## Rajasthan Board RBSE Class 11 Physics Chapter 11 Fluids

### RBSE Class 11 Physics Chapter 11 Textbook Exercises With Solutions

#### RBSE Class 11 Physics Chapter 11 Very Short Answer Type Questions

Question 1.

Why do clouds float in the sky?

Answer:

The final velocity of water drops is zero,so they appear to be floating in the sky.

Question 2.

What is critical velocity?

Answet:

Critical velocity is the limiting velocity above which the fluid flow is turbulent and below which the fluid flow is streamlined.

Question 3.

What is meant by Reynold’s number?

Answer:

It is a pure number that tells about the nature of fluid flow inside the pipe.

Question 4.

Do the two streamlines intersect each other in a streamline flow?

Answer:

A tangent drawn from any point on the streamline indicate the direction of flow of the liquid. As two tangents (different) at a point will indicate two directions of fluid flow which is impossible. Hence, the two streamlines do not cross each other in streamline flow.

Question 5.

What is the effect of temperature on the viscosity of the liquid?

Answer:

Viscosity decreases on increasing the temperature.

Question 6.

At which temperature, the surface tension of liquid is zero?

Answer:

At critical temperature, the surface tension of liquid is zero.

Question 7.

At what temperature, the surface tension of water is maximum?

Answer:

Surface tension of water is maximum at 4 °C because the density of water is maximum at 4°C.

Question 8.

What is the effect on angle of contact on mixing soap solution in water?

Answer:

The angle of contact decreases.

Question 9.

What is the advantage of ploughing the field?

Ans. By ploughing the soil, the small capillaries in the soil break down, hence the water does not evaporate through them. This keeps the field moisturised for a long time which helps in germination of seeds.

Question 10.

What is an ideal liquid?

Answer:

A liquid which is non-compressible and non-viscous is called ideal liquid. In real no liquid is ideal as viscosity and compressibility is not zero for any liquid.

Question 11

What is the angle of contact for pure water and glass?

Answer:

The angle of contact for pure water and glass is zero.

Question 12.

Name the force responsible for surface tension.

Answer:

Cohesive force.

#### RBSE Class 11 Physics Chapter 11 Short Answer Type Questions

Question 1.

How many types of energies are there in a flowing liquid?

Answer:

A flowing liquid has three types of energies :

- Pressure Energy
- Kinetic Energy
- Potential Energy

Question 2.

Define turbulent flow.

Answer:

Turbulent flow is a type of flow of fluid in which the fluid travels in irregular path. In this type of flow, the speed of the fluid at a point undergoes changes continuously in both magnitude and direction. , Most kinds of fluid flow are turbulent. Example : flow of air and water.

Question 3.

There is high pressure on the bottom of sea, then how animals survive?

Answer:

There is high pressure at the bottom of the sea. To balance out this pressure, the fishes and other sea animals have extra air dissolved in their blood. This internal fluid pressure balances the high pressure and animals are able to survive. This is the reason why fishes die when they are brought out of water because outside water the internal fluid pressure exceeds the external atmospheric pressure and their veins burst.

Question 4.

Define streamline flow.

Answer:

Streamline flow (also called flow) is a type of flow of fluid in which fluid travels in regular paths. In streamline flow, the velocity, pressure and other such properties remain constant in fluid at each point.

Question 5.

What is the effect of temperature on the viscosity of gases? Explain it.

Answer:

Viscosity of gases increases with increase in temperature because the velocity of gas molecules increases with temperature and hence collision of molecules increases which decreases the ability of the molecules as a whole to engage in the coordinated move.

Question 6.

A tiny liquid drop is spherical but a larger drop has oval shape?

Answer:

Each free surface of a liquid behaves as a stretcjied membrane under the effect of surface tension and its nature is to bring the surface area to the minimum. Hence, the tiny drop is spherical but a large drop becomes oval under the effect of gravity.

Question 7.

Name the forces which are responsible for shape of liquid surface in a capillary tube.

Answer:

The adhesive force and cohesive force are responsible for shape of liquid surface in a capillary tube.

- When the adhesive force > the cohesive force, the shape of the liquid surface is concave.
- If the adhesive force < the cohesive force, the shape of liquid surface is convex.
- If the adhesive force – the cohesive force, the shape of liquid surface is flat.

Question 8.

Oil is sprinkled on sea waves to calm them. Why?

Answer:

Sprinkling oil reduces the surface tension of sea water due to which a net force acts from lesser surface tension to more surface area and the it calms the sea waves.

Question 9.

Why is it difficult to fill mercury in a glass tube of mercury thermometer?

Answer:

The inner radius of glass tube is very small and due to capillarity the mercury falls in the tube. Hence it is difficult to fill mercury in glass tube of the thermometer.

Question 10.

Why does a small piece of camphor dance about on the water surface?

Answer:

The camphor slowly dissolves in the water lowering the water’s surface tension in the immediate neighbourhood. The strange pull exerted by the uncontaminated portion of water brings about a movement of the surface and the camphor particles are carried along with it. When after sometime the whole surface settles for the reduced tension, the movement of the camphor too stops.

#### RBSE Class 11 Physics Chapter 11 Long Answer Type Questions

Question 1.

State and prove Bernoulli’s theorem.

Answer:

Bernoulli’s Theorem :

According to Bernoulli’s theorem, the sum of the energies possessed by a flowing ideal liquid at a point is constant provided that the liquid is incompressible and non-viseous and flows in streamline.

Potential energy + Kinetic energy + Pressure energy = Constant

where C is a constant.

This relation is called Bernoulli’s theorem.

Dividing eqn. (11.11) by g, we get

where C is another constant.

For horizontal flow, h remains same throughout.

So,

Thus, for horizontal motion, the sum of static and dynamic pressure is constant. If , P_{1},υ_{1} and P_{2}, υ_{2} represent pressure and velocities at two points. Then,

Derivation of Bernoulli’s Theorem :

The energies possessed by a flowing liquid are mutually convertible. When one type of energy increases, the other type of energy decreases and vice-versa. Now, we will derive the Bernoulli’s theorem using the work-energy theorem.

Consider the flow of liquid. Let at any time, the liquid lies between two areas of flowing liquid A_{1} and A_{2}. In time interval ∆t, the liquid displaces from A_{1} by ∆x_{1} = υ_{1}∆t and displaces from A_{2} by ∆ x_{2} = υ_{2} ∆t. Here υ_{1} and υ_{2} are the velocities of the liquid at A_{1} and A_{2}.

The work done on the liquid is P_{1}A_{1}∆x_{1} by the force and P_{2}A_{2}∆x_{2} against the force respectively.

Net work done,

Question 2.

Find out the speed of efflux at a height h from the side of a container both when its top is closed and open. Hence derive Torricelli’s law.

Answer:

Consider the surface of a liquid in a wide container at a height h above the orifice. The velocity of the liquid at the surface is taken zero and the atmospheric pressure at P and at the exit is the same.

Apply Bernoulli’s theorem for the unit mass of the fluid.

At P, only potential energy is there; and, at exit, only kinetic energy is there.

So, υ = \(\sqrt { 2gh }\) …(11.20)

According to Bernoulli’s principle, the sum of total energy and pressure of unit volume of liquid at its free surface and at hole must remain constant.

So P + 0 + ρgh = P + \(\frac { 1 }{ 2 } \) ρυ^{2} + ρg(H – h)

⇒ υ = \(\sqrt { 2gh }\) …..(11.21)

The above expression represents the speed of a freely falling body and is known as Toricelli’s law.

The horizontal range of fluid :

The height (sy) = H – h

The horizontal velocity υ_{x} = \(\sqrt { 2gh }\)

The vertical velocity = υy = 0

The horizontal acceleration (aX) = 0

The vertical acceleration ay = +g

from the vertical motion:

Question 3.

Derive a formula to measure the rate of flow of a liquid through venturi meter.

Answer:

Venturi meter :

This device is used to measure velocities and mass rate of flow of liquid at the different cross-sections of a pipe. It consists of a constriction inserted in a pipe line and have tapers at the inlet and outlet, so that turbulence is avoided. See fig. 11.12.

Let at the cross-sectional areas A_{1} and A_{2}, the velocities are υ_{1} and υ_{2} respectively.

From the equation of continuity,

A_{1}υ_{1}– = A_{2}υ_{2}

Question 4.

Derive an expression for the terminal velocity of a small body falling through a viscous liquid.

Answer:

When a body falls in a viscous medium, it carries with its layers of fluid which are in body’s contact where as the layers of fluid in contact with the stationary surface remain almost at rest. The layers of fluid destroys the relative motion and motion of the body is thus opposed. The viscous drag increases with velocity of the body till viscous drag and upthrust of the body are together equal to the weight of the body which acts downwards.

When there is no net force, the body moves with the uniform velocity. This velocity is called terminal velocity.

Stoke showed that the retarding force F due to viscous drag for a spherical body of radius r that moves with a velocity υ in a fluid, with coefficient of viscosity η, is given as :

F = 6πηrυ

This expression is known as Stoke’s Law.

Derivation with help dimensions :

F ∝ υ, velocity;

F ∝ r, radius of the body;

F ∝η coefficient of viscosity of the fluid.

⇒ F = Aη^{α} r^{b} υ^{c}

where A is a constant with no dimensions. Putting the dimensions,

This is the formula for terminal velocity.

It is apparent from the above expression that the terminal velocity is : (a) directly proportional to the square of the radius of body, (b)directly proportional to the densities of the body and the medium, (conversely proportional to the coefficient of viscosity of the medium.

Variation of viscosity with temperature and pressure effect of temperature on viscosity :

(i) When a liquid is heated then the kinetic energy of it’s molecules increases and the intermolecular force between them is decrease. Hence the viscosity of a liquid decreases with the increase in it’s temperature.

(ii) Viscosity of gases is due to the diffusion of molecules from one moving layer to another. But the rate of diffusion of a gas is directly proportional to the square root of the temperature. So the viscosity of a gas increase with it’s temperature.

Effect of pressure :

(i) Except the water of the viscosity of liquid increases with the increase in pressure. In case of water, the viscosity decrease with increase in pressure.

(ii) The viscosity of gas is independent of pressure.

Question 5.

Explain surface tension on the basis of molecular forces.

Answer:

It is commonly experienced that drops of a liquid are spherical in shape in the absence of external forces. If a small amount of mercury is dropped from a small height, it is spread out in small spherical globules. The water drops falling from a tap are also spherical. Thus, it is clear that for a given volume the surface area of sphere is minimum. So, on the liquid other than the gravitational force another force acts that tries to minimise its surface area. This force is called the surface tension of the liquid.

Question 6.

Explain some examples which illustrate the existence of surface tension.

Answer:

Surface tension can be understood with the help of following examples :

1. When a camel-hair paint brush is dipped in water, its hair spread out. When it is taken out, all hair come closer. This means that the liquid film between the hair is contracting.

2. We make a soap film on the metallic wire ring. Place a moistened cotton thread on it in the form of a loop. It is put on the film as shown in fig. 11.19 (a). This is due to equal and opposite forces acting on each

point of the loop, because soap film lies on both inside the thread and outside the thread. Now, if the film inside the film is picked, it becomes disappeared from the inside region and loop acquires circular form.

We know that a circle encloses the greatest area so that the area of the film is minimum. This shows that the film shrinks to have minimum area.

Question 7.

Derive a relation between surface tension and surface energy.

Answer:

Consider a rectangular frame (see fig. 11.20), having a sliding wire on one of its arms. Dip the frame in a soap solution and take it out. A soap film is formed on the frame and have two surfaces. Both the surfaces are in contact with the sliding wire. So, surface tension acts on the wire due to both the surface.

Let T be the surface tension of the soap solution and L be the length of the wire.

The force exerted by each surface on the wire = T × L

So, total force on the wire = 2TL …(11.23)

Let the surfaces contract by ∆x.

Work done on the film, W = F × ∆x

= 2T L∆x …(11.24)

Here 2L × ∆x → total increase in the area of both of the surfaces of film.

Let 2L × ∆x = ∆A

So, surface tension of a liquid is equal to the work done in increasing the surface area of its free surface by one unit. In other words, surface tension is equal to surface energy per unit area.

Question 8.

Explain some daily life examples based on capillarity.

Answer:

Examples of Capillarity in Daily Life:

- As pen nib is split at the tip to provide the narrow capillary and the ink is drawn upto the point continuously.
- In oil lamps, the oil is drawn up through the capillary of the wick.
- Clay soils are damped as the water rises quickly to the surface through the capillaries.
- Water and minerals rise in the plants through the fine capillaries. .
- Blotting paper absorbs ink through the pores.

Question 9.

Explain shape of meniscus for different liquids in capillary.

Answer:

We know that, the height rises in the capillary tube,

For pure water, the value of angle of θ is very small.

A clean capillary tube of uniform bore is fixed vertically with its lower end dipping into water taken in a beaker. A needle N is also fixed with the capillary tube as shown in the figure. The tube is raised or lowered until the tip of the needle just touches the water surface.

A travelling microscope M is focused on the meniscus of the water in the capillary tube. The reading R_{1} corresponding to the lower meniscus is noted. The microscope is lowered and focused on the tip of the needle and the corresponding reading is taken as R_{2}. The difference between R_{1} and R_{2} gives the capillary rise h.

The radius of the capillary tube is determined using the travelling microscope. If p is the density of water then the surface tension of water is given by

where g is the acceleration due to gravity.

When a liquid rises in a capillary tube, the weight of the liquid column of density p inside the tube is supported by the surface tension (upward) acting around the circumference of the points of contact.

where h → height of the liquid column above the liquid meniscus

p → density of the liquid

r → inner radius of the capillary tube

θ → angle of contact.

Question 10.

What do you understand by the term capillarity? Derive an expression for the rise of liquid in a capillary tube.

Answer:

When a capillary tube of fine bore is dipped in water, then the water rises in the tube (see fig. 11.29).

If the tube is dipped in mercury, there is a depression of mercury level in the tube.

The property of rise or depression of a liquid inside the capillary tube is called capillarity.

Generally, these liquids which wet the glass rise in the tube while those which do not wet the glass gets depressed.

Let r be the radius of a capillary tube and 0 be the angle of contact. From fig. 11.31.

Question 11.

Derive an expression for the excess pressure inside a liquid drop.

Answer:

Suppose that ABC is a half drop of a liquid, whose radius is r and surface tension is T. On the molecules of liquid towards the spart, the cohesive force F_{1} would be downwards.

Thus, the pressure inside the drop is \(\frac { 2T }{ r } \) more than the pressure outside the drop.

Question 12.

What is the effect of contamination and temperature on the surface tension of a liquid.

Answer:

1. Effect of Temperature: On increasing the temperature, the surface tension of the liquid decreases and becomes zero at a definite temperature (Critical temperature).

2. Effect of Contaminations: If there is dust, grease, oil, etc. on the liquid surface, then the surface tension decreases.

3. Effect of Solute: If the solute is more soluble, (sugar, water, salt, lemon), then the surface tension of water increases.

If the solute is less soluble, then the surface tension decreases. For e.g., dissolving soap, petrol, phenol in water, decreases the surface tension. That is why, when we pour petrol in water, the surface tension decreases and hence, the mosquitoes do not sit on water and sinks.

4. Effect of Detergents: On mixing detergents in water, the surface tension decreases and water reaches the small pores of clothes easily. Otherwise, pure water does not enters there due to more surface tension. The adhesive forces of particles and water is more than the cohesive forces. So, the dust comes out of the clothes. The woolen clothes are cleaned using the petrol. Its surface tension is much lesser so ‘the dust particles are easily removed.

#### RBSE Class 11 Physics Chapter 11 Numerical Questions

Question 1.

In a car lift compressed air exerts a force F, on a small piston having a radius of 5 cm. This pressure is transmitted to a second piston of radius 15 cm. If the mass of the car to be lifted is 1500 kg, what is F_{1} ? What is the pressure required to do this task?

Solution:

Question 2.

A bubble of gas whose diameter is 2 cm is travelling in a liquid with constant velocity of 0.90 cm/s. The density of the gas is negligible and density of the liquid is 1.5 gm/cm^{3}. Calculate the value of η.

Solution:

Question 3.

Find out the terminal velocity of rain water drop of radius 1 mm. Viscous coefficient for air is 1.8 × 10^{-5} Ns/m^{2} and density 1.2 kg/m^{3}, density of water is 10^{3} kg/m^{3}, (g = 10 m/s^{2})

Solution:

Question 4.

Water rises in a capillary tube upto a height of 4 cm. If capillary is inclined at an angle of 30° with the vertical, then what will be the position of water in the tube?

Solution:

If the capillary is inclined at an angle of β with the vertical, then the position of water in the tube is

Question 5.

If two soap bubbles of radii R_{1} and R_{2} make a bigger bubble at a constant temperature, then find out the resultant radius of the big bubble.

Solution:

Since, the two bubbles make a bigger bubble at a constant temperature.

∴ Surface energy of the bigger bubble = Sum of surface energies of both the bubbles

Question 6.

A water tank is filled with water upto height H. There is a orifice in the wall of water tank at a depth D from the water surface. If water is emerging from orifice to ground then find out the horizontal range of water on the ground.

Solution:

The velocity of flow of water υ = \(\sqrt { 2gD }\)

After coming out of orifice, time taken (time of flight) to reach the ground.

Question 7.

Two soap bubbles have diameters in the ratio 2 : 3. Compare the excess of pressure inside these bubbles.

Solution:

Excess pressure inside the bubble

Question 8.

Water rises in a capillary tube upto height 10 cm. If surface tension of water is 73 × 10-3 m/s2, then find out the radius of the capillary.

Solution:

Question 9.

Water is flowing through a non-uniform cross-sectional pipe. Where the radius of pipe is 2 cm, velocity of water is 20 cm/s. If another place, where radius of the pipe is 6 cm, find out the velocity of the water.

Solution:

Question 10.

What should be the excess pressure inside a water drop whose radius is 2 mm. Surface tension of water is 0.075 N/m.

Solution:

Given, R = 2 mm = 2 × 10^{-3} m;

T = 0.075N/ m; P_{ex} =?

∵ Since the pressure inside the water drop is more

Question 11.

Work done by a big drop of radius R in spraying n small drops of radius r will be 4π(n^{1/3} – 1) R^{2}T. Prove it, where T is the surface tension

Solution:

Volume of one big drop = Volume of n small drops

Question 12.

An aeroplane is passing through an air tunnel. The velocity of air is 70 m/s and 63 m/s respectively at its upper surface of wing and lower surface of wing. Find out the lifting force. Given that total wing area is 2.5 m^{2} and air density is 1.3 kg/m^{3}.

Solution:

Question 13.

Water is flowing through a non-uniform cross-sectional area pipe. In pipe where the velocity of water is 0.4 m/s, pressure 0.1 m of mercury. At another place, where velocity of water is 0.5 m/s, calculate the value of pressure.

Solution:

Question 14.

1000 small drops of water, each of radius 10^{-7} m are falling down, coalesce to form a bigger drop. Find out the free energy. Surface tension of water is 7 × 10^{-2} N/m.

Solution:

Question 15.

The mass of a spherical glass ball is M.It is falling down through a viscous fluid with the terminal velocity V. Find out the terminal velocity of another spherical glass ball of mass 8M in the same medium.

Solution:

Question 16.

Work done in increasing size 10 × 11cm from 10 cm × 6 cm of a soap film is 3.0 × 10^{-4} Joule. Find out the surface tension of the film.

Solution:

Effective increase in surface area

∆A= 2(A_{2} – A_{1})= 2 [10 × 11 – 10 × 6] cm^{2}

= 2 × 10(11 – 6)cm^{2} = 2 × 10 × 5cm^{2}

= 100cm^{2}

= 100 × 10^{-4} m^{2}

= 1 × 10^{-2} m^{2
}

Question 17.

A pitot tube is dipped in a river, and we receive pressure water column to be 0.05 m. Calculate the rate of flow of water here.

Solution:

Rate of flow of water measured through the pitot tube is

Question 18.

A tank containing water has an orifice 3.5 m below the surface of water in the tank. Calculate the velocity of efflux at the orifice.

Solution:

Given, h = 3.5 m, g = 9.8m^{-2}m/s^{2}, υ = ?

∵ Velocity of efflux,

Question 19.

If work done in blowing a soap bubble of volume V is W. Find out work done in making a bubble of volume 2V.

Solution:

Question 20.

Calculate the minimum pressure in flowing blood from heart to brain (head). Given height of the brain from heart is 0.5 m and density of blood 1040 kg/m^{3}, g = 9,8 m/s^{2}. (Neglecting viscosity)

Solution:

Minimum pressure in flowing blood from heart to brain (head) P = ρgh

= 1040 × 9.8 × 0.5 = 5096

or P = 5.096 × 10^{3} N/m^{2}

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