get, \(T_1=\frac{T_2 \sin \theta_2}{\sin \theta_1}=\frac{5 \sqrt{3} \times 9.8 \times \sin 30^{\circ}}{\sin 60^{\circ}}=49 \mathrm{~N} \text {. }\)

Example 9. A chain of mass m and length l is kept on a horizontal frictionless table, such that \(\frac{1}{4}\) th of the length of chain is the hanging part of the chain onto the table.
Solution:

Mass per unit length of the chain = \(\frac{m}{l}\)

∴ Mass of hanging part of the chain = \(\frac{m}{l}\) x  \(\frac{l}{4}\) =  \(\frac{m}{4}\)

∴ The Centre of gravity of this portion is at \(\frac{l}{4 \times 2}=\frac{l}{8}\) below the top of the table.

∴ Work to be done to lift the hanging part of the chain = \(\frac{m}{4} \times g \times \frac{l}{8}=\frac{m g l}{32}\)

Statics – Equilibrium Of A Body Under Gravity

Applications of Statics in Engineering

Suppose a chair is at rest with its legs resting on the floor. This stationary state of the chair is its equilibrium position. If the chair is tilted a little, it has to be held in that position to prevent it from falling.

• This shows that this position is not the equilibrium position of the chair. On tilting it further the chair falls to the ground on its side and comes to rest again.
• This is another equilibrium position of the chair. In general, under the influence of gravity, a body may have one or more equilibrium states.

Under The Influence Of Gravity, A Body May Stay In Three Types Of Equilibrium States:

1. Stable Equilibrium: If a substance is disturbed from its equilibrium state, and if on release it gets back to the same equilibrium state, then the body is initially in a stable equilibrium.
   • If a book, placed on a table with its largest surface resting on the table, is lifted a little by any one of the edges and then released, it falls back and returns to its initial state of equilibrium.
2. Unstable Equilibrium: If a body, when displaced a little from its equilibrium position, tends to move away further, then the body was in an unstable equilibrium.
   • Unstable Equilibrium Examples:
     • A pencil made to balance on its pointed end
     • An egg is made to stand on its narrow end (it is almost impossible to keep a pencil or an egg in this position by hand). If a slight push is given these fall over.
3. Neutral Equilibrium: Let a body be displaced a little from its state of equilibrium and then released. If it still remains in equilibrium in its new position, then it is said to be in a neutral equilibrium. Let a sphere be placed on a smooth horizontal plane.
   • On being pushed, it rolls a bit and comes to rest in a new position. Hence the sphere is said to be in a neutral equilibrium state. Different equilibrium states of a body can be ascertained by either of the following two methods.

Extension Of The Base: A chair on a floor rests with all four legs in contact with the floor. The quadrilateral obtained by joining the contact points of its legs with the floor is called the base of the chair. A body that has a larger base area compared to the area of its upper regions is always in stable equilibrium if the vertical line through its centre of gravity passes through its base.

From everyday experience, we know that, for an object of a given height, stability increases with the extent of the base area. For a stable object, the line of action of its weight always lies within its base.

Extension Of The Base Stable Equilibrium: Consider two possible ways of placing a brick on a horizontal surface. In the brick is placed on its largest face AB. This is a position of stable equilibrium and the lines of action of the weight W and the normal reaction R coincide. If we tilt the brick to stand it on the edge through point A, as shown, the lines of action of W and R do not lie on the same straight line.

• Consequently, when we release the brick, the couple created by W and R brings it back on the original base AB. Note that the line of action of the weight W still passes through the base AB when the brick is tilted. So the initial state of the brick is a state of stable equilibrium.
• To topple the brick, it needs to be tilted further about the edge through A so that the line of action of W goes outside the base AB. The brick is on the verge of toppling.
• The same brick is shown to be standing on one of its smaller faces like BC. In this case, it will be easier to topple the brick because the line of action of W goes outside the base BC on turning the brick through a smellier angle. But this initial state is also a stable equilibrium state. Thus, we can say that stability increases with the extent of the base area.

Unstable Equilibrium: A cone is shown balanced on its tip. Since the lines of action of W and R coincide, we can say that the cone is in equilibrium.

• However, a minute displacement from this position is enough to make it lose its balance and topple over. Thus, the cone standing on its tip is an example of unstable equilibrium.
• In this state, if the object is disturbed and then released, it does not return to its initial position but seeks a position of stable equilibrium.
• Note that when the cone is displaced slightly, the lines of action of W and R do not lie on the same straight line and the line of action of W does not pass through the base A. The bricks are at state of unstable equilibrium.

Neutral Equilibrium: Consider a ball resting on a horizontal surface as shown. In this situation, the lines of action of W and R coincide. We conclude that the ball is in equilibrium. If we roll the ball to a new position, we find that it is in a similar kind of equilibrium as before.

• As a matter of fact, the lines of action of W and R coincide in any orientation. Such an equilibrium is called a neutral equilibrium. When a body is displaced from a position of neutral equilibrium does not tend to return to its initial position; rather it settles into a new position of neutral equilibrium.
• Thus, if the line of action of its weight does not pass through the base of the body, it cannot remain in equilibrium. Once displaced from its equilibrium position, a body comes to a new equilibrium either by returning to the old base or on a new base.
• A chair is not stable when it stands on two legs, as the line of action of its weight does not pass through the baseline defined by the two legs. Quadrupeds are more stable than animals who walk on two legs.
• Hence, in the latter case, the technique of standing on two legs has to be learned. When a bucket full of water is carried in the right hand, the body is leaned towards the left in order to keep the line of action of the weight of the body and the bucket together within our feet.

Position Of Centre Of Gravity: Objects tend to move towards the centre of the earth due to the gravitational pull of the earth. Since the point of action of this attractive force (i.e., weight) is the centre of gravity of the body, the natural tendency of the centre of gravity is to attain the least height. Therefore, the lower the position of the centre of gravity of a body, the more stable its equilibrium.

If on displacing a body, it’s centre of gravity

1. Goes up, then the body was at a stable equilibrium initially,
2. Goes down, the body attains a more stable equilibrium than before,
3. Remains at the same height, and then the body is said to be in neutral equilibrium.

Position Of Centre Of Gravity Example:

1. A person standing with his feet apart is more stable than when his feet are together. By placing his feet apart, the person increases the base area so he can stand more comfortably.
   • When a person lies down, along with the increase in the base area, there is a considerable lowering of the centre of gravity. Hence, the lying-down position is the most stable equilibrium for a person.
2. Boxes and books, when placed on the floor, are usually kept with their largest surfaces in contact with the ground. This lowers the centre of gravity and the equilibrium becomes most stable.
3. Passengers are not allowed to stand on the upper deck of a double-decker bus in order to keep the centre of gravity low.
4. Hydrometers, lactometers, special types of toys, etc. are loaded with lead in their lower part (base) so that the centre of gravity of the system remains very low. Hence, the vertical position is the most stable equilibrium position in these cases.
   • The toys can be tilted about their bases through large angles, but they will come back to their vertical positions as soon as they are released. Hence, each of these toys has only one state of equilibrium.

Relationship Of The Centre Of Gravity And Gravitational Potential With Different Types Of Equilibrium: Gravitational potential energy is related to the position of the centre of gravity of a body. Potential energy depends on the height of the centre of gravity from the ground.

• Lower the position of the centre of gravity, lower the potential energy and higher the stability of the body.
• Thus, when the centre of gravity of a body or a system attains the lowest possible height and hence has the lowest possible gravitational potential energy, the body achieves stable equilibrium. When the centre of gravity is at its maximum height, potential energy is also the maximum and therefore, the equilibrium becomes unstable.
• The body tends to deviate from that position to a state of lower potential energy.
• For a body in neutral equilibrium, the height of the centre of gravity does not change when it is displaced and the potential energy remains the same. Hence, the body remains in equilibrium even in the displaced state.

Statics Conclusion

Centre Of Mass is a unique point for an extended object or a system of particles such that, any force applied through that point produces only translational motion of the body, but no rotational motion.

1. No object can have more than one centre of mass,
2. The centre of mass of an object can be outside the material of the object.
3. The position of the centre of mass changes as the shape of an object changes.

The vector quantity which tends to rotate a body by the combination of the force acting on the body and the perpendicular distance of the line of action of that force from the axis of rotation is called the moment of the applied force with respect to that axis of rotation.

The magnitude of the moment of force is expressed by the product of the magnitude of the force and the perpendicular distance of the force from the axis of rotation.

The vector form of the relation between the moment of force and the applied force is \(\vec{G}=\vec{r} \times \vec{F}\).

• A number of forces acting in the same direction are called parallel forces. Two forces acting in mutually opposite directions are called unlike parallel forces.
• The straight line along which a force acts is called the line of action of that force.
• The line of action of the resultant of two like parallel forces acting at two different points on a body divides the distance between the like parallel forces in inverse ratio of the magnitudes of the two forces.
• The line of action of the resultant of two unlike parallel forces acting at two different points on a body divides externally the distance between the unlike parallel forces in inverse ratio of the magnitudes of the forces.
• A body is said to be in equilibrium when it is at rest moving with uniform linear velocity moving with a uniform angular speed or moving with both uniform linear and angular velocities.

Conditions of equilibrium of a body under the action of a number of coplanar forces:

1. 1st Condition: The resultant of all coplanar forces acting on the body should be zero. If this condition is fulfilled, the linear acceleration of the body will be zero.
2. 2nd Condition: The algebraic sum of the moments of coplanar forces acting on the body, taken about any point on the plane, should be zero. If this condition is fulfilled, the angular acceleration of the body will be zero.

Both conditions, satisfied simultaneously, keep a body in equilibrium.

• Conditions of equilibrium under the action of two forces: Two forces acting on a body should be equal, collinear and opposite to each other.
• Conditions of equilibrium under the action of three non-parallel forces:
• Three forces must be on the same plane and the resultant of any two should be equal and opposite to the third force and the lines of action of all three forces must pass through the same point.
• The single force that cancels the action of all other forces acting on a body is called an equilibrant.

Centre of gravity is a fixed point of an object through which the weight of the body acts. Its position is independent of the orientation of an object of fixed shape and volume.

1. No object can have more than one centre of gravity.
2. When acceleration due to gravity at all points of an object becomes the same, the centre of mass coincides with the centre of gravity.

At zero gravity, the centre of mass exists but the centre of gravity has no significance.

Under the action of gravity, objects may have three types of equilibrium:

1. Stable,
2. Unstable and
3. Neutral.

A body with a wider base compared to its upper regions, becomes stable. Also if the line drawn vertically through the centre of gravity of the object passes through its base, the object remains in a stable equilibrium.

Statics Useful Relations For Solving Numerical Problems

Let \(\vec{F}_1\) and \(\vec{F}_2\) be two like parallel forces. They act on two different points on x-axis with coordinates x₁ and x₂ respectively. The resultant (\(\vec{R}\)) of the above forces act through a point with coordinate x. Here,

R = \(F_1+F_2 \text { and } x=\frac{F_1 x_1+F_2 x_2}{F_1+F_2}\)

If \(\vec{F}_1\) and \(\vec{F}_2\) are unlike parallel forces then,

R = \(\left|F_1-F_2\right| \text { and } x=\frac{F_1 x_1-F_2 x_2}{F_1-F_2}\)

For n number of parallel forces

R = \(\sum_{i=1}^n F_i \text { and } x=\frac{\sum_{i=1}^n F_i x_i}{\sum_{i=1}^n F_i}\)

For n number of parallel forces \(R=\sum_{i=1}^n F_i \text { and } x=\frac{\sum_{i=1}^n F_i x_i}{\sum_{i=1}^n F_i}\)

For equilibrium og n number of coplanar forces, according to the 1st condition, \(R_x=\sum_{i=1}^n F_{i x}=0 \text { and } R_y=\sum_{i=1}^n F_{i y}=0\)

and according to the 2nd condition, \(\tau=\sum_{i=1}^n F_i x_i=0\)

Mathematical expressions for Lami’s theorem, \(\frac{P}{\sin (Q, R)}=\frac{Q}{\sin (R, P)}=\frac{R}{\sin (P, Q)}\)

Suppose particles of masses m₁, m₂,…..m_n constitute an extended object or a system of particles. The positional coordinates of the particles are (x₁, y₁, z₁, (x₂, y₂, z₂),…..(x_n, y_n, z_n) and that of the centre of mass are (x_cm, y_cm, z_cm), Then

∴ \(x_{\mathrm{cm}}=\frac{\sum_{i=1}^n m_i x_i}{\sum_{i=1}^n m_i}, y_{\mathrm{cm}}=\frac{\sum_{i=1}^n m_i y_i}{\sum_{i=1}^n m_i}, z_{\mathrm{cm}}=\frac{\sum_{i=1}^n m_i z_i}{\sum_{i=1}^n m_i}\)

Statics Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 1 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: The centre of mass of the system will not alter in any direction if the external force is not exerted on it.

Statement 2: If net external force is zero then the linear momentum of the system remains constant.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 2.

Statement 1: The position of the centre of mass of a body is independent of the shape and size of the body.

Statement 2: The centre of mass of a body may be in a position where there is no mass.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 3.

Statement 1: Centre of mass of a rigid body always lies inside the body.

Statement 2: Centre of mass and centre of gravity coincide if gravity is uniform.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 4.

Statement 1: The centre of mass of an electron-proton system, when released, moves faster towards the proton.

Statement 2: Proton is heavier than an electron.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 5.

Statement 1: When a body dropped from a height explodes in mid-air, its centre of mass keeps moving in a vertically downward direction.

Statement 2: Explosion occurs under internal forces only External force is zero.

Answer: 3. Statement 1 is true, statement 2 is false.

Statics Match The Columns

Question 1.

Answer: 1. A, 2. C, 3. D, 4. B

Question 2. The two-block systems shown, match the following.

Answer: 1. C, D, 2. C, D, 3. B, 4. B

Question 3.  If the net force on a system of particles is zero, then match the following columns

Answer: 1. B, 2. C, 3. C, 4. D

Question 4. The arrangement is shown to match the following.

Answer: 1. B, 2. B, 3. B, 4. A

Statics Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A solid cylinder rolls up an inclined plane with an angle of inclination of 30°. At the bottom of the inclined plane, the centre of mass of the cylinder has a speed of 5 m · s^-1.

1. The distance up to which the cylinder goes up the plane is

1. 4m
2. 3.8m
3. 3.6m
4. 3m

Answer: 2. 3.8m

2. The time taken for the cylinder to return to the bottom is

1. 3.5 s
2. 3.7 s
3. 3.0 s
4. 3.8 s

Answer: 3. 3.0 s

Question 2. Four particles of masses 1 kg, 2 kg, 3 kg and 4 kg are placed at the four vertices A, B, C and D of a square of side 1 m.

1. Taking D as origin, x -coordinates of centre of mass is

1. 0.7 m
2. 0.8 m
3. 0.5 m
4. 0.6 m

Answer: 3. 0.5 m

2. Taking D as origin, y -coordinates of the centre of mass is

1. 0.5 m
2. 0.6 m
3. 0.3 m
4. 0.4 m

Answer: 3. 0.3 m

3. Taking C as the origin, the position of the centre of mass of the particles is

1. (0.7 m, 0.6 m)
2. (-0.5 m, 0.3 m)
3. (0.6 m, 0.4 m)
4. (-0.5 m, 0.6 m)

Answer: 2. (-0.5 m, 0.3 m)

Question 3. Two balls having masses m₁ = 4 kg and m₂ = 5 kg have initial velocities of 5m • s^-1 in the directions shown. They collide at the origin. (Take all integer values)

1. Velocity of the centre of mass 3 s before the collision is

1. \((-2 \hat{i}+1 \hat{j}) \mathrm{m} \cdot \mathrm{s}^{-1}\)
2. \((-2 \hat{i}-1 \hat{j}) \mathrm{m} \cdot \mathrm{s}^{-1}\)
3. \((2 \hat{i}+1 \hat{j}) \mathrm{m} \cdot \mathrm{s}^{-1}\)
4. \((2 \hat{i}+1 \hat{j}) \mathrm{m} \cdot \mathrm{s}^{-1}\)

Answer: 1. \((-2 \hat{i}+1 \hat{j}) \mathrm{m} \cdot \mathrm{s}^{-1}\)

2. The position of the centre of mass 2 s after the collision is

1. \(-4 \hat{i}-2 \hat{j}\)
2. \(4 \hat{i}-2 \hat{j}\)
3. \(-4 \hat{i}+2 \hat{j}\)
4. \(4 \hat{i}+2 \hat{j}\)

Answer: 3. \(-4 \hat{i}+2 \hat{j}\)

Question 4. A particle of mass 1 kg has velocity \(\overrightarrow{v_1}=(2 t) \hat{i}\) and another particle of mass 2 kg has velocity \(\overrightarrow{v_2}=\left(t^2\right) \hat{j}\)

1. Net force on the centre of mass at 2 s

1. \(\frac{20}{9}\) unit
2. \(\sqrt{68}\) unit
3. \(\frac{\sqrt{80}}{3}\) unit
4. None of these

Answer: 2. \(\sqrt{68}\) unit

2. Velocity of the centre of mass at 2 s

1. \(\frac{20}{9}\) unit
2. \(\sqrt{68}\) unit
3. \(\frac{\sqrt{80}}{3}\) unit
4. None of these

Answer: 3. \(\frac{\sqrt{80}}{3}\) unit

3. Displacement of the centre of mass in 2 s

1. \(\frac{20}{9}\) unit
2. \(\sqrt{68}\) unit
3. \(\frac{\sqrt{80}}{3}\) unit
4. None of these

Answer: 1. \(\frac{20}{9}\) unit

Statics Integer Answer Type Questions And Answers

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9.

Question 1. Two ice skaters A and B approach each other at right angles. Skater A has a mass of 30 kg and a velocity 1 m · s^-1, and skater B has a mass 20 kg and a velocity 2m · s^-1. They meet and cling together. Find the final velocity (in m · s^-1) of the couple.
Answer: 1

Question 2. Both the blocks as shown in the given arrangement are given a horizontal velocity towards the right together. If a_cm is the subsequent acceleration of the centre of mass of the system of blocks then find the value of a_cm (in m · s^-2).
Answer: 2

Question 3. Blocks A and B shown have equal masses m. The system is released from rest with the spring unstretched. The string between A and the ground is cut when there is a maximum extension in the spring. Find the acceleration of the centre of mass (in terms of how many times of g) of the two blocks.
Answer: 2

 

Newtonian Gravitation And Planetary Motion

WBBSE Class 11 Newtonian Gravitation Notes

Newtonian Gravitation And Planetary Motion Introduction: Since ancient times, scientists have been extremely curious to learn more about the stars and planets.

• Since the time of Copernicus, it is known that planets move around the sun. But to investigate the cause and nature of this motion, it was necessary to know the exact positions of the planets in the sky at different times.
• Astrophysicist Tycho Brahe, for many years, observed the positions of planets without a telescope (Galileo discovered the telescope after the death of Brahe) and published a lot of information about this.
• Kepler analysed the observational findings of Tycho Brahe and arrived at three laws about planetary motion. These laws are known as Kepler’s laws of planetary motion.

Newton’s Law of Universal Gravitation Explained

Read and Learn More: Class 11 Physics Notes

 Newtonian Gravitation And Planetary Motion – Kepler’s Laws

Understanding Kepler’s Laws of Planetary Motion

Kepler’s Laws First Law: Every planet moves in an elliptical orbit with the sun at one of its foci.

Kepler’s Laws Second Law: The line joining the sun and a planet sweep out equal areas in equal intervals of time, i.e., the areal velocity of a planet is constant.

Kepler’s Laws Third Law: The square of the time period of revolution of a planet is directly proportional to the cube of the length of the semi-major axis of its elliptical orbit.

The orbits of planets are known as Keplerian orbits while their motions are known as Keplerian motions.

Proof Of Kepler’s Second Law: Let us consider a planet of mass m moving in an elliptical orbit with the sun at focus S. Also, let \(\vec{r}\) be the position vector of the planet with respect to the sun and \(\vec{F}\) be the required centripetal force for the planet.

The torque exerted on this planet by this force about the sun, \(\)

(\(\vec{r}\) and \(\vec{F}\) are oppositely directed)

But \(\) (\(\vec{L}\) = angular momentum of the planet)

∴ \(\frac{d \vec{L}}{d t}=0 \quad \text { or, } \vec{L}=\text { constant }\)

Now, if the planet moves from position P to P’ in very small time A t, then the area swept out by the radius vector \(\vec{r}\) is,

⇒ \(\Delta \vec{A}=\text { area of triangular region } S P P^{\prime}\)

= \(\frac{1}{2} \vec{r} \times \overrightarrow{P P^{\prime}}\)

Now, \(\quad \overrightarrow{P P^{\prime}} =\Delta \vec{r}=\vec{v} \Delta t=\frac{\vec{p}}{m} \Delta t\)

∴ \(\quad \Delta \vec{A}=\frac{1}{2} \vec{r} \times \frac{\vec{p}}{m} \Delta t\)

or, \(\quad \frac{\Delta \vec{A}}{\Delta t}=\frac{1}{2 m}(\vec{r} \times \vec{p})=\frac{\vec{L}}{2 m}\) (because \(\vec{L}=\vec{r} \times \vec{p}\))

Hence, \(\quad \frac{\Delta \vec{A}}{\Delta t}=\text { constant }\) (because \(\vec{L} \text { and } m\)) are constant

Thus, the areal velocity of the planet remains constant, i.e., the radius vector joining the sun and a planet sweeps out equal areas in equal intervals of time.

Gravitational Attraction Of Extended Bodies

Gravitational Force and Planetary Orbits

Any extended body comprises a large number of particles. Thus, to find the mutual force of attraction between two extended bodies, the force of attraction between each particle of the first body and that of the second body is to be determined individually.

• The resultant of these forces, given by the vector addition method, gives the magnitude and direction of the force of attraction between the two extended bodies. Clearly, this method is complicated and unwieldy. This method can only be used in the case of regular-shaped bodies.
• But in the case of two extended bodies, however large they may be, if the distance separating them is much greater than their sizes, it can be assumed that the entire mass of each body is concentrated at its centre of mass. As a result, two extended bodies can be considered as two particles at their respective centres of mass.
• In this way, the complication involved in applying Newton’s law of gravitation may be removed. This approach can be applied to celestial bodies, like the earth and the moon, even if they are large in size and cannot be considered as particles.
• The distance between them is so large in comparison to their sizes that they may be considered as particles situated at their respective centres of mass for the purpose of measuring the gravitational force between them.

Now taking M = mass of the earth, m = mass of the moon and r = the distance between their centres of mass, the force of gravitation between them = \(\frac{G M m}{r^2}\).

When the distance between the two bodies is small compared to their sizes, this method of calculation fails.

When the gravitational attraction between a homogeneous spherical body and a particle situated outside the body is considered, it is assumed that the entire mass of the spherical body is concentrated at its centre of mass.

Hence, the earth, the sun, the moon and other planets are taken as spherical and the law is applied. A substance is called homogeneous when the physical properties (like density) of its different parts are identical.

Newtonian Gravitation And Planetary Motion – Sun Earth Moon Numerical Examples

Short Answer Questions on Planetary Motion

Example 1. Assuming that the moon moves around the earth in a circular orbit of radius 3.8 x 10⁵ km in 27 days and the earth moves around the sun in a circular orbit of radius 1.5 x 10⁸ km in 365 days, find the ratio of the masses of the sun and the earth.
Solution:

Given

Assuming that the moon moves around the earth in a circular orbit of radius 3.8 x 10⁵ km in 27 days and the earth moves around the sun in a circular orbit of radius 1.5 x 10⁸ km in 365 days,

From Kepler’s law, \(\frac{T^2}{r^3}\) = constant

where T = time period, r = radius of the orbit.

If the mass of the sun is M₀ and the radius of the earth’s orbit around the sun is r₀, the orbital speed,

⇒ \(v_0=\sqrt{\frac{G M_0}{r_0}}\)

∴ Time period T₀ of the earth = \(\frac{2 \pi r_0}{v_0}=2 \pi r_0 \sqrt{\frac{r_0}{G M_0}}=2 \pi \sqrt{\frac{r_0^3}{G M_0}}\)

Similarly, if the mass of the earth is M, the radius of the moon’s orbit is r, time period,

T = \(2 \pi \sqrt{\frac{r^3}{G M}}\)

∴ \(\frac{T}{T_0}=\sqrt{\left(\frac{r}{r_0}\right)^3 \frac{M_0}{M}} \text { or, }\left(\frac{T}{T_0}\right)^2=\left(\frac{r}{r_0}\right)^3 \cdot \frac{M_0}{M}\)

or, \(\frac{M_0}{M}=\left(\frac{T}{T_0}\right)^2 \cdot\left(\frac{r_0}{r}\right)^3=\left(\frac{27}{365}\right)^2 \cdot\left(\frac{1.5 \times 10^8}{3.8 \times 10^5}\right)^3\)

= 3.37 x 10⁵ = 337000

Hence, the sun is 337000 times more massive than the Earth.

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Example 2. Find the mass of the sun considering the orbit of the earth to be circular. Given, a distance of the earth from the sun = 1.49 x 10¹³ cm and G = 6.66 x 10^-8 CGS unit.
Solution:

Given

Given, a distance of the earth from the sun = 1.49 x 10¹³ cm and G = 6.66 x 10^-8 CGS unit.

Let the mass of the sun be M₀, the distance of the earth from the sun r, and the time period of the earth around the sun T.

∴ Mass of the Sun, \(M_0=\frac{4 \pi^2 r^3}{G T^2}\)

= \(\frac{4 \times \pi^2 \times\left(1.49 \times 10^{13}\right)^3}{6.67 \times 10^{-8} \times(365 \times 24 \times 60 \times 60)^2}\)

= \(2 \times 10^{33} \mathrm{~g}=2 \times 10^{30} \mathrm{~kg}\)

Newtonian Gravitation And Planetary Motion – Mass And Average Density Of The Earth Numerical Examples

Example 1. The average density of the earth is 5500 kg · m^-3, the gravitational constant is 6.7 x 10^-11 N · m² · kg^-2 and the radius of the earth is 6400 km. Using the given values, find the magnitude of the acceleration due to gravity on the surface of the earth.
Solution:

Given

The average density of the earth is 5500 kg · m^-3, the gravitational constant is 6.7 x 10^-11 N · m² · kg^-2 and the radius of the earth is 6400 km. Using the given values,

It is known \(\rho=\frac{3 g}{4 \pi R G}\)

or, \(g=\frac{4 \pi \rho R G}{3}\)

g =\(\frac{4 \times 22 \times 5500 \times 6400 \times 10^3 \times 6.7 \times 10^{-11}}{7 \times 3}\)

= \(9.88 \mathrm{~m} \cdot \mathrm{s}^{-2} .\)

Example 2. If the earth is considered as a solid sphere of iron of; radius 6.37 x 10⁶ m and of density 7.86 g · cm^-3, what will be the magnitude of the acceleration due to gravity on the earth’s surface? Gravitational constant = 6.58 x 10^-8 CGS unit.
Solution:

Given

If the earth is considered as a solid sphere of iron of; radius 6.37 x 10⁶ m and of density 7.86 g · cm^-3,

It is known \(\rho=\frac{3 g}{4 \pi R G}\)

∴ g = \(\frac{4 \pi \rho R G}{3}=\frac{4 \times 22 \times 7.86 \times 6.37 \times 10^8 \times 6.58 \times 10^{-8}}{7 \times 3}\)

= \(1380.55 \mathrm{~cm} \cdot \mathrm{s}^{-2}\)

Newtonian Gravitation And Planetary Motion Conclusion

The force with which any two material particles in the universe attract each other is called gravitation.

Newton’s Law Of Gravitation: Any two material particles in the universe attract each other along their line of joining. This force of attraction is directly proportional to the product of the masses of the two particles and inversely proportional to the square of the distance separating them.

Applications of Newtonian Gravitation in Space

• The amount of force with which any two material particles of unit mass kept at a unit distance apart attract each other is called the gravitational constant.
• If a body of unit mass is kept at a point in any gravita¬tional field, the gravitational force acting on that body is called gravitational intensity at that point.
• The amount of work done to bring a body of unit mass from infinity to any point in a gravitational field is called the gravitational potential at that point.
• The force with which anybody on or near the earth’s surface is attracted by the earth is called gravity.
• The acceleration produced in a body falling under the influence of the force of gravity is called the acceleration due to gravity.

Kepler’s laws related to the motion of planets and satellites:

1. First Law: Keeping the sun at one of the foci, each planet revolves in an elliptical path around the sun.
2. Second Law: The line joining the sun and a planet covers equal areas in equal intervals of time.
3. Third Law: The square of the time period of revolution of a planet around the sun is directly proportional to the cube of the semimajor axis of its orbit.

The minimum velocity that should be given to a body so that it can escape from the earth or other planets or sat¬ellites is called its escape velocity.

Some uses of an artificial satellite:

1. Weather information,
2. Communication,
3. Military and defence surveillance,
4. Remote sensing.

If the relative angular velocity of an artificial satellite revolving in the equatorial plane is zero with respect to the diurnal motion of the earth, then, from the earth’s surface, the satellite appears to be at rest at the same place. This kind of satellite is called a geostationary satellite.

Newtonian Gravitation And Planetary Motion Useful Relations For Solving Numerical Problems

If the distance between two particles of masses m₁ and m₂ is r and if the mutual force of attraction between the two particles is F, then according to Newton’s law of gravitation,

F = \(\) where G is the gravitational constant whose value in SI is 6.67 x 10^-11 N · m² · kg^-2.

Gravitational intensity at a point situated at a distance r from the centre of a body of mass M is E = \(\frac{G M}{r^2}\)

Work done in bringing a body of unit mass from infinity to the position r, i.e., the gravitational potential at a distance r from the centre of a body of mass M is V = –\(\frac{G M}{r}\)

The relation between the gravitational field intensity and potential is E = \(\frac{d V}{d r}\)

Acceleration due to gravity at any point on the earth’s surface, g = \(\frac{G M}{R^2}\), where M is the mass and R is the radius of the earth, and G is the gravitational constant.

The relation between the gravitational constant G and the mean density (ρ) of the earth is \(\frac{3 g}{4 \pi R G}\).

1. The relation between the acceleration due to gravity at an altitude h above the surface of the earth (g’) and the acceleration due to gravity on the earth’s surface (g) is g’ = g(1-\(\frac{2h}{R}\)) (this equation is valid for h<<R)
2. The relation between the acceleration due to gravity at a depth h below the earth’s surface (g’) with the acceleration due to gravity on the earth’s surface (g) is g’ = g = (1-\(\frac{h}{R}\)
3. The value of the acceleration due to gravity at θ – latitude (g’) of the earth is \(g^{\prime}=g\left(1-\frac{\omega^2 R}{g} \cos ^2 \theta\right)\)
   1. At the poles, θ = 90° ; so, g’ = g.
   2. At the equator, θ = 0°; so, \(g^{\prime}=g\left(1-\frac{\omega^2 R}{g}\right)\)

If a planet of mass m revolves around the sun of mass M₀ along a circular path of radius r, then the orbital speed of that planet, v = \(v=\sqrt{\frac{G M_0}{r}}.\).

Orbital angular velocity, \(\omega=\sqrt{\frac{G M_0}{r^3}}\)

Time period of revolution, T = \(T=2 \pi \sqrt{\frac{r^3}{G M_0}}\)

The value of the escape velocity of a body from the surface of the earth, \(v_e=\sqrt{2 g R}.\).

If the mass of the earth = M, radius of the earth = R, mass of an artificial satellite = m, the orbital speed of the satellite = v and the height of the orbit above the surface of the earth = h, then the orbital speed of the artificial satellite, v \(=\sqrt{\frac{G M}{R+h}}\) and the time period of revolution of the satellite, T = \(2 \pi \sqrt{\frac{r^3}{8}}\), where r = R+h.

In the case of a satellite revolving very close to the surface of the earth, its orbital speed, \(v=\sqrt{g R}\)

and the time period of revolution, T = \(2 \pi \sqrt{\frac{R}{g}}\)

If the mass of the earth = M, the radius of the earth = R, the mass of a geostationary satellite = m, the distance of the satellite from the centre of the earth = r and its time period of revolution = T, then the orbital speed of the geostationary satellite,

v = \(\sqrt{\frac{g R^2}{r}}\)

and the height of the orbit of the geostationary satellite, r \(=\left(\frac{g R^2 T^2}{4 \pi^2}\right)^{1 / 3}\)

If an artificial satellite of mass m revolves along a circular path of radius r, the kinetic energy of the satellite,

K = \(\frac{G M M}{2 r}\) [where M = mass of the earth]

The potential energy is U = –\(\frac{G M m}{r}\)

Total energy of the satellite, E = –\(\frac{G M m}{2r}\)

Newtonian Gravitation And Planetary Motion Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2 of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, statement 1 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is true; statement 2 is false.

Question 1.

Statement 1: The force of gravitation between a sphere of mass M₁ and a rod of mass M₂ is = \(\frac{G M_1 M_2}{r} \text {. }\)

Statement 2: Newton’s law of gravitation holds correct for point masses.

Answer: 4. Statement 1 is true; statement 2 is false.

Question 2.

Statement 1: An astronaut in an orbiting space station above the earth experiences weightlessness

Statement 2: An object moving around the earth under the influence of the earth’s gravitational force is in a state of free fall.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: If time period of a satellite revolving in a circular orbit in the equatorial plane is 24 h, then it must be a geostationary satellite.

Statement 2: Time period of a geostationary satellite is 24 hours.

Answer: 4. Statement 1 is true; statement 2 is false.

Question 4.

Statement 1: Gravitational force between two masses in the air is F. If they are immersed in the water force will remain F.

Statement 2: Gravitational force does not depend on the medium between the masses.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 5.

Statement 1: The binding energy of a satellite does not depend upon the mass of the satellite.

Statement 2: Binding energy is the negative value of the total energy of the satellite.

Answer: 4. Statement 1 is true; statement 2 is false.

Question 6.

Statement 1: Kepler’s laws for planetary motion are a consequence of Newton’s laws.

Statement 2: Kepler’s laws can be derived by using Newton’s laws.

Answer: 4. Statement 1 is true; statement 2 is false.

Real-Life Examples of Gravitational Effects

Newtonian Gravitation And Planetary Motion Match Column A With Column B.

Question 1.

Answer: 1. A, C, 2. A, B, C, D, 3. A, B, C, D, 4. A, C, D

Question 2. In Column A, four artificial satellites of Earth are shown. In Column B, some statements are given related to motion or other facts about the satellites.

Answer: 1. A, D, 2. B, D, 3. B, D, 4. C

Question 3.

Answer: 1. C, 2. B, 3. A

Question 4. In the elliptical orbit of a planet, as the planet moves from the apogee position to the perigee position, match the following columns.

Answer: 1. C, 2. B, 3. B, 4. A

Newtonian Gravitation And Planetary Motion Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. The PE of the earth-satellite system is shown by a solid line as a function of distance r (the separation between the earth’s centre and satellite). The total energy of two objects which may or may not be bound to earth are shown in the graph by dotted lines,

1. Mark the correct statement(s).

1. The object having total energy E₁ is the bounded one
2. The object having total energy E₂ is the bounded one
3. Both objects are bounded

Answer: 1. The object having total energy E₁ is the bounded one

2. If the object having total energy E₁ is having the same PE curve as shown, then

1. r₀ is the maximum distance of the object from the earth’s centre
2. This object and earth system is a bounded one
3. The KE of the object is zero when r = r₀
4. All of the above

Answer: 4. All of the above

3. If both the objects gave the same PE curve as shown, then

1. For the object having total energy E₂, all values of r are possible
2. For the object having total energy E₂ only values of r < r₀ are possible
3. For the object having total energy E₁ all values of r are possible
4. None of the above

Answer: 1. For the object having total energy E₂, all values of r are possible

Question 2. A uniform metal sphere of radius a and mass M is surrounded by a thin uniform spherical shell of equal mass and radius 4a. The centre of the shell falls on the surface of the inner sphere.

1. Find the gravitational field intensity at point P₁.

1. \(\frac{G M}{16 a^2}\)
2. \(\frac{G M}{8 a^2}\)
3. \(\frac{G M}{2 a^2}\)
4. \(\frac{G M}{4 a^2}\)

Answer: 1. \(\frac{G M}{16 a^2}\)

2. Find the gravitational field intensity at point P₂.

1. \(\frac{21 G M}{900 a^2}\)
2. \(\frac{61 G M}{450 a^2}\)
3. \(\frac{61 G M}{900 a^2}\)
4. \(\frac{61 G M}{1800 a^2}\)

Answer: 3. \(\frac{61 G M}{900 a^2}\)

Question 3. The gravitational field in a region is given by \(\vec{E}=\left(5 \mathrm{~N} \cdot \mathrm{kg}^{-1}\right) \hat{i}+\left(12 \mathrm{~N} \cdot \mathrm{kg}^{-1}\right) \hat{j}\)

1. Find the magnitude of the gravitational force acting on a particle of mass 2 kg placed at the origin.

1. 26 N
2. 30 N
3. 20 N
4. 35 N

Answer: 1. 26 N

2. Find the potential at the points (12 m, 0) and (0,5 m) if the potential at the origin is taken to be zero.

1. -30J · kg^-1_, -30 J · kg^-1
2. -40J · kg^-1_, -30 J · kg^-1
3. -60J · kg^-1_, -60 J · kg^-1
4. -40J · kg^-1_, -50 J · kg^-1

Answer: 3. -60J · kg^-1_, -60 J · kg^-1

3. Find the change in gravitational potential energy if a particle of mass 2 kg is taken from the origin to the point (12 m, 5 m).

1. -225 J
2. -240 J
3. -245 J
4. -250 J

Answer: 2. -240 J

4. Find the change in potential energy if the particle is taken from (12 m, 0) to (0,5 m).

1. -10 J
2. -50 J
3. Zero
4. -60 J

Answer: 3. Zero

Newtonian Gravitation And Planetary Motion Integer Answer Type Questions

In this type, the answer to each of the questions Is a single-digit integer ranging from 0 to 9.

Question 1. Gravitational acceleration on the surface of a planet is (√6/11)g, where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is \(\frac{2}{3}\)times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km · s^-1, the escape speed on the surface of the planet in km · s^-1 will be
Answer: 3

Question 2. A binary star consists of two stars A (mass 2.2 M_s) and B (mass 11 M_s), where M_s is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is
Answer: 6

Question 3. A particle is projected vertically upwards from the surface of the earth of radius R, with a kinetic energy equal to half of the minimum value needed for it to escape. The maximum height to which it rises above the surface of the earth is nR. What should be the value of n?
Answer: 1

Question 4. The time period of a satellite A, whose orbital radius is r₀, is T₀ and the time period of a satellite B having radius 4r₀ is T_B. Find the ratio of T_B and T₀.
Answer: 8

Question 5. Two particles of masses m₁ and m₂ are kept at a separation of r. When a third particle is kept at a distance of \(\frac{d}{3}\) from m, then it does not experience any net force. Determine the ratio \(\frac{m_2}{m_1}\).
Answer: 4

 

Transmission of Heat

Different Mechanisms Of Heat Transfer: When two bodies having different temperatures are brought in contact with each other, heat flows from the hot¬ter body to the colder one.

Again, when there is a temperature difference between different points of a body, heat flows in die body from the point of higher temperature to the point of lower temperature.

The flow of heat from one place to another is called transmission of heat: Heat is transmitted in three different processes:

1. Conduction,
2. Convection and
3. Radiation.

 

1. Conduction: It is the process in which heat energy is transmitted from a hotter region to a colder region of a material without any displacement of molecules. Conduction usually takes place in solids. When one end of an iron rod is placed in fire, the other end becomes so hot that it becomes difficult to hold that end with our hand. In this case heat gets transmitted through the rod from the end held in fire to the other end. This is known as heat conduction.
2. Convection: It is the process in which heat is transmitted from a hotter region to a colder region of a material by the actual movement of heated molecules. This happens only in fluids i.e., in liquids and gases because here molecules can move freely. If we take some liquid in a container and heat it from below, the top part of the liquid gets heated mainly through convection.
3. Radiation: This is the process in which heat is transferred from one place to another in the form of electro-magnetic radiation in the absence of any material medium or without heating a material medium if it is present between the two places. Heat from the sun reaches the earth by radiation.

Comparison among the three modes of transmission of heat

Transmission Of Heat Numerical Examples

Short Answer Questions on Heat Transfer Methods

Example 1. Four metal pieces of the same surface area and of thickness 1 cm, 2 cm, 3cm, and 4cm respectively are connected serially with each other; Coefficients of thermal conductivity of the respective pieces are 0.2 CGS, 0.3 CGS, 0.1 CGS, and 0.4 CGS units. Find the equivalent conductivity of the system.
Solution:

It is known that, \(\frac{x_1+x_2+x_3+x_4}{k}=\frac{x_1}{k_1}+\frac{x_2}{k_2}+\frac{x_3}{k_3}+\frac{x_4}{k_4}\)

or, \(\frac{1+2+3+4}{k}=\frac{1}{0.2}+\frac{2}{0.3}+\frac{3}{0.1}+\frac{4}{0.4} \)

or, \(\frac{10}{k}=51.67 \text { or, } k=0.1935 \text { CGS. }\)

Example 2. A thick composite plate is formed by two plates of equal thickness kept one over the other. If the conductivities of the material of the constituent plates are k₁ and k₂, show that the equivalent conductivity of the thick plate, k = \(\frac{2 k_1 k_2}{k_1+k_2}\)
Solution:

Is is known, \(\frac{x_1+x_2}{k}=\frac{x_1}{k_1}+\frac{x_2}{k_2}\)

Here, \(x_1=x_2=x\) (say)

∴ \(\frac{x+x}{k}=\frac{x}{k_1}+\frac{x}{k_2} or, \frac{2}{k}=\frac{1}{k_1}+\frac{1}{k_2} or, k=\frac{2 k_1 k_2}{k_1+k_2}\).

Example 3. A 75 cm long copper rod and a 125 cm long steel rod are joined face to face. Each rod is of a circular cross-section of diameter 2 cm. Temperatures at the two ends of the composite rod are 100 °C and 0°C and the outer surface of the rod is insulated. Find the temperature at the junction of the two rods What is the rate of conduction of heat through the junction?

1. k for copper = 9.2 x 10^-2 kcal · m^-1 · °C^-1 · s^-1 and
2. k for steel = 1.1 x 10^-2  kcal · m^-1 · °C^-1 · s^-1

Solution: Let the temperature of the junction = θ Conductivity of copper,

⇒ \(k_1=\frac{9.2 \times 10^{-2} \times 10^3}{10^2}=0.92 \text { CGS unit }\)

and that of steel, \(k_2=\frac{1.1 \times 10^{-2} \times 10^3}{10^2}=0.11 \text { CGS unit }\)

Area of cross-section of each rod, \(A=\pi(1)^2=\pi \mathrm{cm}^2\)

∴ Rate of flow of heat through the junction, \(\frac{Q}{t}=\frac{k_1 A\left(\theta_2-\theta\right)}{l_1}=\frac{k_2 A\left(\theta-\theta_1\right)}{l_2}\)

or, \(k_1 l_2(100-\theta)=k_2 l_1 \theta\)

or, \(0.92 \times 125 \times(100-\theta)=0.11 \times 75 \times \theta \text { or, } \theta=93.3^{\circ} \mathrm{C}\)

∴ Rate of conduction, \(\frac{Q}{t}=\frac{k_2 A\left(\theta-\theta_1\right)}{l_2}=\frac{0.11 \times \pi \times 93.3}{125}=0.258 \mathrm{cal} \cdot \mathrm{s}^{-1}\)

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Example 4. The thickness of each metal in a composite bar is 0. 01 m and the temperature of the two external surfaces are 100 °C and 30 °C. If the conductivities of the metals be 0.2 CGS unit and 0.3 CGS unit respectively, find the temperature on the interface.
Solution:

Let the temperature on the interface = θ.

At steady state, rate of flow of heat will be the same through both the plates.

∴ \(\frac{Q}{t}=\frac{k_1 A(100-\theta)}{x_1}=\frac{k_2 A(\theta-30)}{x_2}\)

or, \(\frac{0.2(100-\theta)}{1}=\frac{0.3(\theta-30)}{1} \text { or, } 200-2 \theta=3 \theta-90\)

or, \(5 \theta=290 \text { or, } \theta=58^{\circ} \mathrm{C} .\)

Real-Life Examples of Heat Transmission

Example 5. Three metal rods of the same length and area of cross-section are attached in series. Conductivity of the three metals are k, 2k, and 3k. Free end of the first rod is kept at 200°C while the other end of the combination is kept of 100 °C. Find the temperatures of the two junctions at steady state. Assume that no heat loss occurs due to radiation.
Solution:

Let the required temperatures be θ₁  and θ₂.

At steady state, \(\frac{Q}{t}=\frac{k A\left(200-\theta_1\right)}{l}=\frac{2 k A\left(\theta_1-\theta_2\right)}{l}=\frac{3 k A\left(\theta_2-100\right)}{l}\)

∴ \(200-\theta_1=2\left(\theta_1-\theta_2\right)=3\left(\theta_2-100\right)\)

As \(200-\theta_1=2\left(\theta_1-\theta_2\right)\)

∴ \(3 \theta_1-2 \theta_2=200\)…..(1)

Also, \(200-\theta_1=3\left(\theta_2-100\right)\)

or, \(\theta_1+3 \theta_2=500\)…..(2)

Solving (1) and (2), \(\theta_1=145.45^{\circ} \mathrm{C} \text { and } \theta_2=118.18^{\circ} \mathrm{C} \text {. }\)

Example 6. A composite block is constructed with three plates of equal thickness and of equal cross-sectional area. The coefficients of conductivity of the three plates are k₁, k₂, and k₃ respectively. If the coefficient of conductivity of the composite block is k, then prove that k = \(=\frac{3}{\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}}\)
Solution:

Let the thickness of each plate be d and cross-sectional area be A.

Q amount of heat flows through each plate in time t, The temperatures of the two ends of the composite slab are T₁ and T₂ respectively.

Also, the temperatures of the consecutive junctions

So, \(Q=\frac{k_1 A\left(T_1-T_1{ }^{\prime}\right) t}{d}\)

= \(\frac{k_2 A\left(T_1{ }^{\prime}-T_2{ }^{\prime}\right) t}{d}=\frac{k_3 A\left(T_2{ }^{\prime}-T_2\right) t}{d}\) or, \(\frac{Q d}{A t}=\frac{T_1-T_1{ }^{\prime}}{\frac{1}{k_1}}=\frac{T_1{ }^{\prime}-T_2{ }^{\prime}}{\frac{1}{k_2}}=\frac{T_2{ }^{\prime}-T_2}{\frac{1}{k_3}}\)

or, \(\frac{Q d}{A t}=\frac{T_1-T_2}{\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}}\)…(1)

[by componendo and dividendo process]

Again, since thickness of the composite slab is 3d,

Q = \(\frac{k A\left(T_1-T_2\right) t}{3 d} \text { or, } \frac{Q d}{A t}=\frac{T_1-T_2}{\frac{3}{k}}\)….(2)

From (1) and (2) we get, \(\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}=\frac{3}{k} \text { or, } k=\frac{3}{\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}}\)

Fourier’s Law of Heat Conduction

Example 7. The temperature of a room is kept fixed at 20 °C with the help of an electric heater of resistance 20Ω. The heater is connected to a 200 V main. Temperature is same everywhere in the room. The area of the window through which heat is conducted outside from the room is 1 m² and the thickness of the glass is 0.2 cm. Then calculate the temperature outside. [Given, coefficient of thermal conductivity of glass = 0.002 cal • m^-1 • ° C^-1 • s^-1 and / = 4.2 J/s
Solution:

Heat generated by the heater in 1s,

⇒ \(H_1=\frac{V^2}{R J}=\frac{(200)^2}{20 \times 4.2}=\frac{200 \times 2000}{20 \times 42}=476.19 \mathrm{cal}\)

Heat conducted through the window in 1s,

⇒ \(H_2=\frac{k A\left(\theta_2-\theta_1\right)}{d}=\frac{0.002 \times 10^4 \times\left(\theta_2-\theta_1\right)}{0.2} \mathrm{cal}\)

Here, \(H_2=H_1 \quad \text { or, } \frac{0.002 \times 10^4 \times\left(\theta_2-\theta_1\right)}{0.2}=476.19\)

or, \(\left(\theta_2-\theta_1\right)=4.7619\)

∴ \(\theta_1=\theta_2-4.76=20-4.76=15.24^{\circ} \mathrm{C}\)

Example 8. The temperature of a room is kept fixed at 20°C. With the help of an electric heater when the temperature outside is -10°C. The total area of the walls in the room is 137 m2. The walls have three layers—the innermost layer is made of wood and 2.5 cm thick, the middle layer is made of cement and 1 cm thick, the outermost layer is made of bricks and 25 cm thick. Then calculate the power of the heater. [Given, coefficients of thermal conductivity of wood, cement, and brick are 0.125 W · m^-1 • °C^-1, 1.5 W · m^-1 • °C^-1 and 1.0 W • m^-1 • °C^-1 respectively]
Solution:

If k is the equivalent thermal conductivity of the wall made of three layers of different materials then,

k = \(\frac{x_1+x_2+x_3}{\frac{x_1}{k_1}+\frac{x_2}{k_2}+\frac{x_3}{k_3}}\)

Here, \(x_1=2.5 \mathrm{~cm}, x_2=1.0 \mathrm{~cm}, x_3=25 \mathrm{~cm}\)

⇒ \(k_1=0.125 \mathrm{~W} \cdot \mathrm{m}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}, k_2=1.5 \mathrm{~W} \cdot \mathrm{m}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1},\)

⇒ \(k_3=1 \mathrm{~W} \cdot \mathrm{m}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

k = \(\frac{2.5+1+25}{\frac{2.5}{0.125}+\frac{1}{1.5}+\frac{25}{1}}=0.624 \mathrm{~W} \cdot \mathrm{m}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

Heat conducted outside through the walls per second

= \(\frac{k A\left(\theta_2-\theta_1\right)}{d}=\frac{0.624 \times 137 \times\{20-(-10)\}}{28.5 \times 10^{-2}}\)

= 8998,7 W ≈ 8999 J s^-1

To keep the temperature of the room fixed the healer must generate 8999 J or 8.999 kJ heat per second.

Hence, power of the heater = 9 kW.

Example 9. The temperature gradient at the earth’s surface is 32 C/km and the average thermal conductivity of earth is 0.005 CGS unit. Find the loss of heat per day from the earth surface taking its radius to be  6000 km.
Solution:

We know, Q = \(k \cdot \Lambda \frac{d \theta}{d x} t \quad\left[\frac{d \theta}{d x}=\frac{\theta_2-\theta_1}{d}\right]\)

Here, k = 0.008 CGS unit, A = 4 x (6000 x 105)² cm²

∴ \(\frac{d \theta}{d x} = 32^{\circ} \mathrm{C} / \mathrm{km}=\frac{32}{10^5}{ }^{\circ} \mathrm{C} / \mathrm{cm},\)

t = \(1 \text { day }=24 \times 60 \times 60 \mathrm{~s}\)

∴ Q = \(\frac{0.008 \times 4 \pi \times 36 \times 10^{16} \times 32 \times 24 \times 60 \times 60}{10^5}\)

= \(1.0006 \times 10^{18} \mathrm{cal}\)

Thermal Conductivity and Its Importance

Example 10. Three rods made of material X and three rods of material Y are connected as shown. Each rod has equal length and equal cross-section. If the temperatures of end A and junction E are 60°C and 10°C respectively, find the temperatures of junctions B, C, and D. Coefficients of thermal conductivity of the materials X and Y are 0.92 CGS units and 0.46 CGS units respectively.
Solution:

Let the temperatures of the junctions B, C, and D be θ₁, θ_2, and θ₃ respectively.

Now, heat conducted from A to B

= heat conducted from B to C + heat conducted from B to D

or, 0.46(60 – θ₁) = 0.92(θ₁ -θ₂) + 0.46(θ₁ – θ₃)

or, 60 -θ₁ =2(θ₁ – θ₂) + (θ₁ – θ₃)

or, 4θ₁ -2θ₁-θ₃ = 60 …(1)

Again, heat conducted from B to D

= heat conducted from D to C + heat conducted from D to E

or, \(0.46\left(\theta_1-\theta_3\right)=0.92\left(\theta_3-\theta_2\right)+0.46\left(\theta_3-10\right)\)

or, \(\theta_1-\theta_3=2\left(\theta_3-\theta_1\right)+\left(\theta_3-10\right)\)

or, \(-\theta_1-2 \theta_2+4 \theta_3=10\)….(2)

Similarly, heat conducted from B to C+ heat conducted from D to C= heat conducted from C to E

or, \(0.92\left(\theta_1-\theta_2\right)+0.92\left(\theta_3-\theta_2\right)=0.92\left(\theta_2-10\right)\)

or, \(\theta_1-\theta_2+\theta_3-\theta_2=\theta_2-10\)

or, \(\theta_1-3 \theta_2+\theta_3=-10\)…(3)

Solving (1), (2), and (3), we get, \(\theta_1=30^{\circ} \mathrm{C} ; \theta_2=\theta_3=20^{\circ} \mathrm{C}\)

Example 11. Determine the equivalent thermal conductivity of the system shown. when

1. Heat flows from left to right and
2. Heat flows downwards.

Given, k₁ = 2k₂ and the length and cross section are same for all the slabs.

Solution:

1. Let, the equivalent thermal conductivity is k

Area of cross-section of each plate = A

Temperature difference between the two ends of each Plate =θ

Then heat conducted, \(Q_1=2 \frac{k_1 A \theta t}{l}+2 \frac{k_2 A \theta t}{l}\)

= \(4 \frac{k A \theta t}{l}\)

or, \(\frac{2 A \theta t}{l}\left(k_1+k_2\right)=\frac{4 k A \theta t}{l}\)

or, \(k_1+k_2=2 k\)

or, \(k=\frac{k_1+k_2}{2}=\frac{2 k_1+k_2}{2}=\frac{3 k_2}{2}=1.5 k_2\)

2. In this case if \(k^{\prime}\) be the equivalent thermal conductivity then,

⇒ \(\frac{l}{k^{\prime}} =\frac{\frac{l}{4}}{k_1}+\frac{\frac{l}{4}}{k_2}+\frac{\frac{l}{4}}{k_1}+\frac{\frac{l}{4}}{k_2}\)

or, \(\frac{1}{k^{\prime}}=\frac{1}{2 k_1}+\frac{1}{2 k_2} \frac{1}{4 k_2}+\frac{1}{2 k_2}=\frac{3}{4 k_2}\)

or, \(k^{\prime}=\frac{4 k_2}{3}=1.33 k_2\)

Conduction of Heat through a Slab of Varying Thickness: The thickness of a slab can vary during thermal conduction.

• In cold countries when temperature falls below 0°C, the water on the surface of the lakes and ponds slowly freeze to ice.
• Heat from rest of the mass of water is conducted to the atmosphere through the top frozen layer and the thickness of the ice layer on the surface gradually increases.
• So this is a fine example of conduction of heat through a slab of varying thickness. When the temperature of the atmosphere falls below 0°C, the uppermost surface of water loses its latent heat to the atmosphere and forms a thin sheet of ice.
• In this way, as the thickness of the layer of ice increases, heat will have to be conducted through a thicker layer too.

Let the coefficient of conductivity of ice = k, the density of ice at 0°C = ρ, the temperature of air above the ice surface = -θ °C and the uppermost surface area of the water body = A

After a time t s from the beginning of the formation of ice, let the thickness of ice formed on the water body be x.

If in a small interval of time dt, the increase in thickness of ice is dx, then the volume of ice formed in dt is Adx.

∴ The mass of that ice =Aρdx.

Therefore, the heat lost by water or the heat conducted to air in time dt through ice of thickness x is,

dQ = Aρdx- L [where L = latent heat of ice]

∴ Rate of flow of heat, \(\frac{d Q}{d t}=A \rho L \frac{d x}{d t}=\frac{k A[0-(-\theta)]}{x}=\frac{k A \theta}{x}\)

or, \(xdx  =\frac{k}{\rho L} \theta d t\)…(1)

2. Let us assume that at t = 0, x = x₁ and at t = t₁, x = x₂. This means that in t₁ s time if the thickness of ice slab has increased from x₁ to x₂, integrating the equation (1) we get,

⇒ \(\int_{x_1}^{x_2} x d x=\int_0^{t_1} \frac{k \theta}{\rho L} d t\)

or, \(\frac{1}{2}\left(x_2^2-x_1^2\right)=\frac{k \theta}{\rho L} t_1\)

∴ \(t_1=\frac{\rho L}{2 k \theta}\left(x_2^2-x_1^2\right)\)

1. If x₁ = 0 when t = 0 and x2-x when t = t₁

∴ \(t_1=\frac{\rho L}{2 k \theta} x^2\)….(3)

This equation gives the time (t₁) required for the deposi¬tion of a layer of ice of thickness x.

Note that in both equations (1) and (2), t₁ is independent of the surface area A of the lake. So in the same weather conditions, water bodies of all sizes from ice in this same rate.

Transmission Of Heat Conductivity Of Ice Numerical Examples

Example 1. A 3 cm thick layer of ice is formed over a water reservoir. Temperature over the reservoir is -20°C. In what time, the thickness of the ice layer increase by 1 mm? The conductivity of ice = 0.005 CGS unit; latent heat of fusion of ice = 80 cal · g^-1; density of ice at 0°C = 0.91 g · cm^-3.
Solution:

We shall use the equation, \(t=\frac{\rho L}{2 k \theta}\left(x_2^2-x_1^2\right)\)

where ρ = 0.91 g · cm^-3, L = 80 cal · g^-1,

k = 0.005 CGS unit, θ = 0 – (-20) = 20°C , x₁ = 3cm, x₂ = 3.1 cm

∴ t = \(\frac{0.91 \times 80}{2 \times 0.005 \times 20}\left[(3.1)^2-(3)^2\right]\)

= \(\frac{0.91 \times 80}{2 \times 0.005 \times 20} \times 0.61\)

= 222.04 s = 3 min 42 s

Example 2. The surface of a lake is covered with a layer of ice of a thickness 10 cm. For increase in thickness of the ice layer by 1 mm, time taken is found to be 49 min9s. Conductivity of ice = 0.005 CGS unit, latent heat of fusion of ice = 80 cal · g^-1, and density of ice = 0. 917 g • cm^-3. Find the outside temperature.
Solution:

Let the temperature of air outside = -θ

It is known that, t = \(\frac{\rho L}{2 k \theta}\left(x_2^2-x_1^2\right) \quad \text { or, } \theta=\frac{\rho L}{2 k t}\left(x_2^2-x_1^2\right)\)

In this case, t = 49 min 9 s = 2949 s, x₁ = 10.1 cm, x₂ = 10 cm, ρ = 0.917 g · cm^-3, L = 80 cal · g^-1, k = 0.005 CGS unit

∴ \(\theta =\frac{0.917 \times 80}{2 \times 0.005 \times 2949}\left[(10.1)^2-(10)^2\right]\)

= \(\frac{0.917 \times 80 \times 2.01}{2 \times 0.005 \times 2949}=5\)

∴ The outside temperature is -5°C

Example 3. Water in a tank at 0°G is in contact with a surrounding temperature of -20°C. Prove that the rate of increase of thickness x (in cm) ofice on the surface, is related to time t (in s) as x² = 0.00273t. The density of ice =0.917 g · cm^-3, latent heat of fusion of ice =80 cal • g^-1, and conductivity of ice =0.005 CGS unit.
Solution:

Let, A be the surface area of the tank, x be the initial thickness of ice floating on the tank surface and dx be the increase in thickness in time dt.

Hence, mass of ice formed in the time dt = Adx x ρ [ρ = density of ice]

Heat released by water = Adx x ρ xL [L = latent heat of fusion of ice]

This is the amount of heat that is conducted from water in the tank through the ice of thickness x to the surroundings.

∴ Rate of release of heat, \(\frac{d Q}{d t}=A \rho L \frac{d x}{d t}=\frac{k A[0-(-\theta)]}{x}\)

[-θ°C being the temperature of the surroundings]

or, \(x d x=\frac{k \theta}{\rho L} d t\)

Integrating, \(\int_0^x x d x=\frac{k \theta}{\rho L} \int_0^t d t\)

or, \(x^2=\frac{2 k \theta}{\rho L} t=\frac{2 \times 0.005 \times 20 \times t}{0.917 \times 80}\)

or, \(]y=0.00273 t\) (Proved).

Example 4. A 14.9 cm thick layer of ice floats on a deep lake. The temperature of the upper surface of the ice layers is the same as that of the surrounding air. If this temperature remains constant at -1°C then determine the time required for the layer to increase by 2mm in thickness. Given, latent heat of melting of ice =80 cal · g^-1, density of ice = 0.9 g cm^-3 and coefficient of thermal conductivity of ice = 0.006 CGS unit.
Solution:

We know, \(t=\frac{\rho L}{2 k \theta}\left(x_2^2-x_1^2\right)\)

Here, \(\rho=0.9 \mathrm{~g} \cdot \mathrm{cm}^{-3}, L=80 \mathrm{cal} \cdot \mathrm{g}^{-1}\)

k = \(0.006 \mathrm{CGS} \text { unit }, \theta=1^{\circ} \mathrm{C}\)

∴ \(x_1 =14.9 \mathrm{~cm}, x_2=14.9+0.2=15.1 \mathrm{~cm}\)

t = \(\frac{0.9 \times 80}{2 \times 0.006 \times 1}\left[(15.1)^2-(14.9)^2\right]\)

= \(\frac{0.9 \times 80}{0.012} \times 30 \times 0.2=36000 \mathrm{~s}=10 \mathrm{~h}\)

Energy Distribution In Black Body Radiation Wien’s Displacement Law

Radiation from a hot body consists of electromagnetic waves of different wavelengths (λ). However the energy is not evenly distributed in all wavelengths of radiation.

The intensity of wavelengths in black body radiation, obtained by experimental observations, reveals a pattern as shown by the graph.

The abscissa of the graph denotes the wavelengths A of radiated heat and the ordinate denotes radiation per unit area per unit time (Eλ) for different wavelengths.

Interpretation of the graph:

1. At a temperature T₁, the source

1. Energy distribution is unequal for different wavelengths and
2. The intensity is maximum at a certain wavelength \(\lambda_{m_1}\) as represented by point P on graph A.

2. Radiations from the body at different temperatures T₁, T₂, T_3, and the corresponding intensity distribution can be compared from graphs A, B, C. Clearly

1. As temperature increases the total intensity, represented by the area under the curve, also increases,
2. The wavelength of maximum intensity shifts towards the lower wavelength side, with the increase in temperature of the source.

WBBSE Class 11 Heat Transmission Notes

The observations referred to above led to the formulation of Wien’s displacement law which can be stated as: With the increase in the temperature of a body, the wavelength corresponding to the maximum intensity shifts towards the lower wavelength side.

The mathematical representation of Wien’s displacement law is, λ_mT = b (constant), where λ_m is the wavelength having maximum intensity at a temperature T of the source.

The value of Wien’s constant b, is about 0.0029 m K in SI unit. Thus, if λ_m is known, the temperature of the source T can be calculated. This method is widely used in estimating the temperature of stars. For example, if the temperature of a black body (star) is 1000 K,

∴ \(\lambda_m=\frac{0.0029}{1000}=2.9 \times 10^{-6} \mathrm{~m}\)

= \(29000 \times 10^{-10} \mathrm{~m}=29000\)Å

This wave lies in the infrared region of the electromagnetic spectrum.

Transmission Of Heat Energy Distribution In Black Body Radiation Numerical Examples

Example 1. The wavelength of the radiation of maximum Intensity from the solar surface Is 4.9 x 10^-7 m. From Wien’s displacement law, find the surface temperature of the sun. [ b = 0.0029 m • K]
Solution:

As per Wien’s displacement law, \(\lambda_m T=b=0.0029 \mathrm{~m} \cdot \mathrm{K}\)

Hence, \(T=\frac{0.0029}{4.9 \times 10^{-7}}=5918 \mathrm{~K}\).

Example 2. Assume that a star, which has a surface temperature of 5 x 10⁴ K, is a black body. Calculate the length of maximum intensity In Its radiation [b = 0.0029 m · K]
Solution:

From Wien’s displacement law, \(\lambda_m T=b\); given, \(T=5 \times 10^4 \mathrm{~K}\) and \(b=0.0029 \mathrm{~m} \cdot \mathrm{K}\).

∴ \(\lambda_m=\frac{b}{T}=\frac{0.0029}{5 \times 10^4}=\frac{29}{5} \times 10^{-8}=5.8 \times 10^{-8} \mathrm{~m}\).

 

Transmission Of Heat Long Answer Type Questions

The capability of a material to conduct heat is called its thermal conductivity.

• The quantity of heat conducted per second perpendicularly across the opposite faces of a unit cube, when the difference of temperature between its opposite faces is unity, is called the coefficient of thermal conductivity or simply thermal conductivity (k) of that material.
• In case of an ideal conductor k → ∞ and in case of an ideal insulator k = 0. So in reality, 0 < k < ∞.

Dimension of k: \(\mathrm{MLT}^{-3} \Theta^{-1}\)

• In the pre-steady state, the rate of increase in temperature of any layer of a conducting rod depends on the coefficient of thermal conductivity, the density and the specific heat of the material of the rod.
• In steady state, no layer of the rod has any increase in temperature and hence, the transmission of heat along the rod depends only on the coefficient of thermal conductivity.
• The ratio of the thermal conductivity of a substance to the thermal capacity per unit volume of that substance is called the thermal diffusivity of the substance.

In the radiation process, heat energy spreads all around in the form of electromagnetic waves. This thermal wave is called thermal radiation or radiant heat.

Prevost’s theory of heat exchange: The rise and fall of temperature of a body depends on the heat exchange of the body with its surroundings.

• The body which is a good absorber is also a good radiator of heat. Conversely, a bad absorber is also a bad radiator of heat.
• The body which absorbs all the incident radiations without reflecting or transmitting any part of it is called a perfectly black body. Since a perfectly black body is an ideal absorber of heat, it is also an ideal radiator of heat.

Kirchhoff’s law: At a particular temperature, the ratio of the emissive power and the absorptive power of a substance is always a constant and is equal to the emissive power of a perfectly black body at that temperature.

Stefan’s law: The total radiation of all wavelengths emitted per unit time per unit area of a perfect blackbody is directly proportional to the fourth power of its absolute temperature.

Newton’s law of cooling: For a small difference in temperature, heat lost per unit time by a body is directly proportional to the difference in temperature of the body with its surroundings.

Men’s law: With the increase in temperature of a body, the wavelength of the radiation corresponding to the maximum intensity shifts towards the lower wavelengths.

Transmission Of Heat Useful Relations For Solving Numerical Problems

If the temperature of the two faces of a rectangular plate of cross-sectional area A and of thickness d be θ₂ and θ₁ (θ₂ > θ₁) then the amount of heat conducted in time t from the hot to the cold face of the plate will be

Q = \(\frac{k A\left(\theta_2-\theta_1\right) t}{d}\)

Thermal resistance of conductor = \(\frac{1}{k} \cdot \frac{d}{A} .\)

Thermal resistivity of a substance = \(\frac{1}{k}\)

Thermal diffusivity of a substance, \(h=\frac{k}{\rho s}\)

[k = coefficient of thermal conductivity, ρ = density, s = specific heat]

Understanding Heat Transfer Methods

If the equivalent thermal conductivity of a combined slab is k, then, \(\frac{x_1+x_2+\cdots+x_n}{k}=\frac{x_1}{k_1}+\frac{x_2}{k_2}+\cdots+\frac{x_n}{k_n}\)

where, x_n = thickness of n-th slab, k_n= thermal conductivity of the n-th slab. The temperature of the interface,

⇒ \(\theta=\frac{k_1 \theta_2 x_2+k_2 \theta_1 x_1}{k_2 x_1+k_1 x_2}\)

Emissive power (e) of a surface = amount of radiation emitted at a particular temperature from unit area of the surface in unit time.

Relative emittance of a surface (e) = \(\begin{gathered}
=\frac{\text { the amount of heat radiated by that surface }}{\text { the amount of heat radiated by the same surface area }} \\
\text { of a black body at the same temperature } \\
\text { in an equal interval time }
\end{gathered}\)

Absorptive power (a) of a surface = \(\frac{\text { the amount of heat absorbed }}{\text { the amount of heat incident on the surface }}\)

If the emissive power and absorptive power of a surface at a particular temperature be e and a respectively and the emissive power of a perfectly black body is E, then according to Kirchhoff’s law, \(\frac{e}{a}\) = E.

According to the Stefan’s law, E = σT⁴ (where σ = Stefan’s constant]

If the absolute temperature of an ideal black body and its surroundings be T and T₀, then by Prevosfs theory of heat exchange, the net rate of radiation of heat from the black body, \(E=\sigma\left(T^4-T_0^4\right)\)

Let mass of a body be m, its specific heat s, and the temperature of the surroundings θ₀. If at time t the temperature of the body drops from θ₁ to θ₂, then

t = \(\frac{1}{C} \ln \left(\frac{\theta_1-\theta_0}{\theta_2-\theta_0}\right)\)

According to the Wien’s law, λ_mT= constant.

[where, λ_m = wavelength of the radiation corresponding to the maximum intensity emitted from a black body kept at an absolute temperature T.]

Solar temperature T = \(\left[\left(\frac{R}{r}\right)^2 \times \frac{s}{\sigma}\right]^{1 / 4}\)

(where r = radius of the sun, R = mean distance of the earth from the sun, s = solar constant and σ = Stefan’s constant)

 

Transmission Of Heat Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 Is true, statement 2 Is true; statement 2 Is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 Is false, and statement 2 Is true.

Question 1.

Statement 1: The equivalent thermal conductivity of two plates of same thickness in contact is less than the smaller value of thermal conductivity.

Statement 2: For two plates of equal thickness contact, the equivalent thermal conductivity is given by \(\frac{2}{k}=\frac{1}{k_1}+\frac{1}{k_2} \text {. }\)

Answer: 4. Statement 1 Is false, statement 2 Is true.

Question 2.

Statement 1: If the thermal conductivity of a rod is 5 units, then its thermal resistivity is 0.2 units.

Statement 2: Thermal conductivity = \(\frac{1}{\text { thermal resistivity }}\)

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: When the temperature difference across the two sides of a wall is increased, its thermal conductivity increases.

Statement 2: Thermal conductivity depends on the nature of material of the wall.

Answer: 4. Statement 1 Is false, statement 2 Is true.

Question 4.

Statement 1: If the temperature of a star is doubled then the rate of loss of heat from it becomes 16 times.

Statement 2: The specific heat varies with temperature.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 5.

Statement 1: Radiant heat is an electromagnetic wave.

Statement 2: Heat from the sun reaches the earth by convection.

Answer: 3. Statement 1 is true, statement 2 is false.

Transmission Of Heat Match Column 1 With Column 2

Question 1. On average, the temperature of the earth’s crust increases 1°C for every 30 m of depth. The average thermal conductivity of the earth’s crust is 0.75 J • m^-1 • K^-1 • s^-1. Solar constant is 1.35 kW • m^-2.

Answer: 1. A, 2. 3. 4. E

Question 2. Three rods of equal length of the same material are joined form an equilateral triangle ABC as shown. Area of a cross-section of rod AB is S, of rod BC is 2S and that of AC is S.

Answer: 1. C, 2. D, 3. 1, 4. B

Question 3. A copper rod (initially at room temperature 20°C) of the non-uniform cross section is placed between a steam chamber at 100°C and an ice water chamber at 0°C. A and B are cross sections as shown. Then match the statements in Column 1 with results in Column 2 using comparing only cross sections A and B.

Answer: 1. A, C, 2. D, 3. B

Transmission Of Heat Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A body cools in a surrounding of constant temperature 30°C. Its heat capacity is 2 J • °C^-1. The initial temperature of die body is 40°0. Assume Newton’s law of cooling is valid. The body cools to 38°C in 10 minutes.

1. In further 10 min it will cool from 38 °C to

1. 36°C
2. 36.4 °C
3. 37°C
4. 37.5°C

Answer: 2. 36.4 °C

2. The temperature of the body in T. denoted by θ. The variation of θ versus time t is best denoted as

Answer: 1

3. When the body temperature has reached 38°C, it is heated again so that it reaches 40°C in 10 min. The heat required from a heater by the body is

1. 3.6J
2. 0.364J
3. 8J
4. 4J

Answer: 3. 4

Question 2. Two insulated metal bars each of length 5 cm and rectangular cross-sections with sides 2 cut and 3 cm are wedged between two walls one held at 100°C and the other at 0°C. The bars are lead (Pb) and silver (Ag) k_Pb = 350 W m^-1 K^-1, k_Ag = 425 W m^-1 K^-1

1. Thermal current through lead bar is

1. 210W
2. 420W
3. 510W
4. 930W

Answer: 2. 420W

2. Total thermal current through the two-bar system is

1. 210 W
2. 420W
3. 510 W
4. 930W

Answer: 4. 930W

3. Equivalent thermal resistance of the two-bar system is

1. 0.1 W
2. 0.23 W
3. 0.19 W
4. 0.42W

Answer: 1. 0.1 W

Question 3. Assume that the thermal conductivity of copper is twice that of aluminum and four times that of brass. Three metal rods made of copper, aluminum, and brass are each 15 cm long and 2 cm in diameter. These rods are placed end to end, with aluminum between the other two. The free ends of the copper and brass rods are maintained at 100°C and 0°C, respectively. The system is allowed to reach the steady state condition. Assume there is no loss of heat anywhere,

1. Under steady-state condition, the equilibrium temperature of the copper-aluminum junction will be

1. 86°C
2. 18.8°C
3. 57°C
4. 73°C

Answer: 1. 86°C

2. When steady state condition is reached everywhere,

1. No heat is transmitted across the copper-aluminium or aluminium-brass junction
2. More heat is transmitted across the copper-aluminium junction than across the aluminium-brass junction
3. More heat is transmitted across the aluminium-brass junction than the copper-aluminium junction
4. Equal amount of heat is transmitted at the copper-aluminum and aluminium-brass junction

Answer: 4. Equal amount of heat is transmitted at the copper-aluminium and aluminium-brass junction

3. Under steady-state conditions, the equilibrium temperature of the aluminum-brass junction will be

1. 57°C
2. 35°C
3. 18.8°C
4. 28.5°C

Answer: 1. 57°C

Transmission Of Heat Integer Type Questions And Answers

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9.

Question 1. A metal rod AB of length 10x has its one end A in. ice at 0°C and the other end B in water at 100°C if a point P on the rod is maintained at 400°C, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is 540 cal · g^-1 and the latent heat of melting of ice is 80 cal · g^-1. If the point P is at a distance of λx from the ice end A, find the value of λ. (Neglect any heat loss to the surroundings.)
Answer: 9

Question 2. Two spherical bodies A (radius 6 cm] and B (radius IS cm) are at temperature T₁ and T₂, respectively. The maximum intensity in the emission spectrum of A is at 500 runs and of B is at 1500 nm. Considering them to be blackbodies, what will be the ratio of the rare of total energy radiated by A to that of B?
Answer: 9

Question 3. A liquid takes 5 minutes to cool from 80°C to 50°C- How much time (in min) will it take to cool from 60°C to 30 °C? The temperature of the surroundings is 20°C
Answer: 9

Question 4. The ends of the two rods of different materials with their thermal conductivities, radii of cross-section, and lengths in the ratio 1:2 are maintained at the same temperature difference. If the rate of flow of heat through the larger rod is 4 cal · s^-1, what is the rate of flow of heat (in cal • s^-1) through the shorter rod?
Answer: 1

 

Expansion Of Solid And Liquids Introduction Expansion Of Solids

WBBSE Class 11 Expansion of Solids and Liquids Notes

Usually, all solid substances expand on heating and contract on cooling. For a certain rise in temperature, this change is so small for solids, compared to that of liquids and gases, that the change is not always noticeable.

But with the help of proper experiments, it can be shown that solids expand on heating. This phenomenon of expansion with a change in temperature is called thermal expansion. Expansions in solids by the application of external forces have been discussed in the chapter Elasticity.

Thermal expansion in solids is of three types

1. Linear expansion (a change in length),
2. Surface or superficial expansion (the change in surface area) and
3. Volume or cubical expansion (change in volume).

The thermal expansion of different solids, for the same rise in temperature, is different. For example, copper undergoes a greater thermal expansion than iron, for the same rise in temperature with respect to their initial length, surface or volume.

Generally, with the rise in temperature, a solid expands equally in all directions. But there are exceptions as well. A few crystals expand differently in different directions when heated. Again an alloy of iron and nickel, called invar, practically does not show any expansion with the rise in temperature.

Reason for thermal expansion of solids: From the simple considerations of the structure of a crystalline solid, it can be said that the atoms of the crystal are arranged in a regular array under the elastic force.

Between two adjacent atoms this elastic force behaves like an almost inextensible spring. At a fixed temperature, two atoms thus maintain an average distance between them and vibrate. This average distance increases with rise in temperature. Hence, solids expand with the rise in temperature.

Expansion Of Solid And Liquids – Coefficient Of Surface Or Superficial Expansion

Superficial Expansion Definition: The increase in surface area for a unit rise in temperature for a unit surface area of a solid is called the coefficient of surface expansion of the material of that solid.

Let S₁ and S₂ be the surface areas of a solid at temperatures t₁ and t₂ respectively, where t₂ > t₁

Proceeding in a way similar, we get, the coefficient of surface expansion,

⇒ \(\beta=\frac{S_2-S_1}{S_1\left(t_2-t_1\right)}\)

= \(\frac{\text { increase in area }}{\text { initial area } \times \text { rise in temperature }}\) …….(1)

or, \(S_2-S_1=S_1 \beta\left(t_2-t_1\right)\)

or, \(S_2=S_1\left\{1+\beta\left(t_2-t_1\right)\right\}\) ….(2)

If the initial temperature = 0 and the final temperature = t, we may write, S_t = S₀ {1 + βt}……..(3)

where S₀ = surface area at zero temperature.

1. The coefficient of surface expansion β is not a constant For precise measurements of β, the surface area at 0°C is to be taken as the initial surface area.
2. Value of β does not depend on the unit of surface area,
3. Value of β depends on the unit of temperature.

Unit of β is °C or °F^-1. The change in temperature by 1°F = 5/9°C change in temperature.

∴ \(\beta_F=\frac{5}{9} \beta_C \text {, where } \beta_F\) = value of 0 in Fahrenheit scale, and 0C = value of 0 in Celsius scale.

 

Expansion Of Solid And Liquids – Relation Among The Three Coefficients of Expansion Numerical Examples

Example 1. At 30°C the diameter of a brass disc is 8 cm. What will be the increase in surface area if it is heated to 80°C? a of brass = 18 x 10^-6 °C^-1.
Solution:

Given

At 30°C the diameter of a brass disc is 8 cm.

Increase in surface area = \(S_2-S_1=\beta S_1\left(t_2-t_1\right)\)

Here, \(\beta=2 \alpha=2 \times 18 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1} \text { and } S_1=\pi \times\left(\frac{8}{2}\right)^2 \mathrm{~cm}^2\)

Increase in temperature = t₂– t₁ = 80-30 = 50 °C

∴ Increase in surface area, \(S_2-S_1\)=\(2 \times 18 \times 10^{-6} \times \pi \times\left(\frac{8}{2}\right)^2 \times 50\)

= \(36 \times 10^{-6} \times 16 \pi \times 50=0.0905 \mathrm{~cm}^2 .\)

Example 2. A rectangular copper block measures 20 cm x 12 cm x 3 cm. What will be the change in volume of the block when it is heated from 0°C to 800°C? The coefficient of linear expansion of copper is 0. 16 x 10^-4 °C^-1.
Solution:

Given

A rectangular copper block measures 20 cm x 12 cm x 3 cm.

Initial volume of the block, V₀ =20 X 12 X 3 = 720 cm³, increase in temperature = t₂ – t₁ = 800 – 0 = 800°C.

γ = 3α = 3 x 0.16 x 10^-4 °C^-1

Cubical expansion, \(V_{800}-V_0=V_0 \times \gamma \times(800-0)\)

= 720 x 3 x 0.16 x 10^-4 x 800 = 27.65 cm³.

Example 3. A lead bullet has a volume of 2.5 cm³ at 0°C. Its volume increases by 0.021 cm³ when heated to 98°C. Find the coefficient of linear expansion of lead.
Solution:

Given

A lead bullet has a volume of 2.5 cm³ at 0°C. Its volume increases by 0.021 cm³ when heated to 98°C.

By definition, the coefficient of volume expansion of lead, \(\gamma=\frac{V_t-V_0}{V_0 t}\)

Given, \(V_t-V_0=0.021 \mathrm{~cm}^3, V_0=2.5 \mathrm{~cm}^3 \text { and } t=98^{\circ} \mathrm{C}\)

∴ \(\gamma=\frac{0.021}{2.5 \times 98}=85.7 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\)

∴ Coefficient of linear expansion of lead \(\alpha=\frac{\gamma}{3}=\frac{8.57 \times 10^{-6}}{3}=2.86 \times 10^{-6 \circ} \mathrm{C}^{-1}\)

Example 4. An aluminium sphere of diameter 20 cm Is heated from 0°C to 100°C. What will be its change in volume? Coefficient of linear expansion of aluminium = 23x 10^-6 °C^-1.
Solution:

Given

An aluminium sphere of diameter 20 cm Is heated from 0°C to 100°C.

The initial volume of the aluminium sphere,

= \(\frac{4}{3} \pi\left(\frac{20}{2}\right)^3=\frac{4}{3} \pi(10)^3 \mathrm{~cm}^3\)

Value of γ for aluminium =3 x α =3x23x 10^-6 °C^-1.

Hence, increase in volume, \(V_t-V_0=V_0 \gamma t=\frac{4}{3} \pi \times 10^3 \times 3 \times 23 \times 10^{-6} \times 100\)

= 28.9 cm³.

Example 5. A piece of metal weighs 46 g xg in air. When immersed in a liquid of relative density 1.24, kept at 27°C, its weight is 30 g x g. When the temperature of the liquid is raised to 42°C, the metal piece in it weighs 30.5 g x g. At 42°C, the relative density of the liquid is 1.20. Find the coefficient of linear expansion of the metal.
Solution:

Given

A piece of metal weighs 46 g xg in air. When immersed in a liquid of relative density 1.24, kept at 27°C, its weight is 30 g x g. When the temperature of the liquid is raised to 42°C, the metal piece in it weighs 30.5 g x g. At 42°C, the relative density of the liquid is 1.20.

The apparent loss in weight of the metal at 27°C = weight of an equal volume of the liquid = (46 – 30) g x g;

Thus the volume of the displaced liquid at 27 °C = \(\frac{46-30}{1.24}=\frac{16}{1.24} \mathrm{~cm}^3\) = volume of the metal piece at 27°C(= V₁).

Similarly, the volume of the metal piece at 42 °C (= V₂)

= \(\frac{46-30.5}{1.20}=\frac{15.5}{1.20} \mathrm{~cm}^3\)

∴ Coefficient of volume expansion of the metal,

⇒ \(\gamma=\frac{V_2-V_1}{V_1\left(t_2-t_1\right)}=\frac{1}{\left(t_2-t_1\right)}\left(\frac{V_2}{V_1}-1\right)\)

= \(\frac{1}{42-27}\left(\frac{15.5}{1.2} \times \frac{1.24}{16}-1\right)=\frac{1}{15}\left(\frac{961}{960}-1\right)\)

= \(\approx 6.94 \times 10^{-5 \circ} \mathrm{C}^{-1}\)

∴ The coefficient of linear expansion of the metal piece \(\alpha=\frac{\gamma}{3}=\frac{6.94 \times 10^{-5}}{3}{ }^{\circ} \mathrm{C}^{-1}=23.15 \times 10^{-6{ }^{\circ}} \mathrm{C}^{-1}\)

Change Of Density Of A Solid Due To Change Of Temperature

Understanding Thermal Expansion of Solids and Liquids

It is known that the density of a substance = \(\frac{\text { mass }}{\text { volume }}\).

With the change in temperature, while the mass of a solid remains the same, its volume changes. Hence, with the change in temperature, the density of a solid changes.

With the rise in temperature, volume increases, thus density decreases and with the decrease in temperature, volume decreases, thus density increases.

Let for a solid of mass m at temperature t₁, the volume be V₁ and density be D₁; while at temperature t₂, its volume becomes V₂ and density becomes D₂.

∴ \(D_1=\frac{m}{V_1} \text { and } D_2=\frac{m}{V_2}\)

∴ \(\frac{D_1}{D_2}=\frac{V_2}{V_1}\)

If the coefficient of volume expansion of the solid is γ, then

⇒ \(V_2=V_1\left[1+\gamma\left(t_2-t_1\right)\right] \quad \text { or, } \frac{D_1}{D_2}=\frac{V_1\left[1+\gamma\left(t_2-t_1\right)\right]}{V_1}\)

or, \(D_1=D_2\left[1+\gamma\left(t_2-t_1\right)\right]\)

or, \(D_2=\frac{D_1}{\left[1+\gamma\left(t_2-t_1\right)\right]}=D_1\left[1+\gamma\left(t_2-t_1\right)\right]^{-1}\)

or, \(D_2=D_1\left[1-\gamma\left(t_2-t_1\right)\right]\)

[neglecting higher powers of γ(t₂ – t₂), as γ is very small]

If t₂ > t₁, D₂ < D₁

If D₀ and D_t are the densities of the solid at 0 and t degree temperatures respectively, equation (2) reduces to D_t = D₀(1-γt).

Expansion Of Solid And Liquids – Change Of Density Of A Solid Due To Change Of Temperature Numerical Examples

Example 1. Density of glass at 10°C is 2.6 g · cm^-3 and that at 60°C is 2.596 g • cm^-3. What is the average value of the coefficient of linear expansion of glass between these two temperatures?
Solution:

Given

Density of glass at 10°C is 2.6 g · cm^-3 and that at 60°C is 2.596 g • cm^-3.

Using the equation D₁ = D₂ [1 +γ(t₂ -t₁)], and substituting the given values,

D₁ = 2.6 g · cm^-3, D₁ = 2.596 g · cm^-3, t₁ = 10°C and t₂ = 60°C,

we get, 2.6 = 2.596 [1 + γ(60 – 10)]

or, \(1+50 \gamma=\frac{2.6}{2.596} \quad \text { or, } 50 \gamma=\frac{2.6-2.596}{2.596}\)

or, 50γ = 1.00154 – 1

or, \(\gamma=\frac{0.00154}{50}=30.8 \times 10^{-6 \circ} \mathrm{C}^{-1}\)

∴ \(\alpha=\frac{\gamma}{3}=10.27 \times 10^{-6 \circ} \mathrm{C}^{-1}\)

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Unit 7 Properties Of Bulk Matter Chapter 5 Expansion Of Solid And Liquids Thermal Stress

A change in temperature causes a change in the length of a metal rod. But if the two ends of the rod are rigidly fixed at fixed supports, expansion or contraction of the rod gets obstructed. Hence, a large force is generated along the rod. This force, measured per unit area of the rod is called ther¬mal stress.

Experimental demonstration of thermal stress: A metal rod B is set within the gap of a heavy iron frame Y. One end of the rod B is threaded and two holes P₁ and P₂ are at the other end.

• A screw N is fitted at the threaded end of B. A cast iron pin is introduced through the hole P₁ and the rod B is heated.
• When the rod expands, the pin at P₁ is tightly fitted with the frame by adjusting the screw N.
• If the rod is cooled now, the pin obstructs the contraction of the rod developing a huge force which breaks the pin inserted through P₁.
• Expansion of the rod, when obstructed, also generates a huge force. To demonstrate this, a pin is inserted through P₂ and the rod is fixed rigidly by adjusting the screw N. If now the rod is heated, the pin P₂ gets broken due to the force developed in the rod on expansion.

Magnitude of thermal stress: Let a rod of length l, cross-sectional area A, coefficient of linear expansion α be heated so that the rise in temperature is t. The rod, therefore expands by l∝t.

• Now if the two ends of the rod are rigidly fixed and it is cooled to its original temperature, it tends to contract back to its original length and this contraction is opposed by a force F (say).
• Therefore, the reaction to the force F, which is equal and opposite to F, is the thermal force developed in the rod due to expansion lα t.

From Hooke’s law, the Young’s modulus of the rod,

Y = \(\frac{\text { stress }}{\text { strain }}=\frac{\text { applied force } / \text { area }}{\text { change in length } / \text { initial length }}=\frac{F / A}{l a t / l}\)

∴ F = AYαt

Therefore, the thermal stress = F/A = Yαt

Clearly, thermal stress is independent of length or area of the cross-section of the rod (or a wire).

Expansion Of Solid And Liquids Thermal Stress Numerical Examples

Example 1. Two ends of a steel rod are rigidly fixed with two supports. At 30° C its area of cross-section is 4 cm². How much force will be exerted on the supports by the ends of the rod if the temperature of the rod is raised by 60°C? [Young’s modulus of steel = 2.1 x 10¹² dyn · cm^-2 and its coefficient of linear expansion is 12 x 10^-6 °C^-1
Solution:

Given

Two ends of a steel rod are rigidly fixed with two supports. At 30° C its area of cross-section is 4 cm².

In this case, A = 4 cm², Y = 2.1 x 10¹² dyn · cm^-2, α = 12 x 10^-6 °C^-1 and t = 60 – 30 = 30°C

∴ The force exerted

=AYαt =4×2.1 x 10¹² x 12 x 10^-6 x 30 = 3.024x 10⁹ dyn.

Example 2. Two ends of a wire are rigidly clamped. If its temperature is decreased by 10°C, find the change in the tension of the wire. Area of cross-section of the wire =0.01 cm²; α= 16 x 10 ^-6 °C^-1, Y = 20 x 10¹¹ dyn · cm^-2
Solution:

Given

Two ends of a wire are rigidly clamped. If its temperature is decreased by 10°C,

Area of cross-section of the wire =0.01 cm²; α= 16 x 10 ^-6 °C^-1, Y = 20 x 10¹¹ dyn · cm^-2

Here A = 0.01 cm², Y = 20 x 10¹¹ dyn · cm^-2,α = 16 x 10^-6 °C^-1, t = 10°C

∴ Change in tension

AYαt = 0.01 x 20 x 10¹¹ x 16 x 10^-6 x 10 = 32 x 10⁵ dyn.

 

Unit 7 Properties Of Bulk Matter Chapter 5 Expansion Of Solid And Liquids Apparent And Real Expansion Of Liquids

To heat a liquid, it has to be kept in a container. When heat is applied, the container also expands along with the liquid. As the liquid expands more than the container for the same change in temperature, expansion of the container is sometimes neglected and only the expansion of the liquid is recorded.

• Hence, the recorded expansion of the liquid, ignoring the expansion of the container, is less than the actual expansion of the liquid.
• The expansion of a liquid, ignoring the expansion of the container, is called the apparent expansion of the liquid.
• The sum of the apparent expansion of the liquid and the expansion of the container is called the real expansion of the liquid.

Experiment: Let, surface of the liquid in a flask rest at mark O. If the flask is heated from outside, at first the flask expands and the surface of the liquid comes down at mark A.

Thus the length OA represents expansion of the flask. Then the supplied heat reaches the liquid and surface of the liquid rises up to mark B. Therefore, volume of AB is the real expansion of the liquid.

On the other hand, if the first change of volume is unnoticed, it seems like the surface rises from initial position O to final position B. So, volume of OB is the apparent expansion of the liquid.

∴ For a liquid, real expansion = apparent expansion of the liquid + expansion of container

Coefficients of Apparent and Real Expansion of Liquids

Since liquid expansions are of two types, two separate coefficients of expansion are to be considered:

1. Coefficient of apparent expansion and
2. Coefficient of real expansion.

Coefficient of apparent expansion of a liquid Definition: The apparent expansion of unit volume of a liquid for a temperature rise of 1° is called the coefficient of apparent expansion (γ’) of the liquid.

Expression for γ’: Let the volume of a certain amount of liquid be V₁ at temperature t₁, and its apparent volume be V’₂ at temperature t₂.

∴ For a rise in temperature of (t₂ – t₁), apparent expansion of the liquid of volume \(V1 = (V2′ – Vx).V_1=\frac{V_2^{\prime}-V_1}{\left(t_2-t_1\right)}\)

∴ For a unit rise in temperature, the apparent expansion per unit volume = \(\frac{V_2^{\prime}-V_1}{V_1\left(t_2-t_1\right)}\)

By definition, \(\gamma^{\prime} =\frac{V_2^{\prime}-V_1}{V_1\left(t_2-t_1\right)}\)

= \(\frac{\text { apparent expansion }}{\text { initial volume } \times \text { rise in temperature }}\) ….(1)

From (1), we get, \(V_2^{\prime}=V_1\left\{1+\gamma^{\prime}\left(t_2-t_1\right)\right\}\)…(2)

It is important to note that, the coefficient of apparent expansion of a liquid is not an intrinsic property of the liquid. It depends on the material of the container. Hence, a liquid may have different values of γ’ when heated in containers of different materials.

Coefficient of real expansion of a liquid Definition: The actual or real increase of unit volume of a liquid for a temperature rise of 1° is called the coefficient of real expansion (γ) of the liquid.

Expression for γ: Let the volume of a fixed amount of a liquid at a temperature t₁ be V₁, and at a temperature t₂ be V₂.

Hence, volume increases by (V₂-V₁) for a rise (t₂ – t₁) in temperature.

∴ By definition, \(\gamma=\frac{V_2-V_1}{V_1\left(t_2-t_1\right)}\)

= \(\frac{\text { real expansion }}{\text { initial volume } \times \text { rise in temperature }}\)…(3)

∴ \(V_2 =V_1\left\{1+\gamma\left(t_2-t_1\right)\right\}\) …(4)

The coefficient of real expansion is an intrinsic property of the liquid and does not depend on the material of the container.

1. It is clear from equations (1) and (3) that the values of γ and γ’ are independent of the unit of volume but depend on scale of temperature used.

• For example, the coefficient of real expansion of mercury in the Celsius and the Fahrenheit scales are 18.18 x 10^{-5 °}C^-1 and 10.1 x 10^{-5 °}F^-1 respectively.
• The coefficient of volume expansion of a liquid is the same in the Celsius and the Kelvin scales but in the Fahrenheit scale it is 5/9 times that in Celsius and Kelvin scales.

2. It is assumed during the derivations of (2) and (4) that the value of the coefficient of expansion of a liquid is the same for all ranges of temperature. Precise observations show that the value changes, though the changes are very small.

Hence, the values deduced above are the average values of γ and γ’ for the temperature range between t₁ and t₂. However, in practice, the values of γ and γ’ of a liquid are taken as constants for all temperature ranges.

3. While defining γ or γ’, initial volume at any temperature is taken. But for finer measurements, volume at 0°C should be taken as the initial volume In practice, the difference is ignored.

Values of the coefficient of real expansion of a few liquids are shown in the following table.

Relation between the Coefficients of Apparent and Real Expansions of Liquids

Let the volume of certain mass of a liquid in a container at a temperature t₁ = V₁. At a temperature t₂, the apparent volume of that liquid = V’₂ and its real volume = V₂.

The part of the container, that contained the liquid at t₁, has a volume V at t₂.

∴ Apparent expansion of the liquid = V₂‘ – V₁ and real expansion = V₂ – V₁.

Expansion of the part of the container containing the liquid at t₁ = V- V₁

Since for a liquid, real expansion = apparent expansion + expansion of container, \(V_2-V_1=\left(V_2^{\prime}-V_1\right)+\left(V-V_1\right)\)

Dividing both sides by V₁ t (where t = t₂ – t₁)

⇒ \(\frac{V_2-V_1}{V_1 t}=\frac{V_2^{\prime}-V_1}{V_1 t}+\frac{V-V_1}{V_1 t}\)

or, γ = γ’ + γ_g[γ_g = coefficient of volume expansion of the material of the container]

Hence, the coefficient of real expansion of a liquid = the coefficient of apparent expansion of the liquid + the coefficient of volume expansion of the material of the container.

Coefficient of Linear Expansion in Solids

Relation between Density and Coefficient of Real Expansion of Liquids: The volume of a liquid increases with the increase in temperature. Thus the density decreases. Water between 0°C and 4°C is an exception, and that will be discussed later. Let the mass of some liquid be m, the volume of that liquid be V₁ and the density be ρ₁, at temperature t₁.

At temperature t₁, its volume becomes V₁ and density ρ₂. Consider t₂> t₁.

Hence, \(m_1=V_1 \rho_1=V_2 \rho_2 \quad \text { or, } \frac{\rho_1}{\rho_2}=\frac{V_2}{V_1}\)…(1)

If the coefficient of real expansion of the liquid is γ, then

⇒ \(V_2 =V_1\left\{1+\gamma\left(t_2-t_1\right)\right\}\)

or, \(\frac{V_2}{V_1}=1+\gamma\left(t_2-t_1\right)\)…..(2)

From equations (1) and (2) we get,

⇒ \(\frac{\rho_1}{\rho_2}=\left\{1+\gamma^{\prime}\left(t_2-t_1\right)\right\}\)

or, \(\rho_1=\rho_2\left[1+\left(t_2-t_1\right)\right]\)….(3)

\(\rho_2 =\frac{\rho_1}{1+\gamma\left(t_2-t_1\right)}=\rho_1\left\{1+\gamma\left(t_2-t_1\right)\right\}^{-1}\)

= \(\rho_1\left[1-\gamma\left(t_2-t_1\right)\right]\)

neglecting higher powers of γ, as it is very small.

Hence, the density of a liquid decreases with the increase in temperature.

Equations (3) and (4) both give the relation between the coefficient of real expansion and the density of the liquid.

Equation (4) can be written as \(\gamma=\frac{\rho_1-\rho_2}{\rho_1\left(t_2-t_1\right)}\)

Thus if the densities of a liquid at two different temperatures are known, its coefficient of real expansion can be found out.

 

Unit 7 Properties Of Bulk Matter Chapter 5 Expansion Of Solid And Liquids Apparent Loss In Weight Of A Solid Immersed In A Liquid At Different Temperatures

There is an apparent loss in the weight of a solid when it is immersed in a liquid. This loss in weight is due to the upthrust it receives in a liquid. This upthrust depends directly on the density of the liquid as well as the volume of the immersed portion of the body.

With the change in temperature, density of the liquid and volume of the solid, both change. Hence, the apparent weight of a solid will be different at different temperatures.

Let the weight of a solid in air = W.

Weight of the solid completely immersed in a liquid at temperature t₁ = W₁ and that at temperature t₂ = W₂.

Hence, the apparent loss in weight at temperature t₁ = W- W₁ = M₁g, and that at t₂ = W- W₂ = M₂g,

where M₁ and M₂ are the masses of the liquid displaced by the body.

If V₁ and ρ₁ are the volume and density at t₁, and V₂ and ρ₂ are those at t₂ respectively, then, \(M_1=V_1 \rho_1 \text { and } M_2=V_2 \rho_2\)

If the coefficient of real expansion of the liquid is γ and the coefficient of volume expansion of the material of the solid is γ_s, then,

⇒ \(\rho_1=\rho_2\left[1+\gamma\left(t_2-t_1\right)\right]\)

= \(\rho_2[1+\gamma t]\left[\text { Let } t_2-t_1=t\right]\)

and \(V_2=V_1\left(1+\gamma_s t\right)\)

∴ \(M_2 =V_2 \rho_2=\frac{\rho_1 V_1\left(1+\gamma_s t\right)}{1+\gamma t}\)

= \(\rho_1 V_1\left(1+\gamma_s t\right)(1+\gamma t)^{-1}\)

= \(\rho_1 V_1\left(1+\gamma_s t\right)(1-\gamma t)\)

[neglecting the higher powers of γ]

Coefficient of Volume Expansion in Liquids

= \(M_1\left[1-\left(\gamma-\gamma_s\right) t\right]\) neglecting the term \(\gamma \gamma_s t^2\)

Usually \(\gamma \gg \gamma_s and t=t_2-t_1>0\),

∴ \(M_1>M_2 \text { or, } M_1 g>M_2 g \text { i.e., } W_1<W_2 \text {. }\)

Hence, the apparent weight of a body immersed in a liquid increases with the increase in temperature of the liquid.

Unit 7 Properties Of Bulk Matter Chapter 5 Expansion Of Solid And Liquids Apparent Loss In Weight Of A Solid Immersed In A Liquid At Different Temperatures Numerical Examples

Example 1. A piece of metal weighs 50 g x g in air. It weighs45 g x g when immersed in a liquid at 25°C, and 45.1 g x g at 100°C. If the coefficient of linear expansion of the metal is 12 x 10^-6 °C^-1, find the coefficient of real expansion of the liquid.
Solution:

Given

A piece of metal weighs 50 g x g in air. It weighs45 g x g when immersed in a liquid at 25°C, and 45.1 g x g at 100°C. If the coefficient of linear expansion of the metal is 12 x 10^-6 °C^-1,

Apparent loss in weight at 25°C = M₁g = (50 – 45) gxg = 5gxg Apparent loss in weight at 100°C

= \(M_1 g=(50-45) g \times g=5 g \times g\)

Apparent loss in weight at \(100^{\circ} \mathrm{C}\)

= \(M_2 g=(50-45.1) g \times g=4.9 \mathrm{~g} \times \mathrm{g}\)

As \(M_2=M_1\left[1-\left(\gamma-\gamma_s\right)\left(t_2-t_1\right)\right]\)

∴ 4.9 = \(5\left[1-\left(\gamma-12 \times 10^{-6} \times 3\right)(100-25)\right]\)

or, 4.9 = \(5\left[1-\left(\gamma-12 \times 10^{-6} \times 3\right) \times 75\right]\left[\right. as \left.\gamma_s=3 \times \alpha_s\right]\)

or, \(\left(\gamma-36 \times 10^{-6}\right) \times 75 \times 5=5-4.9\)

or, \(\gamma-36 \times 10^{-6}=\frac{0.1}{75 \times 5}\)

or, \(\gamma=36 \times 10^{-6}+2.67 \times 10^{-4}=3.03 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\).

Example 2. A glass rod weighs 90g x g In air. It weighs 49.6 g x g when immersed In n liquid at 12°C, and 51.9 g x g at 97°C. Wnd the real expansion coefficient of the liquid. Volume expansion coefficient of glass = 2.4 x 10^-5 °C^-1
Solution:

Given

A glass rod weighs 90g x g In air. It weighs 49.6 g x g when immersed In n liquid at 12°C, and 51.9 g x g at 97°C.

let the volume of the glass rod be V₁, at 12°C, and density of the liquid be ρ₁.

The mass of the displaced liquid at that temperature = 90-49.6 =40.4 g

∴ Volume of the glass rod at that temperature, \(V_1=\frac{40.4}{\rho_1}\)……(1)

Again at 97°C, mass of the displaced liquid = 90-51.9 = 38.1 g

Let at 97°C the volume of the glass rod be V₂, and density of the liquid be ρ₂.

∴ \(V_2=\frac{38.1}{\rho_2}\) ….(2)

From (1) and (2), \(\frac{V_2}{V_1}=\frac{\rho_1}{\rho_2} \times \frac{38.1}{40.4}\)….(3)

Now for the glass rod, V₂ = V₁ [1 + 2.4 x 10^-5 x 85]

∴ \(\frac{V_2}{V_1}=1+2.4 \times 10^{-5} \times 85=1.00204\)

In case of liquid, \(\rho_1=\rho_2[1+\gamma \times 85]\)

∴ \(\frac{\rho_1}{\rho_2}=1+\gamma \times 85\)

From equation (3), \(1.00204=(1+\gamma \times 85) \times \frac{38.1}{40.4}\)

or, \(1+85 \gamma=\frac{1.00204 \times 40.4}{38.1}\)

or, \(85 \gamma=\frac{1.00204 \times 40.4}{38.1}-1=\frac{2.382}{38.1}\)

or, \(\gamma=7.35 \times 10^{-4 \circ} \mathrm{C}^{-1} .\)

Example 3. Apparent weights of a solid in a liquid are 50 g x g and 52 g x g at 25°C and 75°C respectively. If the coefficient of linear expansion of the solid Is ag α_s = 6.6 x 10^-6 °C^-1, and γ for the liquid Is 7.3 x 10^-4 °C^-1, what Is the real weight of the solid In air?
Solution:

Given

Apparent weights of a solid in a liquid are 50 g x g and 52 g x g at 25°C and 75°C respectively. If the coefficient of linear expansion of the solid Is ag α_s = 6.6 x 10^-6 °C^-1, and γ for the liquid Is 7.3 x 10^-4 °C^-1,

Let the real weight of the solid in air = M g x g.

Apparent loss in weight at \(225^{\circ} \mathrm{C}:-M_1 g=(M-50) g \times g\)

and apparent loss in weight at \(75^{\circ} \mathrm{C}=M_2 g \circ(M-52) \mathrm{g} \times g\)

∴ \(M_2=M_1\left|1-\left(\gamma-\gamma_s\right) \times t\right|\)

⇒ \((M-52)=(M-50) \mid 1-\left(7.3 \times 10^{-4}\right.\) \(-19.8 \times\left(0^{-6}\right) \times 501\)

= \((M-50)\left\{1-7.102 \times 10^{-4} \times 50\right\}\)

= \((M-50) \times 0.96449\)

M = 0.96449 M-48.2245+52

or, 0.03551 M=3.7755

M = \(\frac{3.7755}{0.03551}=106.32 \mathrm{~g}\)

∴ Real weigth of the solid is 106.32 g x g

Example 4. A sphere of mass 266.5 g and of diameter 7 cm floats on a liquid. When the liquid Is heated to 35°C the sphere sturts sinking In the liquid. If the density of the liquid at 0°C Is 1.527 g • cm^-3, find the coefficient of volume expansion. Neglect the expansion of the sphere.
Solution:

Given

A sphere of mass 266.5 g and of diameter 7 cm floats on a liquid. When the liquid Is heated to 35°C the sphere sturts sinking In the liquid. If the density of the liquid at 0°C Is 1.527 g • cm^-3,

Volume of the sphere = \(\frac{4}{3} \pi\left(\frac{7}{2}\right)^3 \mathrm{~cm}^3\)

∴ Volume of displaced liquid at 35°C = \(\frac{4}{3} \pi\left(\frac{7}{2}\right)^3 \mathrm{~cm}^3\)

If the density of the liquid at 35°C is ρ₃₅, then the mass of the displaced liquid at 35°C = \(\frac{4}{3} \pi\left(\frac{7}{2}\right)^3 \times \rho_{35} g .\)

From the condition of floatation, \(\frac{4}{3} \pi\left(\frac{7}{2}\right)^3 \times \rho_{35}=266.5\)…(1)

Now, \(\rho_{35}=\frac{\rho_0}{1+\gamma \times 35}\)

= \(\frac{1.527}{1+\gamma \times 35}\)

From equations (1) and (2), we get, \(\frac{1.527}{1+\gamma \times 35}=\frac{266.5}{\frac{4}{3} \pi\left(\frac{7}{2}\right)^3}\)

or, \(1+35 \gamma=\frac{4 \times 22 \times(7)^3 \times 1.527}{3 \times 7 \times(2)^3 \times 266.5}\)

or, \(35 \gamma=1.029-1\)

or, \(\gamma=\frac{0.029}{35}=8.28 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\).

Example 5. A piece of metal weighs 46 g x g in air. It weighs 30 g x g in a liquid of specific gravity 1.24 at 27°C. At 42°C, when the specific gravity of the liquid is 1.20, the weight of the piece immersed in it is 30.5 g x g. Find the coefficient of linear expansion (α) of the metal.
Solution:

Given

A piece of metal weighs 46 g x g in air. It weighs 30 g x g in a liquid of specific gravity 1.24 at 27°C. At 42°C, when the specific gravity of the liquid is 1.20, the weight of the piece immersed in it is 30.5 g x g.

Mass of the displaced liquid at 27°C = 46-30 = 16 g

Volume of the displaced liquid at 27°C, \(V_{27}=\frac{16}{1.24} \mathrm{~cm}^3\)

Similarly volume of the displaced liquid at 42 °C, \(V_{42}=\frac{46-30.5}{1.20}=\frac{15.5}{1.20} \mathrm{~cm}^3\)

So the volume of the piece at 27 °C and 42° C are \(\frac{16}{1.24} \mathrm{~cm}^3\) and \(\frac{15.5}{1.20} \mathrm{~cm}^3\) respectively.

Now, \(V_{42}=V_{27}\{1+\gamma(42-27)\}\)

[where γ = coefficient of volume expansion of the metal]

or, \(\frac{15.5}{1.20}=\frac{16}{1.24}\{1+\gamma \times 15\} or, 1+15 \gamma=\frac{15.5 \times 1.24}{1.20 \times 16}\)

or, \(15 \gamma=1.001-1\)

or, \(\gamma=\frac{0.001}{15}=3 \alpha\)

∴ Coefficient of linear expansion of the metal \(\alpha=\frac{\gamma}{3}=\frac{0.001}{45}=2.2 \times 10^{-5 \circ} \mathrm{C}^{-1}\)

Unit 7 Properties Of Bulk Matter Chapter 5 Expansion Of Solid And Liquids Anomalous Expansion Of Water

When a liquid is heated, its volume increases and density decreases with rise in temperature. Exceptions are observed in case of water for a certain range of temperatures.

• When heated from 0°C to 4°C, the volume of water decreases and the density increases. Above 4°C, the volume of water increases again with the increase in temperature.
• Hence, water has a maximum density and a minimum volume at 4°C. Also, on cooling from 4°C to 0°C, the volume of water increases instead of decreasing. This exceptional behaviour of water in respect of expansion within the range of 0°C to 4°C, is called anomalous expansion of water.
• The two graphs below represent the change In volume and density of 1 g of water, with the Increase in temperature.

Conclusions from the two graphs are:

1. Volume of water decreases as its temperature rises from 0°C to 4°C. Hence, expansion of water is anomalous and does not follow the general rule. Consequently, the coefficient of volume expansion of water is negative for this temperature range.
2. Water has the least volume and the maximum density at 4°C.
3. After 4°C, expansion of water takes place following the general rule. This means that, with the increase in tem-perature, the volume also starts increasing. The expan¬sion is no longer anomalous.
4. The slope of the curves for volume or for density, at temperature close to 4°C is almost zero. So, there is practically no change in volume or density for a small variation of temperature at and around 4°C. Hence, the density of water at 4°C is taken as unity.

Effect of Anomalous Expansion of Water on Marine Life: Due to anomalous expansion of water, fishes and various living creatures can survive under frozen lakes, rivers or seas.

• In cold countries, with the fall in atmospheric temperature, upper surface of lakes, seas and various ponds gradully, cool. Water of the upper surface, then being denser and heavier, moves down.
• Water below it, being comparatively warmer and lighter, moves up. This convection process in water continues until the density of the water in the lower part becomes maximum i.e., the temperature of the lower water reaches 4°C.
• As the temperature of the upper surface decreases further below 4°C, density begins to decrease. So water cannot move down further. It then begins to cool further and at last turns into ice. As ice is lighter than water, a thick layer of ice, thus formed, floats over the surface of water.

Both ice and water are bad conductors of heat. So, a negligible amount of heat can be conducted from the lower levels of water to the atmosphere outside. So, the entire vol¬ume of water (top to bottom) in a pond cannot freeze.

Unit 7 Properties Of Bulk Matter Chapter 5 Expansion Of Solid And Liquids Anomalous Expansion Of Water Numerical Examples

Applications of Thermal Expansion in Engineering

Example 1. The area of the cross-section of the capillary tube of a mercury thermometer is A₀ and the volume of the bulb filled with mercury is V₀ at 0°C. Find the length of the mercury column in the capillary tube as the bulb is heated to t°C. The coefficient of linear expansion of glass is α and coefficient of volume expansion of mercury is β.
Solution:

Given

The area of the cross-section of the capillary tube of a mercury thermometer is A₀ and the volume of the bulb filled with mercury is V₀ at 0°C.

The coefficient of linear expansion of glass is α and coefficient of volume expansion of mercury is β.

Let V_g and V_m be the volumes of the bulb and mercury at t°C.

∴ \(V_g=V_0\left(1+\gamma_g t\right)=V_0(1+3 \alpha t)\)

and \(V_m=V_0(1+\beta t)\)

Hence, volume of mercury entering the tube is \(V_m-V_g=V_0(1+\beta t)-V_0(1+3 \alpha t)=V_0 t(\beta-3 \alpha)\)

If the area of the cross-section of the capillary tube at t°C is A_t then A_t = A₀(1+2αt).

If the length of the mercury column in the tube is l, then \(A_t \times l=V_0 t(\beta-3 \alpha)\)

∴ l = \(\frac{V_0 t(\beta-3 \alpha)}{A_0(1+2 \alpha t)}\)….(1)

= \(\frac{V_0 t(\beta-3 \alpha)(1+2 \alpha t)^{-1}}{A_0}=\frac{V_0}{A_0} t(\beta-3 \alpha)(1-2 \alpha t)\)

= \(\frac{V_0}{A_0}(\beta-3 \alpha) t\)…..(2)

[neglecting αβ and α² as they are very small]

Both the equations (1) and (2), indicate the length of mercury column in the capillary tube at t °C.

Example 2. A 1L flask contains some mercury. It is observed that the volume of air in the flask remains unchanged at all temperatures. What is the volume of mercury in the flask? The coefficient of linear expression of the material of the flask = 9 x 10^-6 °C^-1 and coefficient of real expansion of mercury = 1.8 x 10^-4 °C^-1
Solution:

Given

A 1L flask contains some mercury. It is observed that the volume of air in the flask remains unchanged at all temperatures.

The coefficient of linear expression of the material of the flask = 9 x 10^-6 °C^-1 and coefficient of real expansion of mercury = 1.8 x 10^-4 °C^-1

Due to expansion of the scale, the actual barometric reading will be greater than the apparent reading. Let the real reading be h and the apparent reading be H.

∴ h = \(H(1+\alpha t)=75.34\left(1+18 \times 10^{-6} \times 5\right)\)

= \(75.34\left(1+9 \times 10^{-5}\right)\)

h = \(75.34 \times 1.00009 \mathrm{~cm}\)

Again, mercury expands with increase in temperature. Hence, the barometric reading at 0°C will be less than that at 20°C. Let at 0°C, the barometric reading be H₀.

∴ \(H_0=h(1-\gamma t)=75.34 \times 1.00009\left[1-18 \times 10^{-5} \times 25\right]\)

= 75.34 x 1.00009[1 -0.0045]

= 75.34 x 1.00009 x 0.9955 = 75.01 cm.

Example 4. Coefficients of volume expansion of benzene and wood are 1.2×10^-3 °C^-1 and 1.5 x 10^-4 °C^-1 respectively. Their respective densities at 0°C are 900 kg · m^-3 and 880 kg · m^-3. Find the temperature at which wood will just immerse in benzene.
Solution:

Given

Coefficients of volume expansion of benzene and wood are 1.2×10^-3 °C^-1 and 1.5 x 10^-4 °C^-1 respectively. Their respective densities at 0°C are 900 kg · m^-3 and 880 kg · m^-3.

The wood will just immerse in benzene at a temperature for which the densities are equal. Let the required temperature be t°C, when the density of both is ρ.

∴ 900 = \(\rho\left[1+1.2 \times 10^{-3} \times t\right]\)

and \(880=\rho\left[1+1.5 \times 10^{-4} \times t\right]\)

Dividing (1) by (2) we get, \(\frac{900}{880}=\frac{1+1.2 \times 10^{-3} t}{1+1.5 \times 10^{-4} t}\)

or, \(t\left(900 \times 1.5 \times 10^{-4}-880 \times 1.2 \times 10^{-3}\right)=880-900 or, \quad t=21.7^{\circ} \mathrm{C}\).

Example 5. A metal piece of density 8g · m^-3 is suspended from a wooden hook by a weightless string. The tension in the string is 56 g x g. What will be the tension in the string, if the system is Immersed in a liquid at 40°C? The surrounding temperature during the experiment is 20°C. At 20°C the specific gravity of the liquid is 1.24. The coefficients of volume expansion of the liquid and the metal are 4 x 10^-5 °C^-1 and 8 x 10^-4 °C^-1 respectively.
Solution:

Given

A metal piece of density 8g · m^-3 is suspended from a wooden hook by a weightless string. The tension in the string is 56 g x g.

The surrounding temperature during the experiment is 20°C. At 20°C the specific gravity of the liquid is 1.24. The coefficients of volume expansion of the liquid and the metal are 4 x 10^-5 °C^-1 and 8 x 10^-4 °C^-1 respectively.

Volume of the metal piece at 20°C, \(V_{20}=\frac{56}{8}=7 \mathrm{~cm}^3\)

∴ Volume at 40°C, \(V_{40}=V_{20}\left[1+8 \times 10^{-4} \times 20\right]\)

= \(7\left(1+8 \times 10^{-4} \times 20\right) \mathrm{cm}^3\)

Volume of the displaced liquid = V₄₀

∴ Mass of displaced liquid = V₄₀ x ρ₄₀

[ρ₄₀ = density of the liquid at 40°C]

i.e, weight of the displaced liquid = V₄₀ x ρ₄₀ x g

∴ Upthrust = \(V_{40} \times \rho_{40} \times g\)

= \(7\left[1+8 \times 10^{-4} \times 20\right] \times \frac{1.24}{1+20 \times 4 \times 10^{-5}} \times 980\)

= \(7 \times 1.24[1+0.016]\left[1+8 \times 10^{-4}\right]^{-1} \times 980 \)

= \(7 \times 1.24 \times 1.016[1-0.0008] \times 980\)

= \(7 \times 1.24 \times 1.016 \times 0.9992 \times 980=8.81 \times 980 \mathrm{dyn}\)

∴ Tension in the string = \((56-8.81) \times 980=4.625 \times 10^4 \mathrm{dyn} .\)

Example 6. A body, at 4°C, floats with 0.98 part of its volume immersed in water. At what temperature the body will just be immersed in water? Coefficient of real expansion of water = 3.3 x 10^-4 ° C^-1. Neglect expansion of the solid body.
Solution:

Given

A body, at 4°C, floats with 0.98 part of its volume immersed in water.

Coefficient of real expansion of water = 3.3 x 10^-4 ° C^-1.

Let the required temperature be t°C, and the volume of the body = V.

Let the densities of water at 4°C and t°C be d₁ and d₂ respectively.

∴ From the condition of floatation, Vx 0.98 x d₁ = Vxd₂

∴ \(\frac{d_1}{d_2}=\frac{1}{0.98}=\frac{50}{49}\)

As \(d_1=d_2\left\{1+3.3 \times 10^{-4}(t-4)\right\},\)

∴ \(\frac{d_2}{d_2}\left\{1+3.3 \times 10^{-4}(t-4)\right\}=\frac{50}{49}\)

∴ \(3.3 \times 10^{-4} \times(t-4)=\frac{50}{49}-1=\frac{1}{49}\)

∴ t = \(\frac{1}{49 \times 3.3 \times 10^{-4}}+4=61.84+4=65.84^{\circ} \mathrm{C} .\)

Real-Life Examples of Expansion in Solids and Liquids

Example 7. A solid at 0°C floats with 98% of its volume immersed in a liquid. The solid floats completely immersed when the temperature is raised to 25°C. If the coefficient of volume expansion of the solid is 2.6 x 10^-6 °C^-1, find the coefficient of real expansion of the liquid.
Solution:

Given

A solid at 0°C floats with 98% of its volume immersed in a liquid. The solid floats completely immersed when the temperature is raised to 25°C. If the coefficient of volume expansion of the solid is 2.6 x 10^-6 °C^-1

Let at 0°C, volume of the solid = V₀, density of the liquid = ρ₀; at 25°C, volume of the solid = V’ and density of the liquid = ρ’

Now, \(V^{\prime}=V_0\left[1+2.6 \times 10^{-6} \times 25\right]\)

and \(\rho_0=\rho^{\prime}[1+\gamma \times 25]\)

where γ = coefficient of real expansion of’tlre liquid.

From the condition of floatation, \(V_0 \times 0.98 \times \rho_0=V^{\prime} \rho^{\prime}\)

or, \(V_0 \times 0.98 \times \rho^{\prime}[1+\gamma \times 25]\)

= \(V_0\left[1+2.6 \times 10^{-6} \times 25\right] \times \rho^{\prime}\)

or, \(1+25 \gamma=\frac{1+0.000065}{0.98} \quad \text { or, } 25 \gamma=1.02047-1\)

∴ \(\gamma=8.19 \times 10^{-4 \circ \mathrm{C}^{-1} .}\)

Example 8. A mercury thermometer contains 0.4 cm³ of mercury at 0°C. The diameter of the capillary tube of the thermometer is 0.2 mm. What should be the length of the scale to measure temperatures between 0°C to 100°C? The coefficient of apparent expansion of mercury = 1.7 x 10^{-4 °}C^-1.
Solution:

Given

A mercury thermometer contains 0.4 cm³ of mercury at 0°C. The diameter of the capillary tube of the thermometer is 0.2 mm.

The apparent expansion of mercury due to increase in temperature from 0°C to 100°C = 0.4 x 1.7 x 10^-4 x 100 = 0.0068 cm³

The cross-section of the capillary tube = π(0.01)² cm²

∴ The length of the temperature measuring scale = \(\frac{0.0068}{\pi(0.01)^2}=21.6 \mathrm{~cm}\)

Example 9. A and B are two thermometers made of glass and both contain the same liquid. Both thermometers have spherical bulbs. The internal diameter of the bulb of A is 7.5 mm and radius of the capillary tube is 1.25 mm. The corresponding values for B are 6.2 mm and 0.9 mm. Find the ratio of the lengths between two consecutive graduations in thermometers A and B.
Solution:

Given

A and B are two thermometers made of glass and both contain the same liquid. Both thermometers have spherical bulbs. The internal diameter of the bulb of A is 7.5 mm and radius of the capillary tube is 1.25 mm. The corresponding values for B are 6.2 mm and 0.9 mm.

Let the separations between two consecutive graduations for 1° in thermometers A and B be x cm and y cm respectively.

Hence, for the thermometer A, volume expansion of liquid in the bulb due to an increase of 1°C in temperature = vol¬ume of x cm length in the tube,

⇒ \(\frac{4}{3} \pi\left(\frac{0.75}{2}\right)^3 \times \gamma^{\prime} \times 1=x \times \pi(0.125)^2\)…..(1)

where γ’ is the coefficient of apparent expansion of the liquid.

Similarly for the thermometer B, \(\frac{4}{3} \pi\left(\frac{0.62}{2}\right)^3 \times \gamma^{\prime} \times 1=y \times \pi \times(0.09)^2\)….(2)

Dividing equation (1) by (2) we get,

⇒ \(\frac{x \times \pi \times(0.125)^2}{y \times \pi \times(0.09)^2}=\frac{\frac{4}{3} \pi\left(\frac{0.75}{2}\right)^3 \times \gamma^{\prime} \times 1}{\frac{4}{3} \pi\left(\frac{0.62}{2}\right)^3 \times \gamma^{\prime} \times 1}\)

or, \(\frac{x}{y}=\frac{(0.75)^3 \times(0.09)^2}{(0.62)^3 \times(0.125)^2}=0.92 \text { (approx.). }\)

Example 10. A container is filled up to the brim with 500 g of water and 1000 g of mercury. When 21200 cal of heat is supplied to the system, 3.52 g of water flows out of the container. Neglecting the expansion of the container, find the coefficient of real expansion of mercury. Given, the volume expansion coefficient of water =1.5x 10^{-4 °}C^-1, density of mercury ss 13.6 g · cm^-3, density of water 1 g · cm^-3 and speciflc heat capacity of mercury = 0.03 cal · g^-1 · °C^-1. 
Solution:

Given

A container is filled up to the brim with 500 g of water and 1000 g of mercury. When 21200 cal of heat is supplied to the system, 3.52 g of water flows out of the container. Neglecting the expansion of the container,

Given, the volume expansion coefficient of water =1.5x 10^{-4 °}C^-1, density of mercury ss 13.6 g · cm^-3, density of water 1 g · cm^-3 and speciflc heat capacity of mercury = 0.03 cal · g^-1 · °C^-1.

Due to the application of heat, 3.52 g i.e., 3.52 cm³ of water flows out of the container.

∴ The total expansion of mercury and water = 3.52 cm³.

Let the rise in temperature = t°C.

Heat absorbed by mercury + heat absorbed by water = 21200

∴ 500 x 1 x t + 1000 x 0.03 x t= 21200 or, t = 40°C

Expansion of water = 500 x 1.5 x 10^-4 x 40 =3 cm³

∴ Expansion of mercury = (3.52-3) = 0.52 cm3 1000

∴ \(\frac{1000}{13.6} \times \gamma \times 40=0.52\)

[γ = volume expansion coefficient of mercury] 7

∴ γ= 1.768 x 10^-4 C^-1.

Example 11. A glass bulb is filled in at 0°C by 350 g of mercury. When a few steel balls are put in the bulb, it can then hold only 265 g of mercury. When the bulb is heated to 100°C, with the steel balls in mercury, 5 g of mercury flows out. Find the coefficient of linear expansion of steel. Given, the coefficient of real expansion of mercury = 18 x 10^{-5 °}C^-1. Neglect expansion of glass.
Solution:

Given

A glass bulb is filled in at 0°C by 350 g of mercury. When a few steel balls are put in the bulb, it can then hold only 265 g of mercury. When the bulb is heated to 100°C, with the steel balls in mercury, 5 g of mercury flows out.

The coefficient of real expansion of mercury = 18 x 10^{-5 °}C^-1. Neglect expansion of glass.

Let ρ₀ and ρ₁₀₀ be the densities of mercury at 0°C and 100°C respectively.

∴ Volume of mercury in the glass bulb at 0°C = \(\frac{350}{\rho_0}\)

Volume of mercury in the glass bulb after the steel balls are put = \(\frac{265}{\rho_0}\)

∴ Volume occupied by the steel balls at 0°C = \(\frac{350}{\rho_0}-\frac{265}{\rho_0}=\frac{85}{\rho_0}\)

Expansion of mercury at 100°C = \(\frac{265}{\rho_0} \times 18 \times 10^{-5} \times 100=\frac{4.77}{\rho_0}\)

Expansion of the steel balls at 100°C = \(\frac{85}{\rho_0} \times \gamma_s \times 100=\frac{8500 \gamma_s}{\rho_0}\)

[γ_s = coefficient of volume expansion of steel]

Now, expansion of mercury + expansion of the steel balls = volume of mercury expelled

i.e., \(\frac{4.77}{\rho_0}+\frac{8500 \gamma_s}{\rho_0}=\frac{.5}{\rho_{100}}\)

or, \(4.77+8500 \gamma_s=\frac{5 \rho_0}{\rho_{100}}=\frac{5 \rho_{100}\left(1+18 \times 10^{-5} \times 100\right)}{\rho_{100}}\)

= \(5 \times 1.018\)

⇒ \(\left[because \rho_0=\rho_{100}\left(1+18 \times 10^{-5} \times 100\right)\right]\)

or, \(8500 \gamma_s=0.32\) or, \(\gamma_s=\frac{0.32}{8500}=37.65 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\).

∴ Coefficient of linear expansion, \(\alpha_s=\frac{\gamma_s}{3}=\frac{37.65}{3} \times 10^{-6}=12.55 \times 10^{-6{ }^{\circ}} \mathrm{C}^{-1}\)

Example 12. Between two similar thermometers, one is filled with mercury and another with alcohol of the same volume at 0°C. The gap for each degree in the mercury thermometer is l and that in the alcohol thermometer is l’. Show that \(\frac{l}{l^{\prime}}=\frac{\gamma-3 \alpha}{\gamma_1-3 \alpha}\)
coefficient of real expansion of mercury, γ₁ = coefficient of real expansion of alcohol and a = coefficient of linear expansion of glass.
Solution:

Given

Between two similar thermometers, one is filled with mercury and another with alcohol of the same volume at 0°C. The gap for each degree in the mercury thermometer is l and that in the alcohol thermometer is l’.

In both thermometers, let the area of the cross-section of the tube be A and the volume of the bulb be V.

For 1° rise in temperature in mercury thermometer, apparent expansion of mercury = volume of length l of the tube

i.e., \(V(\gamma-3 \alpha) \times 1=l \times A \quad\left[because \gamma^{\prime}=\gamma-\gamma_g\right]\)….(1)

Similarly, for 1° rise in temperature in alcohol thermometer,

∴ \(V\left(\gamma_1-3 \alpha\right) \times 1=l^{\prime} \times A\)….(2)

Dividing (1) by (2) we get,

⇒ \(\frac{l \times A}{l^{\prime} \times A}=\frac{V(\gamma-3 \alpha)}{V\left(\gamma_1-3 \alpha\right)} \text { or, } \frac{l}{l^{\prime}}=\frac{\gamma-3 \alpha}{\gamma_1-3 \alpha}\)

Example 13. A cylindrical container contains some liquid. The coefficient of linear expansion of the material of the container is a. When the container is heated it is observed that the liquid level inside the container remains unchanged. What is the volume expansion coefficient of the liquid?
Solution:

Given

A cylindrical container contains some liquid. The coefficient of linear expansion of the material of the container is a. When the container is heated it is observed that the liquid level inside the container remains unchanged.

Let height of the liquid level inside the container be x and area of the cross-section be A.

Now, the level of liquid surface inside the container will remain unchanged when volume expansion of the container due to an increase in temperature t = volume expansion of the liquid due to increase in temperature t.

∴ xA x 3α x t = xA x γ x t [y = volume expansion coefficient of the liquid]

∴ γ = 3α

Short Answer Questions on Thermal Expansion

Example 14. A body weighs W₀ in air. Its apparent weights in a liquid at t₁ °C and t₂ °C are W₁ and W₂ respectively. If the coefficient of volume expansion of the material of the body is γ, find the coefficient of real expansion of the liquid.
Solution:

Given

A body weighs W₀ in air. Its apparent weights in a liquid at t₁ °C and t₂ °C are W₁ and W₂ respectively. If the coefficient of volume expansion of the material of the body is γ,

Weight of displaced liquid at t₁ °C = W₀-W₁

Let V₁ and V₂ be the volumes of the body and d₁ and d₂ be the densities of the liquid at t₁°C and t₂°C respectively.

We have, \(W_0-W_1=V_1 d_1 g\)….(1)

and \(W_0-W_2=V_2 d_2 g\)

Dividing equation (1) by equation (2) we get,

⇒ \(\frac{W_0-W_1}{W_0-W_2}=\frac{V_1 d_1}{V_2 d_2} .\)

Now \(V_2=V_1\left[1+\gamma\left(t_2-t_1\right)\right] and d_1=d_2\left[1+\gamma_l\left(t_2-t_1\right)\right]\),

where \(\gamma_l\) is the coefficient of real expansion of the liquid.

∴ \(\frac{W_0-W_1}{W_0-W_2}=\frac{1+\gamma_l \times t}{1+\gamma \times t} [where \left.t=\left(t_2-t_1\right)\right]\)

or, \(1+\gamma_l t=\frac{W_0-W_1}{W_0-W_2} \times(1+\gamma t)=\frac{W_0-W_1}{W_0-W_2}+\frac{\gamma t\left(W_0-W_1\right)}{W_0-W_2}\)

or, \(\gamma_l t=\frac{W_2-W_1}{W_0-W_2}+\frac{\gamma t\left(W_0-W_1\right)}{W_0-W_2}\)

or, \(\left(W_0-W_1\right)(1+\gamma t)=\left(W_0-W_2\right)\left(1+\gamma_l \times t\right)\)

or, \(\gamma_l=\frac{W_2-W_1}{\left(W_0-W_2\right) t}+\frac{\gamma\left(W_0-W_1\right)}{W_0-W_2}\)

Example 15. Using two different containers A and B, the coefficients of apparent expansion of a liquid are found to be γ₁  and γ₂ respectively. If the coefficient of linear expansion of the material A is α, find that of the material B.
Solution:

Given

Using two different containers A and B, the coefficients of apparent expansion of a liquid are found to be γ₁  and γ₂ respectively. If the coefficient of linear expansion of the material A is α,

We have, γ = γ’ + γ_g or, γ = γ’ + 3α

[γ’ = Coefficient of apparent expansion of the liquid, γ = coefficient of real expansion of the liquid, γ_g = coefficient of volume expansion of the material of the container.]

For container A, \(\gamma=\gamma+\gamma_g=\gamma_1+3 \alpha\)

and for container B, \(\gamma=\gamma_2+3 \alpha^{\prime}\)

[α’ = coefficient of linear expansion of the material of the container B ]

∴ \(\gamma_2+3 \alpha^{\prime}=\gamma_1+3 \alpha\)

or, \(3 \alpha^{\prime}=\gamma_1+3 \alpha-\gamma_2 \quad or, \alpha^{\prime}=\frac{1}{3}\left(\gamma_1-\gamma_2\right)+\alpha\).

Example 16. A hollow iron ball floats completely immersed in water at 10°C temperature. What will happen if the temperature of both of them is raised to 50°C?
Solution:

Given

A hollow iron ball floats completely immersed in water at 10°C temperature.

The iron ball will sink in water if the temperature of the ball and of water is raised to 50°C from 10°C.

Let dy be the density of water and be the volume of the iron ball at 10°C.

So, the apparent loss in weight of the ball at 10°C = V₁d₁g.

Again, if d₂ is the density of water and V₂, is the volume of the iron ball at 50°C, the apparent loss in weight of the bull at 50°C = V₂d₂g

Now, \(V_2=V_1\left[1+\gamma_i(50-10)\right]\)

[\(\gamma_i=\) coefficient of volume expansion of iron]

and \(d_2=d_1\left[1-\gamma_w(50-10)\right]\)

⇒ \(\gamma_w\). coefficient of volume expansion of water]

∴ \(V_2 d_2=V_1 d_1\left[1+40 \gamma_i\right]\left[1-40 \gamma_w\right]\)

= \(V_1 d_1\left[1-40\left(\gamma_w-\gamma_i\right)\right] \text { (approx.) }\)

Since \(\gamma_w>\gamma_i, V_2 d_2 g<V_1 d_1 g\).

Therefore, the loss in weight of the ball at 50°C temperature is less. Hence its apparent weight increases. So the ball will sink in water.

 

Unit 7 Properties Of Bulk Matter Chapter 5 Expansion Of Solid And Liquids Useful Relations For Solving Numerical Problems

1. If the length of a rod at temperature is t₁ is l and that at temperature t₂ is l₂ then,

coefficient of linear expansion, \(\alpha=\frac{l_2-l_1}{l_1\left(t_2-t_1\right)}\)

2. If the initial length of the rod at temperature 0 is l₀ and the, final length of the rod at temperature t is l_t, then, \(l_t=l_0(1+\alpha t)\)

3. If the values of α in Celsius and Fahrenheit scales are α_C and α_F then, \(\alpha_F=\frac{5}{9} a_C\)

If the surface area of a solid at temperature t₁, is S₁ and that at temperature t₂, is S₂, then,

coefficient of surface expansion, \(\beta=\frac{S_2-S_1}{S_1\left(t_2-t_1\right)}\)

⇒ \(\beta_F=\frac{5}{9} \beta_C\)

If the volume of a solid body at temperature t₁ is V₁ and that at temperature t₂ is V₂ then,

coefficient of volume expansion, \(\gamma=\frac{V_2-V_1}{V_1\left(t_2-t_1\right)}\)

⇒ \(\gamma_F=\frac{5}{9} \gamma_C\)

⇒ \(\alpha=\frac{\beta}{2}=\frac{\gamma}{3}\)

If the density of a solid at temperature t₁ is D₁ and that at temperature t₂ is D₂ then, \(D_2=D_1\left\{1+\gamma\left(t_2-t_1\right)\right\}^{-1} \approx D_1\left\{1-\gamma\left(t_2-t_1\right)\right\}\)

If area of the cross-section of a rod is A, rise in temperature is t, Young’s modulus for the material is Y, coefficient of linear expansion is a and the force developed inside the rod, when its length is kept unchanged, is F then,
thermal stress = \(\frac{F}{A}=Y \alpha t\)

Real expansion of a liquid = apparent expansion of the liquid + expansion of the container.

Coefficient of real expansion of a liquid = coefficient of apparent expansion of the liquid + coefficient of volume expansion of the material of the container, i.e., \(\gamma=\gamma^{\prime}+\gamma_g \)

⇒ \(\gamma=\frac{\rho_1-\rho_2}{\rho_1\left(t_2-t_1\right)}\) [where ρ₁ = density of the liquid at the temperature t₁, ρ₂ = density of the liquid at the temperature t₂]

Apparent loss in weight of a solid body immersed completely in a liquid is M₂ = M₁{1-(γ-γ_s)t}

where M₁g= apparent loss in weight of the body at the temperature t₁, M₂g = apparent loss in weight of the body at the temperature t₂,t = t₂-t₁, γ = coefficient of real expansion of the liquid and γ_s = coefficient of volume expansion of the immersed solid.

 

Unit 7 Properties Of Bulk Matter Chapter 5 Expansion Of Solid And Liquids Very Short Answer Type Questions

Question What is the unit of the coefficient of linear expansion of a substance?
Answer: °C^-1

Question 2. What is the relation between the coefficient of linear expansion (α) and the coefficient of volume expansion (γ) of a solid?
Answer: γ = 3α

Question 3. A bimetallic strip made up of brass and iron remains linear at 20 °C. When the temperature is decreased to 0°C, the strip bends. Which material remains on the convex side of the bent strip?
Answer: Iron

Question 4. Coefficient of linear expansion of platinum is 9 x 10^-6 °C^-1. What will be the value of this coefficient when the temperature is expressed in Fahrenheit unit?
Answer: 5 x10^-6 °F^-1

Question 5. If the coefficient of linear expansion of iron is 0. 0000067 °F^-1, then what is its value in the Celsius scale?
Answer: 0.000012°C^-1

Question 6.Due to rise in temperature, if each side of a copper cube increases by 0.1%, then find out the increase in volume of that cube.
Answer: 0.3%

Question 7.Write down the variation of density of a solid with increase in temperature.
Answer: Decreases

Question 8. Thermal expansion of invar is _______ than that of all other metals or alloys.
Answer: Less

Question 9.Between the coefficient of apparent expansion (γ’) and the coefficient of real expansion (γ) of a liquid which one is a characteristic property of the liquid?
Answer: y

Question 10. A glass vessel is filled with water up to its brim. Now what will happen if temperature is increased?
Answer: Water will overflow

Question 11. At what temperature density of water will be the maximum?
Answer: 4°C

Question 12. How does the density of a liquid change with temperature?
Answer: Density decreases

Question 13. At what temperature under standard atmospheric pressure, the density of pure water will be the maximum?
Answer: 4°C

Question 14. Real expansion of a liquid = apparent expansion of the liquid + volume expansion of the _______.
Answer: vessel

Question 15. Coefficient of real expansion of water from 0°C to 4°C is ___________.
Answer: Negative

Question 16. ‘Usually, the thermal expansion of a liquid is greater than the thermal expansion of an equal volume of a solid.’ State whether the statement is true or false.
Answer: True

Question 17. ‘Usually, the thermal expansion of a liquid is greater than
the thermal expansion of an equal volume of a gas.’ State whether the statement is true or false.
Answer: False

Question 18. What is the density of pure water at 4°C in SI?
Answer: 1000 kg • m^-3

Unit 7 Properties Of Bulk Matter Chapter 5 Expansion Of Solid And Liquids Assertion Reason Type Question And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, statement 2 is true.

Question 1.

Statement 1: A brass disc just fits in a hole in a steel plate. The system must be cooled to loosen the disc from the hole.

Statement 2: The coefficient of linear expansion for brass is greater than the coefficient of linear expansion for steel.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 2.

Statement 1: A beaker is completely filled with water at 4 °C. It will overflow, both when heated or cooled.

Statement 2: There is expansion of water below and above 4°C.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: The coefficient of real expansion of the liquid is independent of the nature of container.

Statement 2: ϒ_r = γ_a + γ_v where γ_v = coefficient of real expansion, γ_a = coefficient of expansion and γ_v = coefficient of apparent expansion of vessel.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 4.

Statement 1: The coefficient of apparent expansion can be negative.

Statement 2: Coefficient of real expansion of a liquid can be less than the coefficient of expansion of vessel.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 5.

Statement 1: A solid and a hollow sphere of same material when heated through the same temperature, wall expand by the same amount.

Statement 2: The change in volume is independent of the original mass but depends on original volume.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Unit 7 Properties Of Bulk Matter Chapter 5 Expansion Of Solid And Liquids Match Column 1 With Column 2.

Question 1. A piece of metal of density ρ₁ floats on mercury of density ρ₂. The coefficients of expansion of the metal and mercury are. γ₁ and γ₂ respectively. The temperatures of both mercury and metal are increased by ΔT.

Answer: 1. C, 2. A, 3. D, 4. B

Question 2.

Answer: 1. B, 2. C, 3. A

Unit 7 Properties Of Bulk Matter Chapter 5 Expansion Of Solid And Liquids Comprehension Type Question And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A copper collar is to fit tightly about a steel shaft that has a diameter of 6 cm at 20 °C. The inside diameter of the copper collar at that temperature is 5.98 cm.

1. To what temperature must the copper collar be raised so that it will just slip on the steel shaft, assuming the steel shaft remains at 20 °C? (α_copper= 17 x 10^-6 K^-1)

1. 324 °C
2. 21.7 °C
3. 217 °C
4. 32.4 °C

Answer: 3. 217 °C

2. The tensile stress in the copper collar when its temperature returns to 20 °C is (Y = 11 x 10¹⁰ N • m^-2)

1. 1.34 x 10⁵ N • m^-2
2. 3.68 x 10^-12 N • m^-2
3. 3.68 x 10⁸ N • m^-2
4. 1.34 x 10^-12 N • m^-2

Answer: 3. 3.68 x 108^-12 N • m^-2

3. If the breaking stress of copper is 230 N • m^-2, at what temperature will the copper collar break as it cools?

1. 20 °C
2. 47 °C
3. 94 °C
4. 217 °C

Answer: 3. 94 °C

Unit 7 Properties Of Bulk Matter Chapter 5 Expansion Of Solid And Liquids Integer Type Question And Answers

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9.

Question 1. A composite rod is made by joining a copper rod, end to end, with a second rod of a different material but of same cross-section. At 25 °C, the composite rod is lm in length of which the length of the copper rod is 30 cm. At 125 °C the length of the composite rod increases by 1.91 mm. The coefficient of linear expansion of copper is α = 1.7 x 10^-5 °C^-1 and that of the second rod is β = n x 10^-5 °C^-1. Find the value of n
Answer: 2

Question 2. The volume of a metal sphere increases by 0.24% when its temperature is raised by 40 °C. The coefficient of linear expansion of the metal is n x 10^-5 °C^-1.Find the value of n.
Answer: 2

Question 3. The coefficient of real expansion of mercury is 18 x 10^-6 °C^-1. A thermometer has a bulb of volume 10^-6 m³ and the cross-section of the stem is 0.002 cm². Assuming the bulb to be filled with mercury at 0 °C, find the length (in cm) of the mercury column at 100 °C.
Answer: 9

Question 4. A thin copper wire of length l increases in length by 1 % when heated from 0 °C to 100 °C. If a thin copper plate of area 2lx l is heated from 0 °C to 100 °C, find the percentage increase in its area.
Answer: 2

 

 

One-Dimensional Motion

One-Dimensional Motion Introduction: Mechanics is the branch of science which deals with the motion of bodies and the effect of force on stationary or moving bodies.

The Branches Of MechanicsAre As Follows:

1. Dynamics: A branch of mechanics in which the motion of a body and the effect of force on the motion of the body are discussed.
2. Kinematics: It is restricted to the study of motion and not the causes of motion.
3. Kinetics: It incorporates the study and analysis of the motion of a body together with the causes of motion. Also, the mass of a body and the effect of force on the mass are studied here.
4. Statics: A branch of mechanics which deals with the equilibrium of a body under the action of a number of forces and studies the conditions of equilibrium.

One-Dimensional, Two-Dimensional And Three-Dimensional Motions: We will be discussing only one-dimensional motion in this chapter. The motion which is confined to a straight line is called one-dimensional motion or rectilinear motion.

• This type of motion can be explained by one-dimensional, two-dimensional or three-dimensional coordinates. The motion of a car moving in a straight line or the motion of a body under gravity are examples of one-dimensional motion.
• The motion which is confined to a plane is called two-dimensional motion or planar motion is called two- dimensional motion can only be described by two-dimensional or three-dimensional coordinates. The motion of a planet around the sun, a body revolving in a circle, a billiard ball moving over the billiard table etc. are examples of two-dimensional motion.
• The motion which is not confined to a plane is called three-dimensional motion. This type of motion can only be explained by three-dimensional coordinates, flic examples of three-dimensional motion are the spiral motion of a particle or the motion of an aeroplane

Read and Learn More: Class 11 Physics Notes

Speed And Velocity Numerical Examples

Example 1. A particle moves in a circular path of radius 7 cm. It covers

1. Half of the circle in 4 s and
2. One complete round in 10 s. In each case find the average speed and average velocity.

Solution:

The circumference of the circle = 2πr = 2 x \(\frac{22}{7}\) x 7 = 44cm

1. The distance travelled by the particle = half of the circumference = \(\frac{44}{2}\) = 22 cm and time taken = 4 s.

∴ Average speed, v = \(\frac{22}{4}\) = 5.5 cm · s^-1.

Displacement = diameter of the circle = 2 x 7 = 14 cm and time taken = 4s.

Hence average velocity, \(\vec{v}\) = \(\frac{14}{4}\) = 3.5 cm · s^-1 along AB.

2. The distance travelled by the particle = circumference of the circle = 44 cm and time taken = 10 s.

∴ Average speed = = 4.4 cm · s^-1.

Displacement in this case is 0 (as initial and final positions are the same.)

Hence, average velocity = \(\frac{\text { displacement }}{\text { time }}\)

Example 2. An aeroplane travels 2000 km to the west It then turns north and moves 2000 km more. Finally, it follows the shortest path to return to its starting point If the speed of the plane is 200 km · h^-1, find its average velocity for the total journey.
Solution:

Since initial and final positions are the same, displacement is zero.

∴ Average velocity = \(\frac{\text { displacement }}{\text { total time }}=\frac{0}{\text { total time }}=0 \text {. }\)

Example 3. Find the speed of the tip of a 3 cm long second’s hand in a clock.
Solution:

The tip of the second’s hand describes an angle of 360° in 60 seconds when it completes the total circular path once.

Hence, distance travelled = circumference of the circle = 2π x 3 = 6π cm; time = 60 s.

∴ Speed = \(=\frac{6 \pi}{60}=\frac{\pi}{10}=0.314 \mathrm{~cm} \cdot \mathrm{s}^{-1} .\)

Example 4. A train travels from station A to station B at a constant speed of 40 km · h^-1 and returns from B to A at 60 km · h^-1. Find the average speed and average velocity of the train.
Solution:

Let the distance between stations A and B be x km.

Time taken by the train to move from A to B = \(\frac{x}{40}\)h and time taken to move from B to A = \(\frac{x}{60}\)h.

Total distance travelled = 2x km.

∴ Average speed = \(\frac{2 x}{\left(\frac{x}{40}+\frac{x}{60}\right)}=\frac{2 \times 40 \times 60}{100}=48 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Initial and final positions are the same so the total displacement becomes zero and hence average velocity is zero.

Example 5. The motion of a particle, along the x-axis, follows the relation x = 8t – 3t². Here x and t are expressed in metre and second respectively. Find

1. The average velocity of the particle in time interval 0 to 1 s and
2. Its instantaneous velocity at t = 1s.

Solution:

1. Let at t = 0, x = x₁ and at t = 1, x = x₂.

∴ x₁ = 8 x 0 -3 x 0² = 0 and x₂ = 8 x 1 -3 x 1² = 5m

∴ Displacement = x₂ – x₁ =5 m and time taken is 1 s

∴ Average velocity = \(\frac{5 m}{1 s}\) = 5 m · s^-1.

2. Here instantaneous velocity,

v = \(\frac{d x}{d t}=\frac{d}{d t}\left(8 t-3 t^2\right)=8-2 \times 3 t=8-6 t\)

At t = \(1 \mathrm{~s}, \frac{d x}{d t}=8-6 \times 1=2 \mathrm{~m} \cdot \mathrm{s}^{-1}\).

WBBSE Class 11 One Dimensional Motion Notes

Acceleration: When the velocity of a particle increases with time, the particle is said to be accelerating. So, in case of acceleration, the final velocity is greater than the initial velocity.

Acceleration Definition: The rate of change of velocity with time is called acceleration.

Thus, acceleration (a) = \(\frac{\text { change in velocity }}{\text { time }}\)

= \(\frac{\text { final velocity }- \text { initial velocity }}{\text { time }}\)

Acceleration Example: A train at rest starts from a station and speeds up. In this case, we can say that the train is moving with acceleration.

Sometimes, acceleration is represented by the symbol ‘f’.

Nature Of Acceleration: Acceleration is related to the change in velocity of a body. So, like velocity, acceleration is also a vector quantity. It has to be specified by its magnitude and direction. However, to specify the acceleration vector, we have to use vector algebra to determine the change in the velocity.

This shows that velocity and change in velocity may have different directions, in general. Thus, the direction of the acceleration may or may not be the same as that of velocity.

Units And Dimension: Unit of acceleration = \(\frac{\text { unit of velocity }}{\text { unit of time }} \text {. }\)

CGS System: cm · s^-2

SI: m · s^-2

Dimension of acceleration = \(\frac{\text { dimension of velocity }}{\text { dimension of time }}=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}}=\mathrm{LT}^{-2}\)

Motion With Uniform And Non-Uniform Acceleration: Uniform acceleration corresponds to a motion in which the velocity of a body changes equally in equal intervals of time.

If a particle moves with uniform acceleration, then its acceleration remains the same, both in magnitude and direction, at each point on its path. When a body falls freely from a height under gravity, its velocity increases. But its acceleration is uniform on or near the surface of the earth and is known as the acceleration due to gravity.

Non-uniform Acceleration corresponds to a motion in which the velocity of a body does not change equally in equal intervals of time.

The motion of an oscillating pendulum is an example of motion with non-uniform acceleration. The acceleration of the bob becomes maximum at its maximum displaced position and becomes zero at its mean position.

Average Acceleration: The acceleration of a particle may not always be uniform. Generally, we can find out the average acceleration using the following relation:

(a) = \(\frac{\text { final velocity }- \text { initial velocity }}{\text { time taken }}\)

= \(\frac{\text { change in velocity }}{\text { time }}\)

Instantaneous Acceleration: The acceleration of a particle at any moment is called its instantaneous acceleration.

Instantaneous Acceleration Definition: The instantaneous acceleration of a particle at a given point is the limiting value of the rate of change in velocity with respect to time when the time interval tends to zero.

According to differential calculus, instantaneous acceleration, a = \(\lim _{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}=\frac{d v}{d t}=\frac{d}{d t}\left(\frac{d s}{d t}\right)=\frac{d^2 s}{d t^2}\)

Deceleration Or Retardation: when the velocity of a particle decreases with time, the particle is in a state of deceleration or retardation. Here, the final velocity is less than the initial velocity.

Deceleration Or Retardation Definition: The rate of decrease of velocity with time is called deceleration or retardation.

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Deceleration Or Retardation Example: When a train approaches a station, it slows down and finally stops. During this period the train decelerates to come to a halt.

Deceleration is a special case of acceleration where the final velocity is less than the initial velocity. Thus, deceleration is essentially a negative acceleration. It is not a different physical quantity.

Let us consider a particle moving with a velocity of 10 cm · s^-1. It then slows down to 4 cm · s^-1 in 3 s. Its acceleration is given by,

a = \(\frac{\text { final velocity }- \text { initial velocity }}{\text { time interval }}\)

= \(\frac{4-10}{3}=-2 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

This example clearly shows that deceleration is described mathematically as negative acceleration.

Distinction Between Acceleration And Retardation

Acceleration Due To Gravity: The earth attracts other bodies towards itself because of its gravity. A body released near the earth’s surface falls freely under the action of the force due to gravity. Whenever a body is allowed to fall freely, it undergoes an acceleration directed towards the earth. This is referred to as the acceleration due to gravity and is denoted by g. Its characteristics are:

1. It is always directed towards the centre of the earth, i.e., vertically downwards.
2. Its value is the greatest on the earth’s surface. It decreases slightly as we go away from the surface in such a way that at an altitude of 3.2 km, its value decreases only by 0.1%. Thus, for practical purposes, we may consider its value to be constant and equal to its value on the earth’s surface.
3. It is independent of the characteristics of an object, such as mass, density or shape.

Bodies that move vertically downwards undergo constant acceleration, while bodies moving vertically upwards undergo constant retardation. In this discussion, we have neglected the small variations in g at different locations and the air resistance faced by freely falling bodies.

The value of g is usually taken as, g = 980 cm · s^-2 = 9.8 m · s^-2

Understanding Displacement in One Dimensional Motion

Motion In A Straight Line

When a body moves without changing its direction, the motion is naturally along a straight line. It is also known as rectilinear motion.

Vertical fall of a body under gravity, the motion of a car on a straight road, etc. are examples of such motion. The discussions henceforth in this chapter will be restricted to rectilinear motions only.

Representation Of The Physical Quantities Of Motion: For the rectilinear motion of a particle, the straight line of motion itself may be chosen as an axis (say, the x-axis) and a point O on it as the origin. Obviously, the motion is one-dimensional.

• There is a basic difference in the representations of the scalar and the vector quantities related to rectilinear motion. The scalar quantities like distance travelled and speed have magnitudes only, and are always expressed by positive numbers. However, two directions, exactly opposite to each other, exist for the motion in a straight line.
• So a vector quantity is expressed by a positive number for one direction and by a negative number for the exactly opposite direction. The direction towards the right may be taken as positive; then the direction towards the left naturally becomes negative. As an example, we may consider the vertical motion of a particle under gravity.
• For a downward motion, we may take each of displacement, velocity and acceleration to be positive. Then, for an upward motion, displacement and velocity would be negative; but acceleration would still remain positive since the acceleration due to gravity (g) is always directed downwards.
• It is important to note that simple algebraic operations are sufficient for calculations involving any quantity, a scalar or a vector, in a rectilinear motion. It means that vector algebra is not at all necessary even for the calculations involving vector quantities of a rectilinear motion.

For motion along a straight line,

1. Distance travelled by a particle = magnitude of its displacement,
2. Speed of the particle at any point = magnitude of its velocity at that point
3. Displacement, velocity and acceleration vectors are along the same straight line.

Rectilinear Motion with Uniform Velocity: If a particle moves with a uniform velocity, its acceleration is zero. Let v be the uniform velocity of a particle and s be its displacement in time t. Therefore, according to the definition of uniform velocity, the particle moves a distance v x 1 in Is, v x 2 in2s, etc.

∴ In t s its displacement is v x t.

∴ s = vt…..(1)

i.e., displacement = uniform velocity x time .

Motion In A Straight Line Numerical Example

Example 1. A person travels half of a distance at an average velocity of 24 km · h^-1. At what average velocity should he move to cover the second half of the path so that his average velocity for the total path becomes 32 km · h^-1?
Solution:

Let the total length of the path = 2s km.

∴ Time required to cover the first half of the path = \(\frac{s}{24}\) h.

If the man travels the total path with an average velocity of 32 km · h^-1, then the total time taken by him = \(\frac{2 s}{32}\) = \(\frac{s}{16}\) h.

∴ Time required to cover the second half of the path = \(\frac{s}{16}\) – \(\frac{s}{24}\) = \(\frac{s}{48}\) h

∴ Average velocity of second half = \(\frac{\text { distance }}{\text { time }}=\frac{s}{\frac{s}{48}}=48 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Rectilinear Motion With Uniform Acceleration

Rectilinear Motion With Uniform Acceleration: For a particle in motion, let

u = initial velocity

v = final velocity after a time t and

corresponding uniform acceleration = a.

Let the displacement = s in that time.

Then the above variables obey the following equations:

1. v = u+at,
2. s= ut+1/2 at²
3. v² = u² + 2as
4. s_t = M + \(\frac{1}{2}\) a(2t – 1) [where s_t = the displacement in the t th second]

Rectilinear Motion With Uniform Acceleration Derivations:

1. v= u+at

Let the initial velocity of a particle be u and its final velocity after time t be v.

∴ In time t, change in velocity = v – u

∴ Rate of change of velocity with time = \(\frac{v – u}{t}\) = a, by definition.

Hence, at = v – u or, v = u + at….(1)

For a particle starting from rest, u = 0 and thus, v = at ….(2)

In case of retardation, the relationship becomes v = u – at …(3)

2. s = ut + \(\frac{1}{2}\) at²

If a particle with initial velocity u and uniform acceleration a, attains final velocity v after time t, the average velocity, considering the two endpoints, is \(\frac{u + v}{2}\).

The acceleration a implies that the velocity increases by an after every second. Hence, the velocity 1 s after the start of motion = u + a and 1 s before the end of motion = v – a

Velocity and Speed in One Dimensional Motion

∴ Average velocity = \(\frac{u+a+v-a}{2}=\frac{u+v}{2}\)

Hence, the particle can be considered to have travelled a distance s with the average velocity \(\frac{(u+v)}{2}\) in time t.

Hence, displacement, s = \(\frac{u+v}{2} \times t=\frac{u+(u+a t)}{2} \times t\) (because v=u+a t)

or, \(s=\frac{2 u t}{2}+\frac{1}{2} a t^2\) or, \(s=u t+\frac{1}{2} a t^2\)…..(4)

For a particle starting from rest, u=0

So, \(s=\frac{1}{2} a t^2\)…..(5)

For a retarding particle, s = \(u t-\frac{1}{2} a t^2\)…..(6)

3. \(v^2=u^2+2 a s\)

From equation (1) we have, v=u+a t

or, \(v^2=(u+a t)^2=u^2+2 u a t+a^2 t^2\)

= \(u^2+2 a\left(u t+\frac{1}{2} a t^2\right)\)

or, \(v^2=u^2+2\) as [using equation (4)]…….(7)

Hence, for a particle starting from rest \(\nu^2=2 a s\)……(8)

and in case of retardation \(v^2=u^2-2 a s\)……(9)

4. \(s_t=u+\frac{1}{2} a(2 t-1)\)

Displacement in t seconds, s = \(u t+\frac{1}{2} a t^2\), from equation (4).

Displacement in (t-1) seconds, \(s^{\prime}=u(t-1)+\frac{1}{2} a(t-1)^2\)

Hence, the displacement in the t th second, \(s_t=s-s^{\prime}\)

or, \(s_t=u t+\frac{1}{2} a t^2-\left\{u(t-1)+\frac{1}{2}(t-1)^2 a\right\}\)

= \(u t+\frac{1}{2} a t^2-u t+u-\frac{1}{2} a t^2+\frac{1}{2} \cdot 2 t a-\frac{a}{2}\)

= \(u+a t-\frac{a}{2}=u+\frac{1}{2} a(2 t-1)\)…….(10)

For a particle starting from rest \(s_t=\frac{1}{2} a(2 t-1)\)……(11)

and in case of retardation \(s_t=u-\frac{1}{2} a(2 t-1)\)…..(12)

Rectilinear Motion With Uniform Acceleration Numerical Examples

Example 1. A velocity of 60 km · h^-1 of a train is reduced by the application of brakes. A retardation of 40 cm · s^-2 is produced. After how much time will the train stop? What will be the velocity of the train after 20 seconds?
Solution:

Given, \(u=60 \mathrm{~km} \cdot \mathrm{h}^{-1}=\frac{60 \times 1000}{60 \times 60}=\frac{50}{3} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

a = 40 cm · s^-2 = 0.4 m · s^-2 and v = 0

Hence, from the relation v = u- at, we get \(0=\frac{50}{3}-0.4 \times t \quad \text { or, } t=\frac{50}{3 \times 0.4}=41.7 \mathrm{~s}\)

∴ The train will stop after 41.7.

The velocity after 20 s, v = \(\frac{50}{3}-0.4 \times 20\)

= \(\frac{26}{3} \mathrm{~m} \cdot \mathrm{s}^{-1}=8.7 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

= \(\frac{8.7 \times 60 \times 60}{1000} \mathrm{~km} \cdot \mathrm{h}^{-1}\)

= \(31.3 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Example 2. A body covers 200 cm in the first 2 s of motion and 220 cm in the next 4 s. Calculate the velocity 7 s after the start.
Solution:

We know, s = ut + \(\frac{1}{2}\) at²

∴ Putting t = 2 s

200 = u x 2 + \(\frac{1}{2}\) a x 4 = 2u + 2a

or, u + a = 100….(1)

The displacement in (4 + 2) or 6 s = 200 + 220 = 420 cm

∴ 420 = u x 6 + \(\frac{1}{2}\) a x 36 = 6u + 18a

or, u + 3a = 70…..(2)

From equations (1) and (2) we get, a = -15 cm · s^-2 and u = 115 cm · s^-1

Velocity after 7 s of motion, v = u + at = 115 – 15 x 7 = 115 – 105 = 10cm · s^-1

Example 3. A man is 9 m behind a train at rest. The train starts with an acceleration of 2 m · s^-2 and simultaneously the man starts running. He is able to board the train somehow after 3 seconds. Find the acceleration of the man.
Solution:

While boarding the train, the positions of the man and of the train must be the same.

Let the acceleration of the man be a and the distance traversed by the train in 3 s be x.

We know, s = ut + \(\frac{1}{2}\)at².

Here u = 0 as the train as well as the man starts from rest.

So, for the train, x = \(\frac{1}{2}\) x 2 x (3)² or, x = 9 m

Thus, the distance traversed by the man in this time = 9 + 9 = 18 m

Then, s = 18 m , u = 0 and t = 3 s

∴ a = \(\frac{2 s}{t^2}\) = \(\frac{2}{3 ^2}\) x 18 = 4 m s

Example 4. A particle moves with a uniform acceleration along a straight line. It covers 41 cm and 49 cm in the 6th and the 10th seconds respectively. What will be the distance covered by the particle in 15 seconds?
Solution:

We know, distance covered in the nth second, s = u + \(\frac{1}{2}\)a(2n- 1)

According to the question, s₀ = u + \(\frac{1}{2}\)a(2 x 6-1) or, 41 = u + \(\frac{11}{2}\) a….(1)

ans s₁₀ = u + \(\frac{1}{2}\)a(2 x 10- 1) or, 49 = u + \(\frac{19}{2}\)a…..(2)

By solving equations (1) and (2) we get, u = 30 and a = 2

Now, putting t = 15 in s = ut + \(\frac{1}{2}\) at² we get,

s = 30 x 15 = \(\frac{1}{2}\) x 2 x (15)² = 450 + 225 = 675 cm

∴ The particle traverses 675 cm in 15 s.

Example 5. A train begins its journey from station A and stops at station B after 45 min. C is a certain point between A and B where the train attains its maximum velocity of 50 km · h^-1. If the train travels from A to C with a uniform acceleration and from C to B with a uniform retardation, calculate the distance between A and B.
Solution:

Let the train start from A with a uniform acceleration a₁ and reach C in time t₁. From here it travels to B in time t₂ with a uniform retardation a₂.

The maximum velocity of A the train is v at point C.

Let AC = s₁ and CB = s₂

So for the motion from A to C, v = \(a_1 t_1 \text { and } s_1=\frac{1}{2} a_1 t_1^2\)

and for the motion from C to B, \(0=v-a_2 t_2 \quad \text { or, } v=a_2 t_2\)

and \(s_2=v t_2-\frac{1}{2} a_2 t_2^2=a_2 t_2^2-\frac{1}{2} a_2 t_2^2=\frac{1}{2} a_2 t_2^2\)

So the distance between A and B,

s = \(A B=A C+C B=s_1+s_2\)

= \(\frac{1}{2} a_1 t_1^2+\frac{1}{2} a_2 t_2^2\)

= \(\frac{1}{2} v t_1+\frac{1}{2} v t_2\) (because \(v=a_1 t_1=a_2 t_2\))

= \(\frac{1}{2} v\left(t_1+t_2\right)=\frac{1}{2} \times 50 \times \frac{3}{4}\)

(\(v=50 \mathrm{~km} \cdot \mathrm{h}^{-1}, t_1+t_2=45 \mathrm{~min}=\frac{3}{4} \mathrm{~h}\))

=18.75 km

Example 6. A train moving with constant acceleration crosses an observer standing on the platform. The first and the second compartments, each 15 m long, cross the observer in 2 s and 2.5 s, respectively. Find the velocity of the train when its first compartment just crosses the observer and also find its acceleration.
Solution:

Let the velocity and the acceleration of the train as its 1st compartment just reaches the observer be u and a, respectively.

Hence, displacement in 2 s = length of the 1st compartment = 15 m

and displacement in (2 + 2.5) or, 4.5 s = total length of the two compartments = 2 x 15 = 30 m

Now from the equation s = ut + \(\frac{1}{2}\) at² we get,

∴ 15 = 2u + \(\frac{1}{2}\)a(2)²

or, u + a = \(\frac{15}{2}\)….(1)

and 30 = 4.5 u + \(\frac{1}{2}\) a(4.5)²

or, 36u + 81a = 240….(2)

Solving equations (1) and (2), we get, u = \(\frac{49}{6}\) and a = –\(\frac{2}{3}\)

The 1st compartment crosses the observer in 2 s; the velocity at that moment,

v = u + at = \(\frac{49}{6}\) x 2 = \(\frac{41}{6}\)

Therefore, the velocity and acceleration of the train as its 1st compartment just crosses the observer are \(\frac{41}{6}\) m · s^-1 and –\(\frac{2}{3}\)m · s^-2 respectively.

Example 7. A bullet with an initial velocity u penetrates a target. After penetrating a distance x, its velocity decreases by \(\frac{u}{n}\). How much farther will the bullet move through the target before it comes to rest?
Solution:

Let us assume that the retardation of the bullet inside the target is a and it is uniform.

After penetrating a distance x, the velocity, v = u – \(\frac{u}{n}\) = \(\frac{u(n-1)}{n}\)

∴ From \(v^2=u^2-2 a s\), we get, \(\frac{u^2(n-1)^2}{n^2}=u^2-2 a x\)

or, \(2 a x=u^2-\frac{u^2(n-1)^2}{n^2}\)

= \(u^2\left(1-\frac{n^2-2 n+1}{n^2}\right)=u^2 \cdot \frac{2 n-1}{n^2}\)

or,  a = \(\frac{u^2(2 n-1)}{2 x n^2}\)

Let the bullet travel an additional distance y before it comes to rest.

∴ \((0)^2=u^2-2 a(x+y) \quad \text { or, } x+y=\frac{u^2}{2 a}\)

or, \(y=\frac{u^2}{2 a}-x=\frac{u^2 \cdot 2 x n^2}{2 u^2(2 n-1)}-x=\frac{x n^2}{2 n-1}-x\)

= \(x\left(\frac{n^2-2 n+1}{2 n-1}\right)=\frac{x(n-1)^2}{2 n-1}\)

Example 8. Starting from rest, a train travels a certain distance with a uniform acceleration α. Then it travels with a uniform retardation β and finally comes to rest again. If the total time of motion is t, find

1. The maximum velocity attained and
2. The total distance travelled by the train.

Solution:

1. Let t₁ be the time taken to travel a distance s₁ with acceleration α, and t₂ be the time taken to travel a farther distance s₂ with retardation β. Let the maximum velocity attained by the train be v.

Here t = t₁ + t₂

∴ For the motion of the train with acceleration α,

v = \(\alpha t_1 \quad \text { or, } t_1=\frac{\nu}{\alpha}\)…..(1)

and \(\nu^2=2 \alpha s_1 or, s_1=\frac{\nu^2}{2 \alpha}\)…..(2)

Similarly, for the motion of the train with retardation β,

0 = \(\nu-\beta t_2\)

or, \(t_2=\frac{v}{\beta}\) and \(0^2=v^2-2 \beta s_2\) or, \(s_2=\frac{\nu^2}{2 \beta}\)

From (1) and (3) we get, \(t=t_1+t_2=v\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \quad \text { or, } \quad v=\frac{\alpha \beta t}{\alpha+\beta}\)

2. From (2) and (4), the total distance travelled,

s = \(s_1+s_2=\frac{\nu^2}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \)

= \(\left(\frac{\alpha \beta t}{\alpha+\beta}\right)^2 \times \frac{1}{2}\left(\frac{\alpha+\beta}{\alpha \beta}\right)=\frac{\alpha \beta t^2}{2(\alpha+\beta)} .\)

Example 9. A particle travelling with uniform acceleration along a straight line has average velocities v₁, v₂ and v₃ in successive time intervals t₁, t₂ and t₃, respectively. Prove that, \(\frac{v_2-v_1}{v_3-v_2}=\frac{t_1+t_2}{t_2+t_3}\)
Solution:

Let the initial velocity of the particle be w and its acceleration be a. Also, x, y and z are the velocities after the respective time intervals t₁, t₂ and t₃.

∴ \(x=u+a t_1, y=u+a\left(t_1+t_2\right), z=u+a\left(t_1+t_2+t_3\right) \)

Again, \(v_1=\frac{u+x}{2}, v_2=\frac{x+y}{2}, v_3=\frac{y+z}{2}\)

∴ \(\frac{v_2-v_1}{v_3-v_2}=\frac{\frac{1}{2}(x+y)-\frac{1}{2}(u+x)}{\frac{1}{2}(y+z)-\frac{1}{2}(x+y)}\)

= \(\frac{y-u}{z-x}=\frac{u+a\left(t_1+t_2\right)-u}{u+a\left(t_1+t_2+t_3\right)-\left(u+a t_1\right)}\)

= \(\frac{t_1+t_2}{t_2+t_3}\)

Example 10. A bullet, moving with a velocity of 200 m • s-1 can just go through a 4 cm thick plank. What should be the velocity of a bullet for just going through a 10 cm thick identical plank?
Solution:

The retardation (a) should be the same inside both the planks. The final velocity of the bullet in both cases is zero.

Let the initial velocities of the bullet in the two cases be u₁ and u₂, respectively.

Then 0 = \(u_1^2-2 a s_1 or, u_1^2=2 a s_1\)

Similarly, \(u_2^2=2 a s_2\)

∴ \(\frac{u_1^2}{u_2^2}=\frac{s_1}{s_2}\)

or, \(u_2=u_1 \sqrt{\frac{s_2}{s_1}}=200 \times \sqrt{\frac{10}{4}}\)

= \(100 \sqrt{10}=316.2 \mathrm{~m} \cdot \mathrm{s}^{-1}\).

Example 11. The speed of a train drops from 48 km h^-1 to 24 km h^-1 after moving through a distance of 108 m with uniform retardation. How much farther would it move with the same retardation before coming to rest?
Solution:

⇒ \(v^2=u^2+2\) as; \(s=108 \mathrm{~m}=0.108 \mathrm{~km}\)

∴ \(a=\frac{v^2-u^2}{2 s}\)

= \(\frac{(24)^2-(48)^2}{2 \times 0.108}=-8000 \mathrm{~km} \cdot \mathrm{h}^{-2}\)

For the second part of the motion, u = \(24 \mathrm{~km} \cdot \mathrm{h}^{-1} \text { and } v=0\)

∴ \(s=\frac{v^2-u^2}{2 a}\)

= \(\frac{0-(24)^2}{2 \times(-8000)}=0.036 \mathrm{~km}=36 \mathrm{~m}\)

Example 12. A particle starts with a velocity u with a uniform acceleration f. In the p-th, q-th and r-th seconds, it moves through distances a, b and c respectively. Prove that, a(q- r) + b(r- p) + c(p- q) = 0.
Solution:

Distance travelled in the n-th second, \(s_n=u+\frac{1}{2} f(2 n-1)\)

∴ a = \(u+\frac{1}{2} f(2 p-1)\)

= \(u+f p-\frac{f}{2}=\left(u-\frac{f}{2}\right)+f p\)…..(1)

Similarly, b = \(\left(u-\frac{f}{2}\right)+f q\)…..(2)

and c = \(\left(u-\frac{f}{2}\right)+f r\)……(3)

Multiplying (1), (2) and (3) by (q-r),(r-p) and (p-q) respectively, and then adding, we get a(q-r)+b(r-p)+c(p-q)

= \(\left(u-\frac{f}{2}\right)[(q-r)+(r-p)+(p-q)]\) + \(f[p(q-r)+q(r-p)+r(p-q)]\)

= \(\left(u-\frac{f}{2}\right) \cdot 0+f \cdot 0=0\)

Example 13. From two stations A and B, two trains started simultaneously towards each other with velocities v₁ and v₂ respectively. After they crossed each other, the first train reached B in time t₁ and the second train reached A in time t₂. Show that v₁: v₂ = √t₂:√t₁

Solution:

Let the two trains cross each other at point O after time t.

So, AO = v₁t and BO = v₂t, i.e., \(\frac{A O}{B O}=\frac{v_1}{v_2}\)

Again, for the first train, OB = v₁t₁

and for the second train, OA = v₂t₂

∴ \(\frac{O A}{O B}=\frac{A O}{B O}=\frac{v_2 t_2}{v_1 t_1}\)

Then, \(\frac{v_1}{v_2}=\frac{v_2 t_2}{v_1 t_1} \quad or, \frac{v_1^2}{v_2^2}=\frac{t_2}{t_1} \quad or, \frac{v_1}{v_2}=\frac{\sqrt{t_2}}{\sqrt{t_1}}\)

Example 14. A train attains a velocity v after starting from rest with a uniform acceleration α. Then the train travels for some time with uniform velocity, and at last, comes to rest with a uniform retardation β. If the overall displacement is s in time t, show that \(t=\frac{s}{v}+\frac{v}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\).
Solution:

For the accelerated motion, \(\nu=0+\alpha t_1 \quad \text { or, } t_1=\frac{\nu}{\alpha}\)

and \(v^2=0+2 \alpha s_1 \quad or, s_1=\frac{v^2}{2 \alpha}\)

For the uniform motion, \(s_2=v t_2 or, t_2=\frac{s_2}{v}\)

For the retarded motion, \(0=v-\beta t_3 \quad or, t_3=\frac{\nu}{\beta}\)

and 0 = \(v^2-2 \beta s_3 \quad or, s_3=\frac{v^2}{2 \beta}\)

∴ \(s=s_1+s_2+s_3=\frac{\nu^2}{2 \alpha}+s_2+\frac{\nu^2}{2 \beta}\)

or, \(s_2=s-\frac{v^2}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\)

Then, t = \(t_1+t_2+t_3=\frac{\nu}{\alpha}+\frac{s_2}{\nu}+\frac{\nu}{\beta}=\frac{v}{\alpha}+\frac{1}{\nu}\left[s-\frac{v^2}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\right]+\frac{\nu}{\beta}\)

= \(v\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+\frac{s}{v}-\frac{\nu}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=\frac{s}{v}+\frac{\nu}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\)

Graphical Proof Of The Equations Of Motion: AB represents the velocity-time graph of a particle starting with an initial velocity u, attaining a final velocity v in time t, moving with a uniform acceleration a.

Along the time and the velocity axes, OC = t, OA = u and CB = v.

∴ a = slope of AB

= tan θ= \(\tan \theta=\frac{D B}{A D}=\frac{C B-C D}{A D}=\frac{C B-O A}{O C}=\frac{\nu-u}{t} .\)

Proof Of v = u + at: We know the equation of a straight line of slope m with an intercept c on the y-axis is y = mx + c …..(1)

Comparing the corresponding values for AB, we get, y = v, m = a, x = t and c = u.

∴ The graph AB follows the equation v = u + at….(2)

Proof Of s = ut + \(\frac{1}{2}\)at²: The area under the graph AB and the time-axis, gives the displacement of the particle in time t.

The area under AB is the area of the trapezium OABC, which is the sum of the areas of the rectangle OADC and the triangle ABD.

Hence, the displacement of the particle in time t,

s = \(O A \times O C+\frac{1}{2} A D \times D B=O A \times O C+\frac{1}{2} \times \frac{D B}{A D} \times A D^2\)

∴ \(s=u t+\frac{1}{2} a t^2\) (because A D=O C=t) ……(3)

Proof Of v² = u² + 2as: Slope of AB = a = tanθ = \(\frac{D B}{A D}\)=\(\frac{DB}{O C}\)

∴ as = slope of AB X area of the trapezium OABC

= \(\frac{B D}{O C} \times \frac{1}{2}(O A+C B) \times O C\)

= \(\frac{1}{2} B D(O A+C B)=\frac{1}{2}(C B-C D)(O A+C B)\)

∴ as = \(\frac{1}{2}(v-u)(v+u)=\frac{1}{2}\left(v^2-u^2\right)\)

∴ 2as =\(v^2-u^2\)

or, \(v^2=u^2+2 a s\)

Acceleration in One Dimensional Motion Explained

Proof Of s_t = u + \(\frac{1}{2}\)a(2t-1): Let AP represent the velocity-time graph of a particle starting with an initial velocity u and moving with a uniform acceleration a.

Let points D and E represent the times (t – 1) and t respectively in the motion. Hence, the area of the trapezium BCED denotes the displacement s_t in the t th second.

∴ s_t = \(\frac{1}{2}\) (DB + EC) x DE

DE = OE- OD = t- (t- 1) = 1 s

DB = the velocity attained in time (t- 1) = u+ a(t- 1)

EC = velocity attained in time t = u+ at.

∴ s_t = \(\frac{1}{2}\)[{u+ a(t—1)} + (u+ at)] x 1

= u + \(\frac{1}{2}\) a(2t- 1)….(5)

Derivation Of The Equations Of Motion Using Calculus

Derivation of v = u + at: From definition, acceleration is the rate of change of velocity,

i.e., a = \(\frac{dv}{dt}\) or, dv = a dt

For motion with a uniform acceleration, a = constant.

∴ ∫dv = a∫dt

or, v = at + A [A = integration constant]……(1)

At t = 0, v = u. Then from equation (1), we get,

u = a·0 + A or, A = u

Hence, equation (1) becomes, v = u + at

Derivation of s = ut + \(\frac{1}{2}\) at²: We know the velocity of a particle is the rate of change of its displacement with time.

∴ v = \(\frac{ds}{dt}\) or, \(\frac{ds}{dt}\) = u + at

or, ds = u dt+ at dt, a and u are constants

∫ds = u∫dt+ a∫t dt

or, s = ut + \(\frac{1}{2}\) at² + A [A = integration constant]……(2)

From initial condition, at t = 0 , s = 0

∴ A = 0

∴ Equation (2) becomes, s = ut + \(\frac{1}{2}\) at²

Derivation Of v² = u² + 2as: Velocity of the particle,

v = \(\frac{ds}{dt}\) and acceleration, a = \(\frac{dv}{dt}\) \(\frac{ds}{dt}\) = v\(\frac{dv}{ds}\)

or, a ds = v dv, where a is a constant, a∫ds = ∫v dv

or, as = \(\frac{v^2}{2}\) + B [B = integration constant]…..(3)

At the start of motion, v = u and s = 0

∴ 0 = \(\frac{u^2}{dt}\) + B or, B = –\(\frac{u^2}{2}\)

Substituting this in equation (3), we get,

as = \(\frac{v^2}{2}\) – \(\frac{u^2}{2}\)

or, v² = u² + 2as…..(4)

Derivation Of The Equations Of Motion Using Calculus Numerical Examples

Example 1. s-t graph for a particle, moving with a constant acceleration, subtends 45° angle with the time axis at time t. That angle becomes 60° 1 s later. Find the acceleration of the particle.
Solution:

Let the velocity of the particle at P₁ and P₂ be u and v respectively.

∴ u = tanθ₁ and v = tanθ₂

∴ Acceleration of the particle

a = \(\frac{\nu-u}{t_2-t_1}\)

= \(\frac{\tan \theta_2-\tan \theta_1}{t_2-t_1}\)

= \(\frac{\tan 60^{\circ}-\tan 45^{\circ}}{1}\)

= \(\sqrt{3}-1=0.732 \text { unit. } \mathrm{s}^{-2} .\)

Equations of Motion for One Dimensional Movement

Example 2. Displacement x and time t, in a rectilinear motion of a particle, are related as t= √x+3. Here x is measured in metres and t in seconds. Find the displacement of the particle when its velocity is zero.
Solution:

Given, t = √x + 3 or, √x = t-3 or, x = (t-3)².

Velocity, v = \(\frac{dx}{dt}\) = 2(t – 3)

If velocity =0, 2(t – 3) = 0 or, t = 3 s

∴ At t = 3 s , the displacement, x = (3 -3)² = 0

Hence at zero velocity, the displacement is also zero

Example 3. A body starts from rest and moves with an acceleration proportional to time,

1. Find its velocity n s after starting,
2. What distance will it travel in ns?

Solution:

According to the problem, a ∝ t

or, a = kt, where k is the constant of proportionality.

Now, a = \(\frac{dv}{dt}\)

dv = kt dt or, ∫dv = k∫t dt

∴ v = \(\frac{k t^2}{2}\) = A(where A = integration constant) ……(1)

Also, as v = 0 at t = 0 , we get A = 0

∴ From equation (1 ) we get, v = \(\frac{k t^2}{2}\)….(2)

Again, if the displacement of the object is s, then v = \(\frac{ds}{dt}\)

∴ From equation (2), \(\)

On integration we get,

s = \(\frac{k t^3}{6}\) + B [where B = integration constant]

Also, at t = 0 , s = 0

∴ B = 0

∴ s = \(\frac{k t^3}{6}\)….(3)

At t = ns, from equations(2) and (3) we get,

1. v = \(\frac{k n^2}{2}\)
2. s = \(\frac{k n^2}{6}\)

Example 4. The velocity of a moving particle v decreases with its displacement. Given, v = v₁ – ax where v₀ = initial velocity, x = displacement and a is a constant. How long will the particle take to reach point B on the x-axis at a distance x_m from the origin?
Solution:

Given, v = v₀ – ax.

At the starting point, v = v₀.

∴ v₀ = v₀– ax or, x= 0 [a = constant]

Hence, the particle was initially at the origin.

Now, \(v=\frac{d x}{d t}=v_0-\alpha x\)

or, \(\frac{d x}{v_0-\alpha x}=d t or, \frac{-\alpha d x}{v_0-\alpha x}=-\alpha d t\)

Then, \(\int \frac{-\alpha d x}{v_0-\alpha x}=-\alpha \int d t or, \log _e\left(\nu_0-\alpha x\right)=-\alpha t+c\); [c is the integration constant]

At t=0 (initially), x=0

∴ \(c=\log _e v_0\)

Therefore, \(\log _e\left(v_0-\alpha x\right)=-\alpha t+\log _e v_0\)

or, \(t=\frac{1}{\alpha} \log _e \frac{v_0}{v_0-\alpha x}\)

At the point \(B, x=x_m\).

Let the corresponding time be \(t_m\).

∴ \(t_m=\frac{1}{\alpha} \log _e \frac{v_0}{v_0-\alpha x_m} \text {. }\)

Example 5. The relation between the time taken and the displacement of a moving body is s = 2t²- 3t² + 4t³, where the unit of s is in metres and that of t is in seconds. Find out the displacement, velocity and acceleration of the body 2 s after initiation of the journey.
Solution:

Here, s = 2t – 3t³ + 4t³……(1)

∴ In 2 s the displacement of the body is,

s = (2 x 2)- [3 x (2)²] + [4 x (2)³]

= (4-12 + 32) = 24 m

Now, the velocity is, v = \(\frac{ds}{dt}\) = 2 – 6t + 12t²

∴ After 2s, v = 2- (6 x 2) +[12 x (2)²] = 38 m · s^-1

Again, the acceleration is, a = \(\frac{dv}{dt}\) -6 + 24t

∴ Acceleration after 2 s, a = – 6 +.24 X 2 = -6 + 48 = 42 m · s^-2.

Example 6. For a particle travelling along a straight line, the equation of motion is s = 16t + 5t². Show that it will always travel with uniform acceleration.
Solution:

Here, s = 16t + 5t²

∴ Velocity, v = \(\frac{ds}{dt}\) = 16 + 10t

Again acceleration, a = \(\frac{dv}{dt}\) = 10 = constant

∴ The particle will always travel with uniform acceleration.

Example 7. If a, b and c are constants of motion and s = at² + bt + c, then prove that 4a(s – c) = v² – b².
Solution:

Here, s = at² + bt+ c

∴ v = \(\frac{ds}{dt}\) = 2at+ b

∴ v² – b² =  (2at+ b)²- b² = 4a²t² + 4abt

= 4a(at² + bt) = 4a(at² + bt + c – c)

or, v² – b² = 4a(s-c) (Proved)

Example 8. The retardation of a particle in rectilinear motion is proportional to the square root of its velocity v. Assume that the constant A of proportionality Is positive. The initial velocity of the particle is v₀. How far would the particle move before coming to rest? What would be the time required to travel that distance?
Solution:

⇒ \(a \propto-\sqrt{v}\)

∴ \(a=-A \sqrt{v}=-A \nu^{1 / 2}\)

or, \(\frac{d \nu}{d t}=-A v^{1 / 2}\)

or, \(v^{-1 / 2} d v=-A d t\)

∴ \(\int v^{-1 / 2} d v=-A \int d t+c\) [c=integration constant]

or, \(2 v^{1 / 2}=-A t+c\)…..(1)

Given, at t=0, v=\(v_0\). Putting in (1), \(c=2 v_0^{1 / 2}\)

∴ \(2 v^{1 / 2}=-A t+2 v_0^{1 / 2}\)

or, \(2\left(\sqrt{v_0}-\sqrt{v}\right)\)= At…..(2)

When the particle comes to rest after a time T, we have v=0 at t=T.

From (2), \(2 \sqrt{v_0}=A T \quad \text { or, } T=\frac{2}{A} \sqrt{v_0}
\)

Now, \(a=\frac{d v}{d t}=\frac{d v}{d x} \frac{d x}{d t}=\frac{d v}{d x} v\)

or, \(d x=\frac{1}{a} v d v=-\frac{1}{A \sqrt{v}} v d v=-\frac{1}{A} v^{1 / 2} d v\)

∴ \(\int d x=-\frac{1}{A} \int v^{1 / 2} d v+k\) [k = integration constant]

or, \(x=-\frac{1}{A} \cdot \frac{2}{3} \nu^{3 / 2}+k=-\frac{2}{3 A} v^{3 / 2}+k\)….(3)

At start, x=0 and \(v=v_0\), Putting in (3),

0 = \(-\frac{2}{3 A} v_0^{3 / 2}+k \quad \text { or, } k=\frac{2}{3 A} v_0^{3 / 2}\)

So, equation (3) becomes, \(x=\frac{2}{3 A}\left(v_0^{\frac{3}{2}}-v^{\frac{3}{2}}\right)\)……..(4)

When the particle comes to rest, v=0. Then the total distance travelled is, \(x_0=\frac{2}{3 A} v_0^{3 / 2}\)

Example 9. The acceleration-time graph of a particle starting from rest is given. Draw the corresponding velocity-time graph and hence find out the displacement in 6s.

Graphical Representation of One Dimensional Motion

Solution:

In the intervals (0 →1), (2 → 3) and (4 → 5) seconds, acceleration is zero, i.e., velocity = constant.

Again, in the intervals (1 → 2), (3 → 4) and (5 → 6) seconds, the velocity increases uniformly and rises to 1 m/s, 2 m/s and 3 m/s respectively, because the uniform acceleration in each interval is 1 m/s²

∴ The velocity-time graph of the motion is ABCDEF

Displacement in 6 s = area under ABCDEF

= area of 6 unit squares + area of 3 triangles

= 6 x (1 x 1) + 3 x (\(\frac{1}{2}\) x 1 x 1) = 7.5 m

Vertical Motion Under Gravity

Acceleration Due To Gravity: when an object is released from a certain height above the earth’s surface, it moves with vertically downward acceleration. Again, when an object is thrown vertically upwards from the ground, it moves up with a deceleration. An upward deceleration is equivalent to a downward acceleration.

Actually, the acceleration is always downward, and its magnitude is the same for both downward and upward motions.

So, the equations of vertical motion are,

v = u + at ….(1)

h = ut + \(\frac{1}{2}\)at² ……(2)

and v² = u² + 2ah ……(3)

Here, initial velocity is u, velocity after time r is v, vertical displacement in time t is h, and acceleration is a.

This acceleration a is called the acceleration due to gravity or free fall acceleration and is represented by the letter g.

The direction of g is always vertically downwards.

If the downward direction is taken as positive, i.e., a = g, then we get the following equations of motion:

v = u + gt …..(4)

h = ut + \(\frac{1}{2}\)gt² …..(5)

and v² = u² + 2 gh …(6))

If the upward direction is taken as positive, i.e., a = we get the following equations: -g, then

v = u – gt …….(7)

h = ut – \(\frac{1}{2}\)gt² …..(8)

and v² = u² – 2gh …..(9)

Maximum Height Attained: When a body is thrown vertically upwards with a velocity u, it momentarily comes to rest on attaining the maximum height and then again starts falling vertically downwards.

Hence, at maximum height H, the velocity of the body, v = 0.

∴ 0 = u² – 2gH [from equation (9)]

or, H = \(\frac{u^2}{2g}\)…….(10)

Time To Reach The Maximum Height: Let the time required to reach the maximum height be T.

∴ 0 = u² – gT [from equation (7)]

or, T= \(\frac{u}{g}\)……(11)

Time Taken To Fall From The Maximum Height: if T₁ is the time taken by the body to fall from the maximum height to the initial position then, using equation (5),

H = \(\frac{1}{2} g T_1^2\)[as at maximum height, u = 0]

or, \(T_1^2=\frac{2 H}{g}=\frac{2}{g} \cdot \frac{u^2}{2 g}=\frac{u^2}{g^2}\)

∴ \(T_1=\frac{u}{g}\)….(12)

Hence, the time to reach the maximum height is equal to the time to return to the starting point.

Time Of Flight: it is the total time required for upward and downward motions

T’ = T + T₁ = \(\frac{u}{g}\) + \(\frac{u}{g}\) = \(\frac{2u}{g}\)

Time Of Flight Alternative Method: After completion of the upward and downward motions, the displacement becomes zero. Thus, using equation (8),

0 = uT’ – \(\frac{1}{2}\)gT’² or, u = \(\frac{1}{2}\)gT’ or, T’ = \(\frac{2u}{g}\).

Time Taken To Reach A Certain Height: Let the object reach a height h at time f. The initial upward velocity = u.

∴ h = \(u t-\frac{1}{2} g t^2\)

or, \(\frac{1}{2} g t^2-u t+h=0\)

or, \(t=\frac{u}{g} \pm \frac{\sqrt{u^2-2 g h}}{g}\)

From this equation, two different values of t are obtained. This is because the object crosses the point at a height h twice, first during the upward motion and then during the downward motion.

Velocity At Any Height: From the equation \(v^2=u^2-2 g h, \text { we get, } v= \pm \sqrt{u^2-2 g h} \text {. }\).

The positive (+) sign is applicable for upward motion and the negative (-) sign is applicable for downward motion. So, an object crosses any point with the same magnitude of velocity in its upward and downward motions.

Velocity Of Projection And Velocity Of Return: Let u = velocity of projection, v = velocity of return and total time of flight = T’.

∴ From equation (7), we can write,

v = u – gT’ = u – g · \(\frac{2u}{g}\) = -u (T’ = \(\frac{2u}{g}\))

Hence, the upward velocity of projection is equal in magnitude to the downward velocity with which an object hits the ground.

Vertical Motion Under Gravity Numerical Examples

Example 1. A stone is dropped from a height of 19.6 m. What is the time taken by the stone to travel the last metre of the path?
Solution:

In this case u = 0

∴ \(h=\frac{1}{2} g t^2 \quad \text { or, } t=\sqrt{\frac{2 h}{g}}\)

Let t₁ and t₂ be the time taken by the stone to travel (19.6-1) = 18.6 m and 19.6 m, respectively.

∴ \(t_1=\sqrt{\frac{2 \times 18.6}{9.8}} \text { and } t_2=\sqrt{\frac{2 \times 19.6}{9.8}}\)

or, \(t_1=1.948 \mathrm{~s} \) and \(t_2=2 \mathrm{~s}\)

Hence, the time taken to travel the last metre, \(t_2-t_1=2-1.948=0.052 \mathrm{~s}\)

Example 2. An object is thrown vertically upwards with an initial velocity of 40 m · s^-1.

1. How long will the object move upwards?
2. What will be the maximum height attained?
3. How much time will it take to reach the ground?
4. When will the object be at a height of 25 m from the ground?
5. What will be its velocity after 2 s ? [g = 9.8 m · s^-2]

Solution:

1. Let the time taken for upward motion be t s. At maximum height, its velocity is zero. From equation v = u – gt, we get, 0 = 40 – 9.8 x t or, t = 4.1s.

2. Let the maximum height attained be h.

From equation v² = u² – 2gh, we get,

0 = (40)² – 2 x 9.8h or, h = 81.6 m .

3. Let the time taken to reach the ground be t₁ starting from the time of projection. Considering both upward and downward motions of the body and using the equation h = ut – \(\frac{1}{2}\)gt², we get,

0 = \(40 t_1-\frac{1}{2} \times 9.8 \times t_1^2\) [total displacement is zero in this case]

∴ t₁ = 8.2 s

4. Let the time after which the body is at a height of 25 m be x.

Hence, from h = ut – \(\frac{1}{2}\)gt² we get,

25 = 40 x – \(\frac{1}{2}\) · 9.8x² or, 49x² – 400x + 250 = 0

or, \(x=\frac{400 \pm \sqrt{(400)^2-4 \cdot 49 \cdot 250}}{2 \cdot 49}=\frac{400 \pm 10 \sqrt{1110}}{2 \cdot 49}\)

or, x = 0.682 s and x = 7.481 s

Two values of x signify that the object will be at a height of 25 m twice during its flight, once (x = 0.682 s) while moving upwards and the next (x = 7.481 s) during its downward motion.

5. Let the velocity acquired 2 s after the projection be v.

∴ v = 40 – 9.8 x 2 = 20.4 m · s^-1.

Example 3. A ball falls freely on a perfectly elastic plate from a height of 3 m. At the instant t = 0, the velocity of the ball is zero. Draw a velocity-time graph for the motion of the ball, [g = 9.8 m · s^-2]
Solution:

In this case, h = \(\frac{1}{2}\) gt²

The time taken by the ball to fall through 3 m is given by t = \(\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 3}{9.8}}=0.783 \mathrm{~s}\)

According to the equation, v² = 2gh, the velocity of the ball just before striking the plate is given by

v = \(\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 3.0}=7.67 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Since the plate is perfectly elastic, the ball after striking it will rebound with the same velocity (7.67 m · s^-1) and its velocity will become zero after the same time (0.78 s). This motion will be repeated again and again as shown in the above graph.

Example 4. A body is thrown vertically upwards. After attaining half of its maximum height its velocity becomes 14 m · s^-1.

1. How high will the body rise?
2. What will be the velocity of the body 1 s and 3 s after its projection?
3. What is the average velocity of the body in the first half second?

Solution:

1. Let the velocity of projection be u.

Hence, maximum height attained by the body, h = \(\frac{u^2}{2g}\) [using v² = u² – 2 gh]

or, u² = 2 gh

For half the maximum height, i.e., \(\frac{h}{2}\), we get

(14)² = u² – 2g\(\frac{h}{2}\) = 2gh –\(\frac{h}{2}\)(2gh) = gh

∴ h = 20 m [where g = 9.8 m · s^-2]

∴ The body will rise up to a height of 20 m. nil Here, the velocity of projection,

2. Here, the velocity of projection, \(u=\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 20}=19.8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ The velocity of the body 1 s after projection is, v₁ = u – g ·1 = 19.8 – 9.8 = 10 m · s^-1.

Velocity 3 s after projection, v₂ = u – g x 3 = 19.8 – 9.8 x 3 = -9.6 m · s^-1

(Negative sign indicates downward motion of the body.)

3. Velocity after  \(\frac{1}{2}\)s

v’ = u – g x \(\frac{1}{2}\) = 19.8 – 9.8 x \(\frac{1}{2}\) = 149 m · s^-1

Hence, average velocity during the given period = \(\frac{u+v^{\prime}}{2}=\frac{(19.8+14.9)}{2}=17.35 \mathrm{~m} \cdot \mathrm{s}^{-1} \text {. }\)

Short Answer Questions on One Dimensional Motion

Example 5. A piece of stone was dropped from a stationary balloon. The stone covered 13.9 m during the last 1/7 s of its descent. Find the height of the balloon and the velocity of the stone when it strikes the ground, [g = 9.8 m · s^-2]
Solution:

Let h = height of the balloon, t = total time of fall of the stone, h’ = downward displacement in time (t – \(\frac{1}{7}\))s.

Hence, \(h-h^{\prime}=\frac{1}{2} g t^2-\frac{1}{2} g\left(t-\frac{1}{7}\right)^2\)

or, \(13.9=\frac{1}{2} \times 9.8 \times t^2-\frac{1}{2} \times 9.8\left(t-\frac{1}{7}\right)^2\)

or, \(13.9=4.9 t^2-4.9 t^2+1.4 t-0.1\)

or, \(1.4 t=14\) or, t=10

∴ Height of the balloon, \(h=\frac{1}{2} g t^2=\frac{1}{2} \times 9.8 \times(10)^2=490 \mathrm{~m}\)

The velocity of the stone when it strikes the ground is, \(v=u+g t=0+9.8 \times 10=98 \mathrm{~m} \cdot \mathrm{s}^{-1} \text {. }\)

Example 6. A stone is dropped from the top of a tower 400 m high. At the same time, another stone is thrown upwards from the ground with a velocity of 100 m · s^-1. When and where will they meet each other? (g = 9.78 m · s^-2).
Solution:

Let the two stones meet after a time t at a distance h from the top of the tower.

h = \(\frac{1}{2}\) gt² = \(\frac{1}{2}\) x 9.8 t² ……(1)

Considering the downward motion of the 1st stone 400 – h = 100t – \(\frac{1}{2}\) x 9.8 t² …..(2)

From equations (1) and (2) we get, 400 – \(\frac{1}{2}\)x 9.8t² = 100t – \(\frac{1}{2}\) x 9.8t²

or, 100t = 400 or, t = 4 s

Hence, from equation (1) we get, h = \(\frac{1}{2}\) x 9.8 x (4)² = \(\frac{1}{2}\) x 9.8 x 16 = 78.4 m

Hence, the two stones meet at 78.4 m below the top of the tower after 4 s.

Example 7. A stone is dropped from the top of a vertical pillar. When the stone has fallen through a height x, another stone is dropped from height y below the top of the pillar. Both the stones touch the ground at the same time. Prove that the height of the pillar should be \(\frac{(x+y)^2}{4 x}\).
Solution:

Let the height of the pillar be h and the velocity of the stone at x below the top of the pillar be v.

∴ \(v^2=2 g x \text { or, } v=\sqrt{2 g x}\)…..(1)

Let the first stone take \(t \mathrm{~s}\) to cover the distance (h-x).

∴ \(h-x=v t+\frac{1}{2} g t^2\)….(2)

According to the problem, the second stone is dropped from a height of (h-y) and this stone takes time t to cross that distance.

∴ \(h-y=\frac{1}{2} g t^2\)…..(3)

From equations (2) and (3) we get, \(h-x=\nu t+h-y \text { or, } y-x=v t \text { or, } y-x=t \sqrt{2 g x}\)

∴ t = \(\frac{y-x}{\sqrt{2 g x}}\)

From equation (3) we get, \(h-y=\frac{1}{2} g \frac{(y-x)^2}{2 g x}=\frac{(y-x)^2}{4 x}\)

or, \((h-y) \cdot 4 x=(y-x)^2 or, 4 x h-4 x y=y^2-2 y x+x^2\)

or, \(h=\frac{(x+y)^2}{4 x}\) (Proved).

Example 8. A, B, C and D are four points on a vertical line such that AB = BC = CD. A body is allowed to fall freely from A. Prove that the respective times required by the body to cross the distances AB, BC, and CD should be in the ratio 1: (√2- 1): (√3-√2).
Solution:

Let AB – BC = CD = x and the time taken by the body to cover these distances be t₁, t₂ and t₃ respectively.

Now, \(x=\frac{1}{2} g t_1^2 \text { or, } t_1=\sqrt{\frac{2 x}{g}}\)…..(1)

2x = \(\frac{1}{2} g\left(t_1+t_2\right)^2 \text { or, } t_1+t_2=\sqrt{\frac{4 x}{g}}\)…..(2)

3 x = \(\frac{1}{2} g\left(t_1+t_2+t_3\right)^2 \text { or, } t_1+t_2+t_3=\sqrt{\frac{6 x}{g}}\)…..(3)

From (1) and (2) we get, \(t_2=\sqrt{\frac{4 x}{g}}-\sqrt{\frac{2 x}{g}}=\sqrt{\frac{2 x}{g}}(\sqrt{2}-1)\)

From (2) and (3) we get, \(t_3=\sqrt{\frac{6 x}{g}}-\sqrt{\frac{4 x}{g}}=\sqrt{\frac{2 x}{g}}(\sqrt{3}-\sqrt{2})\)

∴ \(t_1: t_2: t_3=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})\) (Proved).

Example 9. A rubber ball is thrown vertically downwards from the top of a tower with an Initial velocity of 14 m · s^-1. A second ball is dropped 1 s later from the same place. In 2 s the first ball reaches the ground and rebounds upwards with the same velocity. When will they collide with each other?
Solution:

Height of the tower, h = distance covered by the first ball in 2s = 14 x 2 + \(\frac{1}{2}\)  x 9.8 x (2)² = 47.6 m

(h = ut + \(\frac{1}{2}\)gt²)

The velocity of the first ball just before touching the ground is v = 14 + 9.8 x 2 = 33.6 m · s^-1

Hence, its velocity just after bouncing = 33.6 m · s^-1

Downward displacement of the second ball in 1 s, x = \(\frac{1}{2}\) x 9.8 x (1)² = 4.9 m

Velocity of second ball after 1 s = 9.8 x 1 = 9.8 m · s^-1

Hence, the distance between the two balls, 2 s after the projection of the first ball = 47.6- 4.9 = 42.7 m.

Let the two balls collide with each other ts after the first ball bounces off the ground.

Upward displacement of the first ball in ts, x₁ = 33.6t – \(\frac{1}{2}\) X 9.8 x t² = 33.6t- 4.9t²

Downward displacement of the second ball in t s, x₂ = 9.8t+ \(\frac{1}{2}\) x 9.8 x t² = 9.8t + 4.9t²

Now, x₁ + x₂ = 42.7

or, 33.6t – 4.9t + 9.8t+ 4.9 t³ = 42.7 or, 43.4t = 42.7

∴ t = \(\frac{42.7}{43.78}\) = 0.98 s

Example 10. A lift starts to move up with a constant acceleration of 2 m · s^-2 from the earth’s surface. A piece of stone is dropped outside from the lift 4 s after the start of the lift When will the stone reach the earth’s surface?
Solution:

Initial velocity of the lift, u = 0; acceleration, a = 2 m · s^-2

Let the rise of the lift in 4 s be s and its velocity at that point be v.

Hence, from equation v = u+ at we get,

v = 0 + 2 x 4 = 8 m · s^-1

Also from equation s = ut + \(\frac{1}{2}\) at² we get

s = 0 x 4 + \(\frac{1}{2}\) x 2 (4)² = 16 m

∴ The stone piece was dropped with an initial upward velocity of 8 m · s^-1 and was at a height of 16 m from the ground. If t is the time taken by the stone to reach the ground, then from equation h = ut + \(\frac{1}{2}\)gt²,

16 = \(-8 t+\frac{1}{2} \times 9.8 t^2\)

(because for the stone, \(u=-8 \mathrm{~m} \cdot \mathrm{s}^{-1}, g=9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\), h=16 m)

or, \(4.9 t^2-8 t-16=0\)

or, \( t=\frac{8 \pm \sqrt{(8)^2-4 \times 4.9 \times(-16)}}{2 \times 4.9}=\frac{8 \pm 19.4}{9.8}\)

Since time cannot be negative, \(t=\frac{8+19.4}{9.8}=2.8 \mathrm{~s}\).

Example 11. Two bodies released from different heights fall freely and reach the ground at the same time. The first body takes a time, t₁ = 2 s and the second body takes a time, t₂ = 1 s. What was the height of the first body at the time of the release of the second body?
Solution:

Let the height from which the first body is released be h₁ above the ground. When it is at a height h₂, the second body is released.

So, the first body falls through a height (h₁ – h₂) in time (t₁– t₂) before the release of the second body.

According to the problem, \(h_1=\frac{1}{2} g t_1^2 \text { and } h_1-h_2=\frac{1}{2} g\left(t_1-t_2\right)^2\)

or, \(\frac{1}{2} g t_1^2-h_2=\frac{1}{2} g\left(t_1-t_2\right)^2\)

or, \(h_2=\frac{1}{2} g\left(2 t_1-t_2\right) t_2\)

= \(\frac{1}{2} \times 9.8(2 \times 2-1) \times 1=14.7 \mathrm{~m} .\)

Example 12. A balloon moves vertically upwards with a uniform velocity v₀. A weight is tied to the balloon with a rope. When the balloon attains a height of h₀, the rope snaps. How much time will the weight take to reach the ground?
Solution:

Let the time taken by the weight, after the rope snaps, to reach the ground be t. From the question, the weight ascends with the same velocity as that of the balloon, i.e., v₀.

So, when the rope snaps the initial velocity of the weight =- v₀ [the negative sign comes as the velocity of the weight is directed upwards just when the rope snaps].

∴ For the free fall of the weight, \(h_0=-v_0 t+\frac{1}{2} g t^2 \text { or, } \frac{1}{2} g t^2-v_0 t-h_0=0\)

∴ \(t=\frac{v_0 \pm \sqrt{v_0^2-4 \times \frac{1}{2} g\left(-h_0\right)}}{2 \cdot \frac{1}{2} g}=\frac{v_0 \pm \sqrt{v_0^2+2 g h_0}}{g}\)

So, the time taken by the weight to reach the ground is \(\frac{v_0+\sqrt{v_0^2+2 g h_0}}{g}\) (becasue t>0).

Example 13. According to the, three cars P, Q and R are at three points along the x-axis at a given moment. Now car P starts its motion towards P₁ parallel to the y-axis with a uniform velocity v. Again, R is in motion parallel to the y-axis along RRX with uniform acceleration a. If car Q too moves parallel to the y-axis then under what condition will all of them remain collinear? Given PQ = QR

Solution:

Let the time be t after which P and R are at and respectively.

PP₁ = vt and RR₁ = \(\frac{1}{2}\)at²

According to the question, after time t, Q will be at Q₁.

From the figure we have, \(\frac{P P_1}{P M}=\frac{Q Q_1}{Q M}=\frac{R R_1}{R M}=k\) (say)

∴ \(P P_1=k \cdot P M=k(P Q+Q M)\)

⇒ \(Q Q_1=k \cdot Q M\)

and \(R R_1=k \cdot R M=k(Q R-Q M)\)

∴ \(P P_1-R R_1=k \cdot(Q M+Q M)\) (because P Q=Q R)

= \(2 k \cdot Q M\)

∴ \(k \cdot Q M=\frac{P P_1-R R_1}{2}\)

∴ \(Q Q_1=k \cdot Q M\)

= \(\frac{P P_1-R R_1}{2}=\frac{P P_1}{2}-\frac{R R_1}{2}=\frac{\nu t}{2}-\frac{1}{4} a t^2\)

∴ \(Q Q_1=\left(\frac{\nu}{2}\right) t-\frac{1}{2}\left(\frac{a}{2}\right) \cdot t^2\)……..(1)

Comparing equation (1) with s = ut- \(\frac{1}{2}\)at², we can say that to remain in the same straight line joining the other two cars, the initial velocity of the car Q must be \(\frac{v}{2}\) along the positive y-axis and it should move with retardation \(\frac{a}{2}\).

 

 One-Dimensional Motion Conclusion

Dynamical quantities related to an object are determined with reference to some other object in the surroundings. This external object forms a frame of reference. A frame of reference is represented in forms like cartesian, polar, spherical etc.

• The change of position of a moving object in a fixed direction is called displacement. Displacement is a vector quantity.
• The rate of Hie distance covered by a body with respect to time is called speed. Speed is a scalar quantity.
• The rate of the distance travelled from a point by a particle with respect to an infinitesimal interval of time is called the instantaneous speed of the particle at that point.
• The rate of displacement of a body with respect to time is called velocity. Velocity is a vector quantity.

Applications of One Dimensional Motion in Real Life

The rate of the displacement of a particle from a certain point with respect to an infinitesimally small time interval is called the instantaneous velocity of the particle at that point.

• An object moving in a straight line with uniform speed is said to be moving with uniform velocity.
• An object with uniform velocity must have uniform speed but an object with uniform speed may not have uniform velocity.
• For a moving object, zero average velocity does not necessarily mean a zero average speed, but zero speed always implies zero velocity.
• For an object moving with a uniform velocity, its average velocity is the same as its instantaneous velocity.

The rate of change of velocity of an object with respect to time is called acceleration. Acceleration is a vector quantity.

• The rate of change in velocity of a body at a point with respect to an infinitesimally small time interval is the instantaneous acceleration of the body at that position.
• On a displacement-time graph of a particle, the inclination of the tangent at a point on the graph gives the instantaneous velocity of the particle at that point.
• In a velocity-time graph for a particle, the inclination of the tangent at any point gives the instantaneous acceleration of the particle at that point.

The area enclosed between the velocity-time graph of a particle and the time axis gives the magnitude of the displacement of the particle.

 One-Dimensional Motion Useful Relations For Solving Problems

Speed, v = \(\frac{l}{t}\), where l = distance covered in time t.

Average speed = \(\frac{\text { total distance }}{\text { total time }}\)

or, \(v=\frac{l_1+l_2+l_3+\cdots+l_n}{t_1+t_2+t_3+\cdots+t_n}\)

Instantaneous speed, \(v_i=\lim _{\Delta t \rightarrow 0} \frac{\Delta l}{\Delta t}=\frac{d l}{d t}\)

Velocity, v = \(\frac{s}{t}\), where s= displacement in time t.

Average velocity = \(\frac{\text { total displacement }}{\text { total time }}\)

or, (v)= \(\frac{s_1+s_2+s_3+\cdots+s_n}{t_1+t_2+t_3+\cdots+t_n}\)

Instantaneous velocity, \(v_i=\lim _{\Delta t \rightarrow 0} \frac{\Delta s}{\Delta t}=\frac{d s}{d t}\)

For A Particle In Motion

• Initial velocity = u
• Acceleration = a
• Distance covered in the n-th second =sn
• Acceleration due to gravity = g
• The final velocity after time t = v
• Displacement in time t = s_n
• Height in time t = h

Acceleration or average acceleration = \(\frac{\text { final velocity }- \text { initial velocity }}{\text { time }}=\frac{\text { change in velocity }}{\text { time }}\)

or, (a) =\(\frac{v-u}{t}\)

Instantaneous acceleration, \(a_i=\lim _{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}=\frac{d v}{d t}=\frac{d^2 s}{d t^2}\)

For a particle in motion with uniform acceleration,

1. v = u + at,
2. s = ut +\(\frac{1}{2}\)at²,
3. v² = u² + 2as,

The displacement in the t th second s_t = u + \(\frac{l}{2}\)a(2t-1) In case of retardation,

1. v – u -at,
2. s_t = ut – \(\frac{1}{2}\)at²,
3. v² = u² – 2as,
4. s_t = u – \(\frac{1}{2}\)a(2t-1)

Equations of vertical motion under gravity (downward direction positive):

1. v = u + gt,
2. h = ut + \(\frac{1}{2}\)gt²,
3. v² = u² + 2gh

(upward direction positive):

1. v = u – gt,
2. h = ut- \(\frac{l}{2}\)gt²,
3. v² = u² – 2gh

 

One-Dimensional Motion Very Short Answer Type Questions

Question 1. A particle is in uniform motion with respect to a reference frame. Is it possible for the particle to be at rest with respect to another frame?
Answer: Yes

Question 2. Retardation is essentially a _______ acceleration
Answer: Negative

Question 3. For a particle, displacement (x) and time (t) are related by the following equation: x = (3t² + 2t+5)m. If time is expressed in seconds, find the initial velocity of the particle.
Answer: 2 m · s^-1

Question 4. The displacement of a moving particle is directly proportional to the square of the time duration. State whether the particle is moving at a constant velocity or at a constant acceleration.
Answer: Constant acceleration

Question 5. When is the average velocity of a particle equal to its instantaneous velocity?
Answer: In motion with uniform velocity

Question 6. Does a particle with a uniform speed in a curved path possess any acceleration?
Answer: Yes

Question 7. How many dimensions are there in the motion of a ship in a turbulent sea?
Answer: Three

Question 8. If the position of a particle at instant t is given by x = t⁴, find the acceleration of the particle.
Answer: 12t²

Question 9.For a moving body, displacement y (in metre) and time t (in second) are related as y = – \(\frac{2}{3}\) t²+16t+2. When will the body stop?
Answer: After 12s

Question 10. The displacement equation for a particle moving in a straight line is x = αt³ + βt² + γt+ δ. The ratio of the initial acceleration to the initial velocity depends only on _______
Answer: β and γ

Question 11. The motion of an artificial satellite around the earth is a _________ dimensional motion.
Answer: Two

Question 12. An athlete runs with a velocity of 18 km · h^-1. How much distance will he cover in 10 min?
Answer: 3 km

Question 13. What is the nature of the time-displacement graph for a particle moving with a constant velocity?
Answer: A straight line

Question 14. What does the slope of a position-time graph represent?
Answer: Velocity of a body

Question 15. The area under v-t graph =?
Answer: Displacement of the particle

Question 16. In the same displacement-time graph, two motions are represented by two straight lines having slopes of 45° and 60° respectively. Which line represents a higher velocity and what is the ratio between the first and the second velocities?
Answer: Second: 1: √3

Question 17. Velocity (in m · s^-1 unit)-time (in s unit) graph, of a particle moving in a straight line, is a straight line and it is inclined at an angle 45 with the time axis. What is the acceleration of the particle?
Answer: 1 m s^-2

One-Dimensional Motion Assertion Reason Type Questions And Answers

These questions have statements 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: The average velocity of a particle may be equal to its instantaneous velocity.

Statement 2: For a given time interval of a given motion, average velocity is single-valued while average speed can have many values.

Answer: 3. Statement 1 is false, statement 2 is true.

Question 2.

Statement 1: A scooter moves towards the north and then moves towards the south with the same speed. There will be no change in the velocity of the scooter.

Statement 2: Velocity is a vector quantity.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 3.

Statement 1: An object can possess acceleration even when it has a uniform speed.

Statement 2: When the direction of motion of an object keeps changing, its velocity also changes with time.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 4.

Statement 1: Acceleration of a moving particle can change its direction without any change in the direction of velocity.

Statement 2: If the direction of the change in the velocity vector changes, the direction of the acceleration vector also changes.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 5.

Statement 1: The distance between two particles moving with constant velocities always remains constant.

Statement 2: The relative motion between two particles moving with a constant is velocities.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 6.

Statement 1: A particle with zero velocity may have a non-zero acceleration.

Statement 2: A particle comes to rest at the instant of reversing its direction of motion.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 7.

Statement 1: A particle moving with uniform accelera¬tion has its displacement proportional to the square of time.

Statement 2: If the motion of a particle is represented by a straight line on the velocity-time graph, its acceleration is uniform.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 8.

Statement 1: A freely falling body travels through distances in this ratio 1: 3: 5: 7:…… in successive equal intervals of time (Galileo’s law of odd integers).

Statement 2: In one-dimensional motion, a particle with zero speed may have a non-zero velocity.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 9.

Statement 1: If the average velocity of a body is equal in two successive time intervals, its velocity is a constant.

Statement 2: When a body travels with constant velocity, its displacement is proportional to time.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 10.

Statement 1: If two bodies of different masses one dropped simultaneously from the same height, then they touch the ground simultaneously.

Statement 2: The time of flight of a freely falling body is independent of its mass.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 11.

Statement 1: The distance between two bodies does not change if they move in the same direction with the same constant acceleration.

Statement 2: Two bodies moving with the same velocity are at rest relative to each other.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 12.

Statement 1: A body is dropped from a height h and another body is thrown simultaneously from the ground with a velocity u in the vertically upward direction. They meet after a time of \(\frac{h}{u}\).

Statement 2: For a body projected in the vertically upward direction, the ascent in the last second is always 4.9 m, whatever the velocity of projection.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

One-Dimensional Motion Match Column A And Column B

Question 1. In column 1 velocity-time graphs of a particle moving along a straight line are given and in column 2 the corresponding acceleration-time graphs are given.

Answer: 1. A, 2. E, 3. D, 4. C

Question 2. In column 1 some statements related to various physical quantities are given while in column 2 information about the motion of a particle moving along a straight line are given.

Answer: 1. C, 2. B, 3. D, 4. A

Question 3. The displacement-time curve of a moving particle is given.

Answer: 1. A, 2. C, 3. B, 4. D

Question 4.

Answer: 1. A, B, 2. A, B, 3. C, 4. A, B

Question 5.

Answer: 1. A, 2. B, E, 3. C, 4. D

Question 6.

Answer: 1. A, B, 2. C 3. A, B, D, 4. C

Question 7.

Answer: 1. C, 2. B, 3. D, 4. A

Question 8. For a moving particle, displacement and velocity at time t are s and v, respectively.

Answer: 1. A, 2. A, 3. D, 4. B

Question 9. Two cars start from the origin and move along the x-axis. Their displacements (in m) are related to time (in s) as, x_A = 4t+ t² and x_B = 2t² + 2t³. Then

Answer: 1. D, 2. B, 3. C, 4. A

One-Dimensional Motion Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. The vertical rise or fall of a particle under gravity is governed by the equations:

1. v = u + gt,
2. h = ut + \(\frac{1}{2}\)gt², and
3. v² = u2² + 2gh, the symbols having their usual meanings. Then for a particle dropped from the top of a tower and falling freely, choose the correct options:

1. The distance covered by it after n seconds is directly proportional to

1. n¹
2. n
3. 2n-1
4. 2n²-1

Answer: 1. n²

2. The distance covered in the nth second is proportional to

1. n²
2. n
3. 2n-1
4. 2n²-1

Answer: 3. 2n-1

3. The velocity of the body after n seconds is proportional to

1. n²
2. n
3. 2n-1
4. 2n²-1

Answer: 2. n

Question 2. Shows the speed-time graph of two cars A and B which are travelling in the same direction over a period of 40 s. Car A travelling at a constant speed of 40 m · s^-1, overtakes car B at t = 0. In order to catch up with car A, car B immediately accelerates uniformly for 20 s.

1. Distance travelled by car B in 20 s is

1. 800 m
2. 750 m
3. 1000 m
4. 500 m

Answer: 2. 750m

2. What is the acceleration of car B for the first 20 s?

1. 1.25 m · s^-2
2. 3.75 m · s^-2
3. 2.5 m · s^-2
4. 5 m · s^-2

Answer: 1. 1.25 m · s^-2

3. At what time car B overtake car A?

1. 12 s
2. 50 s
3. 18 s
4. 25 s

Answer: 4. 25 s

4. What is the distance travelled by car A before car B overtakes it?

1. 480 m
2. 1000 m
3. 800 m
4. 1200 m

Answer: 2. 1000 m

5. What is the maximum separation between the two cars during the 40 s interval?

1. 150 m
2. 480 m
3. 390 m
4. None of these

Answer: 1. 150 m

Question 3. A ball is thrown up with velocity u = 50 m · s^-1 as shown. The origin and positive and negative directions are also indicated in the figure. Neglect air resistance and take g = 10 m · s^-2.

1. How much time does the ball take to reach half the maximum height?

1. 1.46 s
2. 2.5 s
3. 3s
4. 1.82s

Answer: 1. 1.46 s

2. After how much time will it again cross the same position?

1. 7.07 s
2. 7.5 s
3. 8.18 s
4. Would not cross again

Answer: 1. 7.07 s

3. Determine the distance travelled by the ball in the 3rd second of its motion.

1. Zero
2. 45 m
3. 25 m
4. 80 m

Answer: 3. 25

One-Dimensional Motion Integer Answer Type Questions

In this type, the answer to each of the questions is a single-digit integer between 0 and 9.

Question 1. The displacement x of a particle moving in one dimension, under the action of a constant force, is related to time t by the equation t = √x + 3, where x is in meters and t is in seconds. Find the displacement (in metres) of the particle when its velocity is zero.
Answer: 0

Question 2. The motion of a body is defined by \(\frac{dv(t)}{dt}\) = 6-3v(t) where v(t) is the velocity (in m/s) of the body at time t (in seconds). If the body was at rest at t = 0, find its velocity (in m/s) when the acceleration is half the initial value.
Answer: 1

Question 3. A balloon is at a height of 40 m and is ascending with a velocity of 10 m · s^-1. A bag of 5 kg weight is dropped from it. When will the bag reach the surface of the earth? Given g = 10 m · s^-2.
Answer: 4

Question 4. A bike, initially at rest, travels the first 20 m in 4s along a straight line with constant acceleration. Determine the acceleration of the bike in m · s^-2. Consider the bike as a particle.
Answer: 5

Question 5. Starting from rest, a particle moving along a straight line attains a speed of 2 m · s^-1 in 1.5 s. What is the particle’s speed after an additional 3 s has elapsed assuming that the particle is moving with constant acceleration?
Answer: 6