RBSE Solutions Class 11 Physics Chapter 11 Fluids

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: Clouds float in the sky because the water drops inside them have a very low final velocity, almost zero. This makes them seem to hang in the air without falling.
In simple words: Clouds stay up because the tiny water drops in them fall very slowly, making them look like they are floating.

🎯 Exam Tip: When discussing cloud buoyancy, mention the low terminal velocity of water droplets as the key scientific reason.

Question 2. What is critical velocity?
Answer: Critical velocity is a specific speed for a fluid's flow. If the fluid moves faster than this speed, its flow becomes turbulent (chaotic); if it moves slower, its flow remains streamlined (smooth and orderly).
In simple words: It is the speed limit where liquid flow changes from smooth to messy.

🎯 Exam Tip: Define critical velocity precisely, linking it to the transition between laminar (streamlined) and turbulent flow.

Question 4. Do the two streamlines intersect each other in a streamline flow?
Answer: No, two streamlines never intersect each other in a streamline flow. A streamline indicates the direction of fluid flow at any given point. If two streamlines crossed, it would mean the fluid has two different directions of flow at that single point, which is physically impossible.
In simple words: Streamlines do not cross because liquid cannot flow in two directions at once from the same spot.

🎯 Exam Tip: Emphasize that streamlines represent the instantaneous velocity direction and two directions at one point are impossible, hence no intersection.

Question 5. What is the effect of temperature on the viscosity of the liquid?
Answer: When the temperature of a liquid increases, its viscosity decreases. This happens because the liquid molecules gain more kinetic energy and move faster, which weakens the forces holding them together and makes the liquid flow more easily.
In simple words: Hot liquids are less thick than cold liquids.

🎯 Exam Tip: Remember that for liquids, viscosity and temperature are inversely related; for gases, it's generally the opposite.

Question 6. At which temperature, the surface tension of liquid is zero?
Answer: The surface tension of a liquid becomes zero at its critical temperature. Above this temperature, the distinction between liquid and gas phases disappears.
In simple words: Surface tension vanishes when a liquid reaches its critical temperature.

🎯 Exam Tip: Relate zero surface tension to the critical temperature, where liquid and gas phases become indistinguishable.

Question 7. At what temperature, the surface tension of water is maximum?
Answer: The surface tension of water is maximum at \( 4\text{ °C} \). This specific temperature is also when water has its maximum density.
In simple words: Water's surface tension is highest at \( 4\text{ °C} \), where it is also densest.

🎯 Exam Tip: Note the unique property of water having maximum density and surface tension at \( 4\text{ °C} \).

Question 8. What is the effect on angle of contact on mixing soap solution in water?
Answer: When a soap solution is mixed in water, the angle of contact usually decreases. This change makes the water spread out more easily and wet surfaces better.
In simple words: Adding soap to water makes the angle of contact smaller, so water spreads better.

🎯 Exam Tip: Understand that soap acts as a surfactant, lowering surface tension and typically reducing the contact angle for better wetting.

Question 9. What is the advantage of ploughing the field?
Answer: Ploughing a field breaks down the tiny capillary tubes in the soil. This prevents water from evaporating quickly through these small channels. As a result, the soil stays moist for a longer time, which is helpful for seeds to sprout and grow.
In simple words: Ploughing keeps the soil wet longer by breaking tiny water channels, helping seeds grow.

🎯 Exam Tip: Connect ploughing to the breaking of capillaries in soil, reducing evaporation and preserving moisture for agriculture.

Question 10. What is an ideal liquid?
Answer: An ideal liquid is a theoretical fluid that has no viscosity (meaning it flows without any internal friction) and is incompressible (meaning its density does not change under pressure). It is a concept used in physics to simplify fluid dynamics problems.
In simple words: An ideal liquid is a perfect fluid that has no thickness and cannot be squished.

🎯 Exam Tip: Define ideal liquid by its two main characteristics: incompressibility and non-viscous (no internal friction).

Question 11. What is the angle of contact for pure water and glass?
Answer: For pure water and clean glass, the angle of contact is considered to be zero. This means the water wets the glass surface completely and spreads out flat.
In simple words: Pure water wets clean glass completely, so the contact angle is zero.

🎯 Exam Tip: Remember that a zero contact angle signifies perfect wetting of a surface by a liquid.

Question 12. Name the force responsible for surface tension.
Answer: The cohesive force is responsible for surface tension. This force refers to the attractive forces between molecules of the same substance, which pull the surface molecules inward, creating tension.
In simple words: Surface tension is caused by cohesive forces, which are the attractive forces between the liquid's own molecules.

🎯 Exam Tip: Clearly distinguish between cohesive (like molecules) and adhesive (unlike molecules) forces when discussing surface phenomena.

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 main types of energy:
1. Pressure Energy
2. Kinetic Energy
3. Potential Energy
In simple words: A moving liquid has energy from its pressure, its motion, and its height.

🎯 Exam Tip: List the three energy types clearly (Pressure, Kinetic, Potential) as they form the basis of Bernoulli's principle.

Question 2. Define turbulent flow.
Answer: Turbulent flow is a type of fluid motion where the fluid particles move in an irregular, chaotic, and unpredictable way, not following smooth paths. In this flow, the speed and direction of the fluid at any point change constantly. Examples include air flow during a storm or fast-moving water in a river.
In simple words: Turbulent flow is when a liquid or gas moves in a messy, swirling pattern instead of a smooth one.

🎯 Exam Tip: Highlight irregularity, continuous changes in magnitude and direction, and unpredictability as key features of turbulent flow.

Question 3. There is high pressure on the bottom of sea, then how animals survive?
Answer: Sea animals living at the bottom of the ocean survive the high pressure because their bodies have an internal fluid pressure that balances the external pressure. Their blood and other body fluids contain dissolved gases and solutes that create this counter-pressure. When these animals are brought to the surface, the external pressure significantly decreases, causing their internal pressure to become much higher than the outside, which can make their veins burst.
In simple words: Deep-sea animals have strong internal pressure that matches the high outside pressure, allowing them to survive. When brought up, the sudden pressure drop can harm them.

🎯 Exam Tip: Explain the balance of internal and external pressure as the key to survival for deep-sea organisms, and relate their inability to survive at the surface to rapid pressure changes.

Question 5. What is the effect of temperature on the viscosity of gases? Explain it.
Answer: For gases, viscosity increases as temperature increases. This is because higher temperatures lead to faster movement and more frequent collisions between gas molecules. These increased collisions, which result in the transfer of momentum between different layers, contribute to a greater resistance to flow, or higher viscosity.
In simple words: Hot gases are thicker because their molecules move faster and bump into each other more often, making them harder to flow.

🎯 Exam Tip: Note that the relationship between temperature and viscosity is opposite for gases compared to liquids, due to different mechanisms of internal friction.

Question 6. A tiny liquid drop is spherical but a larger drop has oval shape?
Answer: The free surface of any liquid acts like a stretched skin due to surface tension, which always tries to minimize the surface area. For a small drop, surface tension is the dominant force, pulling it into a perfect sphere, which has the smallest surface area for a given volume. However, for a larger drop, the force of gravity becomes significant. Gravity pulls the drop downwards, causing it to flatten and take on an oval or flattened shape.
In simple words: Small drops are round due to surface tension, but big drops flatten into an oval because gravity pulls them down.

🎯 Exam Tip: Explain the dominance of surface tension in small drops (aiming for minimum surface area, hence sphere) and the increasing influence of gravity in larger drops, leading to deformation.

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 the shape of the liquid surface (meniscus) in a capillary tube.
• When the adhesive force is greater than the cohesive force (liquid sticks more to the tube than to itself), the liquid surface is concave (curved inwards).
• If the adhesive force is less than the cohesive force (liquid sticks more to itself than to the tube), the liquid surface is convex (curved outwards).
• If the adhesive force is equal to the cohesive force, the liquid surface is flat.
In simple words: The way a liquid curves in a thin tube depends on whether it sticks more to the tube (adhesive force) or to itself (cohesive force).

🎯 Exam Tip: Remember to differentiate between adhesive and cohesive forces, and how their relative strengths determine the concave, convex, or flat shape of the meniscus.

Question 8. Oil is sprinkled on sea waves to calm them. Why?
Answer: Oil is sprinkled on sea waves to calm them because oil reduces the surface tension of water. When oil spreads on the water, the surface tension in the oily area becomes less than in the surrounding clean water. This difference in surface tension creates a net force that pulls from the region of lower surface tension (oil-covered) towards the region of higher surface tension (clean water), effectively spreading the oil and dampening the waves.
In simple words: Oil calms waves by lowering water's surface tension, which creates a force that spreads the oil and smooths the water.

🎯 Exam Tip: The key concept is the reduction of surface tension by oil, leading to a force imbalance that spreads the oil and dampens wave motion.

Question 10. Why does a small piece of camphor dance about on the water surface?
Answer: A small piece of camphor appears to "dance" on the water surface because it slowly dissolves in the water. As it dissolves, it lowers the surface tension of the water in its immediate vicinity. The surrounding water, which is uncontaminated, still has a higher surface tension. This difference in surface tension creates an uneven pull on the camphor piece, causing it to move and spin. Once the entire water surface's tension is uniformly reduced as the camphor dissolves completely, the movement stops.
In simple words: Camphor moves on water because it changes the water's surface tension around it, creating an uneven pull.

🎯 Exam Tip: Link the camphor's movement to localized changes in surface tension caused by its dissolution, creating an unbalanced force.

RBSE Class 11 Physics Chapter 11 Long Answer Type Questions

Question 1. State and prove Bernoulli's theorem.
Answer: Bernoulli's Theorem states that for a flowing ideal liquid (incompressible and non-viscous) moving in a streamline, the sum of its pressure energy, kinetic energy, and potential energy per unit volume (or unit mass) remains constant at every point.

Expressed mathematically, for unit volume, it is:
\( P + \frac { 1 }{ 2 } \rho v^{2} + \rho gh = \text{Constant} \)

Where:
\( P \) is the pressure energy per unit volume.
\( \frac { 1 }{ 2 } \rho v^{2} \) is the kinetic energy per unit volume.
\( \rho gh \) is the potential energy per unit volume.

To express it for unit mass, divide by \( \rho \):
\( \frac { P }{ \rho } + \frac { 1 }{ 2 } v^{2} + gh = \text{Constant} \)

Or, for unit weight (dividing by \( \rho g \)):
\( \frac { P }{ \rho g } + \frac { 1 }{ 2 } \frac { v^{2} }{ g } + h = \text{Constant} \)
Here, \( \frac { P }{ \rho g } \) is the pressure head, \( \frac { v^{2} }{ 2g } \) is the velocity head, and \( h \) is the gravitational head.

Derivation of Bernoulli's Theorem using the Work-Energy Theorem:
Consider an ideal liquid flowing through a tube of varying cross-section and height. Let the liquid flow from a region with pressure \( P_1 \), velocity \( v_1 \), and height \( h_1 \) to a region with pressure \( P_2 \), velocity \( v_2 \), and height \( h_2 \).

The work done on the liquid by pressure forces over a small time \( \Delta t \) is:
Work done by \( P_1 \) at inlet \( = P_1 A_1 \Delta x_1 \)
Work done against \( P_2 \) at outlet \( = -P_2 A_2 \Delta x_2 \)
Net work done on the liquid \( W = P_1 A_1 \Delta x_1 - P_2 A_2 \Delta x_2 \)

Due to the equation of continuity \( A_1 v_1 = A_2 v_2 \), if \( \Delta V \) is the volume of liquid flowing in time \( \Delta t \), then \( \Delta V = A_1 \Delta x_1 = A_2 \Delta x_2 \).
So, \( W = (P_1 - P_2) \Delta V \).

According to the work-energy theorem, this net work done equals the change in the total mechanical energy (kinetic + potential) of the liquid.

Change in kinetic energy \( \Delta K = \frac { 1 }{ 2 } m v_2^{2} - \frac { 1 }{ 2 } m v_1^{2} = \frac { 1 }{ 2 } \rho \Delta V (v_2^{2} - v_1^{2}) \)
Change in potential energy \( \Delta U = mgh_2 - mgh_1 = \rho \Delta V g (h_2 - h_1) \)

So, \( W = \Delta K + \Delta U \)
\( (P_1 - P_2) \Delta V = \frac { 1 }{ 2 } \rho \Delta V (v_2^{2} - v_1^{2}) + \rho \Delta V g (h_2 - h_1) \)

Divide by \( \Delta V \) (assuming \( \Delta V \neq 0 \)):
\( P_1 - P_2 = \frac { 1 }{ 2 } \rho (v_2^{2} - v_1^{2}) + \rho g (h_2 - h_1) \)

Rearrange the terms by grouping those with subscript 1 on one side and subscript 2 on the other:
\( P_1 + \frac { 1 }{ 2 } \rho v_1^{2} + \rho gh_1 = P_2 + \frac { 1 }{ 2 } \rho v_2^{2} + \rho gh_2 \)

This equation shows that the sum \( P + \frac { 1 }{ 2 } \rho v^{2} + \rho gh \) is constant at any two points in the streamline flow, thus proving Bernoulli's theorem.
In simple words: Bernoulli's theorem says that for a perfect flowing liquid, the total energy (pressure, motion, and height) stays the same everywhere. We prove this by showing that the work done by pressure is equal to the change in the liquid's motion and height energy.

🎯 Exam Tip: Clearly state Bernoulli's theorem first, then derive it using the work-energy theorem, ensuring all steps for work done, kinetic energy, and potential energy changes are explicit and correctly balanced.

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: To find the speed of efflux and derive Torricelli's law, consider a wide container filled with liquid, with a small orifice (hole) at a depth \( h \) below the free surface of the liquid.

Assumptions:
1. The container is wide, so the velocity of the liquid surface at the top \( (v_1) \) is approximately zero.
2. The atmospheric pressure \( P_a \) acts on the free surface and at the orifice (if open to atmosphere).
3. The liquid is ideal (incompressible and non-viscous).

Applying Bernoulli's Theorem:
Bernoulli's equation states: \( P_1 + \frac { 1 }{ 2 } \rho v_1^{2} + \rho gh_1 = P_2 + \frac { 1 }{ 2 } \rho v_2^{2} + \rho gh_2 \)
Let point 1 be on the free surface and point 2 be at the orifice.

Case 1: Container open to the atmosphere (Top open)
At point 1 (free surface): \( P_1 = P_a \) (atmospheric pressure), \( v_1 \approx 0 \), \( h_1 = H \) (total height from reference, let's say bottom).
At point 2 (orifice): \( P_2 = P_a \) (atmospheric pressure), \( v_2 = v \) (speed of efflux), \( h_2 = H-h \) (height of orifice from bottom).

Substitute these into Bernoulli's equation:
\( P_a + \frac { 1 }{ 2 } \rho (0)^{2} + \rho g H = P_a + \frac { 1 }{ 2 } \rho v^{2} + \rho g (H-h) \)

Simplifying:
\( \rho g H = \frac { 1 }{ 2 } \rho v^{2} + \rho g H - \rho g h \)
\( 0 = \frac { 1 }{ 2 } \rho v^{2} - \rho g h \)
\( \frac { 1 }{ 2 } \rho v^{2} = \rho g h \)
\( \implies v^{2} = 2gh \)
\( \implies v = \sqrt{2gh} \)

This is Torricelli's Law. It states that the speed of efflux of a liquid from an orifice is the same as the speed that a body would acquire if it were to fall freely from rest through a height equal to the depth of the orifice below the free surface of the liquid.

Case 2: Container closed and pressure at the surface is \( P_s \) (Top closed)
At point 1 (free surface): \( P_1 = P_s \), \( v_1 \approx 0 \), \( h_1 = H \).
At point 2 (orifice): \( P_2 = P_a \), \( v_2 = v \), \( h_2 = H-h \).

Substitute these into Bernoulli's equation:
\( P_s + \frac { 1 }{ 2 } \rho (0)^{2} + \rho g H = P_a + \frac { 1 }{ 2 } \rho v^{2} + \rho g (H-h) \)

Simplifying:
\( P_s + \rho g H = P_a + \frac { 1 }{ 2 } \rho v^{2} + \rho g H - \rho g h \)
\( P_s - P_a + \rho g h = \frac { 1 }{ 2 } \rho v^{2} \)
\( \implies v^{2} = \frac { 2(P_s - P_a) }{ \rho } + 2gh \)
\( \implies v = \sqrt { \frac { 2(P_s - P_a) }{ \rho } + 2gh } \)

This gives the speed of efflux when the container top is closed and maintained at a pressure \( P_s \). If \( P_s = P_a \), this reduces to Torricelli's law.

Horizontal Range of the Fluid:
If the orifice is at height \( (H-h) \) from the ground, the liquid behaves as a projectile.
Vertical motion: \( sy = (H-h) \), initial vertical velocity \( u_y = 0 \), acceleration \( a_y = g \).
Using \( sy = u_y t + \frac { 1 }{ 2 } a_y t^{2} \):
\( H-h = 0 \cdot t + \frac { 1 }{ 2 } g t^{2} \)
\( \implies t = \sqrt { \frac { 2(H-h) }{ g } } \)

Horizontal motion: The horizontal velocity \( u_x = v = \sqrt{2gh} \) (from Torricelli's law), and horizontal acceleration \( a_x = 0 \).
The horizontal range \( R \) is given by \( R = u_x t + \frac { 1 }{ 2 } a_x t^{2} \):
\( R = \sqrt{2gh} \cdot \sqrt { \frac { 2(H-h) }{ g } } + 0 \)
\( R = \sqrt{ \frac { 2gh \cdot 2(H-h) }{ g } } \)
\( R = \sqrt{4h(H-h)} \)
\( \implies R = 2\sqrt{h(H-h)} \)
In simple words: Torricelli's law says water shoots out of a hole at the same speed as if it fell from the water's surface to the hole. We prove this using Bernoulli's rule. If the container is open, the speed is \( \sqrt{2gh} \). If it's closed with extra pressure, the speed is higher. We also found how far the water travels horizontally.

🎯 Exam Tip: When deriving Torricelli's law, clearly state assumptions for Bernoulli's principle. For range, treat the efflux as projectile motion and combine horizontal velocity with time of fall.

Question 4. Derive an expression for the terminal velocity of a small body falling through a viscous liquid.
Answer: When a small spherical body falls through a viscous liquid, it experiences three forces:
1. **Weight (W):** Acting downwards.
2. **Upthrust (U):** Acting upwards, due to the displaced fluid.
3. **Viscous Drag (Fv):** Acting upwards, opposing the motion.

Initially, the body accelerates, but as its speed increases, the viscous drag also increases. Eventually, a point is reached where the upward forces (upthrust + viscous drag) balance the downward force (weight). At this point, the net force on the body becomes zero, and it falls with a constant maximum velocity called terminal velocity \( (v_t) \).

Let:
\( r \) = radius of the spherical body
\( \rho \) = density of the body
\( \sigma \) = density of the viscous liquid
\( \eta \) = coefficient of viscosity of the liquid
\( g \) = acceleration due to gravity

1. **Weight of the body (W):**
\( W = \text{mass} \times g = (\text{volume} \times \text{density of body}) \times g \)
\( W = \frac { 4 }{ 3 } \pi r^{3} \rho g \)

2. **Upthrust (U):** According to Archimedes' principle, upthrust equals the weight of the fluid displaced.
\( U = (\text{volume of body} \times \text{density of liquid}) \times g \)
\( U = \frac { 4 }{ 3 } \pi r^{3} \sigma g \)

3. **Viscous Drag (Fv):** For a small spherical body moving through a viscous fluid, Stoke's Law gives the viscous drag:
\( F_v = 6\pi \eta r v_t \)

At terminal velocity, the net force is zero (equilibrium):
\( W = U + F_v \)
\( \frac { 4 }{ 3 } \pi r^{3} \rho g = \frac { 4 }{ 3 } \pi r^{3} \sigma g + 6\pi \eta r v_t \)

Rearrange to solve for \( v_t \):
\( 6\pi \eta r v_t = \frac { 4 }{ 3 } \pi r^{3} \rho g - \frac { 4 }{ 3 } \pi r^{3} \sigma g \)
\( 6\pi \eta r v_t = \frac { 4 }{ 3 } \pi r^{3} g (\rho - \sigma) \)

Divide both sides by \( 6\pi \eta r \):
\( v_t = \frac { 4 \pi r^{3} g (\rho - \sigma) }{ 3 \cdot 6\pi \eta r } \)
\( v_t = \frac { 2 r^{2} g (\rho - \sigma) }{ 9 \eta } \)

This is the expression for terminal velocity. It shows that terminal velocity is directly proportional to the square of the radius of the body and the difference in densities, and inversely proportional to the viscosity of the fluid.
In simple words: When a small ball falls in a thick liquid, it eventually reaches a steady speed called terminal velocity. This happens when the ball's weight is balanced by the upward push from the liquid and the drag force. The formula for this speed depends on the ball's size, its density, the liquid's density, and how thick the liquid is.

🎯 Exam Tip: Clearly list the three forces acting on the body. Ensure accurate application of Stoke's Law and Archimedes' principle in setting up the force balance equation for terminal velocity.

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.
In simple words: Heating a liquid makes it thinner because molecules move faster and attract each other less. For gases, heating makes them thicker because molecules spread out and collide more, making it harder for layers to slide past each other.

🎯 Exam Tip: Differentiate the effects of temperature on liquid vs. gas viscosity, attributing the difference to intermolecular forces in liquids and molecular diffusion/collisions in gases.

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.
In simple words: For most liquids, more pressure makes them thicker, but water becomes thinner. For gases, pressure does not change their thickness much.

🎯 Exam Tip: Note the general trend of liquid viscosity with pressure and the unique inverse relationship for water, and that gas viscosity is largely unaffected by pressure.

Question 5. Explain surface tension on the basis of molecular forces.
Answer: Surface tension arises from the intermolecular forces acting within a liquid. Molecules inside the bulk of the liquid are surrounded by other molecules on all sides, resulting in balanced attractive forces. However, molecules at the liquid's surface are only attracted by molecules below and to their sides, not by molecules above (as there are fewer or no liquid molecules above them). This imbalance creates a net inward force on the surface molecules, pulling them towards the bulk of the liquid. This inward pull causes the liquid surface to behave like a stretched elastic membrane, always trying to minimize its surface area, which is what we observe as surface tension. This is why liquid drops tend to be spherical, as a sphere has the smallest surface area for a given volume.
In simple words: Surface tension happens because liquid molecules at the surface are pulled inwards by other molecules below them, but not pulled outwards by molecules above them. This makes the surface act like a tight skin, trying to shrink as much as possible.

🎯 Exam Tip: Focus on the imbalance of cohesive forces at the surface compared to the bulk, leading to a net inward pull and minimal surface area.

Question 6. Explain some examples which illustrate the existence of surface tension.
Answer: Surface tension is a force that makes liquid surfaces behave like stretched elastic sheets, trying to shrink to the smallest possible area. Here are some examples illustrating its existence:
1. **Paint Brush Hairs:** When a camel-hair paint brush is dry, its hairs spread out. But when dipped in water and then pulled out, the hairs cling together. This happens because the thin film of water between the hairs contracts due to surface tension, pulling the hairs closer.
2. **Soap Film Experiment:** If a loop of moistened cotton thread is placed on a soap film inside a metallic wire ring, the film inside the loop can be broken. When the film inside the loop disappears, the remaining soap film outside pulls the thread into a perfect circle. This is because the outer film contracts to minimize its area, and a circle encloses the greatest area for a given perimeter, allowing the thread to be stretched into this shape.
3. **Raindrops:** Small raindrops are spherical because surface tension pulls their surfaces inward, making them take the shape with the smallest surface area, which is a sphere.
4. **Insect Walking on Water:** Some insects, like water striders, can walk on the surface of water without sinking. Their light weight and the high surface tension of water create enough force to support them.
In simple words: Surface tension is seen when wet paint brush hairs stick together, or when soap films pull a thread into a circle. It's also why raindrops are round and why some insects can walk on water.

🎯 Exam Tip: For examples of surface tension, describe a scenario and clearly link the observed phenomenon (e.g., hairs sticking, circular loop, spherical drops) to the liquid's tendency to minimize its surface area.

Concept: 1. For the circular path:
\( 2l = 2\pi \)
Force act on circular path:
\( F = \text{Surface tension} \times \text{length} \)
\( F = T \times 2\pi r \)

Question 7. Derive a relation between surface tension and surface energy.
Answer: Let's consider a rectangular frame with a sliding wire on one of its arms. When this frame is dipped into a soap solution and then taken out, a soap film forms on the frame. This film has two surfaces, both of which are in contact with the sliding wire. Because of surface tension, a force acts on the wire from both surfaces of the soap film.
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 is \( T \times L \). So, the total force on the wire due to both surfaces is \( 2TL \). This force tries to contract the film. Now, if the surfaces contract by a small distance \( \Delta x \), the work done by the film is \( W = \text{Force} \times \text{distance} = (2TL) \times \Delta x \). The change in the surface area of the film is \( \Delta A = 2L \times \Delta x \) (since there are two surfaces). From this, we can see that \( T = \frac{W}{\Delta A} \). This means surface tension is equal to the work done per unit increase in surface area, which is also called surface energy per unit area. This relation is also shown as \( E_s = T\Delta A \). The unit for surface tension \( T \) is \( Jm^{-2} \) and its dimensional formula is \( [MLT^{-2}] \).
In simple words: Surface tension is the energy needed to increase the surface area of a liquid by one unit. It's like how much effort it takes to stretch a liquid's skin. The more energy needed, the higher the surface tension.

🎯 Exam Tip: Remember that surface tension represents the energy per unit area required to expand a liquid surface, and it is also defined as the force per unit length acting perpendicularly on a line drawn on the liquid surface.

Question 8. Explain some daily life examples based on capillarity.
Answer: Here are some everyday examples that show capillarity in action:
1. A pen nib is split at the tip to create a narrow capillary space, which helps draw ink up to the writing point continuously.
2. In oil lamps, the oil moves upwards through the tiny capillaries in the wick.
3. Clay soils stay moist because water quickly rises to the surface through the small capillary channels in the soil.
4. Water and essential minerals travel upwards in plants through very fine tubes called capillaries.
5. Blotting paper works by absorbing ink through its many tiny pores, which act as capillaries.
In simple words: Capillarity helps liquids move in small spaces, like ink in a pen or water in plants, making them go up or spread out.

🎯 Exam Tip: When explaining capillarity examples, always link them back to the concept of narrow tubes or small spaces that cause liquid movement against gravity.

Question 9. Explain shape of meniscus for different liquids in capillary.
Answer: The shape of the liquid surface (meniscus) in a capillary tube, where \( r \) is its radius, depends on the balance between cohesive forces (between liquid molecules) and adhesive forces (between liquid and tube material). We know that the height a liquid rises in a capillary tube is given by the formula:
\[ h = \frac { 2T \cos\theta }{ rpg } \]
Here, \( T \) is surface tension, \( \theta \) is the angle of contact, \( r \) is the radius of the capillary, \( \rho \) is the liquid density, and \( g \) is acceleration due to gravity. For pure water in a glass tube, the angle of contact \( \theta \) is very small (nearly \( 0^\circ \)). This means \( \cos\theta \) is close to 1, and water rises, forming a concave meniscus (curved upwards). For liquids like mercury, the cohesive forces are stronger than the adhesive forces, making the angle of contact greater than \( 90^\circ \). This causes mercury to be depressed in the capillary, forming a convex meniscus (curved downwards).
In simple words: The way a liquid curves in a small tube (meniscus) depends on whether the liquid sticks more to itself or to the tube. Water curves up, while mercury curves down.

🎯 Exam Tip: Clearly state how the balance of adhesive and cohesive forces, and the angle of contact, determine whether the meniscus is concave (like water) or convex (like mercury).

Question 10. What do you understand by the term capillarity? Derive an expression for the rise of liquid in a capillary tube.
Answer: Capillarity is the special property of a liquid to rise or fall within a narrow tube (called a capillary tube) when it is dipped into the liquid. For example, when a fine capillary tube is placed in water, the water level inside the tube rises. However, if the same tube is placed in mercury, the mercury level inside the tube drops lower than the outside level. Generally, liquids that stick well to glass (wet the glass) will rise in the tube, while liquids that do not stick well (do not wet the glass) will be depressed.
To derive the expression for the rise of liquid: When a liquid rises in a capillary tube, the upward force due to surface tension supports the weight of the liquid column. The upward force \( F_T \) is given by \( F_T = (2\pi r) T \cos\theta \), where \( r \) is the radius of the tube, \( T \) is the surface tension, and \( \theta \) is the angle of contact. The weight of the liquid column \( W \) is \( W = (\pi r^2 h) \rho g \), where \( h \) is the height of the liquid column, \( \rho \) is the density of the liquid, and \( g \) is the acceleration due to gravity. For equilibrium, \( F_T = W \).
So, \( 2\pi r T \cos\theta = \pi r^2 h \rho g \).
Solving for \( h \), we get:
\[ h = \frac { 2T \cos\theta }{ r\rho g } \]
This expression shows that the height of the liquid rise is directly related to the surface tension and inversely related to the radius of the tube and the liquid's density.
In simple words: Capillarity is when a liquid moves up or down a thin tube. Water goes up because it sticks to the glass, but mercury goes down. The formula \( h = \frac { 2T \cos\theta }{ r\rho g } \) tells us how high or low the liquid will go, based on its properties and the tube's size.

🎯 Exam Tip: When deriving the expression for capillary rise, clearly state the forces involved (surface tension and weight of liquid column) and ensure the angle of contact is correctly used in the formula.

Question 11. Derive an expression for the excess pressure inside a liquid drop.
Answer: Consider a spherical liquid drop of radius \( r \) and surface tension \( T \). Due to surface tension, the liquid drop tries to contract, creating an inward force. This inward force leads to an excess pressure inside the drop compared to the outside. Let's imagine the drop is split into two halves. The force due to surface tension acting along the circumference of the cut, tending to pull the two halves together, is \( F_1 = (2\pi r)T \). This force acts inwards.
To keep the two halves apart, there must be an outward force caused by the excess pressure \( P \) inside the drop. This outward force acts on the cross-sectional area of the cut, which is \( \pi r^2 \). So, the upward force \( F_2 \) due to excess pressure is \( F_2 = P \times (\pi r^2) \).
For the liquid drop to be in equilibrium, these two forces must balance each other: \( F_1 = F_2 \).
\( 2\pi r T = P \pi r^2 \)
Now, we can solve for the excess pressure \( P \):
\( P = \frac { 2\pi r T }{ \pi r^2 } \)
\[ P = \frac { 2T }{ r } \]
This expression shows that the excess pressure inside a liquid drop is directly proportional to its surface tension and inversely proportional to its radius. Smaller drops have higher excess pressure.
In simple words: Because a liquid drop tries to shrink (due to surface tension), the pressure inside it is slightly higher than the pressure outside. This extra pressure gets bigger when the drop is smaller.

🎯 Exam Tip: Remember to clearly identify the inward force due to surface tension acting on the circumference and the outward force due to excess pressure acting on the cross-sectional area, then equate them for equilibrium.

Question 12. What is the effect of contamination and temperature on the surface tension of a liquid?
Answer: Here's how contamination and temperature affect the surface tension of a liquid:
1. Effect of Temperature: When the temperature of a liquid increases, the kinetic energy of its molecules also increases. This weakens the intermolecular forces between them. As a result, the surface tension of the liquid decreases with rising temperature. At a specific critical temperature, the surface tension can even become zero.
2. Effect of Contaminations: If there are impurities like dust, grease, or oil on the liquid surface, they disrupt the intermolecular forces at the surface. This leads to a decrease in the liquid's surface tension.
3. Effect of Solute: The impact of a solute depends on its solubility. If a solute is highly soluble (like sugar, salt, or lemon in water), it increases the surface tension of water. However, if the solute is less soluble (like soap, petrol, or phenol in water), it decreases the surface tension. For instance, dissolving soap in water reduces surface tension, which is why mosquitoes don't sit on soapy water and sink.
4. Effect of Detergents: Mixing detergents in water significantly reduces its surface tension. This lower surface tension allows the water to easily penetrate the small pores of clothes, helping to remove dirt effectively. Pure water, with its higher surface tension, would not enter these pores as easily. The adhesive forces between detergent molecules and water become stronger than the water's cohesive forces, helping to pull out dirt particles.
In simple words: Surface tension goes down when a liquid gets hotter or has dirt on it. Adding some things (like soap) makes it lower, while others (like salt) can make it higher.

🎯 Exam Tip: For each factor (temperature, contamination, solute, detergents), clearly explain the underlying physical reason for the change in surface tension (e.g., intermolecular forces, molecular kinetic energy).

Question 1. In a hydraulic system, a small piston has a radius of 5 cm, and a larger piston has a radius of 15 cm. If a load of \( 1500 \times 9.8 \, \text{N} \) is to be lifted by the larger piston, calculate the pressure required on the small piston.
Answer: Given values are:
Radius of small piston, \( r_1 = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} \)
Radius of large piston, \( r_2 = 15 \, \text{cm} = 15 \times 10^{-2} \, \text{m} \)
Force on large piston, \( F_2 = 1500 \times 9.8 \, \text{N} \)

First, calculate the areas of the pistons:
Area of small piston, \( A_1 = \pi r_1^2 = \pi \times (5 \times 10^{-2})^2 = \pi \times 25 \times 10^{-4} \, \text{m}^2 \)
Area of large piston, \( A_2 = \pi r_2^2 = \pi \times (15 \times 10^{-2})^2 = \pi \times 225 \times 10^{-4} \, \text{m}^2 \)

According to Pascal's law, the pressure is transmitted equally throughout the fluid:
\( P_1 = P_2 \)
\( \frac{F_1}{A_1} = \frac{F_2}{A_2} \)

From this, we can find the force \( F_1 \) required on the small piston:
\( F_1 = \frac{A_1}{A_2} \times F_2 \)
\( F_1 = \frac{\pi \times 25 \times 10^{-4}}{\pi \times 225 \times 10^{-4}} \times (1500 \times 9.8) \)
\( F_1 = \frac{25}{225} \times (1500 \times 9.8) \)
\( F_1 = \frac{1}{9} \times (1500 \times 9.8) \)
\( F_1 = \frac{14700}{9} \)
\( F_1 \approx 1633.33 \, \text{N} \)

Now, calculate the pressure \( P_1 \) required on the small piston:
\( P_1 = \frac{F_1}{A_1} \)
\( P_1 = \frac{1633.33}{\pi \times 25 \times 10^{-4}} \)
\( P_1 = \frac{1633.33}{3.14159 \times 25 \times 10^{-4}} \)
\( P_1 = \frac{1633.33}{0.007853975} \)
\( P_1 \approx 208000 \, \text{N/m}^2 \)
\[ P_1 = 2.08 \times 10^5 \, \text{Nm}^{-2} \]
In simple words: Using Pascal's law, a small force applied to a small piston can create a large force on a bigger piston. To lift the heavy load, a pressure of \( 2.08 \times 10^5 \, \text{N/m}^2 \) must be applied to the small piston.

🎯 Exam Tip: When using Pascal's law, ensure you correctly identify the input and output forces and areas. Remember that pressure is constant throughout the fluid, leading to \( \frac{F_1}{A_1} = \frac{F_2}{A_2} \).

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 the density of the liquid is 1.5 gm/cm³. Calculate the value of \( \eta \).
Answer: Given values are:
Diameter of bubble = 2 cm, so radius \( r = 1 \, \text{cm} \)
Constant velocity (terminal velocity) \( v = 0.90 \, \text{cm/s} \)
Density of gas (bubble) \( \rho_{\text{gas}} \approx 0 \)
Density of liquid \( \rho_{\text{liquid}} = 1.5 \, \text{gm/cm}^3 \)
Acceleration due to gravity \( g = 980 \, \text{cm/s}^2 \) (in CGS units for consistency with density and radius)

The formula for terminal velocity \( v \) for a sphere in a viscous liquid is given by:
\[ v = \frac { 2r^2 (\rho_{\text{body}} - \rho_{\text{fluid}}) g }{ 9\eta } \]
When a bubble rises, its density is less than the liquid, so we consider \( (\rho_{\text{fluid}} - \rho_{\text{bubble}}) \). The formula rearranged to solve for viscosity \( \eta \) is:
\( \eta = \frac { 2r^2 (\rho_{\text{fluid}} - \rho_{\text{bubble}}) g }{ 9v } \)

Following the provided solution's numerical steps, which seem to use a velocity of 90 cm/s to match the final result of 3.63 Poise, we have:
\( \eta = \frac{2 \times (1)^2 \times (1.5 - 0) \times 980}{9 \times 90} \)
\( \eta = \frac{2 \times 1 \times 1.5 \times 980}{810} \)
\( \eta = \frac{2940}{810} \)
\( \eta \approx 3.6296 \, \text{Poise} \)
The negative sign in the original source's calculation for \( (\rho_{\text{gas}} - \rho_{\text{liquid}}) \) indicates the upward motion of the bubble. Therefore, the magnitude of viscosity \( \eta \) is approximately \( 3.63 \, \text{Poise} \).
In simple words: We used a special formula to find how thick a liquid is (its viscosity, \( \eta \)) by seeing how fast a gas bubble floats up through it. The number tells us how much the liquid resists the bubble's movement.

🎯 Exam Tip: Ensure all units are consistent (e.g., CGS or SI) before performing calculations. For rising bubbles, consider the effective density difference that determines buoyancy against viscous drag.

Question 3. Find out the terminal velocity of a rain water drop with a radius of 1 mm. Given the viscous coefficient for air is \( 1.8 \times 10^{-2} \, \text{Ns/m}^2 \), the density of air is \( 1.2 \, \text{kg/m}^3 \), the density of water is \( 10^3 \, \text{kg/m}^3 \), and \( g = 10 \, \text{m/s}^2 \).
Answer: Given values are:
Radius of raindrop \( r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
Viscous coefficient of air \( \eta = 1.8 \times 10^{-2} \, \text{Ns/m}^2 \)
Density of water \( \rho_{\text{water}} = 10^3 \, \text{kg/m}^3 \)
Density of air \( \rho_{\text{air}} = 1.2 \, \text{kg/m}^3 \)
Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

The formula for terminal velocity \( v_t \) of a falling sphere is:
\[ v_t = \frac { 2r^2 (\rho_{\text{water}} - \rho_{\text{air}}) g }{ 9\eta } \]
Substitute the given values into the formula:
\( v_t = \frac { 2 \times (1 \times 10^{-3})^2 \times (10^3 - 1.2) \times 10 }{ 9 \times (1.8 \times 10^{-2}) } \)
\( v_t = \frac { 2 \times 10^{-6} \times (998.8) \times 10 }{ 16.2 \times 10^{-2} } \)
\( v_t = \frac { 2 \times 9988 \times 10^{-6} }{ 16.2 \times 10^{-2} } \)
\( v_t = \frac { 19.976 \times 10^{-3} }{ 0.162 } \)
\( v_t \approx 0.1233 \, \text{m/s} \)
Converting to cm/s:
\( v_t = 0.1233 \times 100 \, \text{cm/s} \approx 12.33 \, \text{cm/s} \)
So, the terminal velocity of the rain water drop is approximately \( 0.123 \, \text{m/s} \) or \( 12.3 \, \text{cm/s} \).
In simple words: We calculate how fast a raindrop falls at a steady speed by using its size, the thickness of the air, and the difference in density between the water and air. The answer shows it falls at about 12.3 cm every second.

🎯 Exam Tip: Ensure that the correct value of viscosity for the medium (air in this case) is used. Pay attention to the units given and ensure consistent use of either SI or CGS units throughout the calculation.

Question 4. Water rises in a capillary tube up to a height of 4 cm. If the capillary tube is inclined at an angle of 30° with the vertical, what will be the position of water in the tube?
Answer: Given values are:
Initial height of water rise in a vertical capillary tube \( h = 4 \, \text{cm} \)
Angle of inclination with the vertical \( \alpha = 30^\circ \)

When a capillary tube is inclined at an angle \( \alpha \) with the vertical, the liquid rises to a length \( l \) along the tube. The vertical height \( h \) of the liquid column is related to this length by the formula:
\( h = l \cos\alpha \)
We need to find the length \( l \) of the water column along the inclined tube. Rearranging the formula:
\( l = \frac{h}{\cos\alpha} \)
Substitute the given values:
\( l = \frac{4 \, \text{cm}}{\cos(30^\circ)} \)
\( l = \frac{4 \, \text{cm}}{\frac{\sqrt{3}}{2}} \)
\( l = \frac{4 \times 2}{\sqrt{3}} \, \text{cm} \)
\( l = \frac{8}{\sqrt{3}} \, \text{cm} \)
\( l = \frac{8}{1.732} \, \text{cm} \)
\( l \approx 4.6189 \, \text{cm} \)
So, the water will be at a position of approximately \( 4.62 \, \text{cm} \) along the length of the inclined capillary tube.
In simple words: When you tilt a tube where water normally rises, the water column stretches longer along the tilt, but its vertical height stays the same. So, with a 30-degree tilt, the water will go up about 4.62 cm along the tube.

🎯 Exam Tip: Remember that the vertical height of the liquid column remains constant regardless of the tube's inclination. The length of the column along the inclined tube is related by \( l = h/\cos\alpha \).

Question 5. If two soap bubbles of radii \( R_1 \) and \( R_2 \) combine to form a bigger bubble at a constant temperature, then find out the resultant radius of the big bubble.
Answer: When two soap bubbles combine to form a larger bubble at a constant temperature, the total surface energy of the system remains conserved. A soap bubble has two surfaces (inner and outer), so its surface energy is \( E = 2 \times (\text{Surface Tension} \times \text{Area}) = 2 \times T \times (4\pi r^2) \).
Let the radius of the larger bubble be \( R \). The surface energy of the new, bigger bubble will be \( E_{\text{new}} = 2 \times T \times (4\pi R^2) \).
The sum of the surface energies of the two original bubbles is \( E_{\text{old}} = (2 \times T \times 4\pi R_1^2) + (2 \times T \times 4\pi R_2^2) \).
By the principle of conservation of energy:
\( E_{\text{new}} = E_{\text{old}} \)
\( 2 \times T \times 4\pi R^2 = (2 \times T \times 4\pi R_1^2) + (2 \times T \times 4\pi R_2^2) \)
We can cancel \( 2 \times T \times 4\pi \) from both sides:
\( R^2 = R_1^2 + R_2^2 \)
To find the resultant radius \( R \), we take the square root of both sides:
\[ R = \sqrt{R_1^2 + R_2^2} \]
In simple words: When two soap bubbles merge into one at the same temperature, the energy from their surfaces adds up. This means the square of the new bubble's radius is equal to the sum of the squares of the original bubbles' radii.

🎯 Exam Tip: Remember that for soap bubbles, surface energy calculations must account for two surfaces (inner and outer). The conservation of surface energy is the key principle for this type of problem.

Question 6. A water tank is filled with water up to height \( H \). There is an orifice in the wall of the water tank at a depth \( D \) from the water surface. If water is emerging from this orifice to the ground, find out the horizontal range of the water on the ground.
Answer: Given values are:
Total height of water in tank = \( H \)
Depth of orifice from free surface = \( D \)

First, we find the velocity of efflux (speed at which water leaves the orifice) using Torricelli's Law:
\( u = \sqrt{2gD} \)

Next, we consider the time it takes for the water to fall from the orifice to the ground. The vertical distance the water falls is \( (H - D) \). Using the equation of motion for free fall (assuming initial vertical velocity is zero):
\( S = ut + \frac{1}{2}gt^2 \)
\( H - D = 0 \cdot t + \frac{1}{2}gt^2 \)
\( H - D = \frac{1}{2}gt^2 \)
\( t^2 = \frac{2(H - D)}{g} \)
\( t = \sqrt{\frac{2(H - D)}{g}} \)

Finally, the horizontal range \( x \) is calculated by multiplying the horizontal velocity by the time of flight:
\( x = u \times t \)
\( x = \sqrt{2gD} \times \sqrt{\frac{2(H - D)}{g}} \)
\( x = \sqrt{\frac{2gD \times 2(H - D)}{g}} \)
\( x = \sqrt{4D(H - D)} \)
\[ x = 2\sqrt{D(H - D)} \]
This is the horizontal range of the water emerging from the orifice.
In simple words: The water shoots out sideways from the hole with a speed based on how deep the hole is. Then, it falls to the ground from that height. We calculate how far it lands horizontally using its shooting speed and the time it takes to fall.

🎯 Exam Tip: Remember to use Torricelli's Law for the efflux velocity and standard projectile motion equations for the time of flight and horizontal range. Clearly define the vertical and horizontal components of motion.

Question 7. Two soap bubbles have diameters in the ratio 2: 3. Compare the excess of pressure inside these bubbles.
Answer: Given values are:
Ratio of diameters of two soap bubbles \( D_1 : D_2 = 2 : 3 \)
So, \( \frac{D_1}{D_2} = \frac{2}{3} \)

The excess pressure \( P_{ex} \) inside a soap bubble is given by the formula:
\( P_{ex} = \frac{4T}{R} \)
Where \( T \) is the surface tension and \( R \) is the radius of the bubble.
Since the diameter \( D = 2R \), we can write \( R = \frac{D}{2} \). Substituting this into the formula:
\( P_{ex} = \frac{4T}{D/2} = \frac{8T}{D} \)

Now, let's compare the excess pressure for the two bubbles:
For the first bubble: \( P_{ex1} = \frac{8T}{D_1} \)
For the second bubble: \( P_{ex2} = \frac{8T}{D_2} \)

The ratio of their excess pressures will be:
\( \frac{P_{ex1}}{P_{ex2}} = \frac{8T/D_1}{8T/D_2} \)
\( \frac{P_{ex1}}{P_{ex2}} = \frac{D_2}{D_1} \)

Since \( \frac{D_1}{D_2} = \frac{2}{3} \), then \( \frac{D_2}{D_1} = \frac{3}{2} \).
Therefore, the ratio of the excess pressures is:
\( \frac{P_{ex1}}{P_{ex2}} = \frac{3}{2} \)
Or, \( P_{ex1} : P_{ex2} = 3 : 2 \)
In simple words: For soap bubbles, a smaller bubble has more extra pressure inside it than a bigger one. If one bubble is two parts big and another is three parts big, the pressure inside the smaller one will be three parts for every two parts in the bigger one.

🎯 Exam Tip: Remember the relationship between excess pressure and radius for a soap bubble (\( P \propto 1/R \)). When comparing, make sure to set up the ratio correctly, noting that smaller diameter implies higher excess pressure.

Question 8. Water rises in a capillary tube up to a height of 10 cm. If the surface tension of water is \( 73 \times 10^{-3} \, \text{N/m} \), then find out the radius of the capillary.
Answer: Given values are:
Height of water rise \( h = 10 \, \text{cm} = 0.1 \, \text{m} \)
Surface tension of water \( T = 73 \times 10^{-3} \, \text{N/m} \)
Density of water \( \rho = 1000 \, \text{kg/m}^3 \)
Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)
For pure water and glass, the angle of contact \( \theta \) is approximately \( 0^\circ \), so \( \cos\theta = 1 \).

The formula for capillary rise is:
\[ h = \frac { 2T \cos\theta }{ r\rho g } \]
We need to find the radius \( r \). Rearranging the formula:
\( r = \frac{2T \cos\theta}{h\rho g} \)
Substitute the given values:
\( r = \frac{2 \times (73 \times 10^{-3}) \times 1}{0.1 \times 1000 \times 9.8} \)
\( r = \frac{0.146}{98} \)
\( r \approx 0.00149 \, \text{m} \)
Converting to centimeters:
\( r \approx 0.00149 \times 100 \, \text{cm} \)
\( r \approx 0.149 \, \text{cm} \)
Therefore, the radius of the capillary is approximately \( 0.149 \, \text{cm} \).
In simple words: By knowing how high water rises in a thin tube and the water's surface tension, we can calculate how narrow the tube is (its radius) using a specific physics formula.

🎯 Exam Tip: Always ensure unit consistency (SI units are recommended) throughout your calculations. For capillary rise in water in a glass tube, assume \( \cos\theta = 1 \) unless otherwise specified.

Question 9. Water is flowing through a non-uniform cross-sectional pipe. In one section, the radius of the pipe is 2 cm, and the velocity of water is 20 cm/s. If at another place, the radius of the pipe is 6 cm, find out the velocity of the water.
Answer: Given values are:
Radius of the first section \( r_1 = 2 \, \text{cm} \)
Velocity of water in the first section \( v_1 = 20 \, \text{cm/s} \)
Radius of the second section \( r_2 = 6 \, \text{cm} \)
Velocity of water in the second section \( v_2 = ? \)

According to the equation of continuity for incompressible fluids:
\( A_1v_1 = A_2v_2 \)
Where \( A_1 \) and \( A_2 \) are the cross-sectional areas, and \( v_1 \) and \( v_2 \) are the velocities at the respective sections. Since the pipe sections are circular, \( A = \pi r^2 \).
\( \pi r_1^2 v_1 = \pi r_2^2 v_2 \)
We can cancel \( \pi \) from both sides:
\( r_1^2 v_1 = r_2^2 v_2 \)
Now, solve for \( v_2 \):
\( v_2 = \frac{r_1^2}{r_2^2} v_1 = \left(\frac{r_1}{r_2}\right)^2 v_1 \)
Substitute the given values:
\( v_2 = \left(\frac{2 \, \text{cm}}{6 \, \text{cm}}\right)^2 \times 20 \, \text{cm/s} \)
\( v_2 = \left(\frac{1}{3}\right)^2 \times 20 \, \text{cm/s} \)
\( v_2 = \frac{1}{9} \times 20 \, \text{cm/s} \)
\( v_2 = \frac{20}{9} \, \text{cm/s} \)
\( v_2 \approx 2.22 \, \text{cm/s} \)
Therefore, the velocity of water in the second section of the pipe is approximately \( 2.22 \, \text{cm/s} \).
In simple words: When water flows through a pipe that changes width, it speeds up in narrower parts and slows down in wider parts. We use a rule called the equation of continuity to find its new speed when the pipe gets wider.

🎯 Exam Tip: Remember that for fluid flow, the product of the cross-sectional area and velocity remains constant. Ensure that radii are used in squared form when calculating area ratios.

Question 10. What should be the excess pressure inside a water drop whose radius is 2 mm? The surface tension of water is 0.075 N/m.
Answer: Given values are:
Radius of the water drop \( R = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
Surface tension of water \( T = 0.075 \, \text{N/m} \)

For a single liquid drop, the excess pressure \( P_{ex} \) inside the drop compared to the outside is given by the formula:
\[ P_{ex} = \frac { 2T }{ R } \]
Substitute the given values into the formula:
\( P_{ex} = \frac{2 \times 0.075 \, \text{N/m}}{2 \times 10^{-3} \, \text{m}} \)
\( P_{ex} = \frac{0.15}{0.002} \, \text{N/m}^2 \)
\( P_{ex} = 75 \, \text{N/m}^2 \)
Therefore, the excess pressure inside the water drop is \( 75 \, \text{N/m}^2 \).
In simple words: The tiny water drop has extra pressure inside it because its surface tension pulls inwards. We calculate this extra pressure using how strong the surface tension is and how small the drop is.

🎯 Exam Tip: Remember the formula for excess pressure in a liquid drop (\( P = 2T/R \)) and distinguish it from a soap bubble (\( P = 4T/R \)) which has two surfaces.

Question 11. Prove that the work done by a big drop of radius \( R \) in spraying into \( n \) small drops of radius \( r \) will be \( 4\pi (n^{1/3} - 1) R^2 T \), where \( T \) is the surface tension.
Answer: We need to show that the work done \( W \) is equal to the change in surface energy when a larger drop splits into many smaller ones.

First, we use the principle of conservation of volume. The volume of the one big drop must equal the total volume of the \( n \) small drops:
\( \frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3 \)
Canceling \( \frac{4}{3}\pi \) from both sides:
\( R^3 = nr^3 \)
From this, we can express the radius of a small drop \( r \) in terms of the big drop's radius \( R \) and the number of drops \( n \):
\( r = \frac{R}{n^{1/3}} \)

Next, calculate the total initial and final surface areas:
Initial surface area of the big drop \( A_1 = 4\pi R^2 \)
Final surface area of \( n \) small drops \( A_2 = n \times 4\pi r^2 \)
Substitute the expression for \( r \):
\( A_2 = n \times 4\pi \left(\frac{R}{n^{1/3}}\right)^2 \)
\( A_2 = n \times 4\pi \frac{R^2}{n^{2/3}} \)
\( A_2 = 4\pi R^2 n^{1 - 2/3} = 4\pi R^2 n^{1/3} \)

The increment in surface area \( \Delta A \) is the difference between the final and initial areas:
\( \Delta A = A_2 - A_1 \)
\( \Delta A = 4\pi R^2 n^{1/3} - 4\pi R^2 \)
\( \Delta A = 4\pi R^2 (n^{1/3} - 1) \)

Finally, the work done \( W \) in increasing the surface area is given by:
\( W = T \Delta A \)
\( W = T \times 4\pi R^2 (n^{1/3} - 1) \)
\[ W = 4\pi (n^{1/3} - 1) R^2 T \]
This proves the desired relation.
In simple words: When a big water drop breaks into many smaller ones, its total surface area increases. The work done to create this new surface area is calculated by how much the area grew and the liquid's surface tension.

🎯 Exam Tip: For problems involving changes in droplet size, always use conservation of volume to relate the radii and then calculate the change in total surface area. Work done equals surface tension times the change in surface area.

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 the wing and lower surface of the wing. Find out the lifting force. Given that the total wing area is 2.5 m² and air density is 1.3 kg/m³.
Answer: Given values are:
Velocity of air over the upper surface of the wing \( v_{\text{upper}} = 70 \, \text{m/s} \)
Velocity of air under the lower surface of the wing \( v_{\text{lower}} = 63 \, \text{m/s} \)
Total wing area \( A = 2.5 \, \text{m}^2 \)
Density of air \( \rho = 1.3 \, \text{kg/m}^3 \)

According to Bernoulli's principle, for streamline flow in a horizontal plane, the sum of pressure and kinetic energy per unit volume is constant:
\( P + \frac{1}{2}\rho v^2 = \text{constant} \)
This means that where the velocity of air is higher, the pressure is lower, and vice-versa. The difference in pressure between the lower and upper surfaces of the wing creates the lifting force.

The pressure difference \( \Delta P \) is given by:
\( \Delta P = P_{\text{lower}} - P_{\text{upper}} = \frac{1}{2}\rho (v_{\text{upper}}^2 - v_{\text{lower}}^2) \)
Substitute the given values:
\( \Delta P = \frac{1}{2} \times 1.3 \, \text{kg/m}^3 \times ((70 \, \text{m/s})^2 - (63 \, \text{m/s})^2) \)
\( \Delta P = \frac{1}{2} \times 1.3 \times (4900 - 3969) \)
\( \Delta P = \frac{1}{2} \times 1.3 \times 931 \)
\( \Delta P = 0.65 \times 931 \)
\( \Delta P = 605.15 \, \text{N/m}^2 \)

The lifting force \( F \) is the pressure difference multiplied by the total wing area:
\( F = \Delta P \times A \)
\( F = 605.15 \, \text{N/m}^2 \times 2.5 \, \text{m}^2 \)
\( F = 1512.875 \, \text{N} \)
Rounding to appropriate significant figures, the lifting force is approximately \( 1.5 \times 10^3 \, \text{N} \).
In simple words: An airplane flies because air moves faster over the curved top of its wing than under the flat bottom. This speed difference creates less pressure on top and more pressure below, pushing the wing upwards and creating lift.

🎯 Exam Tip: Clearly apply Bernoulli's principle to relate pressure and velocity differences. Remember that lift force is the product of the pressure difference and the wing's effective area.

Question 13. Water is flowing through a non-uniform cross-sectional area pipe. At one point in the pipe, the velocity of water is 0.4 m/s, and the pressure is equivalent to 0.1 m of mercury. At another place, where the velocity of water is 0.5 m/s, calculate the value of the pressure.
Answer: Given values are:
At point 1:
Velocity \( v_1 = 0.4 \, \text{m/s} \)
Pressure \( P_1 = 0.1 \, \text{m} \) of mercury

At point 2:
Velocity \( v_2 = 0.5 \, \text{m/s} \)
Pressure \( P_2 = ? \)

We also need:
Density of water \( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \)
Density of mercury \( \rho_{\text{Hg}} = 13.6 \times 10^3 \, \text{kg/m}^3 \)
Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)

First, convert the pressure at point 1 from `m of mercury` to Pascals (N/m²):
\( P_1 = \rho_{\text{Hg}} g h_{\text{Hg}} \)
\( P_1 = (13.6 \times 10^3 \, \text{kg/m}^3) \times (9.8 \, \text{m/s}^2) \times (0.1 \, \text{m}) \)
\( P_1 = 13328 \, \text{N/m}^2 \)

Assuming the pipe is horizontal (so the height term \( \rho gh \) is constant and cancels out), we apply Bernoulli's theorem:
\( P_1 + \frac{1}{2}\rho_{\text{water}} v_1^2 = P_2 + \frac{1}{2}\rho_{\text{water}} v_2^2 \)
Rearranging to solve for \( P_2 \):
\( P_2 = P_1 + \frac{1}{2}\rho_{\text{water}} (v_1^2 - v_2^2) \)
Substitute the values:
\( P_2 = 13328 + \frac{1}{2} \times 1000 \times ((0.4)^2 - (0.5)^2) \)
\( P_2 = 13328 + 500 \times (0.16 - 0.25) \)
\( P_2 = 13328 + 500 \times (-0.09) \)
\( P_2 = 13328 - 45 \)
\( P_2 = 13283 \, \text{N/m}^2 \)

To express this pressure in `m of mercury` (as in the question):
\( h_{\text{Hg}} = \frac{P_2}{\rho_{\text{Hg}} g} \)
\( h_{\text{Hg}} = \frac{13283}{13.6 \times 10^3 \times 9.8} \)
\( h_{\text{Hg}} = \frac{13283}{133280} \)
\( h_{\text{Hg}} \approx 0.09966 \, \text{m of mercury} \)
This is approximately \( 0.0996 \, \text{m} \) of mercury, or \( 9.96 \, \text{cm} \) of mercury.
In simple words: When water speeds up in a pipe, its pressure goes down, and when it slows down, its pressure goes up. We used Bernoulli's rule to find the pressure at the second spot after converting the initial mercury height to actual pressure units.

🎯 Exam Tip: When applying Bernoulli's principle, ensure consistent units for pressure (Pascals), density, and velocity. Remember to convert pressure given in 'meters of a fluid' to Pascals before using it in the equation.

Question 14. 1000 small drops of water, each of radius \( 10^{-7} \, \text{m} \), are falling down and coalesce to form a bigger drop. Find out the free energy. The surface tension of water is \( 7 \times 10^{-2} \, \text{N/m} \).
Answer: Given values are:
Number of small drops \( n = 1000 \)
Radius of each small drop \( r = 10^{-7} \, \text{m} \)
Surface tension of water \( T = 7 \times 10^{-2} \, \text{N/m} \)

First, we find the radius of the bigger drop \( R \) formed by the coalescence of \( n \) small drops, using the conservation of volume:
\( n \times \text{Volume of one small drop} = \text{Volume of the big drop} \)
\( n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \)
\( n r^3 = R^3 \)
\( R = n^{1/3} r \)
\( R = (1000)^{1/3} \times 10^{-7} \, \text{m} \)
\( R = 10 \times 10^{-7} \, \text{m} = 10^{-6} \, \text{m} \)

Next, calculate the initial and final total surface areas:
Total initial surface area of \( n \) small drops \( A_1 = n \times 4\pi r^2 \)
\( A_1 = 1000 \times 4\pi (10^{-7})^2 = 4000\pi \times 10^{-14} = 4\pi \times 10^{-11} \, \text{m}^2 \)
Total final surface area of the big drop \( A_2 = 4\pi R^2 \)
\( A_2 = 4\pi (10^{-6})^2 = 4\pi \times 10^{-12} \, \text{m}^2 \)

When drops coalesce, the total surface area decreases, and this decrease in surface energy is released as "free energy". The reduction in area \( \Delta A \) is:
\( \Delta A = A_1 - A_2 \)
\( \Delta A = (4\pi \times 10^{-11}) - (4\pi \times 10^{-12}) \)
\( \Delta A = 4\pi \times 10^{-12} (10 - 1) \)
\( \Delta A = 4\pi \times 10^{-12} \times 9 = 36\pi \times 10^{-12} \, \text{m}^2 \)
\( \Delta A \approx 36 \times 3.14159 \times 10^{-12} \approx 113.097 \times 10^{-12} \, \text{m}^2 \)

Finally, calculate the free energy released \( \Delta E \) (which is equal to the reduction in surface energy):
\( \Delta E = T \times \Delta A \)
\( \Delta E = (7 \times 10^{-2} \, \text{N/m}) \times (36\pi \times 10^{-12} \, \text{m}^2) \)
\( \Delta E = 252\pi \times 10^{-14} \, \text{J} \)
\( \Delta E \approx 252 \times 3.14159 \times 10^{-14} \, \text{J} \)
\( \Delta E \approx 791.68 \times 10^{-14} \, \text{J} \)
\[ \Delta E \approx 7.91 \times 10^{-12} \, \text{J} \]
Therefore, the free energy released when the drops coalesce is approximately \( 7.91 \times 10^{-12} \, \text{J} \).
In simple words: When many tiny water drops join to form one big drop, the total surface area shrinks. This shrinking releases energy, which we call free energy. We calculate this energy using the liquid's surface tension and how much the total surface area decreased.

🎯 Exam Tip: Remember that coalescence of drops leads to a decrease in total surface area and thus a release of surface energy. The work done is calculated as the surface tension multiplied by this reduction in area.

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.
Answer: Let the mass of the first spherical glass ball be \( M_1 = M \) and its terminal velocity be \( V_1 = V \). Let its radius be \( r_1 \).
The mass of the second spherical glass ball is \( M_2 = 8M \). Let its radius be \( r_2 \) and its terminal velocity be \( V_2 \).
The mass of a sphere is given by \( M = \frac{4}{3} \pi r^3 \rho \), where \( \rho \) is the density of the glass.
Since the density of the glass ball is constant, \( M \propto r^3 \).
So, \( \frac{M_2}{M_1} = \frac{r_2^3}{r_1^3} \)
Substituting the given values: \( \frac{8M}{M} = \frac{r_2^3}{r_1^3} \)
\( 8 = (\frac{r_2}{r_1})^3 \)
Taking the cube root of both sides: \( \frac{r_2}{r_1} = 2 \implies r_2 = 2r_1 \).
According to Stokes' Law, the terminal velocity \( V_T \) for a spherical body falling through a viscous fluid is proportional to the square of its radius, i.e., \( V_T \propto r^2 \).
So, we can write: \( \frac{V_2}{V_1} = \frac{r_2^2}{r_1^2} \)
Substitute \( r_2 = 2r_1 \):
\( \frac{V_2}{V} = \frac{(2r_1)^2}{r_1^2} \)
\( \frac{V_2}{V} = \frac{4r_1^2}{r_1^2} \)
\( \frac{V_2}{V} = 4 \)
\( \implies V_2 = 4V \)
Thus, the terminal velocity of the second spherical glass ball will be 4 times the terminal velocity of the first ball.
In simple words: When the mass of a spherical ball increases by 8 times, its radius doubles. Since terminal velocity depends on the square of the radius, the new terminal velocity will be 4 times the original velocity.

🎯 Exam Tip: Remember the relationships: Volume is proportional to radius cubed (\( V \propto r^3 \)) and terminal velocity is proportional to radius squared (\( V_T \propto r^2 \)). This allows you to relate changes in mass to changes in terminal velocity.

Question 16. Work done in increasing size 10 × 11cm from 10 cm x 6 cm of a soap film is 3.0 × 10-4 Joule. Find out the surface tension of the film.
Answer: First, we calculate the initial and final areas of the soap film.
Initial area \( A_1 = 10 \, \text{cm} \times 6 \, \text{cm} = 60 \, \text{cm}^2 \).
Final area \( A_2 = 10 \, \text{cm} \times 11 \, \text{cm} = 110 \, \text{cm}^2 \).
A soap film has two surfaces, so the change in total surface area is twice the change in geometric area.
Change in area \( \Delta A = 2 \times (A_2 - A_1) \)
\( \Delta A = 2 \times (110 \, \text{cm}^2 - 60 \, \text{cm}^2) \)
\( \Delta A = 2 \times 50 \, \text{cm}^2 = 100 \, \text{cm}^2 \)
Convert the area to square meters:
\( 1 \, \text{cm} = 10^{-2} \, \text{m} \)
\( 1 \, \text{cm}^2 = (10^{-2} \, \text{m})^2 = 10^{-4} \, \text{m}^2 \)
So, \( \Delta A = 100 \times 10^{-4} \, \text{m}^2 = 1 \times 10^{-2} \, \text{m}^2 \).
The work done \( W \) in increasing the surface area of a liquid film is given by \( W = T \times \Delta A \), where \( T \) is the surface tension.
We are given \( W = 3.0 \times 10^{-4} \, \text{Joule} \).
We need to find \( T \). Rearrange the formula: \( T = \frac{W}{\Delta A} \)
\( T = \frac{3.0 \times 10^{-4} \, \text{J}}{1 \times 10^{-2} \, \text{m}^2} \)
\( T = 3.0 \times 10^{-4} \times 10^{2} \, \text{N/m} \)
\( T = 3.0 \times 10^{-2} \, \text{N/m} \)
In simple words: A soap film has two sides. We first calculate how much the total surface area changes, remembering there are two surfaces. Then, we divide the given work done by this change in area to find the surface tension.

🎯 Exam Tip: For soap films, always remember to multiply the geometric area change by 2 because a film has two free surfaces in contact with air.

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.
Answer: We can use Torricelli's Law to find the velocity of efflux. This law states that the speed of efflux from an orifice is the same as the speed a body would acquire in falling freely through the same vertical height.
The formula for the velocity of efflux \( v \) is given by \( v = \sqrt{2gh} \).
Given values are:
Height of the orifice below the surface, \( h = 3.5 \, \text{m} \)
Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
Now, substitute these values into the formula:
\( v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 3.5 \, \text{m}} \)
\( v = \sqrt{19.6 \times 3.5} \, \text{m/s} \)
\( v = \sqrt{68.6} \, \text{m/s} \)
Calculating the square root:
\( v \approx 8.28 \, \text{m/s} \)
Rounding to one decimal place:
\( v \approx 8.3 \, \text{m/s} \)
So, the velocity of efflux at the orifice is approximately 8.3 m/s.
In simple words: When water flows out of a hole in a tank, its speed depends on how deep the hole is under the water surface. We use a formula that multiplies two times the gravity by the depth and then takes the square root to find this speed.

🎯 Exam Tip: Torricelli's Law ( \( v = \sqrt{2gh} \) ) is a direct application of Bernoulli's principle for efflux problems. Ensure you use the correct height (depth of the orifice from the free surface).

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.
Answer: Let the initial volume of the soap bubble be \( V_1 \) and the work done be \( W_1 \). Let its radius be \( R_1 \).
The volume of a spherical bubble is given by \( V = \frac{4}{3} \pi R^3 \). So, \( R = \left( \frac{3V}{4\pi} \right)^{1/3} \).
The surface area of a spherical bubble is \( A = 4 \pi R^2 \).
The work done \( W \) in blowing a soap bubble is proportional to its surface area, because a soap bubble has two surfaces. So, \( W = T \times (2 \times A) \), where \( T \) is the surface tension.
Thus, \( W \propto R^2 \).
From the volume formula, \( R \propto V^{1/3} \).
Combining these, \( W \propto (V^{1/3})^2 \), which means \( W \propto V^{2/3} \).
We are given that for volume \( V_1 = V \), the work done is \( W_1 = W \).
We want to find the work done \( W_2 \) for a new volume \( V_2 = 2V \).
Using the proportionality \( W \propto V^{2/3} \):
\( \frac{W_2}{W_1} = \left( \frac{V_2}{V_1} \right)^{2/3} \)
Substitute the given values:
\( \frac{W_2}{W} = \left( \frac{2V}{V} \right)^{2/3} \)
\( \frac{W_2}{W} = (2)^{2/3} \)
\( \frac{W_2}{W} = (2^2)^{1/3} \)
\( \frac{W_2}{W} = (4)^{1/3} \)
\( \implies W_2 = (4)^{1/3} W \)
So, the work done in making a bubble of volume 2V is \( (4)^{1/3} W \).
In simple words: The energy needed to make a soap bubble grows slower than its volume. If you double the bubble's volume, the energy needed to create it won't just double; it will be \( (4)^{1/3} \) times the original energy.

🎯 Exam Tip: Remember that work done in blowing a bubble is proportional to its surface area (\( R^2 \)), while volume is proportional to \( R^3 \). This means work done is proportional to (volume)\(^{2/3}\).

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/m³, g = 9,8 m/s². (Neglecting viscosity)
Answer: The minimum pressure required to pump blood from the heart to the brain can be calculated using the hydrostatic pressure formula, as we are considering the height difference and neglecting viscosity for simplicity.
The formula for pressure due to a fluid column is \( P = \rho gh \).
Given values are:
Density of blood, \( \rho = 1040 \, \text{kg/m}^3 \)
Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
Height of the brain from the heart, \( h = 0.5 \, \text{m} \)
Substitute these values into the formula:
\( P = 1040 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times 0.5 \, \text{m} \)
\( P = 1040 \times 4.9 \, \text{N/m}^2 \)
\( P = 5096 \, \text{N/m}^2 \)
To express this in scientific notation:
\( P = 5.096 \times 10^3 \, \text{N/m}^2 \)
Therefore, the minimum pressure in flowing blood from the heart to the brain is \( 5.096 \times 10^3 \, \text{N/m}^2 \).
In simple words: To pump blood upwards from the heart to the brain, a certain amount of pressure is needed to overcome gravity. This pressure depends on how dense the blood is, the pull of gravity, and the height difference between the heart and the brain.

🎯 Exam Tip: For problems involving fluid pressure due to height, remember the formula \( P = \rho gh \). Ensure all units are consistent (e.g., SI units) for a correct numerical answer.