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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.

Read and Learn More: Class 11 Physics Notes

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.

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• 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 Long Answer Type Question And Answers

Hydrostatic Pressure Questions Explained

Question 1. How can a body be cut easily with the sharp edge of a knife, but not with its blunt edge?
Answer:

A body can be cut easily with the sharp edge of a knife, but not with its blunt edge

The working of a knife depends on the pressure, i. e., the force per unit area. We know that pressure (p) = \(\frac{\text { force }(F)}{\text { area }(A)}\); so keeping F constant, if we decrease A,p  will increase.

Now, the surface area of the blunt edge of a knife is more than that of the sharp edge. As a result, for the same force, the blunt edge produces less pressure, but the sharp edge creates a much higher pressure. For this reason, a body can be cut easily with the sharp edge of a knife.

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Question 2. During the construction of a dam, why is the bottom of the dam wall made thicker than the top?
Answer:

During the construction of a dam the bottom of the dam wall made thicker than the top because

The lateral pressure of water depends on the depth of water. With increase in the depth, the lateral pressure increases. The depth of water at the top of a dam is less and hence the lateral pressure of water is also less.

But due to the maximum depth of water at the bottom of a dam, the lateral pressure of water is the maximum there. To withstand such enormous pressure, the bottom of the dam wall is made thicker.

Question 3. Some liquid is kept in a container inside an artificial satellite revolving in a circular orbit around the Earth. What will be the pressure at a point inside the liquid?
Answer:

Some liquid is kept in a container inside an artificial satellite revolving in a circular orbit around the Earth.

We know that a body becomes weightless when placed inside an artificial satellite revolving around the earth. For this reason, the liquid inside the satellite has no weight, and hence no pressure develops at any point within the liquid. However, there is the gravitational attraction of the satellite itself. This force on the liquid is so low that the pressure due to this is negligible.

Step-by-Step Solutions to Hydrostatics Problems

Question 3. A cylindrical vessel is filled with a liquid such that the thrust on the bottom of the vessel and that on Its side wall become equal. Prove that the height of the liquid column in the cylinder is numerically equal to the radius of the vessel.
Answer:

Given

A cylindrical vessel is filled with a liquid such that the thrust on the bottom of the vessel and that on Its side wall become equal.

Let the height of the liquid column in the vessel = h, the radius of the cylindrical vessel = r, and the density of the liquid = ρ.

∴ The weight of the liquid in the cylinder = πr²h x ρ x g

∴ Thrust on the base of the cylinder = weight of the liquid in the cylinder = πr²hρg ……(1)

Now, the average pressure on the wall of the cylinder
=1/2 x hρg, area of the curved surface of the cylinder = 2 πrh.

Total thrust on the wall of the cylinder

= \(\frac{1}{2} h \rho g \times 2 \pi r h=\pi r h^2 \rho g\)

According to the problem, from equations (1) and (2), we get

∴ \(\pi r^2 h \rho g=\pi r h^2 \rho g \quad \text { or, } r=h\)

i. e., the radius of the cylinder will be equal to the height of the liquid column.

Real-Life Applications of Hydrostatics Questions

Question 5. A large shallow wooden container is filled with water. But the container does not crack. A hole is then made on the surface of the container and a long narrow tube is inserted vertically into the container through this hole. The tube Is now filled with water. It is seen that the container now cracks. Explain why.
Answer:

Given

A large shallow wooden container is filled with water. But the container does not crack. A hole is then made on the surface of the container and a long narrow tube is inserted vertically into the container through this hole. The tube Is now filled with water.

The thrust exerted by the liquid on the bottom of the container depends on the base area and the depth of the liquid in it, but not on the quantity of liquid kept in that vessel.

• In the present problem, initially, the depth (h) of water in the container was not large before the inclusion of the tube, although the amount of water in it was large enough. It produced less thrust on the bottom of the container hence the container did not crack.
• When the narrow tube is filled with water, up to the height (h + H), the water column increases considerably. It exerts a large thrust on the base of the container, and so the container cracks.

Question 6. A vessel full of liquid is descending with an acceleration a [a < g]. Find the relation between the liquid pressure with the depth of the liquid in the vessel.
Answer:

Given

A vessel full of liquid is descending with an acceleration a [a < g].

Let the density of the liquid = ρ, area of crosssection of the vessel = α, mass of the liquid =m, and pressure at a depth h inside the liquid = p.

The upward reaction force exerted on the liquid in part A by the liquid in part B = pα.

So, considering the motion of the liquid in part A, we get mg- pa = ma

or, pa = mg-ma = m(g-a) = ahp(g-a)

or, p = hpρ(g- a).

Question 7. A cylindrical vessel contains a liquid up to a height h. A hole is made on the wall of the vessel. Water I comes out through the hole and falls at a distance x from the base of the vessel. At what depth should the hole be created so that x becomes maximum?
Answer:

Given

A cylindrical vessel contains a liquid up to a height h. A hole is made on the wall of the vessel. Water I comes out through the hole and falls at a distance x from the base of the vessel.

Suppose a hole is made at a depth y from the free surface of the liquid and the horizontal velocity of efflux of water through the hole is v.

∴ ν = √2gy

Suppose the liquid takes r second to fall at a distance x from the base of the vessel. Considering the vertical motion of the ejected liquid, we get

h-y = \(\frac{1}{2} g t^2 \text { or, } t^2=\frac{2(h-y)}{g}\)

Considering the horizontal motion of the ejected liquid, we get

x = νt or, x = √2gyt

or, \(x^2=2 g y t^2=2 g y \cdot \frac{2(h-y)}{g}=4 y(h-y)\)…….(1)

Now, for the maximum value of x, x² is also maximum, and hence \(\frac{d}{d y}\left(x^2\right)=0\).

So, differentiating equation (1) with respect to y. we get

⇒ \(\frac{d}{d y}\left(x^2\right)=4(h-y)+4 y(-1)=4(h-2 y)\)

∴ \(h-2 y=0\) or, \(y=\frac{h}{2}\)

Hence, for the maximum range of the water coming out of the hole, the hole should be made at a depth of h/2.

Question 8. Prove that In static equilibrium the pressure exerted by a fluid decreases with its height.
Answer:

Let us assume two layers at heights h and h + dh from the base inside a fluid. Let the density of the fluid = ρ. A part having unit cross-sectional area and thickness dh is considered.

So, the mass of the part = ρdh.

Let the upward pressure on this part be p and the downward pressure = p + dp.

∴ In equilibrium, (p+dp)-p + (ρdh·g) = 0 or, \(\frac{d p}{d h}=-\rho g\)

or, \(\frac{d p}{d h}\) is negative, which indicates that with increase in h, the dh value of p decreases.

So, it is seen that the pressure exerted by the fluid decreases with its height from the base.

Question 9. Two holes are made on the wall and at the bottom of a cylindrical vessel. Now the two holes are dosed with two corks and the vessel is filled with water. If the vessel is now allowed to fall under gravity and the two holes are opened during its flight, then state what will happen.
Answer:

Given

Two holes are made on the wall and at the bottom of a cylindrical vessel. Now the two holes are dosed with two corks and the vessel is filled with water. If the vessel is now allowed to fall under gravity and the two holes are opened during its flight,

We know that a body falling freely under gravity experiences weightlessness. So, the vessel filled with water remains weightless while it is falling under gravity. Then, the upper layers of the liquid exert no pressure on the lower layers and hence there is no pressure difference in the vessel. As a result, no water will come out through the two holes.

Question 10. State whether pressure applied on any part of a confined liquid is transmitted instantaneously to different parts of the liquid. Also, state whether Pascal’s law is applicable.
Answer:

The pressure applied on any part of a confined liq¬uid does not reach different points of the liquid at once. The applied pressure is transmitted through compression and rarefaction within the liquid, and hence, with the velocity of sound. During the transmission of pressure, Pascal’s law is inapplicable. Pascal’s law is valid once the system comes to an equilibrium.

Question 11. Does a hydraulic press work, if the liquid used in it is replaced by a gas?
Answer:

The thrust applied will no doubt be multiplied if the liquid used in a hydraulic press is replaced by a gas. But, due to the greater compressibility of a gas, the thrust developed on the larger piston will not be large enough. For this reason, the press cannot be used for efficient pressing.

Question 12. Does the density have any significance in the weight-less state?
Answer:

Density = \(\frac{\text { mass }}{\text { volume }} \); the quantity of matter contained in a body is called its mass. This mass remains the same even in the weightless state of the body and hence the density of the substance remains unchanged.

Question 13. A long cylinder is fitted with a tap on its wall at its lower end. Keeping the tap closed, the cylinder is filled with water. Now the cylinder is made to float by placing it over a cork and then the tap is opened. State what will happen.
Answer:

Given

A long cylinder is fitted with a tap on its wall at its lower end. Keeping the tap closed, the cylinder is filled with water. Now the cylinder is made to float by placing it over a cork and then the tap is opened.

Due to the lateral pressure of water, it will be seen that a jet of water comes out through the opening of the tap. If the opening of the tap is in the horizontal direction, then the water jet will come out horizontally. Due to an equal but opposite reaction of this thrust, the cylinder (floating on the cork) will move in the opposite direction.

Question 14. A cubical box is completely filled with water. Prove that the total thrust exerted by water on one of the vertical walls is equal to half the weight of water kept in the box.
Answer:

Given

A cubical box is completely filled with water.

Let each side of the cubical box be of length x. Mass of water kept in the box, m = x³ x 1 = x³

∴ Thrust exerted by water on any one of the vertical walls = lateral pressure x area

= \(\frac{x}{2} \times 1 \times g \times x^2=\frac{1}{2} x^3 g=\frac{1}{2} m g\)

= 1/2 x weight of water kept in the box.

Question 15. 1. Prove that the density of the mixture of two substances with densities ρ₁ and ρ₂ of equal mass will be \(\frac{2 \rho_1 \rho_2}{\rho_1+\rho_2}\)

2. Prove that if the two substance with densities ρ₁ and ρ₂ are mixed in equal volumes, then the density of the mixture thus formed will be \(\frac{1}{2}\left(\rho_1+\rho_2\right)\)
Answer:

1. Let the mass of each of the substances be m.

∴ Mass of the mixture = m + m = 2 m

Volume of the mixture = \(\left(\frac{m}{\rho_1}+\frac{m}{\rho_2}\right)\)

∴ Density of the mixture = \(\frac{2 m}{\frac{m}{\rho_1}+\frac{m}{\rho_2}}=\frac{2 \rho_1 \rho_2}{\rho_1+\rho_2}\)

2. Let the volume of each of the substances be V.

∴ Total mass of the mixture = (Vρ₁ + Vρ₂)

Total volume of the mixture = (V+ V) = 2 V

∴ Density of the mature = \(\frac{V \rho_1+V \rho_2}{2 V}=\frac{1}{2}\left(\rho_1+\rho_2\right)\)

Question 16. State with reason whether Pascal’s law is applicable to the water in a pond.
Answer:

Pascal’s law is not applicable to the water in a pond. The law is applicable only in the case of confined fluids. As a pond’s water is not confined, the law cannot be applied here.

Question 17. A piece of iron (specific gravity = 7.8) sinks in water, but floats on mercury (specific gravity = 13.6)— explain why.
Answer:

Given

A piece of iron (specific gravity = 7.8) sinks in water, but floats on mercury (specific gravity = 13.6)

The density of iron is 7.8 g · cm^-3 and the density of water is 1 g · cm^-3. The density of iron is greater than that of water and hence the weight of a piece of iron is greater than the upthrust of water exerted on it.

• As a result, the piece of iron sinks in water due to the action of the resultant downward force. But the density of mercury is 13.6 g • cm^-3 which is greater than the density of iron.
• Hence, the weight of the piece of iron is less than the upthrust exerted on it by mercury. As a result, the piece of iron floats on mercury due to the action of the resultant upward force.

Question 18. Why is it easier to swim in sea water than in a river?
Answer:

Due to the presence of salt dissolved in sea water, its density is more than that of river water. Due to the greater density of sea water it exerts a stronger buoyant force on a swimmer than what is exerted by river water. Hence, it is easier to swim in sea water than in a river.

Question 19. Explain whether Archimedes’ principle is applicable in the case of a freely falling body under gravity.
Answer:

A freely falling body has no weight. Hence, the weight of a freely falling body and the weight of the liquid displaced by it—both will be zero so the body will not experience any buoyant force. It means that the application of Archimedes’ principle in this case is unnecessary.

Question 20. An empty balloon (or a soft plastic bag) weighs the same as what it does when filled with air at atmospheric pressure—explain why.
Answer:

An empty balloon (or a soft plastic bag) weighs the same as what it does when filled with air at atmospheric pressure

An empty balloon or a soft plastic bag displaces a very small quantity of air. The balloon gets inflated when filled with air at atmospheric pressure and displaces a larger amount of air. More the amount of air inserted into the balloon, the larger the volume of air displaced by it.

1. Hence, the increase in weight of the inflated balloon due to the inclusion of air into it becomes just equal to the apparent decrease in the weight of the balloon due to the displaced air.
2. For this reason, an empty balloon (or a plastic bag) weighs the same as what it does when filled with air at atmospheric pressure.

Question 21. A wooden block is floating on water in a closed vessel. What will happen

1. if the air above the water is compressed,
2. if all the air above the water is removed from the closed vessel?

Answer:

1. If the air above the water in the closed vessel is compressed, the block will float upwards some more.

Here, weight of the wooden block = weight of displaced water + weight of displaced air.

If the air is compressed, then its density increases, and hence the weight of the displaced air increases while the weight of the displaced water decreases. Hence, the block will float upwards some more.

2. If all the air above the water is removed from the closed vessel, then the wooden block sinks more. This happens because when the air is removed, the buoyancy due to air is absent. Consequently, the apparent weight of the body increases, and the block sinks some more.

Question 22. Explain why it is sometimes safer for a ship while floating on water to load more goods than to get rid of them.
Answer:

For stable equilibrium of a floating body, its centre of gravity should lie below its metacentre. So, if the metacentre lies below the centre of gravity of a floating ship, then, to lower the position of its centre of gravity below the metacenter, more goods should be loaded in the hold of the ship.

In that case, the ship will float under stable equilibrium and hence it is safer for a ship floating on water to be loaded with more goods.

Question 23. A piece of ice is floating on a liquid of density 1.5 g · cm^-3 kept in a beaker. Will there be any change in the level of the liquid in the beaker as the piece of ice melts completely?
Answer:

Given

A piece of ice is floating on a liquid of density 1.5 g · cm^-3 kept in a beaker.

If the piece of ice melts completely, then the level of the liquid in the beaker will rise.

Let the mass of the piece of ice be m. So, the mass of the liquid displaced is also m.

∴ Volume of displaced liquid, \(V_1=\frac{m}{\rho_l}\left[\rho_l=\text { density of the liquid }\right]\)

As the piece of ice melts completely, the mass of water formed = m.

But the volume of that water, \(V_2=\frac{m}{\rho_w}=m\left[\rho_w=\text { density of water }\right]\)

According to question, \(\rho_l>\rho_w \text {, so } V_2>V_1 \text {. }\)

Hence, the level of liquid in the beaker will rise. If the liquid is immiscible in water, then water will float above the liquid.

Question 24. A wooden block is floating on water at 0°C keeping a portion V of its volume outside water. If the temperature of the water is increased from 0°C to 20°C, then what change of V will be observed?
Answer:

Given

A wooden block is floating on water at 0°C keeping a portion V of its volume outside water. If the temperature of the water is increased from 0°C to 20°C,

The density of water increases with the increase in temperature from 0°C to 4°C. With the increase in density of water, the upthrust on the block will increase. As a result, the block will gradually move up in water, i.e., the value of V will increase.

During the rise in temperature from 4°C to 20°C, the density of water decreases and hence the upthrust on the block will decrease. As a result, the wooden block will gradually sink in water, i.e., the value of V will decrease.

Question 25. A piece of ice with a cork in it is floating on water in a beaker. Will the level of water in the beaker change after complete melting of the ice?
Answer:

Given

A piece of ice with a cork in it is floating on water in a beaker.

The level of water in the beaker will not change.

Let the density of water be ρ, the weight of the piece of ice be W and the weight of the cork inside the piece of ice be w.

Total weight = (W+ w) = weight of displaced water

∴ Volume of displaced water  = \(\frac{W+w}{\rho g}\)

Suppose the ice melts completely. In this case, the weight of water thus formed = W and the volume of that amount of water = \(\frac{W}{\rho g}\).

In this situation, only the cork will float on water. Weight of water displaced by the cork = weight of the cork = w.

∴ Volume of water displaced by the cork = \(\frac{W}{\rho g}\)

Total volume of water formed due to melting of ice and also due to the water displaced by the cork = \(\frac{W + W}{\rho g}\)

Therefore, even after the ice melts completely, the level of water in the beaker does not change.

Question 26. A piece of ice with a stone in it is floating on water in a beaker. Will the level of water in the beaker change after complete melting of the ice?
Answer:

Given

A piece of ice with a stone in it is floating on water in a beaker.

If ice melts completely, the level of water in the beaker goes down.

Let the density of water be ρ, the weight of ice be W and the weight of stone be w.

∴ Weight of displaced water = weight of ice + weight of stone = (W+ w)

∴ Volume of displaced water = V = \(\frac{W+w}{\rho g}\)

Let the density of the stone be ρ_s.

Suppose the ice melts completely. In this case, mass of water thus formed due to melting of ice = W and volume of this water = \(\frac{W}{\rho g}\)

In this situation, the piece of stone will sink into water. Volume of water displaced by the stone = \(\frac{w}{\rho_s g}\)

∴ Total volume of water formed due to melting of ice and also due to displacement by stone

= \(V_2=\frac{W}{\rho g}+\frac{w}{\rho_s g}\)

∴ \(\rho_s>1 \quad therefore V_2<V_1\)

So, when all the ice melts, the level of water in the beaker goes down.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 27. A man carries a bucket containing water in one hand 1 and a live fish in the other. If he releases the fish into the bucket of water, will he carry less weight?
Answer:

A man carries a bucket containing water in one hand 1 and a live fish in the other. If he releases the fish into the bucket of water,

The man will carry the same weight when he releases the live fish into the bucket of water. When the fish is released into the water of the bucket, the displaced water exerts an upthrust and hence the apparent weight of the fish will decrease.

• According to Newton’s third law of motion, an equal reaction force will act downwards at the bottom of the bucket. So, the decrease in the weight of the fish becomes equal to the increase in the thrust at the bottom of the bucket.
• As a result, the net weight of the bucket will remain the same and hence the man will carry the same weight as before.

Question 28. A bird is sitting at the bottom of a cage and, in this condition, the cage is weighed with the help of a spring balance. Now, the bird starts flying inside the cage. Will the reading of the balance change?
Answer:

A bird is sitting at the bottom of a cage and, in this condition, the cage is weighed with the help of a spring balance. Now, the bird starts flying inside the cage.

The reading of the spring balance depends on the design of the bottom of the cage. Suppose the bottom of the cage is totally covered with a metal sheet. As soon as the bird starts flying, it flaps its wings.

• Hence, due to an extra thrust developed in each flapping acting on the bottom of the cage, the reading of the spring balance fluctuates.
• But, if the bottom of the cage is covered with a net, then the reading of the spring balance will decrease. Because the spring balance will then give the reading of the weight of the cage only.

Question 29. Two balloons of the same volume are filled with two gases at the same pressure, one with hydrogen and the other with helium. Which of the two experiences a greater upward force?
Answer:

Two balloons of the same volume are filled with two gases at the same pressure, one with hydrogen and the other with helium.

The upward force for the hydrogen-filled balloon will be greater.

Here, resultant upward force = upthrust – the weight of the gas-filled balloon.

• Since the volumes of both balloons are the same, the total weight of the air displaced by them is also the same. So the upthrust exerted by air on both of lime is equal.
• But, under the same conditions of temperature and pressure, the density of helium is more than that of hydrogen. So, due to the lower weight of the hydrogen-filled balloon, the upward force on it will be greater.

Question 30. An egg sinks in freshwater, but it floats when a suitable quantity of salt is mixed in that water. Why?
Answer:

An egg sinks in freshwater, but it floats when a suitable quantity of salt is mixed in that water.

An egg is heavier than the weight of an equal vol¬ume of water and hence it sinks in water. But when a suitable quantity of salt is mixed in water, the density of this saline water becomes greater than the density of the egg and so it floats.

Question 31. A solid sphere and a hollow sphere having the same mass and external radius are immersed in the same liquid. Which one will feel heavier?
Answer:

A solid sphere and a hollow sphere having the same mass and external radius are immersed in the same liquid.

Since the radii of both spheres are the same, their external volumes are equal. So, both of them will displace an equal weight of the liquid.

According to Archimedes’ principle, the apparent loss of weight of a body in a liquid = weight of liquid displaced by the body. So, loss of weight suffered by both spheres will be the same and, hence, their apparent weight of them will be the same.

Question 32. A glass of water is placed on one of the pans of a balance. On the other pan, another similar glass of water is placed with a piece of wood floating on it Water is at the same level in both the glasses. Which of the two glasses is heavier? Explain your answer.
Answer:

A glass of water is placed on one of the pans of a balance. On the other pan, another similar glass of water is placed with a piece of wood floating on it Water is at the same level in both the glasses.

The masses of both the glasses will be the same.The amount of water contained in the second glass is less than that in the first glass because the submerged part of the floating piece of wood has displaced an equal volume of water.

But we know that the mass of this displaced water is the same as the mass of the floating piece of wood and hence the masses of the two glasses will be the same.

Question 33. A flat disc, a solid cube, and a solid sphere of equal mass made of the same material are completely Immersed in water. Which one of them will experience the minimum and which one will experience the maximum buoyant force?
Answer:

A flat disc, a solid cube, and a solid sphere of equal mass made of the same material are completely Immersed in water.

When a body is immersed completely in water, then the buoyant force becomes equal to the weight of the water displaced by the body. Here, the masses of the three bodies are equal and they are made of the same material (i.e., of the same density) hence their volumes are equal. Hence, in each case, the weight of the displaced water becomes equal and the same buoyant force is experienced.

Question 34. A hollow glass sphere is balanced by counterpoising weights made of brass in a common balance. The whole system is then covered with an air-tight bell jar and the jar is evacuated. What will be the result?
Answer:

A hollow glass sphere is balanced by counterpoising weights made of brass in a common balance. The whole system is then covered with an air-tight bell jar and the jar is evacuated.

Since the density of glass is less than the density of brass, the volume of the glass sphere is greater than the vol¬ume of the same mass of brass weights. So, in air, the glass sphere experiences a larger buoyant force than the brass weights.

Now, when the bell jar is evacuated, this buoyant force ceases to act and hence the weight of the glass sphere becomes more than that of the brass weights. Hence, the end of the balance beam containing the glass sphere will move downwards.

Question 35. A common balance has a beaker of water and a piece of stone in one of the pans and is balanced. Now, the stone is dipped into the beaker of water. Is any change in reading of the balance observed?
Answer:

A common balance has a beaker of water and a piece of stone in one of the pans and is balanced. Now, the stone is dipped into the beaker of water.

When the piece of stone is immersed in the water of the beaker, the balance beam remains in its equilibrium condition. Due to the upthrust exerted by water, the piece of stone loses a part of its weight.

According to Newton’s third law of motion, the reaction of the upthrust of water will act vertically downwards on the pan of the common balance. As the apparent loss in weight of the piece of stone becomes equal to the downward reaction force on the balance pan, the net weight remains the same and hence the balance beam remains in its equilibrium condition.

Question 36. What will happen if the upper part of a ship Is made heavier than its lower part?
Answer:

If the upper part of a ship is made heavier than its lower part, then the centre of gravity of the ship lies above the centre of gravity of the water displaced by the ship. As a result, if the ship is tilted a little, then, due to the couple that develops due to the weight of the ship and the buoyant force, the ship tilts more, and hence it may turn over.

Question 37. Determine the change in potential energy of a body when it is raised through a height h, inside water. Volume of the body is V, its density is ρ and the density of water is ρ₀. Give your answer for the situations when

1. ρ > ρ₀ and
2. ρ < ρ₀.

Solution:

Two forces act on a body when it is immersed in water. One is its weight (acting downwards) and another is the buoyant force (acting upwards). Force due to weight = Vρg, and upthrust due to buoyancy = Vρ₀g

1. If ρ > ρ₀, the weight of the body becomes greater than the upthrust.

Hence, resultant downward force = \(V \rho g-V \rho_0 g=V g\left(\rho-\rho_0\right) .\)

If the body is raised to a height h against this force, then work done = \(V g\left(\rho-\rho_0\right) h\) = increase in potential energy.

2. When ρ > ρ₀, the upthrust becomes greater than the weight. Then resultant upward force = \(V g\left(\rho-\rho_0\right)\). To raise the body through a height h, work done = \(V g\left(\rho-\rho_0\right) h\) = decrease in potential energy.

 

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Useful Relations For Solving Numerical Problems

If the mass of volume V of a substance is M, then the density of that substance, D = M/V.

• Specific gravity, S = \(\frac{\text { density of the substance }}{\text { denisity of water at } 4^{\circ} \mathrm{C}}\)
• In the CGS system, the numerical value of the density of a substance = its specific gravity.
• In SI, the numerical value of the density of a substance = 1000 x its specific gravity.
• Pressure (p) = \(\frac{\text { normal applied force }(F)}{{area}(A)} \text {. }\)
• Thrust exerted by a liquid (F) = pressure of the liquid (p) x area (A).

1. The pressure of a liquid of density p at a depth h, or gauge pressure, p = hρg.
2. The pressure at a point in a liquid at a depth h, when the liquid is exposed to free air, or absolute pressure, p = B + hρg

[where B = atmospheric pressure].

The lateral thrust exerted on a rectangular lamina dipped vertically in a liquid

= average lateral pressure x area = \(\left(h+\frac{1}{2} b\right) \rho g \times a b\)

[where a = length of the lamina, b = breadth of the lamina, h = depth of the upper edge of the lamina from the free surface of the liquid].

In a hydraulic press, if the cross-sectional area of the smaller piston is a, the cross-sectional area of the larger piston is b, and the force applied on the smaller piston  is F₁, then the thrust developed on the larger piston, \(F_2=F_1 \times \frac{b}{a}\)

Apparent weight of a body immersed in a liquid = real weight of the body – buoyant force.

Apparent loss in weight of a body immersed in a liquid = weight of liquid displaced by the body = weight of an equal volume of liquid = buoyant force.

In the case of floatation of a body partly immersed in a liquid,

⇒ \(\frac{\text { volume immersed part of the body in the liquid }}{\text { total volume of the body }}\)

= \(\frac{\text { density of the body }}{\text { density of the liquid }}\)

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Assertion Reason Type Questions And Answers

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

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

Question 1.

Statement 1: If a barometer is accelerated upwards, the level of mercury in the tube of the barometer will decrease.

Statement 2: The effective value of g will increase, so upthrust will increase.

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

Question 2.

Statement 1: A hydrogen-filled balloon stops rising after it has attained a certain height in the sky.

Statement 2: The atmospheric pressure decreases with height and becomes zero when maximum height is attained.

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

Question 3.

Statement 1: A solid sphere and a hollow sphere ot same material are floating in a liquid. Radius of both spheres are same. The percentage of volume immersed in both the spheres will be same.

Statement 2: Upthrust acts on volume of liquid displaced. It has nothing to do with whether the body is solid or hollow.

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

Question 4.

Statement 1: For a floating body to be in stable equilibrium, its centre of buoyancy must be located above the centre of gravity.

Statement 2: The torque developed by the weight of the body and the upthrust will restore the body back to its normal position, after the body is disturbed.

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

Question 5.

Statement 1: The blood pressure in humans is greater at the feet than that at the brain.

Statement 2: Pressure of liquid at any point is proportional to height, density of liquid, and acceleration due to gravity.

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

Question 6.

Statement 1: A cylinder fitted with a movable piston contains a certain amount of liquid in equilibrium with its vapour. The temperature of the system is kept constant with the help of a thermostat. When the volume of the vapour is decreased by moving the piston inwards, the vapour pressure does not increase.

Statement 2: Vapour in equilibrium with its liquid, at a constant temperature, does not obey Boyle’s law.

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

Question 7.

Statement 1: A body floats in a liquid with a fraction n of its volume above the surface of the liquid. If the system is taken to a planet where the acceleration due to gravity is greater than that on earth, the fraction n will decrease.

Statement 2: For flotation, the weight of the body is equal to the weight of the liquid displaced.

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

 

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Very Short Answer Type Questions

Question 1. On what other factor does the pressure at a point in a liquid depend besides the depth of that point and the acceleration due to gravity at that place?
Answer: Density of liquid

Question 2. ‘The pressure at all points on the same horizontal plane inside a liquid are equal’—is the statement true or false?
Answer: True

Question 3. If two immiscible liquids in a U-tube are in equilibrium, then how are the heights of the two liquids from the surface of separation-related with the densities of the two liquids?
Answer: Inversely proportional

Question 4. Besides the volume of the immersed part of a body in a liquid and acceleration due to gravity, on what factor does the buoyancy acting on a body depend?
Answer: Density of the liquid

Question 5. For a body floating partly immersed in a liquid, state whether the density of the body will be greater or less than the density of the liquid. 
Answer: Less

Question 6. As a liquid can flow even under the influence of a minute tangential force, it is called a _________
Answer: Fluid

Question 7. ‘A liquid does not undergo change in density due to change in pressure.’ State whether the statement is true or false.
Answer: True

Question 8. Density of a substance = specific gravity of a substance x density of ________
Answer: Water at 4°C

Question 9. In all systems of units, the value of specific gravity is _____
Answer: Equal

Question 10. ‘The free surface of a liquid at rest is always horizontal’—Correct or incorrect?
Answer: Correct

Question 11. The density of the material of a body in its weightless
condition remains ______.
Answer: Unaltered

Question 12. ‘If the density of a body is more than the density of a liquid, then the body sinks in that liquid.’—State whether the statement is true or false.
Answer: True

Question 13. The weight of a body in air is 100 g and its weight in water is 40 g. Find the volume of the body.
Answer: 60 cm^-3

Question 14. Which quantity represents the normal force applied on unit area of a surface?
Answer: Pressure

Question 15. Is pressure a vector quantity?
Answer: No

Question 16. Write down the variation of the pressure at a point in a liquid to its density.
Answer: Directly proportional

Question 17. ‘During the construction of a dam, the base of the dam wall is usually made wider.’—Correct or incorrect?
Answer: Correct

Question 18. Due to the upthrust of a liquid, a body immersed in the liquid suffers an apparent ________ in weight.
Answer: Loss

Question 19. Standard atmospheric pressure is equal to pressure exerted by __________ cm of mercury column.
Answer: 76

Question 20. What is the height of homogeneous column of air?
Answer: 8 km (approx.)

Question 21. Does a hydraulic press work if the liquid in it is replaced by a gas?
Answer: Yes

Question 22. What is the name of the upward force exerted by displaced liquid or gas on a body when it is partly or totally immersed in it?
Answer: Buoyant force

Question 23. Does buoyancy depend on the depth up to which a body is completely immersed in a liquid?
Answer: No

Question 24. What fraction of floating ice remains below the surface of water?
Answer: 10/11 part

Question 25. In which direction does the buoyant force act?
Answer: In the direction opposite to the weight of the body

Question 26. What is the dimension of buoyancy?
Answer: [MLT^-2]

Question 27. Name the principle which is effective in a hydraulic press.
Answer: Multiplication of thrust

Question 28. Thrust = pressure x __________
Answer: Area

Question 29. Pascal’s law obeys the principle of conservation of _________
Answer: Energy

Question 30. In the tilted position of a floating body, the point at which the vertical line drawn through the centre of buoyancy cuts the central line, name the point of the body.
Answer: Metacentre

Question 31. Due to the upthrust of a liquid, the weight of a body immersed in that liquid decreases or increase?
Answer: Decreases

Question 32. 1 torr = ____________ dyn • cm^-2
Answer: 1332.8

Question 33. Is Archimedes’ principle applicable in the case of a freely falling body?
Answer: No

Question 34. Is Archimedes’ principle applicable inside an artificial satellite revolving around the Earth?
Answer: No

Question 35. A boat floating in a pond is carrying a number of stones. If the stones are dropped into the pond, will the water level rise or fall?
Answer: Fail

Question 36. Does Archimedes’ principle hold good in the case of a gas?
Answer: Yes

Question 37. State whether a lump of iron floats or sinks in mercury.
Answer: Floats

Question 38. Where does the centre of buoyancy lie in a displaced liquid.
Answer: At the centre of gravity

Question 39. If the weight of a body is greater than the weight of the liquid displaced by it, then the body _______ in that liq-uid.
Answer: Sinks

Question 40. What is the apparent weight of a floating body?
Answer: Zero

Question 41. For a floating body in unstable equilibrium, the meta-centre lies __________ the centre of gravity of the body.
Answer: Below

Question 42. When a ship enters a river from the sea. What will you find?
Answer: It sinks more

Question 43. A wooden block is floating on water in a closed vessel. If the air in the vessel is compressed, then the block will ________ more. [Fill in the blank]
Answer: Float up

Question 44. State whether the weight of a body measured in air will be more or less than its weight in vacuum.
Answer: Less

Question 45. For the stable equilibrium of a body, state whether its metacentre should lie above or below its centre of gravity.
Answer: Above

Question 46. A body of mass m is falling freely under gravity. What will be its weight in this condition?
Answer: Zero

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Match Column 1 With Column 2

Question 1. There are two points A and B inside a liquid as shown in Fig. Now the vessel starts moving upwards with an acceleration a.

Answer: 1. A, 2. A, 3. A

Question 2. A cube is floating in a liquid as shown in Fig.

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

Question 3. A tube is inverted in a mercury vessel as shown in Fig. If the pressure p is increased.

Answer: 1. 2, 2. C, 3. C

Question 4.

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

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Comprehension Type Question And Answers

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

Question 1. A small spherical ball of radius r is released from its completely submerged position in a liquid whose density varies with height h as \(\rho_L=\rho_0\left[4-\frac{3 h}{h_0}\right]\). The density of the ball is \(\frac{5}{2} \rho_0\). The height of the vessel is \(h_0=\frac{12}{\pi^2}\). Consider r <<h₀ and g = 10 m • s^-2

1. Where will the ball be in the equilibrium condition?

1. at a depth \(\frac{h_0}{2}\) from top
2. at the bottom of the vessel
3. at a depth \(\frac{3h_0}{4}\) from top
4. The ball will never be at equilibrium

Answer: 1. at a depth \(\frac{h_0}{2}\) from top

2. The motion of the ball in the vessel is

1. Oscillatory but not SHM
2. SHM with time period 2s
3. SHM with time period 1 s
4. The motion is not oscillatory

Answer: 2. SHM with time period 2s

3. Time taken by the ball to reach the bottom of the vessel is

1. 1s
2. 2s
3. 0.5s
4. The ball will not reach the bottom of the vessel.

Answer: 1. 1s

Question 2. A block of mass 1 kg and density 0.8 g · cm^-3 is held stationary with the help of a string as shown in Fig. The tank is accelerating vertically upwards with an acceleration a = 1.0 m · s^-2. Take g = 10 m • s^-2 and density of water = 10³ kg • m^-3.

1. What is the tension in the string?

1. 2.60 N
2. 2.85 N
3. 2.75 N
4. 3.10 N

Answer: 3. 2.75 N

2. If the string is now cut, find the acceleration of the block

1. 3.95 m • s^-2
2. 3.75 m • s^-2
3. 4.25 m • s^-2
4. 3.29 m • s^-2

Answer: 2. 3.75 m • s^-2

Hydrostatics Q&A for Students

Question 3. An engineering firm is assigned the job to design the cylindrical pressured water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 m • s^-2. The pressure at the surface of the wafer will be 130 kPa and depth of the water will be 14.2 m- The pressure of the air in the building outside the tank will be 93 kPa.

1. Find the net downward force on the tank’s flat bottom of area 2 m²

1. 179.4 kN
2. 365.4 kN
3. 105.36 kN
4. None

Answer: 1. 179.4 kN

2. What is the buoyant force on a wooden block of mass 2 kg and relative density 0.8, height 10 cm?

1. 5.94 N
2. 7.42 N
3. 1.48 N
4. None

Answer: 2. 7.42 N

Question 4. The vessel shown in Fig. has two sections of areas of cross-section A₁ and A₂. A liquid of density ρ fills both the sections, up to a height h in each. Neglect the atmospheric pressure.

1. What is the pressure at the base of the vessel?

1. 1/2 hρg
2. 2hρg
3. hρgA₂
4. None

Answer: 2. 2hρg

2. What is the force exerted by the liquid on the base of the vessel?

1. 2hρg A₁A₂
2. 1/2 hρg A₂
3. 2hρg A₂
4. hρg x (A₁+A₂)

Answer: 3. 2hpg A₂

3. Find the downward force exerted by the walls of the vessel at the level B.

1. 2hρg x (A₂-A₁)
2. 1/2hρg x (A₁ + A₂)
3. hρgA₂
4. hρg x (A₂-A₁)

Answer: 4. hρg x (A₂-A₁)

Question 5. A tank has a cylindrical hole H of diameter 2r at its bottom as shown in Fig. A cylindrical block B of diameter 4r and height h is placed on the hole H to prevent the flow of liquid through the hole. The liquid in the tank stands at a height h₁ above the top face of the block. The density of liquid is ρ and that of the block is \(\frac{\rho}{3} .\).

1. If the liquid is gradually taken out of the tank, the height h₁ of the liquid surface above the top face of the block for which the block just begins to rise is

1. 2h/4
2. 3h/4
3. 5h/3
4. 2h

Answer: 3. 5h/3

2. If the liquid level is further lowered so that it stands at a depth h₂ above the bottom face of the block as shown in the figure, then the maximum value of  h₂ so that the block does not move is

1. 4h/9
2. h/3
3. 2h/3
4. 5h/9

Answer: 1. 4h/9

3. If the liquid level is lowered below h₂, then

1. The block will never rise
2. The block will start rising if h₂ = h/3
3. The block will start rising if h₂ = h/4
4. The block will start rising if h₂ = h/5

Answer: 1. The block will never rise

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Integer Type Question And Answers

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

Question 1. A candle of diameter d is floating on a liquid in a cylindrical container of diameter D(D >>d) as shown in Fig It is burning at the rate of 2 cm • h^-1. At what rate will the top of the candle fall?

Answer: 1

Question 2. A cylinder has radius 8 cm. Up to what height (in cm) should it be filled with water so that the thrust on its walls is equal to that on its bottom?
Answer: 8

Question 3. Mass of a balloon with its contents is 1.5 kg. It is descending with an acceleration equal to half of the acceleration due to gravity. If it is to go up with the same acceleration keeping the volume same, what amount of mass (in kg) should be decreased?
Answer: 1

Question 4. A ball whose density is 0.4 x 10³ kg · m^-3 falls into water from a height of 9 cm. To what depth does the ball sink?
Answer: 6

Question 5. A water tank is 20 m deep. If a water barometer reads 10 m at that place, then what is the pressure (in SI unit) at the bottom of the tank in the atmosphere?
Answer: 3

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Short Answer Type Questions

Question 1. The density of a body is d and that of air is ρ. If the body weighs w in air, what will be its actual weight?
Answer:

Given

The density of a body is d and that of air is ρ. If the body weighs w in air,

Let the actual weight of the body = w₀

∴ buoyant force of air, B = \(\frac{w_0}{d} \rho\)

So, the weight of the body in air,

⇒ \(w=w_0-B=w_0-\frac{w_0}{d} \rho=w_0\left(\frac{d-\rho}{d}\right)\)

∴ \(w_0=\frac{d}{d-\rho} w\)

Question 2. A drop of oil rises through water with an acceleration ag. If a is a constant quantity and g is the acceleration due to gravity, find the specific gravity of the oil. Neglect the friction of water.
Answer:

Given

A drop of oil rises through water with an acceleration ag. If a is a constant quantity and g is the acceleration due to gravity,

Let d and D be the density of water and oil respectively, and the mass of the oil drop is m.

∴ Volume of the oil drop = m/D

= volume of the displaced water by the oil drop or, the mass of the displaced water

= density of water x volume of the displaced water

= d x m/D

Now, buoyancy = weight of the displaced water = \(\frac{d m g}{D}\)

∴ Net upward force acting on the oil drop

= buoyant force – weight of the oil drop

= \(\frac{d m g}{D}-m g=\left(\frac{d}{D}-1\right) m g\)

∴ Acceleration of the oil drop inside water, \(\alpha g=\frac{\left(\frac{d}{D}-1\right) m g}{m}=\left(\frac{d}{D}-1\right) g\)

∴ Specific gravity of oil = \(/frac{D}{d}=\frac{1}{1+\alpha}\)

Question 3. A uniform rod is suspended horizontally from its mid-point. A piece of metal whose weight is W is suspended at a distance l from the mid-point. Another weight W₁ is suspended on the other side at a distance l₁ from the mid-point to bring the rod to a horizontal position. When W is completely immersed in water, W₁ needs to be kept at a distance l₂ from the mid-point to get the rod back into horizontal position. The specific gravity of the metal piece is

1. \(\frac{W}{W_1}\)
2. \(\frac{W l_1}{W l-W_1 l_2}\)
3. \(\frac{l_1}{l_1-l_2}\)
4. \(\frac{l_1}{l_2}\)

Answer:

Given

A uniform rod is suspended horizontally from its mid-point. A piece of metal whose weight is W is suspended at a distance l from the mid-point. Another weight W₁ is suspended on the other side at a distance l₁ from the mid-point to bring the rod to a horizontal position. When W is completely immersed in water, W₁ needs to be kept at a distance l₂ from the mid-point to get the rod back into horizontal position.

In equilibrium condition of rod in air, Wl = W₁l₁

In equilibrium condition of rod immersed in water, \(\left(W-F_B\right) l=W_1 l_2\)

Where, F_B = buoyant force on the body of weight W = \(\frac{W}{\rho}\)

So, \(\left(W-\frac{W}{\rho}\right) l=W_1 l_2=W_{\frac{l}{l}}^{l_1} l_2 \quad \text { or, } 1-\frac{1}{\rho}=\frac{l_2}{l_1}\)

∴ \(\rho=\frac{l_1}{l_1-l_2}\)

The option 3 is correct.

Question 4. A wooden block is floating on water kept in a beaker. 40% of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to g/2. The block will then

1. Sink
2. Float with 10% above the water surface
3. Float with 40% above the water surface
4. Float with 70% above the water surface

Answer:

Given

A wooden block is floating on water kept in a beaker. 40% of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to g/2.

Weight of the body and the buoyancy of the liquid both depend on the effective acceleration. Hence there will be no change in the floating portion of the block.

The option 3 is correct.

Question 5. To determine the composition of a bimetallic alloy, a sample is first weighed in air and then in water. These weights are found to be w₁ and w₂ respectively. If the densities of the two constituent metals are p₁ and p₂ respectively, then the weight of the first metal in the sample is (where p_w is the density of water)

1. \(\frac{\rho_1}{\rho_{w^{(}}\left(\rho_2-\rho_1\right)}\left[w_1\left(\rho_2-\rho_w\right)-w_2 \rho_2\right]\)
2. \(\frac{\rho_1}{\rho_w\left(\rho_2+\rho_1\right)}\left[w_1\left(\rho_2-\rho_w\right)+w_2 \rho_2\right]\)
3. \(\frac{\rho_1}{\rho_w^{(}\left(\rho_2-\rho_1\right)}\left[w_1\left(\rho_2+\rho_w\right)-w_2 \rho_1\right]\)
4. \(\frac{\rho_1}{\rho_w\left(\rho_2-\rho_1\right)}\left[w_1\left(\rho_2-\rho_w\right)-w_2 \rho_1\right]\)

Answer:

⇒ \(w_1-w_2=V \rho_w g=\left(V_1+V_2\right) \rho_w g=\left[\frac{x}{\rho_1}+\frac{w_1-x}{\rho_2}\right] \rho_w\)

Hence, required weight,

x = \(\frac{\rho_1}{\rho_w\left(\rho_2-\rho_1\right)}\left[w_1\left(\rho_2-\rho_w\right)-w_2 \rho_2\right]\)

The option 1 is correct.

Question 6. A cylinder of height h is filled with water and is kept on a block of height h/2. The level of water in the cylinder is kept constant. Four holes numbered 1, 2, 3, and 4 are at the side of the cylinder and at heights 0, h/4 , h/2, and 3h/4 respectively. When all four holes are opened together, the hole from which water will reach farthest distance on the plane PQ is the hole number

1. 1
2. 2
3. 3
4. 4

Answer:

Net height of the block and the cylinder = \(\frac{h}{2}+h=\frac{3 h}{2}\)

If we consider a hole at a depth x from the top, height of that hole from the plane PQ, H = 3h/2 – x

Velocity of water flowing out of that hole horizontally, v = √2gx

Now, vertical velocity of water flowing out of that hole = 0

Hence, if the water takes t time to reach PQ plane,

H = \(\frac{1}{2} g t^2 \quad \text { or, } t=\sqrt{\frac{2 H}{g}}\)

For this time t, water will flow horizontally with the same velocity.

Hence, distance travelled on the PQ plane, \(D=v t=\sqrt{2 g x} \cdot \sqrt{\frac{2 H}{g}}=\sqrt{2 x \cdot 2 H}=\sqrt{2 x(3 h-2 x)}\)

For hole no. \(1, x=h ; D_1=\sqrt{2 h(3 h-3 h)}=h \sqrt{2}\)

For hole no. \(2, x=\frac{3 h}{4} ; D_2=\sqrt{\frac{3 h}{2}\left(3 h-\frac{3 h}{2}\right)}=h \cdot \frac{3}{2}\)

For hole no. \(3, x=\frac{h}{2}, D_3=\sqrt{h(3 h-h)}=h \sqrt{2}\)

For hole no. \(4, x=\frac{h}{4} ; D_4=\sqrt{\frac{5 h}{2}\left(3 h-\frac{5 h}{2}\right)}=h \sqrt{\frac{5}{4}}\)

∴ \(D_2>D_1=D_3>D_4\)

The option 2 is correct.

Question 7. There is a circular tube in a vertical plane. Two liquids that do not mix and of densities d₁ and d₂ are Filled in the tube. Each liquid subtends 90° angle at centre. Radius joining their interface makes an angle a vertical. Ratio d₁/d₂ is

1. \(\frac{1+\sin \alpha}{1-\sin \alpha}\)
2. \(\frac{1+\cos \alpha}{1-\cos \alpha}\)
3. \(\frac{1+\tan \alpha}{1-\tan \alpha}\)
4. \(\frac{1+\sin \alpha}{1-\cos \alpha}\)

Answer:

Equating pressure at A, (Rcosa + Rsina)d₂g (Rcosa – Rsina)d₁g

⇒ \(\frac{d_1}{d_2}=\frac{\cos \alpha+\sin \alpha}{\cos \alpha-\sin \alpha}=\frac{1+\tan \alpha}{1-\tan \alpha}\)

The option 3 is correct.

Question 8. An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)

1. 16cm
2. 22 cm
3. 38 cm
4. 6 cm

Solution:

⇒ \(p+x=p_0\)

or, \(p=p_0-x\)

or, \(8 \times A \times 76=(76-x) \times A \times(54-x)\)

∴ x=38

So, length of air column =54-38 = 16 cm.

The option 1 is correct.

Question 9. Two non-mixing liquids of densities p and np(n > 1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL(p < 1) in the denser liquid. The density d is equal to

1. {2 + (n+ 1)p}ρ
2. {2 + (n- 1)p}ρ
3. {1 +(n-1)p}ρ
4. {1 + (n+1)p}ρ

Answer:

Let, the area of the cross-secton of the cylinder be A.

Weight of the cylinder = LAgd

The buoyant force on the cylinder due to the liquid of higher density = pLAnρg

and buoyant force on the cylinder due to the liquid of lower density = (1- p)LAρg

In equilibrium,

LAgd = (1 – p)LAρg+ pLAnρg

or, d = (1 -ρ)ρ + pnρ = ρ-pρ + pnρ

∴ d = [1 +(n- l)p]ρ

The option 3 is correct.