Class 11 Physics

Properties Of Bulk Matter

Expansion Of Solid And Liquids Short Question And Answers

Short Answer Questions on Thermal Expansion

Question 1. The liquid with a co-efficient of volume expansion (γ) is filled in a container of a material having the coefficient of linear expansion a. If the liquid overflows on heating, then

1. γ = 3α
2. γ > 3α
3. γ <3α
4. γ > 3α³

Answer:

Given

The liquid with a co-efficient of volume expansion (γ) is filled in a container of a material having the coefficient of linear expansion a.

Apparent coefficient of volume expansion, γ’= γ-3α

Now, γ’ is positive when the liquid overflows on heating.

So the condition is γ – 3α > 0 or, ϒ > 3α

The option 2 is correct.

Question 2. When a copper sphere is heated, which physical quantity of the sphere will show maximum percentage change?
Answer:

When a copper sphere is heated, its volume will show maximum change in percentage.

Question 3. A metal rod is fixed rigidly at two ends so as to prevent its thermal expansion. If L, a and Y respectively denote the length of the rod, coefficient of linear thermal expansion and Young’s modulus of its material, then for an increase in temperature of the rod by ΔT, the longitudinal stress developed in the rod is

1. Inversely proportional to α
2. Inversely proportional to Y
3. Directly proportional to ΔT/Y
4. Independent of L

Answer:

Given

A metal rod is fixed rigidly at two ends so as to prevent its thermal expansion. If L, a and Y respectively denote the length of the rod, coefficient of linear thermal expansion and Young’s modulus of its material, then for an increase in temperature of the rod by ΔT,

Longitudinal thermal stress = YαΔT.

The option 4 is correct.

Key Concepts in Solid and Liquid Expansion

Question 4. A solid rectangular sheet has two different coefficients of linear expansion α₁ and α₂ along its length and breadth respectively. The coefficient of surface expansion is (for α₁ t<<1 1, α₂ t<<l)

1. \(\frac{\alpha_1+\alpha_2}{2}\)
2. \(2\left(\alpha_1+\alpha_2\right)\)
3. \(\frac{4 \alpha_1 \alpha_2}{\alpha_1+\alpha_2}\)
4. \(\alpha_1+\alpha_2\)

Answer:

Given

A solid rectangular sheet has two different coefficients of linear expansion α₁ and α₂ along its length and breadth respectively.

Let, initial length and breadth of rectangular sheet are a₁ and b₁ respectively.

If the final length and breadth are a₂ and b₂ respectively in increase in temperature t° C, then

⇒ \(a_2=a_1\left(1+a_1 t\right) \text { and } b_2=b_1\left(1+\alpha_2 t\right)\)

∴ \(a_2 b_2=a_1 b_1\left(1+\alpha_1 t\right)\left(1+\alpha_2 t\right)\)

or, \(a_1 b_1(1+\beta t)=a_1 b_1\left\{1+\left(\alpha_1+\alpha_2\right) t+\alpha_1 \alpha_2 t^2\right\}\)

or, \(1+\beta t=1+\left(\alpha_1+\alpha_2\right) t\)

[neglecting the term \(\alpha_1 \alpha_2 t^2\)]

∴ \(\beta=\alpha_1+\alpha_2\)

The option 4 is correct

Real-Life Applications of Thermal Expansion

Question 5. The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100 °C is (For steel, Young’s modulus is 2 x 10¹¹ N • m^-2 and coefficient of linear expansion is 1.1 x 10^-5 K^-1)

1. 2.2x 10⁸ Pa
2. 2.2x 10⁹ Pa
3. 2.2 x 10⁷ Pa
4. 2.2 x 10⁶ Pa

Answer: As length is constant, strain =ΔL/L = αΔQ

Now, pressure = stress = Yx strain

= 2x 10¹¹ x 1.1 x 10^-5 x 100

= 2.2 x 10⁸ Pa

The option 1 is correct.

Question 6. A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the coefficient of linear expansion (α) of the metal of the pendulum shaft are respectively

1. 25°C and 1.85 x 10^-5 l °C
2. 60°C and 1.85 x 10^-4 l °C
3. 30°C and 1.85 x 10^-3 l°C
4. 55°C and 1.85 x 10^-2 l °C

Answer:

Given

A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time,

Increase or decrease in time for change in temperature ΔT,

⇒ \(\Delta t =\frac{1}{2} \alpha \Delta T t\)

∴ 12 = \(\frac{1}{2} \alpha\left(40-T_0\right) \times 1\)

T₀ = temperature at which clock gives accurate breading, t = 1 day]

and 4 = \(\frac{1}{2} \alpha\left(T_0-20\right) \times 1\)….(2)

(1)+(2) we get,

3 = \(\frac{40-T_0}{T_0-20} \quad \text { or, } 3 T_0-60=40-T_0\)

or, \(T_0=\frac{100}{4}=25^{\circ} \mathrm{C}\)

Putting the value of T₀ in equation (1) we get,

⇒ \(15 \alpha=\frac{24}{24 \times 60 \times 60}\)

or, \(\alpha=\frac{1}{15 \times 60 \times 60}=1.85 \times 10^{-5} /{ }^{\circ} \mathrm{C}\)

The option 1 is correct.

Coefficient of Expansion Short Answer Questions

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Question 7. An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by:

1. \(3 P K \alpha\)
2. \(\frac{P}{3 \alpha K}\)
3. \(\frac{P}{\alpha K}\)
4. \(\frac{3 a}{P K}\)

Answer:

Given

An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating

Bulk modulus \(K=\frac{P}{\left(\frac{\Delta V}{V}\right)}\)

∴ \(\frac{\Delta V}{V}=\frac{P}{K}[\Delta V= change in volume ]\)

If increase in temperature Δt brings the cube to its original size, then

∴ \(\Delta V=V \cdot \gamma \Delta t \quad \text { or, } \Delta t=\frac{\Delta V}{V} \cdot \frac{1}{\gamma}=\frac{\Delta V}{V} \cdot \frac{1}{3 \alpha}=\frac{P}{3 K \alpha}\)

The option 2 is correct.

Question 8. Coefficient of linear expansion of brass and steel rods are α₁ and α₂. Lengths of brass and steel rods are l₁ and l₂ respectively. If (l₂ – l₁) is maintained same at all temperatures, which one of the following relations holds good?

1. \(\alpha_1 l_2^2=\alpha_2 l_1^2\)
2. \(\alpha_1^2 l_2=\alpha_2^2 l_1\)
3. \(\alpha_1 l_1=\alpha_2 l_2\)
4. \(\alpha_1 l_2=\alpha_2 l_1\)

Answer:

Given

Coefficient of linear expansion of brass and steel rods are α₁ and α₂. Lengths of brass and steel rods are l₁ and l₂ respectively. If (l₂ – l₁) is maintained same at all temperatures,

Let, lengths of brass rod at temperatures t₁ and t₂ are l₂ and l’₂ respectively.

∴ \(l_1^{\prime}=l_1\left\{1+\alpha_1\left(t_2-t_1\right)\right\}\)

Again, let lengths of steel rod at temperatures t₁ and t₂ are l₂ and l’₂ respectively.

∴ \(l_2^{\prime}=l_2\left\{1+\alpha_2\left(t_2-t_1\right)\right\}\)

According to the question, \(l_2-l_1=l_2^{\prime}-l_1^{\prime}\)

or, \(l_2-l_1=l_2\left\{1+\alpha_2\left(t_2-t_1\right)\right\}-l_1\left\{1+\alpha_1\left(t_2-t_1\right)\right\}\)

or, \(l_2-l_1=l_2+l_2 a_2\left(t_2-t_1\right)-l_1-l_1 a_1\left(t_2-t_1\right)\)

or, \(l_2-l_1=\left(l_2-l_1\right)+\left(l_2-t_1\right)\left(l_2 a_2-l_1 a_1\right)\)

or, \(\left(t_2-t_1\right)\left(l_2 a_2-l_1 a_1\right)=0\)

as \(\left(t_2-t_1\right) \neq 0\)

∴ \(\left(l_2 \alpha_2-l_1 \alpha_1\right)=0 \quad or, l_1 \alpha_1=l_2 a_2\)

The option 3 is correct.

Thermal Expansion Problems with Solutions

Question 9. Show that the coefficient of superficial expansion of a rectangular sheet of the solid is twice its coefficient of linear expansion.
Answer:

Let a₁,b₁ be the length and breadth, respectively, of the rectangular sheet at the initial temperature t₁.

Then, S₁ = a₁b₁ = initial area

Now, the temperature is raised to t₂, where t₂  – t₁ =t

The new values of the length and breadth are, respectively, a₂ and b₂; area S₂ = a₂b₂.

If α be the coefficient of linear expansion, then

⇒ \(a_2=a_1(1+\alpha t) \text { and } b_2=b_1(1+\alpha t)\)

∴ \(S_2=a_2 b_2=a_1 b_1(1+\alpha t)^2=S_1(1+2 \alpha t)\)

(neglecting α²t² as α<<1)

On the other hand, if β be the coefficient of superficial expansion, then

S₂ = S₁(1 +βt)

On comparison, we have β = 2α.

Properties Of Bulk Matter Thermometry

Thermometry deals with the measurement of temperature.
Generally, on touching an object we have some sort of sensation. This sensation is known as thermal sensation which indicates the thermal condition of the object.

1. This thermal sensation enables us to understand
2. Whether a body is hot or cold, and
3. How hot or how cold of a body is compared to other bodies. For example, thermal sensation gives the feeling that the water of a pond is hotter than ice, but is colder than water that has been heated for some time.

Again, if a cold body is in contact with a hot body, the cold one gets gradually hotter and the hot one gets gradually colder. After some time both of the bodies produce the same thermal sensation.

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But thermal sensation often is not completely dependable because,

1. It is unsafe to touch very hot or very cold bodies.
2. The thermal sensation is different for objects that are in contact with one another or in the same environment for a long time. For instance, during the winter a piece of iron feels colder than a piece of wood.
3. Thermal sensation is subjective and may differ from person to person.

• Thus, it is essential to establish a measurement system that can determine the thermal condition of a body accurately and that is independent of a person, place, or environment.
• To achieve this, a law, based on experiences and experimental results, was formulated. This law is the zeroth law of thermodynamics.
• Before discussing the zeroth law of thermodynamics, we should gain a basic notion of two widely used terms in thermal science—thermal contact and thermal equilibrium.

Comparative Analysis of Temperature Scales

Thermal contact: Thermal contact may exist between two objects even when they do not touch each other. For example, a cup of hot tea kept on a table for quite some time ceases to remain hot.

This happens because the cup of tea and the other objects in the room have thermal contact among them and after a while they produce the same thermal sensation. In the study of thermal science, thermal contact is abbreviated as contact.

Thermal equilibrium: In general, on touching different objects and feeling the same thermal sensation, we say that these objects are in thermal equilibrium. For example, the objects kept in the same room are in thermal equilibrium most of the time.

• Let us consider various objects that are not in thermal equilibrium. If these objects are kept in thermal contact of one another, they attain thermal equilibrium on their own.
• For example, a cup of hot tea kept on a table for quite some time is seen to attain thermal equilibrium with other objects in the room.

Properties Of Bulk Matter – The Zeroth Law Of Thermodynamics

Thermodynamics is the branch of physics that discusses the mutual conversion between heat and work and the changes in the physical properties of different bodies caused by this type of conversion.

• Thermodynamics is based on two fundamental laws—the first and the second laws of thermodynamics. After the two laws were formulated and put in use, the need for the identification of a fundamental property of matter was felt. This property is the temperature of a body.
• Hence, a law was to be formulated that would define temperature. It was reasonable that this law should precede the already existing first and second laws of thermodynamics. Thus this law was named the zeroth law of thermodynamics.

Statement of Zeroth law: If two bodies are sepa¬rately in thermal equilibrium with a third body, then the first two bodies are also in thermal equilibrium with each other.

• For example, if A and B are separately in thermal equilibrium with C, then according to the zeroth law, A and B are also in thermal equilibrium with each other.
• This may appear to be an obvious truth and we are tempted to believe that the statement need not be considered as a separate law. But it is not true our experiences dictate that this type of argument does not always hold. For example,

1. if A and B are two iron pieces and C is a magnet, then both A and B will be attracted by C, but A and B will not attract each other.
2. When each of two straight lines is per¬pendicular to a third straight line, the first two are not neces¬sarily perpendicular to each other.

Thus, we can understand that zeroth law is not an obvious truth, but has been formulated as a ‘law’ on the basis of experiments and experiences.

Analogical examples of the zeroth law:

1. a, b and c are three line segments. Out of these lines, if a is equal to c and b is also equal to c, we know from our experience that a and b are equal to each other.

There exists a property whose value is the same for all three line segments. This property is the length of a line segment. Hence, equality in length is the condition for the three line segments to be equal.

2. a, b, and c are three straight lines on a plane. a is parallel to c and b is also parallel to c. We know that a and b will also be parallel to each other, a, b, and c must have the same gradient or slope, and thus, equality in slope is the condition for the lines a, b, and c to be parallel to one another.

WBBSE Class 11 Thermometry Study Notes

Significance Of The Zeroth law: From the analogies discussed above, it can be concluded that there must exist a characteristic property of every body, whose equality dictates thermal equilibrium among the bodies. This characteristic property is the temperature of a body.

Significance Of The Zeroth Law Definition: Temperature is a physical property of everybody, whose equality is the necessary and sufficient condition for thermal equilibrium among the bodies.

According to this definition,

1. All substances kept in a room for a long time are at the same temperature,
2. Temperature of hot water is different from that of cold water,
3. A hot body and a cold body, kept in thermal contact, attain the same temperature after some time, etc.

Properties Of Bulk Matter – Thermometer

An instrument that measures the temperature of a body is called a thermometer.

From zeroth law we see that, to determine the thermal equilibrium between A and B, a third body C can be used as a reference body. In this case C acts as a thermometer.

Temperature scale: A scale of temperature is needed to measure the temperature of a body accurately. To draw up a scale we follow the norm that a hot body is at a higher temperature while a cold body is at a lower temperature.

It implies that when a hot and a cold body are in contact, the temperature of the hot body will decrease and that of the cold body will increase. When the temperatures of both bodies become equal, the bodies attain thermal equilibrium.

Fixed point: To set up a temperature scale, one or more conveniently reproducible, well-established temperatures are chosen as standard temperatures. These fixed tempera¬tures are called the fixed points.

Primary thermometer: There are several kinds of thermometers for practical use. There are liquid thermometers (mercury or alcohol), ideal gas thermometers, platinum-resis-tance thermometers etc. Among these, there is a special kind of thermometer that is used to deduce the accurate value of different fixed points.

• This is known as a primary thermometer. Using these fixed points, other thermometers are calibrated. The universally accepted primary thermometer is the ideal gas thermometer.
• The thermometer contains a gas, under ideal conditions, kept at a constant volume. The property of the gas, that changes with temperature, is its pressure.

As such, pressure is called the thermometric property of the gas. The thermometer is also known as a constant-volume gas thermometer. To fix the ideal gas temperature scale, the temperature (T) is assumed to be proportional to the pres¬sure (p) of the gas. So,

T ∝ p, or, T = kp, where k is an unknown constant.

• To know the value of k, one fixed point should be chosen and its temperature allotted with a definite value. This fixed point, known as the fundamental fixed point, is the triple point of water (discussed below).
• The value of its temperature is universally accepted as 273.16. Later, Kelvin introduced a thermodynamic temperature scale, which exactly coincides with the ideal gas scale.

So the unit of temperature is chosen as kelvin or K; then the temperature of the triple point of water is T₀ = 273.16 K.

Triple point of water: The triple point of water is the state at which ice, water, and water vapour can coexist in thermal equilibirum. At this state, the pressure and the temperature are fixed so it is a fixed point.

• The value of pressure at this fixed point is 4.58 mm of Hg the value of temperature is assigned as 273.16 K in the ideal gas scale (see the chapter ‘Change of State of Matter’).
• Unless a very accurate value of temperature is required, we may use the number 273 in place of 273.16 for the triple point of water. In that case, C = K – 273; so practically there is no difference between the values of the triple point of water and ice point (see the table below).

Ideal gas scale: Scaling of the ideal gas thermometer is done considering the triple point of water as 273.16. Then temperatures of a few more fixed points are measured using this scale. The scale obtained from this is called ideal gas scale. Nowadays, Kelvin (K) is used as the unit instead of degree kelvin in ideal gas scale.

Thermometry Applications in Real Life

The following table lists some of the important fixed points and their temperatures measured and ascertained by an ideal gas thermometer. All fixed points other than the triple point of water are called secondary fixed points.

Secondary thermometer: All thermometers, other than the ideal gas thermometer, are secondary thermome¬ters. They are called secondary because they are calibrated according to the values of temperatures of fixed points already determined by an ideal gas thermometer.

• So all secondary thermometers actually obey the ideal gas temperature scale. However, it should be noted that secondary thermometers should never be regarded as less reliable or less efficient.
• Rather, they are often highly accurate and very easy to use. The most important of them are liquid ther¬mometer and resistance thermometer.

The four scales of temperature—Celcius, Fahrenheit, Reaumur, and Kelvin Fixed points and symbols:

Equivalence of the temperatures recorded in the four different scales mentioned above: Let C, F, R, and K be the temperature of a body as recorded in three different scales viz., Celsius, Fahrenheit, Reaumur, and Kelvin respectively. Then,

⇒ \(\frac{C-0}{100-0}=\frac{F-32}{212-32}=\frac{R-0}{80-0}=\frac{K-273}{373-273}\)

or, \(\frac{C}{5}=\frac{F-32}{9}=\frac{R}{4}=\frac{K-273}{5}\)

Equivalence of the temperature recorded in any two different scales: Let the temperature of a body in scale A be p and that in scale B be q. Then,

⇒ \(\frac{p \text {-lower fixed point in scale } A}{\text { upper fixed point in } A \text {-lower fixed point in } A}\)

= \(\frac{q \text {-lower fixed point in scale } B}{\text { upper fixed point in } B \text {-lower fixed point in } B}\)

 Properties Of Bulk Matter Heat

Heat Definition: The energy transferred from one body to another only because of a difference in temperature 1 between them is called heat.

Heat Discussion:

• It is to be noted that in thermodynamics, temperature has been define, first. Then heat energy is defined based on the difference in temperature.
• Hence, a statement like, “Heat is the cause and temperature is the effect” is not applicable in thermodynamics. In fact, there is no real need for such a simplification.
• The accepted convention for the direction of flow of heat is that, it flows from a body at a higher temperature to a body at a lower temperature.
• The statement, ‘only because of a difference in tempera-ture’, has been used in the definition of heat because difference in other physical properties, between two bodies may cause a flow of other forms of energy. For example, a difference in pressure between two bodies brought in contact causes the transfer of mechanical energy.
• While temperature is an intrinsic property of a body, ‘heat’ is not ‘Heat of a body’ is a meaningless concept. Heat energy manifests itself only when it is transferred from one body to another.
• Hence, heat is energy in transit. The statement, ‘temperature of a body is 20 °C ’, is meaningful. But the ‘heat of a body is 200 cal ’ is meaningless. Instead ‘heat transferred from A to B is 200 cal ’ is a correct statement.
• The amount of heat contained in a body can never be measured. What we measure in calorimetry is the amount of heat absorbed or liberated by a body not the ‘heat content’ of the body.

Thermal equilibrium: When two or more bodies are in thermal contact and the temperature of every one of them is the same, then no heat is exchanged among them. This state is called thermal equilibrium.

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Unit 7 Properties Of Bulk Matter Chapter 4 Thermometry Heat Numerical Examples

Example 1. What is the temperature which has the same value in Celsius and in Fahrenheit scales?
Solution:

The temperature which has the same value in Celsius and in Fahrenheit scales

Let the temperature be x degree. As per the question C = F = x

From the relation for equivalence of temperature scales, we
know \(\frac{C}{5}=\frac{F-32}{9}\)

∴ \(\frac{x}{5}=\frac{x-32}{9}\) or, 9x = 5x- 160

or, 4x = -160

∴ x = -40

∴ -40 °C = -40 °F

Example 2. A thermometer has its lower fixed point and upper fixed point marked as 0.5 and 101 respectively. What is the reading on this thermometer at 30 °C?
Solution:

A thermometer has its lower fixed point and upper fixed point marked as 0.5 and 101 respectively.

Let the reading be t degree.

∴ \(\frac{t-0.5}{101-0.5}=\frac{C}{100} \text { or, } \frac{t-0.5}{100.5}=\frac{30}{100}\)

or, 10t-5 = 301.5 or, t = 30.65 degree

Hence, the faulty thermometer reads 30.65 degree.

Step-by-Step Solutions to Thermometry Problems

Example 3. A faulty thermometer reads -0.5 °C in melting ice and 99 °C in boiling water at the pressure of 747 mm of Hg. What is the correct temperature when the faulty thermometer reads 45 °C? The actual boiling point of water is 99 °C at 734 mm of Hg.
Solution:

A faulty thermometer reads -0.5 °C in melting ice and 99 °C in boiling water at the pressure of 747 mm of Hg.

Actual boiling point of water at the pressure of 760 mm of Hg is 100 °C.

Now, a decrease in pressure by (760 – 734) or 26 mm of Hg decreases the boiling point of water by (100 – 99) = 1 °C.

∴ The decrease in pressure by (760-747) or, 13 mm of Hg decreases the boiling point of water by 1/26 x 13 = 0.5 °C.

Hence, the boiling point of water at 747 mm of Hg should be (100-0.5) = 99.5 °C.

Let the correct temperature be x °C, when the faulty ther-mometer reads 45 °C.

Hence, \(\frac{x}{99.5}=\frac{45-(-0.5)}{99-(-0.5)}=\frac{45.5}{99.5} \quad therefore x=45.5^{\circ} \mathrm{C} \text {. }\)

Temperature of the freezing mixture = -23°C

Example 4. A centimetre scale is attached with a thermometer of uniform bore. The thermometer reads 7.3 cm in melting ice, 23.8 cm in boiling water and 3.5 cm in a freezing mixture. What is the temperature of this freezing mixture in °C?
Solution:

A centimetre scale is attached with a thermometer of uniform bore. The thermometer reads 7.3 cm in melting ice, 23.8 cm in boiling water and 3.5 cm in a freezing mixture.

The lower and the upper fixed points correspond to readings of 7.3 cm and 23.8 cm respectively. The tem­perature of the freezing mixture in this scale corresponds to a scale reading of 3.5 cm.

Let C be the freezing mixture’s temperature in degree Cel­sius.

Temperature of the freezing mixture = – 23.03 °C.

Alternative solution: When the temperature increases from 0 °C to 100 °C, the corresponding change in scale reading = 23.8 – 7.3 = 16.5 cm.

So when the temperature changes by 1 °C, the corresponding changes in scale reading = 16.85/100 = 0.165 cm

Let the temperature of the freezing mixture =-x °C

So change in temperature in Celcius scale = 0 – (-x) = x °C

The corresponding change in scale reading = 0.165x cm.

According to the question,

0. 165X = 7.3-3.5 or, x = 38/0.165 = 23.03

So the temperature of the freezing mixture is -23.03 °C.

Example 5. A substance is heated from 30 °C to 75 °C. What is the change in its temperature on the Fahrenheit scale and on the Kelvin scale?
Solution:

A substance is heated from 30 °C to 75 °C.

Let a temperature be C on the Celsius scale, F on the Fahrenheit scale and T on the Kelvin scale. We can write,

F = 9/5C + 32 ….(1)

and T = C+273 ….(2)

Differentiating equation (1) we get,

⇒ \(\Delta F=\frac{9}{5} \Delta C\)

Here, ΔC = 75 – 30 = 45

∴ ΔF = 9/5 x 45 = 81.

Similarly, by differentiating equation (2) we get, ΔT = ΔC = 45.

Common Thermometry Questions and Answers

Example 6. The graph between Celcius and Fahrenheit temperature of a body is shown. Show that the angle made by the graph with Celsius axis is \(\sin ^{-1} \frac{9}{\sqrt{106}}\)

Solution:

The graph between Celcius and Fahrenheit temperature of a body is shown.

We know, \(\frac{C}{5}=\frac{F-32}{9} \text { or, } C=\frac{5}{9}(F-32) \text { or, } F=\frac{9}{5} C+32\)

This is the equation of a straight line, Here the slope of the line, tanθ = 9/5

∴ OA = \(\sqrt{5^2+9^2}=\sqrt{106}\)

∴ \(\sin \theta=\frac{9}{\sqrt{106}}\)

or, \(\theta=\sin ^{-1} \frac{9}{\sqrt{106}}\)

 

Unit 7 Properties Of Bulk Matter Chapter 4 Thermometry Synopsis

Zeroth law of thermodynamics: If two bodies are separately, in thermal equilibrium with a third body, then the first two bodies are also in thermal equilibrium with each other.

• Temperature is a physical property of any system, whose equality indicates thermal equilibrium between different systems.
• The instrument that measures the temperature of a body is called a thermometer.
• The energy transferred from one body to another due to the difference of temperature only is called heat.
• Thermal equilibrium: When two or more bodies are in thermal contact and the temperature of each body is the same, then no heat is exchanged among them. This state is called thermal equilibrium.

Unit 7 Properties Of Bulk Matter Chapter 4 Thermometry Useful Relations For Solving Numerical Problems

If any temperature in Celsius, Fahrenheit and Kelvin scales be C, F and K respectively, then

⇒ \(\frac{C}{5}=\frac{F-32}{9}=\frac{K-273}{5}\)

Relation of temperature readings in any two scales: If any temperature on one scale be p and that on some other scale be q then,

⇒ \(\frac{p-\text { lower fixed point in the first scale }}{\text { upper fixed point in that scale }- \text { lower fixed point in that scale }}\)

= \(\frac{q-\text { lower fixed point in the second scale }}{\text { upper fixed point in that scale }- \text { lower fixed point in that scale }}\)

Unit 7 Properties Of Bulk Matter Chapter 4 Thermometry Match Column 1 And Column 2

Question 1. Ranges of different thermometers are given below.

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

Unit 7 Properties Of Bulk Matter Chapter 4 Thermometry Comprehension Type Questions

Read the following passage carefully and answer the questions at the end of it.

Question 1. Perhaps the highest temperature material you will ever see is the sun’s outer atmosphere or corona. At a temperature of about 2 x 10⁶ °C or 3.6 x 10⁶ °F, the corona glows with a light that is literally unearthly. But because corona is also very thin, its light is rather faint. You can only see the corona during a total solar eclipse when the sun’s disk is covered by the moon.

1. Is it accurate to say that the corona contains heat?

1. Yes
2. No
3. In particular, conditions, say during solar eclipse, it contain heat
4. None of these

Answer: 1. Yes

2. What is the highest temperature which can be created on earth for a sufficiently long time?

1. 1500°C
2. 2000°C
3. 2500°C
4. 3000 K

Answer: 4. 3000 K

3. To measure high temperatures > 2500°C we use

1. Constant volume gas thermometer
2. Thermocouple
3. Resistance thermometer
4. Pyrometer

Answer: 4. Pyrometer

Unit 7 Properties Of Bulk Matter Chapter 4 Thermometry Integer Type Questions

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

1. A Celsius and a Fahrenheit thermometer are put in a hot bath. The reading of the Fahrenheit thermometer is just 29/5 times the reading on the Celsius thermometer. What is the temperature of the bath in Celsius?
Answer: 8

Chapter 4 Thermometry Multiple Choice Questions

Question 1. At the triple point of water the magnitude of pressure is

1. 4.58 mm of Hg
2. 4.57 mm of Hg
3. 4.59 mm of Hg
4. 4.56 mm of Hg

Answer: 1. 4.58 mm of Hg

Question 2. The temperature at the triple point of water is

1. 273.16 K
2. 273.16 °F
3. 273.16 °C
4. 273 K

Answer: 1. 273.16 K

Question 3. The universally accepted primary thermometer is

1. Liquid thermometer
2. Platinum resistance thermometer
3. Ideal gas thermometer
4. Alcohol thermometer

Answer: 3. Ideal gas thermometer

Step-by-Step Solutions to Thermometry MCQs

Question 4. If the difference between two temperatures in the Kelvin scale is ΔT and that in the Celsius scale is Δt, then—

1. ΔT = Δt
2. ΔT = At+ 273
3. ΔT = At-273
4. None of the above

Answer: 1. AT = At

Question 5. A centigrade and a Fahrenheit thermometer are dipped in boiling water. Now this water is cooled and the reading on the Fahrenheit scale is 140 °F. The decrease in temperature on the Centigrade thermometer is

1. 30 °C
2. 40 °C
3. 60 °C
4. 80 °C

Answer: 2. 40 °C

WBBSE Class 11 Thermometry MCQs

Question 6. A gas thermometer is more sensitive than a liquid thermometer because

1. The expansibility of gas is more than that of a liquid
2. Gas is easily available
3. Gas is comparatively lighter
4. Gas thermometer is the primary thermometer

Answer: 1. Expansibility of gas is more than that of a liquid

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Question 7. A constant volume gas thermometer shows pressure readings of 50cm of Hg and 90cm of Hg at 0°C and 100° C respectively. When the pressure reading is 60cm of Hg, the temperature is

1. 25°C
2. 40°C
3. 15°C
4. 12.5°C

Answer: 1. 25°C

Question 8. On which of the following scales of temperature, the temperature is never negative

1. Celsius
2. Fahrenheit
3. Reaumur
4. Kelvin

Answer: 4. Kelvin

Short Notes on Temperature Scales with MCQs

Question 9. ‘Stem correction’ in platinum resistance thermometers are eliminated by the use of

1. Cells
2. Electrodes
3. Compensating leads
4. None of the above

Answer: 3. Compensating leads

Question 10. The temperature of the sun is measured with

1. Platinum thermometer
2. Gas thermometer
3. Pyrometer
4. Vapour pressure thermometer

Answer: 3. Pyrometer

In these type of questions, more than one options are correct.

Question 11. To measure the temperature say around 400°C. Which of the following thermometers can be used most conveniently?

1. Gas thermometer
2. Mercury thermometer
3. Platinum resistance thermometer
4. Thermocouple thermometer

Answer:

1. Gas thermometer

3. Platinum resistance thermometer

4. Thermocouple thermometer

Thermal Expansion and Temperature Measurement MCQs

Question 12. Reading of temperature may be same on

1. Celsius and Kelvin scale
2. Fahrenheit and Kelvin scale
3. Celsius and Fahrenheit scale
4. All the three scales

Answer:

2. Fahrenheit and Kelvin scale

3. Celsius and Fahrenheit scale

Question 13. Which of the following statements are not true?

1. Size of degree is the smallest on celsius scale
2. Size of degree is smallest on Fahrenheit scale
3. Size of degree is equal on Fahrenheit and kelvin scale
4. Size of degree is equal on celsius and kelvin scale

Answer:

1. Size of degree is the smallest on celsius scale

3. Size of degree is equal on Fahrenheit and kelvin scale

Chapter 4 Thermometry Long Answer Type Questions

Question 1. If a person enters a room at 25 °C, will thermal equilibrium be established?
Answer:

A person enters a room at 25 °C:

Normal body temperature of a human being is 37 °C. The human body maintains this temperature, generating energy through food intake. Hence, thermal equilibrium will not be established unless the temperature of the room rises to 37 °C due to some external reasons.

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Question 2. In inter blankets and quilts warm up after being wrapped around a body. Why?
Answer:

In inter blankets and quilts warm up after being wrapped around a body

The normal temperature of the body is 37 °C. Blankets, quilts etc. are at a lower temperature in winter. After being wrapped, they warm up by taking heat from the body, and after some time thermal equilibrium is established. This means that their temperature also increases to 37 °C.

Question 3. When would two chairs, one made of wood and the other of iron, feel equally hot or cold?
Answer:

When the temperature of the chairs are equal to our body temperature, thermal equilibrium will exist between the body and the two chairs, preventing any heat flow. Then they will appear equally hot or cold on touch.

Question 4. How can a thermometer be used to find out whether the atmospheric pressure is above or below its normal value?
Answer:

Pure water boils at 100 °C when its superincumbent pressure is equal to the normal atmospheric pressure. With the decrease or increase in the superincumbent pressure, its boiling point decreases or increases.

A thermometer determines the boiling point. Boiling point, higher or lower than 100 °C, indicates atmospheric pressure as above or below the normal value (water surface should be open to the atmosphere).

WBBSE Class 11 Thermometry Q&A

Question 5. Is there any thermal equilibrium in the solar system?
Answer:

As the temperatures of the sun, the planets and their satellites are not the same, there is no thermal equilibrium in the solar system.

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Question 6. What happens when water at 80 °C is mixed with water at 20 °C?
Answer:

Heat will be transferred from water at 80 °C to that at 20 °C till thermal equilibrium is achieved. Then both samples would attain an equal temperature somewhere between 20 °C and 80 °C.

Question 7. There are two thermometers in a room. One reads the temperature as 25 degrees and the other as 77 degree. Why is a difference?
Answer:

There are two thermometers in a room. One reads the temperature as 25 degrees and the other as 77 degree

The two thermometers must be measuring the temperatures on two different scales.

As 25 °C = 77 °F, the reading of 25 degree should be in a Celsius thermometer and 77 degrees in a Fahrenheit thermometer.

Chapter 4 Thermometry Very Short Answer Type Questions

Question 1. What is the value of -273°C in Kelvin scale?
Answer: [0 K]

Question 2. What is the value of absolute zero temperature on Fahrenheit scale?
Answer: – 459.4 ° F

Question 3. Write down the physical property of a substance that is defined from the zeroth law of thermodynamics.
Answer: Temperature

Question 4. If two bodies are in thermal equilibrium, then their temperature must be equal’. State whether the statement is true or false.
Answer: True

Question 5. What is the name of the temperature-measuring instrument?
Answer: Thermometer

Question 6. Due to temperature difference only, the energy transferred from one body to another is called _______
Answer: Heat

Thermal Expansion and Temperature Measurement Questions

Chapter 4 Thermometry 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 n is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: Fahrenheit is the smallest unit of measuring temperature.

Statement 2: Fahrenheit was the first temperature scale used for measuring temperature.

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

Question 2.

Statement 1: The temperature at which Centigrade and Fahrenheit thermometers read the same is -40°.

Statement 2: There is no relation between Fahrenheit and Centigrade temperature.

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

Long Answers on Temperature Scales with Q&A

Question 3.

Statement 1: Degree Fahrenheit is the smallest unit for measuring temperature.

Statement 2: Fahrenheit was the first temperature scale used for measuring temperature.

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

Question 4.

Statement 1: Two bodies at different temperatures, if brought in contact do not necessary settle to the mean temperature.

Statement 2: The two bodies may have different thermal capacities.

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

Question 5.

Statement 1: Water is considered unsuitable for use in thermometers.

Statement 2: This is due to small specific heat and high thermal conductivity.

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

Unit 1 Physical World And Measurement

• Chapter 1 Measurement and Dimensions Of Physical Quantity

Unit 2 Kinematics

• Chapter 1 One Dimensional Motion
• Chapter 2 Vector

Unit 3 Laws of Motion

• Newton’s Laws of Motion
• Friction
• Circular Motion

Unit 4 Work, Energy and Power

• Chapter 1 Work and Energy

Unit 5 Motion of System of Particles And Rigid Body

• Chapter 1 Statics
• Chapter 2 Rotation of Rigid Bodies

Unit 6 Gravitation

• Newtonian Gravitation and Planetary Motion

Unit 7 Properties of Bulk Matter

• Chpater 1 Elasticity
  • Elasticity Questions and Answers
• Chapter 2 Hydrostatics
  • Hydrostatics Question And Answers
• Chapter 3 Viscosity and Surface Tension
  • Surface Tension Viscosity Definition and Examples
  • Bernoulli’s Principle: Formula, Derivation and Application
• Chapter 4 Thermometry
  • Thermometry Question and Answers
• Chapter 5 Expansion of Solids and Liquids
• Chapter 6 Expansion of Gases
• Chapter 7 Calorimetry
• Chapter 8 Change of State of Matter
• Chapter 9 Transmission of Heat

Unit 8 Thermodynamics

• Chapter 1 First and Second Law of Thermodynamics

Unit 9 Behaviour of Perfect Gas and Kinetic Theory

• Chapter 1 Kinetic Theory of Gases

Unit 10 Oscillation and Waves

• Chapter 1 Simple Harmonic Motion
• Chapter 2 Nature of Vibration
• Chapter 3 Wave Motion
• Chapter 4 Superposition of Waves
• Chapter 5 Doppler Effect in Sound

Viscosity And Surface Tension Rate Of Flow Of A Liquid And Continuity

Rate of flow Of liquid: For streamline flow of a perfectly incompressible liquid, the amount of liquid flowing through any cross section of a tube in a given time interval remains constant.

• The rate of flow of a liquid through a tube means the volume of liquid flowing through any cross-section of the tube per second.
• Suppose a liquid flows through a tube of cross-sectional area a with a uniform velocity ν. The volume of liquid flowing through any cross-section of the tube per second is equal to the volume of a cylinder of length v and cross-sectional area α.

∴ Volume of liquid flowing per second = the rate of flow of the liquid = velocity of flow x area of cross section of the tube = νa

Bernoulli’s Principle Formula

Therefore, the mass of liquid flowing per second = velocity of flow x area of cross-section of the tube x density of the liquid = ναρ [ρ = density of the liquid]

Continuity of flow: For a streamlined flow of a fluid (liquid or gas) through a tube, the mass of the fluid flowing per second through any cross-section of the tube remains constant. This is known as the continuity of flow.

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Equation Of continuity: Let us consider two sections A and B of a tube having cross-sectional areas a₁ and a₂ respectively. The velocities of the fluid at sections A and B are ν₁ and ν₂, and its densities are ρ₁ and ρ₂ respectively.

The mass of fluid flowing through section A per second = \(v_1 \alpha_1 \rho_1\) and the mass of fluid flowing through section B per second = \(v_2 \alpha_2 \rho_2\)

For streamline flow, the fluid enters through section A and leaves through section B, and does not remain stored in the region between A and B, hence

⇒ \(v_1 \alpha_1 \rho_1\) = \(v_2 \alpha_2 \rho_2\) …..(1)

The product vαρ is the mass flow rate. If the fluid is incompressible (like a liquid), then its density is constant, and in that case ρ₁ = ρ₂.

∴ \(v_1 \alpha_1\) = \(v_2 \alpha_2\)…….(2)

or, να = constant …..(3)

Equations (2) and (3) are known as the equations of continuity of liquid flow.

∴ \(\nu \propto \frac{1}{\alpha},\) which means that the velocity of liquid flow through any cross-section of a tube is inversely proportional to its cross-sectional area.

The equations of continuity essentially express the law of conservation of mass.

Bernoulli’s Principle Formula

Energy of Liquid in Streamline Flow: At any point inside a flowing liquid, there are three forms of energy

1. Kinetic energy,
2. Potential energy and
3. Energy due to pressure.

1. Kinetic energy: If mass m of a liquid flows with a velocity v, then the kinetic energy of that liquid = = \(\frac{1}{2} m v^2\).

Kinetic energy per unit mass = \(\frac{1}{2}v^2\)

Kinetic energy per unit volume = \(\frac{1}{2} \frac{m}{V} v^2\) [volume of the liquid]

= \(\frac{1}{2} \rho v^2\left[\rho=\frac{m}{V}=\text { density of the liquid }\right]\)

2. Potential energy: If mass m of a liquid is at a height h above the surface of the earth, then the potential energy of that liquid = mgh.

Potential energy per unit mass = gh

Potential energy per unit volume = \(\frac{m g h}{V}\) = ρgh.

Bernoulli’s Principle Formula

3. Energy due to pressure: if a liquid is under the action of some applied pressure, then it acquires some energy and this energy is known as energy due to pressure. The liquid can perform work by expending this energy. Let some liquid of density ρ whose free surface is PQ be kept in a container.

A narrow side tube AB of cross sectional area α is attached near the bottom of the container. This tube is fitted with a piston P, which can move freely along the tube. If the pressure of the liquid at rest along the axis of the narrow tube is p, then the force acting on the piston = pα.

If the piston is slowly pushed inside the tube through a distance x, then work done = pαx. As a result, liquid of volume αx or mass αxρ enters the container. Since the piston is moved slowly, the liquid acquires negligible velocity and hence it will possess no kinetic energy.

So the work done pax remains stored as potential energy in mass αxρ of the liquid that has entered the container. This energy is called the energy due to pressure for the liquid.

The energy due to pressure per unit mass of the liquid = = \(\frac{p a x}{a x \rho}=\frac{p}{\rho} .\)

∴ The energy due to pressure per unit volume of the liquid = \(\frac{p a x}{a x}=p\).

Bernoulli’s Theorem: The Swiss mathematician Daniel Bernoulli established a law for the streamline flow of an ideal fluid (which is incompressible and non-viscous). This law is known as Bernoulli’s theorem. It is an important theorem in Hydrodynamics.

Bernoulli’s Principle Formula

Statement Of the theorem: For a streamline flow of an ideal liquid, the sum of the potential energy, kinetic energy, and energy due to pressure per unit volume of the liquid always remains constant at every point on the I streamline.

If the kinetic energy per unit volume of the liquid = \(\frac{1}{2} \rho v^2\); potential energy = pgh and energy due to pressure = p, then

⇒ \(\frac{1}{2} \rho v^2+\rho g h+p\) = constant

or, \(\frac{1}{2} v^2+g h+\frac{p}{\rho}\) = constant …..(1)

This is the mathematical form of Bernoulli’s theorem. Dividing equation (1) by g, we get,

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

This also is a form of Bernoulli’s theorem. Here, \(\frac{v^2}{2 g}\) is called the velocity head, h is the elevation head, and \(\frac{p}{\rho g}\) is the pressure head. Each of these heads has the dimension of length.

So, velocity head + elevation head + pressure head
= constant …..(3)

According to relation (3), Bernoulli’s theorem can also be stated as follows.

For a streamline flow of an ideal liquid, the sum of the velocity head, elevation head and pressure head always remains constant at any point in the liquid.

Bernoulli’s theorem is based on the law of conservation of energy for the streamline motion of an ideal fluid. The theorem states that energy remains conserved along any streamline.

Bernoulli’s Theorem Derivation

When the flow of liquid is horizontal, the height of each point in the liquid is assumed to be the same, i.e., h = constant. We can rewrite equation (2) as,

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

or, \(p+\frac{1}{\mathrm{a}} \rho v^2\) = constant

Hence, in the horizontal flow of a liquid, the sum of pressure and kinetic energy per unit volume of the liquid at any point is constant. This implies that where the velocity of flow is high, the pressure is low and vice-versa.

Applications of Bernoulli’s theorem

1. Velocity of efflux of a liquid and Torricelli’s theorem: If a small hole is present on the wall of a deep container containing liquid, then the velocity with which the liquid comes out through that small hole is called the velocity of efflux of the liquid.

In Fig, a liquid kept in a large container is emerging with velocity ν through the small hole on the wall of the container. The height of the free surface of the liquid above the hole is h and the depth of the liquid below the hole is h₁.

The total depth of the liquid H = h+ h₁. Let us consider a point B just outside the hole and another point A on the surface of the liquid. Atmospheric pressure p acts on A and B.

If the container is large and the hole is very small, then the free surface of the liquid will come down so slowly that the velocity of the free surface of the liquid would seem to be almost zero.

Bernoulli’s Theorem Derivation

If we imagine a tube of flow starting from the free surface of the liquid and ending at the point B and apply Bernoulli’s theorem in that tube of flow, then

⇒ \(0+H+\frac{p}{\rho g}=\frac{v^2}{2 g}+h_1+\frac{p}{\rho g}\)

or,\(\frac{v^2}{2 g}=H-h_1=h\)

or, \(v^2=2 g h\)

or, \(v=\sqrt{2 g h}\)

• This is the velocity of efflux of a liquid through a small hole and it is known as Torricelli’s formula. According to this formula, the velocity of efflux of a liquid is the same as that of a body falling freely under gravity through a height h. So, Torricelli’s theorem can be stated as follows:
• The velocity of efflux of a1 liquid through any small hole or orifice is equal to that acquired by a body falling freely from rest under gravity from the free surface of the liquid to the level of the small hole.
• It should be mentioned that this ideal velocity cannot be attained by any liquid in reality because no liquid is non- viscous. It should be remembered that in Bernoulli’s theorem the effect of viscosity of the liquid has been neglected.

Horizontal range: Let, the first drop of liquid emerged from the orifice touches the ground at a distance x after time t. That means, the vertical displacement of the liquid drop is h₁.

Now, considering the motion of the liquid along the vertical direction,

initial velocity = 0, acceleration = g

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Bernoulli’s Theorem Derivation

From the equation h = ut+1/2gt² we get, h = 0 + 1/2gt²

∴ t = \(\sqrt{\frac{2 h_1}{g}}\)

Again, considering the motion of the liquid aong the horizontal direction, the initial velocity, ν = √2gh, acceleration = 0, time = t.

∴ Horizontal range, \(x=v t=\sqrt{2 g h} \times \sqrt{\frac{2 h_1}{g}}=2 \sqrt{h h_1}\)

2. Venturimeter: A venturimeter is used to measure the rate of flow of liquid through a tube. Its working principle is based on Bernoulli’s theorem.

• Fig shows the action of a venturimeter. The two ends of this tube are equally wide and the middle portion is narrow. Liquid flows through this tube in streamlines. The tube is kept horizontal.
• When a liquid flows through a venturimeter, the velocity of the liquid increases at the narrow part of the tube with consequent decrease in pressure. This decrease in pressure is measured with the help of two vertical tubes attached at the wide and the narrow parts of the venturimeter.

Let the velocity of the liquid at the wider part of the tube be ν₁ and the pressure be p₁ At the narrower part of the tube, the velocity of the liquid is ν₁ and the pressure is p₂.

According to Bernoulli’s theorem, \(\frac{v_1^2}{2 g}+\frac{p_1}{\rho g}=\frac{v_2^2}{2 g}+\frac{p_2}{\rho g}\)

the elevation head, h₁=h₂ since the tube is horizontal.

∴ \(\frac{p_1-p_2}{\rho g}=\frac{1}{2 g}\left(v_2^2-v_1^2\right) \text { or, } p_1-p_2=\frac{\rho}{2}\left(v_2^2-v_1^2\right)\)

or, \(h \rho g=\frac{\rho}{2}\left(v_2^2-v_1^2\right)\)

[h = difference in liquid levels in the vertical tubes attached to the venturimeter]

∴ h = \(\frac{1}{2 g}\left(v_2^2-v_1^2\right)\)

If the cross-sectional areas of the wide and the narrow parts of the venturimeter are α₁ and α₂ respectively, then according to the equation of continuity, we get,

⇒ \(\alpha_1 v_1=\alpha_2 v_2 \text { or, } \frac{v_1}{v_2}=\frac{\alpha_2}{\alpha_1}\)

∴ h = \(\frac{v_2^2}{2 g}\left(1-\frac{v_1^2}{v_2^2}\right)=\frac{v_2^2}{2 g}\left(1-\frac{\alpha_2^2}{\alpha_1^2}\right)\)

or, \(v_2^2=2 g h \cdot \frac{\alpha_1^2}{\alpha_1^2-\alpha_2^2}\)

or, \(v_2=\frac{\alpha_1}{\sqrt{\alpha_1^2-\alpha_2^2}} \cdot \sqrt{2 g h}\) …..(1)

Therefore, the volume of liquid flowing out per second,

Venturimeter Derivation Class 11

V = \(\alpha_2 v_2=\frac{\alpha_1 \alpha_2 \sqrt{2 g h}}{\sqrt{\alpha_1^2-\alpha_2^2}}=\alpha_1 \alpha_2 \sqrt{\frac{2 g h}{\alpha_1^2-\alpha_2^2}}\)…….(2)

So, when α₁ and α₂ are known, by measuring h we can determine the rate of flow of the liquid through the tube with the help of equation (2).

3. Pitot tube: A pitot tube is also used to measure the rate of flow of liquids. Its working principle is similar to that of a venturimeter. Its action also depends on Bernoulli’s theorem.

• In this instrument, two tubes AB and CED, open at both ends are introduced vertically and side by side inside the liquid. The open end B of the tube AB remains parallel to the flow of the liquid. The DE part of the tube CED is so bent that the opening D faces the flowing liquid normally.
• The height of the liquid column in the tube AB expresses the pressure of the liquid at the point B. Since the liquid flow is obstructed at the portion DE of the tube CED, the velocity of flow at point D is zero.

The difference in the liquid levels in the two tubes = h.

Let the velocity of liquid flow be ν.

The points B and D lie on the same horizontal plane; therefore, according to Bernoulli’s theorem,

⇒ \(\frac{1}{2 g} v^2+\frac{p_B}{\rho g}=0+\frac{p_D}{\rho g}\) [ρ= density of the liquid]

or, \(\frac{1}{2 g} v^2=\frac{p_D-p_B}{\rho g}\)

or, \(\frac{1}{2} v^2=\frac{h \rho g}{\rho} v^2=2 g h or, v=\sqrt{2 g h}\)

Pressure and Velocity Relationship in Bernoulli’s Principle

Venturimeter Derivation Class 11

If the cross-section of the pipe where the two tubes are placed is a, then the volume of liquid flowing per second through that section, V = \(\alpha v=\alpha \sqrt{2 g h} .\)

When an aeroplane is in motion, the velocity of air currents can be determined with the help of a pitot tube.

4. Sprayer or atomizer: A sprayer or atomizer is used for spraying water, insecticides, etc. Its action also depends on Bernoulli’s theorem.

• The liquid to be sprayed is kept in a container A and its mouth is closed with the help of a cork or cap. A narrow tube B passes through the cap of the container. C is another tube through which air is blown. The tube C has a narrow tip.
• When air comes out from this narrow tip O with a high velocity, pressure at O decreases. Since O lies just above the open end of the tube B, the liquid rises through the tube B due to this low pressure, and as it meets the high-velocity air coming out of the tube C, it sprays out in the form of fine droplets.

Explanation of Some Phenomena with Bernoulli’s Theorem

1. It is not safe to stand near a fast-moving train: Due to the very high speed of the train, the air near the train also flows at a very high speed. As a result, pressure in that region decreases compared to the air pressure of the surrounding region. This excess surrounding pressure behind a person pushes him towards the train and may cause a serious accident.

2. The tin roof of a house is sometimes blown off during a storm: Since the velocity of the wind above the roof is very high, pressure becomes very low. The air inside the room is still and so the higher pressure from inside pushes the roof upwards and hence the roof may be lifted and blown off with the wind.

3. Two boats or ships moving side by side have a tendency to come closer: The speed of water in the narrow gap between boats or ships is greater than the speed of water on the other sides of the vessels. So, the pressure in that narrow region decreases. As a result, due to higher water pressure on the other sides of the boats or ships, they experience a lateral force and, hence, come closer.

Venturimeter Derivation Class 11

4. Flying in air of typical-shaped objects: Let us take an object moving through air towards right.

• Its lower surface is flat, but the upper surface is oval-shaped. The relative motions of the streamlines of air moving above and below it are shown by arrows.
• Clearly, the upper streamline traverses a greater distance in any fixed interval of time; so its velocity is higher. Then, according to Bernoulli’s theorem, the air pressure above the object is less than that below it.
• As a result, a net upward pressure acts on the object. This helps the object to fly through air, provided its weight is sufficiently low. This is one of the principles utilised to fly an aeroplane.

5. Magnus effect: When a spinning ball is thrown horizontally with a large velocity, it deviates from its usual parabolic path of spin free motion. This deviation can be explained on the basis of Bernoulli’s principle.

• When a ball moves forward, the air ahead the ball, moving with velocity v (say), rushes to fill up the vacant space behind the ball left evacuated by the motion of the ball.
• As the ball spins, the layer of air surrounding the ball also moves with the ball at a velocity u (say). From the fig, it can be stated that the resultant velocity of air above the ball becomes (v+ u) while that below that ball is (v- u).
• This difference in the velocities of air results in the pressure difference between the lower and upper faces of the ball. This pressure difference exerts a net upward force on the ball due to which it moves along a curved path as shown in Fig.
• If the spin of the ball is opposite to that shown in the Fig. a net downward force will act on it, deviating it from its original path. This effect is known as Magnus effect. If the surface of the ball is rough, more air is dragged and the path of the ball becomes more curved.

6. Blood flow and heart attack: An artery may get constricted due to the accumulation of plaque on its inner walls. In order to drive blood through this constriction the speed of the flow of blood is increased.

• This increased velocity lowers the blood pressure in the constricted region and the artery may collapse due to the external pressure.
• As a result, the heart exerts more pressure to open this artery and forces the blood through. As the blood rushes through the opening, the internal pressure once again drops due to some reasons leading to a repeat collapse which results in heart attack.

Venturimeter Derivation Class 11

Rate Of Flow Of A Liquid Numerical Examples

Example 1. The pressure at an orifice situated at the lower side of a vessel filled with a liquid is greater than the atmospheric pressure by 9.8 x 10³ N · m^-2. What is the velocity of efflux if the density of the liquid is 2500 kg · m^-3? [g = 9.8 m · s^-2]
Solution:

Given

The pressure at an orifice situated at the lower side of a vessel filled with a liquid is greater than the atmospheric pressure by 9.8 x 10³ N · m^-2.

Velocity of efflux of the liquid, v = √2gh

According to the problem, hρg = 9.8 x 10³

or, \(g h=\frac{9.8 \times 10^3}{2500}\)

∴ v = \(\sqrt{\frac{2 \times 9.8 \times 10^3}{2500}}=2.8 \mathrm{~m} \cdot \mathrm{s}^{-1} .\)

Step-by-Step Guide to Bernoulli’s Equation

Example 2. Water is flowing through a horizontal tube of unequal cross-section. At a point where the velocity of water is 0.4 m · s^-1, the pressure is 0.1 m Hg. What is the pressure at a point where the velocity of water is 0.8 m · s^-1? The density of mercury = 13.6 x 10³ kg · m^-3.
Solution:

Given

Water is flowing through a horizontal tube of unequal cross-section. At a point where the velocity of water is 0.4 m · s^-1, the pressure is 0.1 m Hg.

According to Bernoulli’s theorem,

⇒ \(\frac{1}{2} v_1^2+\frac{p_1}{\rho}=\frac{1}{2} v_2^2+\frac{p_2}{\rho}\) (since the tube is horizontal)

or, \(p_2=p_1+\frac{1}{2} \rho\left(v_1^2-v_2^2\right)= 0.1 \times\left(13.6 \times 10^3\right) \times 9.8\) + \(\frac{1}{2} \times 10^3 \times\left\{(0.4)^2-(0.8)^2\right\}\)

= \(13088 \mathrm{~Pa}=\frac{13088}{\left(13.6 \times 10^3\right) \times 9.8} \mathrm{~m} \mathrm{Hg}\)

= 0.0982 m Hg

Example 3. A pitot tube is connected to a main pipeline of diameter 16 cm. The difference in height of the water columns in the two arms of the tube is 10 cm. Determine the rate of flow of water through the main pipe.
Solution:

Given

A pitot tube is connected to a main pipeline of diameter 16 cm. The difference in height of the water columns in the two arms of the tube is 10 cm.

Volume of water flowing through the pipe per second,

V = αv [v = velocity of water]

V = \(\alpha \sqrt{2 g h} \text { (Here, } \alpha=\pi(8)^2=64 \pi \mathrm{cm}^2, h=10 \mathrm{~cm} \text { ) }\)

∴ V = \(64 \pi \sqrt{2 \times 980 \times 10}=64 \pi \times 140\)

= 2.8 x 10⁴ cm³ = 0.028 m³.

The rate of flow of water through the main pipe = 0.028 m³.

Venturimeter Derivation Class 11

Example 4. A venturimeter has been connected between two points of a pipe. The radii of the pipe at these two points are 5 cm and 3 cm respectively. The difference in pressure between these points is equal to that of a water column of height 5 cm. Determine the rate of flow of water through the pipe.
Solution:

Given

A venturimeter has been connected between two points of a pipe. The radii of the pipe at these two points are 5 cm and 3 cm respectively. The difference in pressure between these points is equal to that of a water column of height 5 cm.

The rate of flow of water,

V = \(\alpha_1 \alpha_2 \sqrt{\frac{2 g h}{\alpha_1^2-\alpha_2^2}}\)

Here, \(\alpha_1=\pi(5)^2=25 \pi \mathrm{cm}^2, \quad \alpha_2=\pi(3)^2=9 \pi \mathrm{cm}^2\), h=5 \(\mathrm{~cm}, g=980 \mathrm{~cm} \cdot \mathrm{s}^{-2}\)

= \(25 \pi \cdot 9 \pi \sqrt{\frac{2 \times 980 \times 5}{(25 \pi)^2-(9 \pi)^2}} \approx 3000.6 \mathrm{~cm}^3 \cdot \mathrm{s}^{-1} .\)

Example 5. Water from a tap falls vertically with a velocity of 3 m · s^-1. The area of the cross-section of the mouth of the tap is 2.5 cm². If the flow of water is uniform and steady throughout, then determine the cross-sectional area of the tube of flow of water at a depth of 0.8 m from the mouth of the tap. [g = 10 m · s^-2]
Solution:

Given

Water from a tap falls vertically with a velocity of 3 m · s^-1. The area of the cross-section of the mouth of the tap is 2.5 cm². If the flow of water is uniform and steady throughout

The area of cross-section of the mouth of the tap, A₁ = 2.5 cm², and velocity of water flow there, v₁ = 3 m · s^-1.

Let the area of cross-section of the tube of flow of water at a depth of 0.8 m below the tap be A₂ and the velocity of water flow there be v₂

∴ \(v_2^2=v_1^2+2 g hp\)

= (3)² + 2 x 10 x 0.8

= 9 + 16 = 25

or, v₂ = 5 m · s^-1

We know that, A₁ V₁ = A₂ V₂

or, \(A_2=\frac{A_1 v_1}{v_2}=\frac{2.5 \times 3}{5}=1.5 \mathrm{~cm}^2 .\)

Bernoulli’s Principle in Everyday Life

Example 6. Water flows in streamline motion through a vertical tube of non-uniform cross-section. The radius of the tube at point P and at point Q are 1 cm and  0. 5 cm respectively. Point Q is at a distance of 40 cm from point P. The pressure difference between these points is 39.3 cm of water. Determine the rate of flow of water through the tube.

Solution:

Given

Water flows in streamline motion through a vertical tube of non-uniform cross-section. The radius of the tube at point P and at point Q are 1 cm and  0. 5 cm respectively. Point Q is at a distance of 40 cm from point P. The pressure difference between these points is 39.3 cm of water.

Accoring to the equation of continuity,

⇒ \(A_1 v_1=A_2 v_2\)

∴ \(v_2^2=\frac{A_1^2}{A_2^2} v_1^2\)

According to Bernoulli’s theorem,

⇒ \(\frac{v_1^2}{2 g}+h_1+\frac{p_1}{\rho g}=\frac{v_2^2}{2 g}+h_2+\frac{p_2}{\rho g}\)

or, \(\frac{v_2^2-v_1^2}{2 g}=\left(h_1-h_2\right)+\left(\frac{p_1-p_2}{\rho g}\right)\)

or, \(\frac{v_2^2-v_1^2}{2 g}=40+39.3 \frac{\rho g}{\rho g}=79.3 \mathrm{~cm}\)

∴ \(\frac{v_1^2}{2 g}\left[\frac{A_1^2}{A_2^2}-1\right]=79.3\)

or, \(\frac{v_1^2}{2 g}\left[\frac{(\pi)^2}{(0.25 \pi)^2}-1\right]=79.3\)

or, \(v_1=101.79 \mathrm{~cm} / \mathrm{s}\)

The rate of flow of water = \(A_1 v_1=101.79 \times \pi\)

=319.78 cm³

The rate of flow of water through the tube =319.78 cm³

Real-World Applications of Bernoulli’s Principle

Example 7. A liquid of density 1000 kg/m³ is flowing in streamline motion through a tube of the non-uniform cross-section. The tube is inclined with the ground, The area of cross sections at points P and Q of the tube are 5 x 10^-3 m² and 10 x 10^-3 m² respectively. The height of the points P and Q from the ground are 3 m and 6 m respectively. The velocity of the liquid at point P is 1 m/s. Calculate the work done per unit volume due to

1. the pressure and
2. gravitational force for the flow of liquid from point P to point Q.

Solution:

Given

A liquid of density 1000 kg/m³ is flowing in streamline motion through a tube of the non-uniform cross-section. The tube is inclined with the ground, The area of cross sections at points P and Q of the tube are 5 x 10^-3 m² and 10 x 10^-3 m² respectively. The height of the points P and Q from the ground are 3 m and 6 m respectively. The velocity of the liquid at point P is 1 m/s.

From equation  of continuity, \(A_1 v_1=A_2 v_2\)

or, \(v_2=\left(\frac{A_1}{A_2}\right) v_1=\left(\frac{5 \times 10^{-3}}{10 \times 10^{-3}}\right) \cdot(1)=\frac{1}{2} \mathrm{~m} / \mathrm{s}\)

Applying Bernoulli’s theorem we get,

⇒ \(p_1+\frac{1}{2} \rho v_1^2+\rho g h_1=\rho_2+\frac{1}{2} \rho v_2^2+\rho g h_2\)

or, \(\rho_1-\rho_2=\rho g\left(h_2-h_1\right)+\frac{1}{2} \rho\left(v_2^2-v_1^2\right)\)……….(1)

1. Work done per unit volume of the liquid due to the pressure of the streamline flow from P to Q is,

⇒ \(W_p =p_1-p_2\)

⇒ \(W_p =\rho g\left(h_2-h_1\right)+\frac{1}{2} \rho\left(\nu_2^2-v_1^2\right)[\text { from equation (1)] }\)

= \(\left[(1000)(9.8)(6-3)+\frac{1}{2}(1000)\left(\frac{1}{4}-1\right)\right]\)

= \(\left[3 \times 9.8-\frac{3}{8}\right] \times 10^3=29025 \mathrm{~J} / \mathrm{m}^3\)

2. Work done due to gravitational force for the streamline motion from P to Q is,

W_g = ρg(h₁ – h₂) = 1000 x 9.8 x (3 – 6)

= -29400 J/m³

Example 8. A big container is kept on a horizontal surface of uniform area of cross-section, A. Two non-viscous liquids of densities d and 2d which are not mixed with each other and not compressed, are kept in the container. The height of both liquid column is H/2 and the atmospheric pressure on the open surface of the liquid is p₀. A small orifice is created at a height h from the bottom of the container,

1. Calculate the initial velocity of efflux of the liquid through the orifice,
2. Calculate the horizontal distance x at which the first liquid drop emerged from the orifice will reach,
3. What will be the value of h if the value of the horizontal distance x to be maximum x_m? Also, calculate the value of x_m neglecting the air resistance.

Solution:

Given

A big container is kept on a horizontal surface of uniform area of cross-section, A. Two non-viscous liquids of densities d and 2d which are not mixed with each other and not compressed, are kept in the container. The height of both liquid column is H/2 and the atmospheric pressure on the open surface of the liquid is p₀. A small orifice is created at a height h from the bottom of the container,

1. Let, the initial velocity of efflux of the liquid = v.

According to Bernoulli’s theorem,

⇒ \(p_0+d g\left(\frac{H}{2}\right)+2 d g\left(\frac{H}{2}-h\right)=p_0+\frac{1}{2}(2 d) v^2\)

or, \(v^2=\left(\frac{H}{2}+\frac{2 H}{2}-2 h\right) g or, v=\sqrt{(3 H-4 h)_2^g}\)

2. The time required to reach the ground of the first liquid drop emerged from the orifice is,

t = \(\sqrt{\frac{2 h}{g}}\)

∴ The horizontal distance traversed by the liquid is,

x = vt = \(\sqrt{(3 H-4 h)_2^g} \sqrt{\frac{2 h}{g}}\)

= \(\sqrt{h(3 H-4 h)}\)

3. The condition for x to be maximum (x_m):

x = \(\sqrt{h(3 H-4 h)}=\sqrt{-\left(4 h^2-3 h H\right)}\)

= \(\sqrt{\left\{(2 h)^2-2 \cdot 2 h \cdot \frac{3}{4} H+\left(\frac{3}{4} H\right)^2-\left(\frac{3}{4} H\right)^2\right\}}\)

= \(\sqrt{\frac{9}{8} H^2-\left(2 h-\frac{3}{4} H\right)^2}\)

∴ x = \(x_m \text { when } 2 h-\frac{3}{4} H=0\)

Hence, h= \(\frac{3}{8}H\)

∴ \(x_m=\sqrt{\frac{9}{8} H^2}=\frac{3}{2 \sqrt{2}} H\)

Surface Tension Viscosity

Definition of Surface Tension and 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.

Read And Learn More WBCHSE Solutions for Class 11 Physics

• 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.

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

The greater the viscosity of a liquid, the lesser is its mobility. For example, the viscosity of honey is more than that of water and hence honey flows much slower than water. Coal tar has the least mobility.

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. Velocity profile for flow above a horizontal surface is shown in Fig.

Viscosity Measurement Methods

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 centre 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 2D graph).

Coefficient of Viscosity: Let PQ be a solid horizontal surface. A liquid is in streamline 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

Viscosity Explained with Examples

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 streamline 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.

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 coeffcient of viscocity: \(\eta=\frac{F}{A \frac{d v}{d x}}=\frac{F d x}{A d \nu}\)

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.

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Applications of Surface Tension in Industry

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.

• But 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 rise in temperature. 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

Real-Life Applications of Viscosity

Critical velocity: The maximum velocity of a fluid, up to which the flow of the fluid is streamlined and beyond which the flow becomes turbulent, is regarded as the critical velocity for that fluid.

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 Reynolds number is less than 2300. So, if the value of 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 Reynolds number will be greater than 2300. If Reynolds number is greater than 2300, then the flow will be turbulent.

Dimension of Reynolds number: From equation (2) we get, 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 streamline flow of a fluid gradually changes into 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 streamline flow of fluid gradually changes into 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 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:

Given

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.

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. But the viscosity of the fluid opposes this relative motion.
• The opposing force due to viscosity increases with increase in the velocity of the body due to the gravitational acceleration g. If the body is small in size, then after an interval of 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.

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:

Given

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.

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]

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

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

The terminal velocity of the oil drop = \(1.14 \times 10^{-4} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

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:

Given

The density of the liquid is 1.47 g · cm^-3. Ignore the density of air

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

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]

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

The coefficient of viscosity of the liquid = 1524.4 poise. 

Hydrostatics Fluid

Hydrostatics Basics and Definitions

Fluid

Liquids and gases do not have any definite shape of their own and thus can change their shapes easily. As these substances are not rigid, they do not have the capability to restore their original shape.

For this reason, liquids and gases at rest cannot resist the tangential force acting on them, and due to the application of even a slight tangential force, these substances begin to flow. Since liquids and gases can flow easily, they are called fluids.

• The branch of physics in which the characteristic properties of fluids at rest are studied is called hydrostatics.
• A liquid has a volume but no definite shape. It takes the shape of the container in which it is kept. But gases have neither a definite shape nor a volume of their own they usually take the shape and the volume of their container.
• Consequently, a gas can be compressed easily by applying pressure on it, while a liquid is almost incompressible.
• Any substance which has no definite shape and has the ability to flow is called fluid. Thus, both liquids and gases are fluids.

Fluids are everywhere around us. Earth has an envelope of air, and two third of earth’s surface is covered with water. Fluids are a phases of matter and include liquids, gases, plasmas and to some extent, plastic solids.

The fundamental difference between solid and liquid:

1. Shearing stress causes a change in the shape of a solid without changing its volume whereas fluids offer little resistance to shearing stress. The shearing stress of fluids is about million times smaller than that of solids.
2. A fluid can exert or withstand a force in a direction perpendicular to its surface. So a fluid does have a bulk modulus of rigidity.

The fundamental difference between liquid and gas: A liquid is incompressible and has a definite volume and a free surface of its own. However, a gas is compressible and it expands to occupy all the space available to it.

Hydrostatics Archimedes Principle

Hydrostatic Force on Submerged Objects

When a body is totally or partly immersed in a liquid or gas at rest, it appears to lose a part of its weight. This apparent loss of weight is equal to the weight of the liquid or gas displaced by the body. This is Archimedes’ principle.

It is to be noted, that Archimedes’ principle is related to the weight of a body. So, this principle is not applicable for a body in a weightless condition. The weight of the body in an artificial satellite or in a freely falling situation is zero. So in these cases, Archimedes’ principle is not applicable.

Application of Archimedes’ Principle: We can determine the following with the help of Archimedes’ principle:

1. The volume of a solid of any shape
2. The density of a substance
3. The amount of the constituent elements in a piece of alloy made of two elements

For the determination of the above things, the choice of the liquid used should be such that

1. The body should not be soluble in that liquid and
2. The body under experiment should not react chemically with the liquid.

Hydrostatics Principles with Examples

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1. Determination of the volume of a solid of any shape: The method described below is the simplest one to find the volume of a body of any shape.

The body is heavier than the liquid: Let the weight of the body in air be W₁ and its weight when totally immersed in the liquid be W₂.

According to Archimedes’ principle,

W₁ – W₂ = apparent loss in the weight of the body

= weight of the liquid displaced by the body

= weight of the liquid equal to the volume of the body

If the density of the liquid is ρ, then volume of the displaced liquid = \(\frac{W_1-W_2}{\rho g}\)

= volume of the body……(1)

The body is lighter than the liquid: in this case, a sinker is used to immerse the lighter body experiment completely in the liquid.

Let the weogth of the body in air be W₁

the weight of the sinker inside the liquid be W₂ and the weight the body with the sinker inside the liquid = W₃

According to Archimedes’ principle, W₁ – (W₃ – W₂)

= apparent weight loss of the body

= weight of the liquid displaced by the body

= weight of equal volume of the liquid If the density of the liquid is ρ, then the volume of the displaced liquid = \(\frac{W_1-\left(W_3-W_2\right)}{\rho g}\)

∴ volume of the body….. (2)

2. Determination of density of a substance: Let the weight of a body in air be W₁ when it is immersed in a liquid of density ρ, it weighs W₂.

According to Archimedes’ principle, from equation (1) we get the volume of the body, V = \(\frac{W_1-W_2}{\rho g}\)

∴ Density of the material of the body,

D = \(\frac{\text { mass of the body }}{\text { volume of the body }}=\frac{W_1 / g}{V}=\frac{W_1 \rho}{W_1-W_2}\)…..(3)

From the measured value of density, the purity of a metal can be known. If the measured value of the density is equal to the actual density of that metal, then we can say that the metal is pure; otherwise it is impure.

3. Determination of the amounts of constituent elements in a piece of alloy made of two elements: Let us assume that an alloy is made of two metals A and B. Let the mass of the alloy in air be W₁ and its weight when immersed completely in water be W₂.

According to Archimedes’ principle, volume of the alloy V = \(\frac{W_1-W_2}{\rho g}\)

Let us assume that the amount of masses and the densities of metals A and B in the alloy are m_a, m_b, and p_a, p_b respectively.

∴ Volume of metal \(A=\frac{m_a}{\rho_a}\)

and volume of metal B = \(\frac{m_b}{\rho_b}=\frac{\left(W_1 / g\right)-m_a}{\rho_b}\)

∴ \(\frac{m_a}{\rho_a}+\frac{\left(W_1 / g\right)-m_a}{\rho_b}=\frac{W_1-W_2}{\rho g}\)

or, \(m_a\left(\frac{1}{\rho_a}-\frac{1}{\rho_b}\right)=\frac{W_1-W_2}{\rho g}-\frac{W_1}{\rho_b g}\)

or, \(m_a=\frac{1}{g}\left(\frac{W_1-W_2}{\rho}-\frac{W_1}{\rho_b}\right) \frac{\rho_a \rho_b}{\rho_b-\rho_a}\)

So, if the values of W₁ , W₂, ρ, ρ_a, and ρ_b are known, then the amount of metal A can be determined from equation (4), and from this, the amount of metal B can be determined.

Hydrostatics Synopsis

Density Definition:

The mass per unit volume of a substance is called its density.

• The ratio of the density of a solid or a liquid substance to the density of pure water at 4°C is called the specific gravity of that substance.
• , The ratio of the mass of a certain volume of any solid or liquid substance to the mass of an equal volume of pure water at 4°C is called the specific gravity of that substance.
• The force acting normally on unit area of a surface is called pressure.
• The normal force exerted by a liquid on any surface in contact with it is called the thrust of the liquid.

The characteristics of pressure at a point in a liquid at rest are:

1. The pressure at a point in a liquid at rest is the same in all directions.
2. The pressure at all points on the same horizontal level in a liquid at rest is the same.
3. The pressure at a point in a liquid at rest is directly proportional to the depth of that point inside the liquid.

The free surface of a liquid at rest is always horizontal.

If two immiscible liquids in a U-tube are in equilibrium, then the heights of the liquids from the plane of separation are inversely proportional to the densities of the liquids.

The free surface of a liquid at rest in connected vessels remains in the same horizontal plane.

Key Formulas in Hydrostatics

Pascal’s law: The pressure applied at any point of a confined fluid is transmitted with undiminished magnitude in all directions throughout the fluid and acts normally on the surface in contact with the fluid.

• The ability of a liquid or gas at rest to exert an upward force on a body immersed in that fluid is called buoyancy.
• The upward thrust exerted on a body by a liquid or gas immersed partly or totally in it is called the buoyant force.
• If the liquid with the body remains in a weightless state, then no buoyant force acts on the body.
• The point where the centre of gravity of the liquid or gas lies before it is displaced by a body immersed in it is the centre of buoyancy or centre of floatation of the immersed body.

Archimedes’ principle: When a body is immersed partly or totally in a liquid or gas at rest, the body appears to lose a part of its weight. This apparent loss in weight is equal to the weight of the liquid or gas displaced by the body.

In the case of a body immersed in a liquid at rest, let the weight of the body be W₁ and the buoyant force acting on the body be W₂.

If W₁ > W₂, then the body sinks in the liquid.

If W₁ = W₂, then the body remains floating at any position inside the liquid being totally immersed in the liquid.

If W₁ <W₂, the body moves up through the liquid and remains floating partly submerged in the liquid.

A floating body has no apparent weight.

1. Condition of floatation: The weight of a floating body must be equal to the weight of the liquid displaced by the body.
2. Condition of equilibrium: The centre of gravity and the centre of buoyancy must lie on the same vertical plane.

• For a floating body in a tilted position, the vertical line drawn through the centre of buoyancy cuts the central line at a point called the metacentre of the body.
• For stable equilibrium of a floating body, the metacentre of the body should lie above its centre of gravity.