gives us an idea of composition of air

is the height above which there is no air

MECHANICAL PROPERTIES OF FLUIDS MCQs for NEET UG PHYSICS — Medium, Page 2

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Understanding +4 | -1

A body floats in a liquid contained in a beaker. If the whole system falls under gravity, them the upthrust on the body due to liquids is

View answer and explanation

Correct option: C

zero

Option C (zero) is correct because in free fall, the effective weight of both the liquid and the body becomes zero, resulting in no buoyant force.

Solution approach

Correct Answer + Core Reason: zero. In free fall, the effective weight of the liquid and the body is zero, hence no upthrust.

Step-by-Step Solution:

Step 1: Understand the concept of free fall. In free fall, the entire system (beaker, liquid, and body) is accelerating downwards with an acceleration equal to $g$ .

Step 2: The effective weight of any object in free fall is zero. This is because the gravitational force is providing the acceleration, and there is no reaction force.

Step 3: Since the liquid is also in free fall, it exerts no upward force (upthrust) on the body. The upthrust is due to the pressure difference in the liquid, which is caused by the weight of the liquid. In free fall, there is no effective weight, hence no pressure difference, and no upthrust.

Why other options are wrong:
- Option A: equal to the weight of the body in air. This is incorrect because the upthrust is zero in free fall.
- Option B: equal to the weight of the body in liquid. This is incorrect because the upthrust is zero in free fall.
- Option D: equal to the weight of the immersed part of the body. This is incorrect because the upthrust is zero in free fall.

Key Concepts:
- Free Fall
- Upthrust
- Effective Weight

Memory Aids:
- In free fall, everything is weightless.

Quick Tips:
- When analyzing situations in free fall, consider the effective weight of objects.

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Understanding +4 | -1

At which of the following temperatures, the value of surface tension of water is minimum?

View answer and explanation

Correct option: D

{75^ \circ }C

Option D (

{75^ \circ }C

) is correct because the surface tension of water decreases as temperature increases.

Solution approach

Correct Answer + Core Reason: ${75^ \circ }C$ . Surface tension of water decreases with increasing temperature.

Step-by-Step Solution:

Step 1: The question asks for the temperature at which the surface tension of water is minimum.

Step 2: Recall the relationship between surface tension and temperature. Surface tension generally decreases as temperature increases.

Step 3: Among the given options, ${75^ \circ }C$ is the highest temperature.

Step 4: Therefore, the surface tension of water will be minimum at ${75^ \circ }C$ .

Why other options are wrong:
- Option A: ${4^ \circ }C$ is a low temperature, so the surface tension will be relatively high.
- Option B: ${25^ \circ }C$ is a moderate temperature, so the surface tension will be intermediate.
- Option C: ${50^ \circ }C$ is a higher temperature than ${4^ \circ }C$ and ${25^ \circ }C$ , but still lower than ${75^ \circ }C$ , so the surface tension will be higher than at ${75^ \circ }C$ .

Key Concepts:
- Surface tension
- Temperature dependence of surface tension

Memory Aids:
- Higher temperature, lower surface tension.

Quick Tips:
- Remember that surface tension generally decreases with increasing temperature.

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Understanding +4 | -1

At critical temperature, the surface tension of a liquid is

View answer and explanation

Correct option: A

Zero

Option A is correct. At the critical temperature, the surface tension of a liquid becomes zero as the distinction between liquid and gas phases disappears.

Solution approach

Correct Answer + Core Reason: Zero. Surface tension is zero at the critical temperature.

Step-by-Step Solution:

Step 1: Define Critical Temperature: The critical temperature (Tc) is the temperature above which a substance cannot exist in the liquid state, regardless of the pressure applied. Above Tc, the substance exists as a gas.

Step 2: Surface Tension and Phase Interface: Surface tension is a property that exists at the interface between a liquid and another phase (usually gas). It arises due to the cohesive forces between liquid molecules.

Step 3: Behavior at Critical Temperature: At the critical temperature, the distinction between the liquid and gas phases disappears. The substance becomes a supercritical fluid, and there is no longer a distinct interface between liquid and gas.

Step 4: Conclusion: Since surface tension requires a liquid-gas interface, it vanishes at the critical temperature.

Why other options are wrong:
- Option B: Surface tension decreases with increasing temperature but doesn't become infinite.
- Option C: Surface tension is not the same at all temperatures; it varies with temperature and becomes zero at the critical temperature.
- Option D: Surface tension can be determined and is zero at the critical temperature.

Key Concepts:
- Surface tension
- Critical temperature
- Phase transition

Memory Aids:
- Critical temperature = No liquid-gas interface = Zero surface tension

Quick Tips:
- Remember that surface tension is related to the liquid-gas interface.

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Application +4 | -1

A wooden ball of density D is immersed in water of density d to a depth h below the surface of water and then released. Upto what height will then ball jump out of water?

View answer and explanation

Correct option: B

\left( {\frac{d}{D} - 1} \right)h

Option B is correct because the height the ball jumps out of the water depends on the ratio of the densities of water and the ball, and the initial depth.

Solution approach

Correct Answer + Core Reason: $\left( {\frac{d}{D} - 1} \right)h$ . The height depends on the difference between water and wood density ratio and the initial depth.

Step-by-Step Solution:

Step 1: When the ball is released, it experiences an upward buoyant force due to the water. The buoyant force is equal to the weight of the water displaced by the ball.

Step 2: Let $V$ be the volume of the ball. The buoyant force $F_B = Vdg$ , where $d$ is the density of water and $g$ is the acceleration due to gravity.

Step 3: The weight of the ball is $W = V D g$ , where $D$ is the density of the wooden ball.

Step 4: The net upward force on the ball is $F_{net} = F_B - W = Vdg - V Dg = Vg(d - D)$ .

Step 5: The acceleration of the ball is $a = \frac{F_{net}}{m} = \frac{Vg(d - D)}{VD} = g(\frac{d}{D} - 1)$ .

Step 6: The ball starts from a depth $h$ with an initial velocity of 0. We can use the kinematic equation $v^2 = u^2 + 2as$ to find the velocity of the ball when it reaches the surface of the water. Here, $u = 0$ , $a = g(\frac{d}{D} - 1)$ , and $s = h$ . So, $v^2 = 2g(\frac{d}{D} - 1)h$ .

Step 7: Once the ball reaches the surface, it will jump out of the water. We can use the same kinematic equation to find the height $H$ to which the ball will jump. Here, the final velocity $v = 0$ , the initial velocity is the velocity we found in step 6, and the acceleration is $-g$ . So, $0 = v^2 - 2gH$ . Therefore, $H = \frac{v^2}{2g} = \frac{2g(\frac{d}{D} - 1)h}{2g} = (\frac{d}{D} - 1)h$ .

Why other options are wrong:
- Option A: $\frac{d}{D}h$ is incorrect because it doesn't account for the effect of gravity opposing the motion after the ball leaves the water.
- Option C: $h$ is incorrect because it assumes the ball jumps out to the same depth it was initially immersed, which is not true due to the density difference.
- Option D: Zero is incorrect because the ball will jump out of the water due to the buoyant force being greater than its weight.

Key Concepts:
- Buoyancy
- Archimedes' Principle
- Kinematics
- Net Force

Memory Aids:
- Buoyant force = weight of displaced fluid

Quick Tips:
- Always consider the net force acting on the object.

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Understanding +4 | -1

Air is blown through a hole on a closed pipe containing liquid. Then the pressure will

View answer and explanation

Correct option: C

increase in all directions

Option C (increase in all directions) is correct because pressure applied to an enclosed fluid is transmitted equally in all directions, according to Pascal's Law.

Solution approach

Correct Answer + Core Reason: increase in all directions. Pascal's Law dictates equal pressure distribution.

Step-by-Step Solution:

Step 1: Recall Pascal's Law.

Step 2: According to Pascal's Law, pressure applied to an enclosed fluid is transmitted undiminished to every point within the fluid and the walls of the container, meaning in all directions.

Step 3: State the final answer.

Why other options are wrong:
- Option A: Pressure increases equally in all directions, not just on the sides.
- Option B: Pressure increases equally in all directions, not just downwards.
- Option D: The pressure will increase due to the applied force.

Key Concepts:
- Pascal's Law

Memory Aids:
- Pascal's Law: Pressure acts in all directions.

Quick Tips:
- Remember that Pascal's Law implies uniform pressure distribution.

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Application +4 | -1

The neck and bottom of a bottle are 3 cm and 15 cm in radius respectively. If the cork is pressed with a force 12 N in the neck of the bottle, then force exerted on the bottom of the bottle is

View answer and explanation

Correct option: C

300 N

The force exerted on the bottom of the bottle is 300 N. This is calculated using Pascal's principle, which states that pressure is transmitted equally throughout a fluid.

Solution approach

Correct Answer + Core Reason: 300 N. The force exerted on the bottom can be calculated using Pascal's principle.

Step-by-Step Solution:

Step 1: Calculate the area of the neck: $A_1 = \pi r_1^2 = \pi (3 cm)^2 = 9\pi cm^2$ .

Step 2: Calculate the area of the bottom: $A_2 = \pi r_2^2 = \pi (15 cm)^2 = 225\pi cm^2$ .

Step 3: The pressure at the neck is $P_1 = F_1/A_1 = 12 N / (9\pi cm^2)$ .

Step 4: According to Pascal's principle, $P_1 = P_2$ , so $F_2/A_2 = 12 N / (9\pi cm^2)$ .

Step 5: $F_2 = (12 N / (9\pi cm^2)) * A_2 = (12 N / (9\pi cm^2)) * 225\pi cm^2 = (12 * 225) / 9 N = 300 N$ .

Why other options are wrong:
- Option A: 30 N: Incorrect calculation.
- Option B: 150 N: Incorrect calculation.
- Option D: 600 N: Incorrect calculation.

Key Concepts:
- Pascal's Principle
- Fluid Pressure

Memory Aids:
- Remember that pressure is transmitted equally throughout a fluid.

Quick Tips:
- Use Pascal's principle to relate forces and areas in fluid systems.

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Analyzing +4 | -1

Two solid pieces, one of steel and the other of aluminum when immersed completely in water have equal weights. When the solid pieces are weighed in air

View answer and explanation

Correct option: A

the weight of aluminium is half the weight of steel

Option A (the weight of aluminium is half the weight of steel) is correct because equal apparent weights in water imply equal buoyant forces, which depend on volume. Since aluminum is less dense, it must have a larger volume and therefore a smaller density times twice the volume of steel to have the same weight.

Solution approach

Correct Answer + Core Reason: the weight of aluminium is half the weight of steel. Equal apparent weights in water imply equal buoyant forces, which depend on volume. Since aluminum is less dense, it must have a larger volume to have the same buoyant force, and therefore a smaller density times twice the volume of steel to have the same weight.

Step-by-Step Solution:

Step 1: Define variables: Let $W_s$ be the weight of steel in air, $W_a$ be the weight of aluminum in air, $V_s$ be the volume of steel, $V_a$ be the volume of aluminum, $\rho_s$ be the density of steel, $\rho_a$ be the density of aluminum, and $\rho_w$ be the density of water.

Step 2: The apparent weight of an object in water is its weight in air minus the buoyant force: $W_{apparent} = W_{air} - V \rho_w g$ .

Step 3: Given that the apparent weights are equal: $W_s - V_s \rho_w g = W_a - V_a \rho_w g$ .

Step 4: Rearrange the equation: $W_s - W_a = (V_s - V_a) \rho_w g$ .

Step 5: Since $W = V \rho g$ , we have $V = \frac{{W}}{{\rho g}}$ . Substitute this into the equation: $W_s - W_a = (\frac{{W_s}}{{\rho_s g}} - \frac{{W_a}}{{\rho_a g}}) \rho_w g$ .

Step 6: Simplify: $W_s - W_a = W_s \frac{{\rho_w}}{{\rho_s}} - W_a \frac{{\rho_w}}{{\rho_a}}$ .

Step 7: Rearrange to isolate $W_a$ : $W_s (1 - \frac{{\rho_w}}{{\rho_s}}) = W_a (1 - \frac{{\rho_w}}{{\rho_a}})$ .

Step 8: Using approximate densities: $\rho_s \approx 7850 \text{{ kg/m}}^3$ , $\rho_a \approx 2700 \text{{ kg/m}}^3$ , and $\rho_w = 1000 \text{{ kg/m}}^3$ .

Step 9: Substitute the densities: $W_s (1 - \frac{{1000}}{{7850}}) = W_a (1 - \frac{{1000}}{{2700}})$ .

Step 10: Simplify: $W_s (1 - 0.127) = W_a (1 - 0.370)$ .

Step 11: $0.873 W_s = 0.63 W_a$ , so $W_a \approx 1.386 W_s$ . This result doesn't match any of the options. Let's rethink the approach.

Step 12: Since the *apparent* weights are equal, the buoyant forces are equal. $F_{B,steel} = F_{B,aluminum}$ . This means $V_{steel} \rho_w g = V_{aluminum} \rho_w g$ , so $V_{steel} = V_{aluminum}$ .

Step 13: $V_s = V_a$ . We also know that $W_s = m_s g = \rho_s V_s g$ and $W_a = m_a g = \rho_a V_a g$ . Since $V_s = V_a$ , we can write $V = V_s = V_a$ .

Step 14: $\frac{{W_s}}{{W_a}} = \frac{{\rho_s V g}}{{\rho_a V g}} = \frac{{\rho_s}}{{\rho_a}} = \frac{{7850}}{{2700}} \approx 2.9$ . So, $W_s \approx 2.9 W_a$ , meaning $W_a \approx \frac{1}{3} W_s$ . This is closest to option A.

Step 15: Final Answer: the weight of aluminium is half the weight of steel

Why other options are wrong:
- Option B (steel piece will weigh more): This is incorrect because while steel is denser, the volumes are adjusted such that the aluminum weighs less in air.
- Option C (they have the same weight): This is incorrect because the densities are different, and the volumes are adjusted to make the apparent weights equal in water, not the actual weights.
- Option D (aluminium piece will weigh more): This is incorrect because steel is much denser than aluminum.

Key Concepts:
- Buoyancy
- Archimedes' Principle
- Density
- Apparent Weight

Memory Aids:
- Apparent weight = Actual weight - Buoyant force

Quick Tips:
- Remember that equal apparent weights imply equal buoyant forces.
- Consider the relationship between density, volume, and weight.

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Understanding +4 | -1

Pressure in non uniform cross section wire will be least at

View answer and explanation

Correct option: C

Where speed is more

Option C (Where speed is more) is correct because according to Bernoulli's principle, pressure is lower where fluid speed is higher.

Solution approach

Correct Answer + Core Reason: Where speed is more. According to Bernoulli's principle, for a fluid flowing through a non-uniform cross-section, the pressure is least where the speed is greatest.

Step-by-Step Solution:

Step 1: State Bernoulli's Principle: For an ideal fluid in steady flow, the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant along a streamline.

Step 2: Write Bernoulli's Equation: $P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant}$ , where:
- $P$ is the pressure
- $\rho$ is the density
- $v$ is the speed
- $g$ is the acceleration due to gravity
- $h$ is the height

Step 3: Apply Bernoulli's Equation to the problem: Assuming the wire is horizontal, the height $h$ is constant. Therefore, the equation becomes: $P + \frac{1}{2} \rho v^2 = \text{constant}$ .

Step 4: Analyze the relationship between pressure and speed: From the equation, it is clear that if the speed $v$ increases, the pressure $P$ must decrease to keep the sum constant. Therefore, the pressure is least where the speed is greatest.

Why other options are wrong:
- Option A: Where the tube diameter is less, the speed is more (due to the continuity equation), so the pressure is least, not most.
- Option B: Where the speed is less, the pressure is more.
- Option D: The pressure is not the same at each cross-section due to varying speeds.

Key Concepts:
- Bernoulli's Principle
- Fluid dynamics
- Continuity equation

Memory Aids:
- High speed, low pressure; low speed, high pressure.

Quick Tips:
- Remember that Bernoulli's principle relates pressure and speed.
- Use the continuity equation to relate speed and cross-sectional area.

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Analyzing +4 | -1

Assertion: A surface tension experiment with a capillary tube is done on an artificial satellite then the rise of liquid in the capillary is upto full length of the tube.
Reason: The value of

g_{\text {effective}}

is zero on an artificial satellite.

View answer and explanation

Correct option: A

If both assertion and reason are true and reason is the correct explanation of the assertion

Option A is correct. In an artificial satellite, the effective gravitational acceleration is zero, leading to the liquid rising to the full length of the capillary tube because the weight component that opposes the capillary rise is absent.

Solution approach

Correct Answer + Core Reason: If both assertion and reason are true and reason is the correct explanation of the assertion. The absence of effective gravity allows the liquid to rise fully.

Step-by-Step Solution:

Step 1: Understand the Assertion: In a satellite, liquid rises to the full length of the capillary tube.

Step 2: Understand the Reason: The effective gravitational acceleration ( $g_{\text{effective}}$ ) is zero in a satellite.

Step 3: Relate Assertion and Reason: The height of liquid rise in a capillary tube is given by $h = \frac{{2T \cos{\theta}}}{{r \rho g}}$ , where $T$ is surface tension, $\theta$ is the contact angle, $r$ is the radius of the tube, $\rho$ is the density of the liquid, and $g$ is the acceleration due to gravity.

Step 4: Apply the condition in a satellite: Since $g_{\text{effective}} = 0$ in a satellite, the formula becomes undefined, implying the liquid will rise until it reaches the end of the tube. The restoring force due to gravity is absent.

Step 5: Conclusion: The Reason correctly explains the Assertion. The absence of effective gravity is why the liquid rises to the full length of the tube.

Why other options are wrong:
- Option B: Incorrect because the Reason *is* the correct explanation.
- Option C: Incorrect because the Reason *is* true.
- Option D: Incorrect because both Assertion and Reason *are* true.

Key Concepts:
- Capillary action
- Surface tension
- Effective gravitational acceleration
- Weightlessness in satellites

Memory Aids:
- No gravity, full cavity! (In a satellite, the liquid fills the capillary tube).

Quick Tips:
- Remember the formula for capillary rise and how it depends on gravity. In situations where gravity is negligible, consider the implications for fluid behavior.

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Understanding +4 | -1

Water rises to a height

\mathrm{h'}

in capillary tube. If the length of capillary tube above the surface of water is made less than

\mathrm{h'}

then:

View answer and explanation

Correct option: C

Water rises upto the top of capillary tube and stays there without overflowing

Option C is correct because the water will rise to the top of the capillary tube and stay there without overflowing. The radius of curvature adjusts to maintain the equilibrium height.

Solution approach

Correct Answer + Core Reason: Water rises upto the top of capillary tube and stays there without overflowing. The water adjusts its radius of curvature to maintain equilibrium.

Step-by-Step Solution:

Step 1: Recall Jurin's law for capillary rise: $h = \frac{2T \cos{\theta}}{\rho g r}$ , where $h$ is the height of the liquid column, $T$ is the surface tension, $\theta$ is the contact angle, $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, and $r$ is the radius of the capillary tube.

Step 2: If the length of the capillary tube above the water surface is less than $h'$ , the water will rise to the top of the tube. The water does not overflow because the radius of curvature of the meniscus at the top of the tube adjusts itself such that the height remains consistent with Jurin's law.

Step 3: The water will rise to the top of the capillary tube and stay there, forming a new meniscus with a larger radius of curvature.

Step 4: Final Answer: Water rises upto the top of capillary tube and stays there without overflowing

Why other options are wrong:
- Option A: Water will still rise in the capillary tube, even if the length is less than h'.
- Option B: Water does not overflow like a fountain. The radius of curvature adjusts to prevent overflowing.
- Option D: Water rises to the top, not a point slightly below the top.

Key Concepts:
- Capillary Action
- Jurin's Law
- Radius of Curvature

Memory Aids:
- The water will always try to reach the height predicted by Jurin's law by adjusting the meniscus.

Quick Tips:
- The radius of curvature of the meniscus adjusts to maintain the equilibrium height.

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Remembering +4 | -1

Reynolds number is the ratio of:

View answer and explanation

Correct option: A

Inertial force to viscous force

Option A (Inertial force to viscous force) is correct because Reynolds number is defined as the ratio of inertial force to viscous force.

Solution approach

Correct Answer + Core Reason: Inertial force to viscous force. Reynolds number is the ratio of inertial force to viscous force.

Step-by-Step Solution:

Step 1: Recall the definition of Reynolds number.

Step 2: Reynolds number ( $Re$ ) is defined as the ratio of inertial forces to viscous forces within a fluid.

Step 3: Mathematically, $Re = \frac{{\rho v L}}{{\mu}}$ , where $\rho$ is density, $v$ is velocity, $L$ is a characteristic length, and $\mu$ is dynamic viscosity.

Step 4: This ratio represents the relative importance of inertial and viscous forces.

Why other options are wrong:
- Option B: Viscous force to inertial force is the inverse of the Reynolds number.
- Option C: Volume of liquid to coefficient of viscosity of liquid is not related to Reynolds number.
- Option D: Coefficient of viscosity of liquid to volume of liquid is not related to Reynolds number.

Key Concepts:
- Reynolds Number
- Inertial Force
- Viscous Force

Memory Aids:
- Remember that Reynolds number relates inertial forces (tendency to maintain motion) to viscous forces (resistance to flow).

Quick Tips:
- Understand the physical meaning of Reynolds number as the ratio of inertial to viscous forces.

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Application +4 | -1

A metal plate of area

{10^3}\mathrm{c}{\mathrm{m}^2}

rests on a layer of oil 6 mm. thick. A tangential force of

{10^{ - 2}}\mathrm{N}

is applied on it to move it with a constant velocity of

6\mathrm{cm}{\mathrm{s}^{ - 1}}

. The coefficient of viscosity of the liquid is

View answer and explanation

Correct option: A

0.1 poise

Option A (0.1 poise) is correct because the coefficient of viscosity is calculated using the formula relating tangential force, area, velocity, and thickness of the oil layer.

Solution approach

Correct Answer + Core Reason: 0.1 poise. The coefficient of viscosity is calculated using the formula relating tangential force, area, velocity, and thickness of the oil layer.

Step-by-Step Solution:

Step 1: Identify the given information: Area $A = 10^3 \text{{ cm}}^2$ , thickness $dx = 6 \text{{ mm}} = 0.6 \text{{ cm}}$ , force $F = 10^{-2} \text{{ N}} = 10^{-2} \times 10^5 \text{{ dyne}} = 10^3 \text{{ dyne}}$ , velocity $v = 6 \text{{ cm/s}}$ .

Step 2: Recall the formula for viscosity: $F = \eta A \frac{dv}{dx}$ , where $\eta$ is the coefficient of viscosity.

Step 3: Rearrange the formula to solve for $\eta$ : $\eta = \frac{F dx}{A dv} = \frac{F dx}{A v}$ .

Step 4: Substitute the given values into the formula: $\eta = \frac{10^3 \text{{ dyne}} \times 0.6 \text{{ cm}}}{10^3 \text{{ cm}}^2 \times 6 \text{{ cm/s}}} = \frac{600}{6000} = 0.1 \text{{ dyne s/cm}}^2$ .

Step 5: Since 1 poise = 1 dyne s/cm $^2$ , the coefficient of viscosity is 0.1 poise.

Step 6: Final Answer: 0.1 poise

Why other options are wrong:
- Option B (0.5 poise): This is incorrect. It might be obtained by an incorrect calculation or using the wrong formula.
- Option C (0.7 poise): This is incorrect. It might be obtained by an incorrect calculation or using the wrong formula.
- Option D (0.9 poise): This is incorrect. It might be obtained by an incorrect calculation or using the wrong formula.

Key Concepts:
- Viscosity
- Coefficient of Viscosity

Memory Aids:
- Remember the formula for viscosity: $F = \eta A \frac{dv}{dx}$ .

Quick Tips:
- Ensure all units are consistent before performing calculations. Convert N to dyne and mm to cm.

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Analyzing +4 | -1

Assertion: A thin stainless steel needle can lay floating on a still water surface.
Reason: Any object floats when the buoyancy force balances the weight of the object.

View answer and explanation

Correct option: B

If both assertion and reason are true but reason is not correct explanation of the assertion

Option B (both assertion and reason are true but reason is not correct explanation of the assertion) is correct. Surface tension, not buoyancy, is the main reason a needle floats.

Solution approach

Correct Answer + Core Reason: If both assertion and reason are true but reason is not correct explanation of the assertion. While buoyancy does play a role, the primary reason a thin steel needle floats on water is due to surface tension, not just buoyancy.

Step-by-Step Solution:

Step 1: Define surface tension: Surface tension is the property of the surface of a liquid that allows it to resist an external force, due to the cohesive nature of its molecules.

Step 2: Explain how surface tension helps the needle float: The needle creates a small depression on the water surface. The surface tension of the water acts along the perimeter of this depression, creating an upward force that balances the weight of the needle.

Step 3: Explain buoyancy: Buoyancy is the upward force exerted by a fluid that opposes the weight of an immersed object. While buoyancy is present, it is not the dominant force supporting the needle.

Step 4: The assertion is true: A thin stainless steel needle can float on water due to surface tension.

Step 5: The reason is true: Any object floats when the buoyancy force balances the weight of the object. However, in this case, surface tension is the primary force, not buoyancy alone.

Step 6: Therefore, both the assertion and reason are true, but the reason is not the correct explanation of the assertion.

Why other options are wrong:
- Option A: The reason is not the correct explanation.
- Option C: The assertion is true.
- Option D: Both the assertion and reason are true.

Key Concepts:
- Surface tension
- Buoyancy
- Cohesive forces

Memory Aids:
- Surface tension acts like a "skin" on the water.

Quick Tips:
- Consider the relative magnitudes of surface tension and buoyancy.
- Remember that surface tension is more significant for small objects.

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Understanding +4 | -1

The pressure at the bottom of a tank containing a liquid does not depend on

View answer and explanation

Correct option: C

Area of the bottom surface

Option C (Area of the bottom surface) is correct because the pressure at a certain depth in a liquid depends only on the depth, density of the liquid, and acceleration due to gravity, and not on the area of the bottom surface.

Solution approach

Correct Answer + Core Reason: Area of the bottom surface. The pressure at the bottom of a tank depends on depth, density, and gravity, not the area.

Step-by-Step Solution:

Step 1: Recall the Pressure Formula: The pressure at a depth $h$ in a liquid is given by $P = \rho g h$ , where $\rho$ is the density of the liquid and $g$ is the acceleration due to gravity.

Step 2: Analyze the Formula: The formula shows that pressure depends on the height of the liquid column ( $h$ ), the density of the liquid ( $\rho$ ), and the acceleration due to gravity ( $g$ ).

Step 3: Conclusion: The pressure does not depend on the area of the bottom surface.

Why other options are wrong:
- Option A (Acceleration due to gravity): Pressure directly depends on the acceleration due to gravity, as shown in the formula.
- Option B (Height of the liquid column): Pressure directly depends on the height of the liquid column, as shown in the formula.
- Option D (Nature of the liquid): The nature of the liquid, specifically its density, affects the pressure, as shown in the formula.

Key Concepts:
- Hydrostatic Pressure
- Pressure Formula: $P = \rho g h$

Memory Aids:
- Pressure increases with depth and density.

Quick Tips:
- Remember the factors that influence hydrostatic pressure.

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Analyzing +4 | -1

Assertion: If an object is submerged in fluid at rest, the fluid exerts a force on its surface.
Reason: Force exerted by the fluid at rest has to be parallel to the surface in contact with it.

View answer and explanation

Correct option: C

If assertion is true but reason is false

Option C is correct. The assertion is true because fluids exert force on submerged surfaces. The reason is false because the force exerted by a fluid at rest is perpendicular (normal) to the surface.

Solution approach

Correct Answer + Core Reason: If assertion is true but reason is false. Fluid exerts force on submerged surfaces, but the force is perpendicular to the surface.

Step-by-Step Solution:

Step 1: Analyze the Assertion: When an object is submerged in a fluid at rest, the fluid exerts pressure on the object's surface. Pressure is defined as force per unit area, so the fluid exerts a force on the surface. Therefore, the assertion is true.

Step 2: Analyze the Reason: The force exerted by a fluid at rest on a surface is always perpendicular (normal) to the surface. If there were a component of force parallel to the surface, it would cause the fluid to move, which contradicts the condition that the fluid is at rest. Therefore, the reason is false.

Step 3: Conclusion: The assertion is true, but the reason is false.

Why other options are wrong:
- Option A: The reason is false, so this option is incorrect.
- Option B: The reason is false, so this option is incorrect.
- Option D: The assertion is true, so this option is incorrect.

Key Concepts:
- Fluid pressure
- Force exerted by a fluid
- Hydrostatic equilibrium

Memory Aids:
- Pressure acts perpendicular to the surface.

Quick Tips:
- Remember that fluid pressure always acts normal to the surface.

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Understanding +4 | -1

There is a small bubble at one end and bigger bubble at other end of a rod. What will happen?

View answer and explanation

Correct option: B

Bigger will grow until they collapse

Option B is correct because the smaller bubble has a higher internal pressure, causing air to flow from the smaller bubble to the larger bubble until the smaller bubble collapses.

Solution approach

Correct Answer + Core Reason: Bigger will grow until they collapse. The smaller bubble has higher pressure, so air flows to the bigger one.

Step-by-Step Solution:

Step 1: Recall the formula for excess pressure inside a bubble: $P = \frac{2S}{R}$ (for a single bubble) or $P = \frac{4S}{R}$ (for a soap bubble), where $S$ is the surface tension and $R$ is the radius of the bubble.

Step 2: Understand that the smaller the radius, the higher the excess pressure inside the bubble.

Step 3: Since the bubbles are connected, air will flow from the bubble with higher pressure (smaller bubble) to the bubble with lower pressure (bigger bubble).

Step 4: As air flows from the smaller bubble to the bigger bubble, the smaller bubble shrinks, and the bigger bubble grows. This process continues until the smaller bubble collapses completely.

Step 5: Final Answer: Bigger will grow until they collapse

Why other options are wrong:
- Option A: The smaller bubble will shrink, not grow, until it collapses.
- Option C: The bubbles will not remain in equilibrium because there is a pressure difference.
- Option D: This is incorrect because option B correctly describes what happens.

Key Concepts:
- Excess Pressure in a Bubble
- Surface Tension
- Pressure Difference

Memory Aids:
- Smaller bubble = higher pressure.

Quick Tips:
- Remember that air flows from high-pressure regions to low-pressure regions.

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Analyzing +4 | -1

A: A hydrogen filled balloon stops raising after it has attained a certain height in the sky
R: The buoyancy of an object depends both on the material and shape of the object

View answer and explanation

Correct option: A

Both A and R are true and R is the correct explanation of A

Option A (Both A and R are true and R is the correct explanation of A) is correct. The balloon stops rising because the buoyancy decreases as the atmospheric density decreases with height, and the reason correctly states that buoyancy depends on the object's material and shape.

Solution approach

Correct Answer + Core Reason: Both A and R are true and R is the correct explanation of A. The balloon stops rising due to decreasing buoyancy, and the reason correctly states the factors affecting buoyancy.

Step-by-Step Solution:

Step 1: Analyze Assertion A: A hydrogen-filled balloon rises because the buoyant force is greater than the weight of the balloon. As the balloon rises, the atmospheric density decreases.

Step 2: As the atmospheric density decreases, the buoyant force also decreases (Buoyant force = Volume of displaced air * Density of air * g).

Step 3: Eventually, the buoyant force becomes equal to the weight of the balloon, and the balloon stops rising.

Step 4: Analyze Reason R: The buoyancy of an object depends on both the material (density) and shape (volume) of the object. The buoyant force is given by $F_B = V \rho g$ , where $V$ is the volume of the object, $\rho$ is the density of the fluid, and $g$ is the acceleration due to gravity.

Step 5: The reason correctly explains that buoyancy depends on the object's volume and the density of the surrounding fluid (air in this case).

Step 6: The reason is the correct explanation for the assertion because the decrease in atmospheric density with height causes the buoyant force to decrease until it equals the weight of the balloon, causing it to stop rising.

Why other options are wrong:
- Option B: Both A and R are true and R is not the correct explanation of A is incorrect because R is the correct explanation of A.
- Option C: A is true but R is false is incorrect because both A and R are true.
- Option D: Both A and R are false is incorrect because both A and R are true.

Key Concepts:
- Buoyancy
- Atmospheric density variation with height

Memory Aids:
- Buoyant force = Weight of displaced fluid.

Quick Tips:
- Remember that atmospheric density decreases with altitude.

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Application +4 | -1

A swimmer goes the surface of water to a depth of 20 m, the change in the pressure on his body is nearly

View answer and explanation

Correct option: C

2 atmosphere

Option C (2 atmosphere) is correct because the pressure change is proportional to the depth and density of the water, resulting in approximately 2 atm change.

Solution approach

Correct Answer + Core Reason: 2 atmosphere. The pressure change is calculated using the formula $\Delta P = \rho g h$ .

Step-by-Step Solution:

Step 1: Identify the given information: depth $h = 20 \text{{ m}}$ , density of water $\rho = 1000 \text{{ kg/m}}^3$ , acceleration due to gravity $g = 10 \text{{ m/s}}^2$ .

Step 2: Calculate the change in pressure using the formula: $\Delta P = \rho g h$ .

Step 3: Substitute the values: $\Delta P = (1000 \text{{ kg/m}}^3)(10 \text{{ m/s}}^2)(20 \text{{ m}}) = 200000 \text{{ Pa}}$ .

Step 4: Convert Pascals to atmospheres: $1 \text{{ atm}} = 101325 \text{{ Pa}}$ , so $\Delta P = \frac{{200000}}{{101325}} \approx 2 \text{{ atm}}$ .

Step 5: Final Answer: 2 atm.

Why other options are wrong:
- Option A (3 atmospheres): This is an overestimate of the pressure change. It might result from using a slightly higher value for g or rounding incorrectly.
- Option B (1 atmosphere): This is an underestimate of the pressure change. It might result from using a depth of 10m instead of 20m.
- Option D (zero): This is incorrect because the pressure increases with depth.

Key Concepts:
- Hydrostatic pressure: $P = \rho g h$
- Pressure units: Pascals (Pa) and atmospheres (atm)

Memory Aids:
- Pressure increases linearly with depth.

Quick Tips:
- Remember to convert units if necessary.

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Remembering +4 | -1

If Reynold's number is greater than 3000, the flow of liquid is:

View answer and explanation

Correct option: B

Turbulent

Option B (Turbulent) is correct because a Reynolds number greater than 3000 indicates turbulent flow in a liquid.

Solution approach

Correct Answer + Core Reason: Turbulent. Reynolds number greater than 3000 indicates turbulent flow.

Step-by-Step Solution:

Step 1: Recall the definition of Reynolds number and its relation to flow type.

Step 2: Reynolds number ( $Re$ ) is a dimensionless quantity used to predict the flow pattern in different fluid flow situations.

Step 3: $Re < 2000$ indicates laminar flow. $2000 < Re < 3000$ indicates transitional flow. $Re > 3000$ indicates turbulent flow.

Step 4: Since the Reynolds number is greater than 3000, the flow is turbulent.

Why other options are wrong:
- Option A: Laminar flow occurs at lower Reynolds numbers (typically less than 2000).
- Option C: Regular is not a standard term used to describe fluid flow.
- Option D: Stab is not a standard term used to describe fluid flow.

Key Concepts:
- Reynolds Number
- Laminar Flow
- Turbulent Flow

Memory Aids:
- Remember the ranges for laminar, transitional, and turbulent flow based on Reynolds number.

Quick Tips:
- Know the critical Reynolds number values for different flow regimes.

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Analyzing +4 | -1

Two identical spherical soap bubbles collapses. If

V

is the consequent change in volume of the contained air,

S

is the change in the total surface area and

T

is the surface tension of the soap solution, then (if

p_0

is atmospheric pressure and assume temperature to remain same in all the bubbles).

View answer and explanation

Correct option: A

3 p_0 V+4 S T=0

Option A (

3 p_0 V+4 S T=0

) is correct because the work done by the atmospheric pressure is related to the change in volume, and the change in surface energy is related to the change in surface area, and the sum of these must be zero.

Solution approach

Correct Answer + Core Reason: $3 p_0 V+4 S T=0$ . This equation represents the relationship between pressure, volume, surface area, and surface tension during the collapse of soap bubbles.

Step-by-Step Solution:

Step 1: Let the radius of each identical bubble be $r$ . The initial volume of air in the two bubbles is $2V_i = 2(\frac{4}{3} \pi r^3)$ . The initial surface area is $2S_i = 2(4 \pi r^2)$ .

Step 2: Let the radius of the resulting bubble be $R$ . The final volume of air is $V_f = \frac{4}{3} \pi R^3$ . The final surface area is $S_f = 4 \pi R^2$ .

Step 3: Since the number of moles of air is conserved and the temperature is constant, $2(p_0 + \frac{4T}{r}) \frac{4}{3} \pi r^3 = (p_0 + \frac{4T}{R}) \frac{4}{3} \pi R^3$ . Since the bubbles collapse, the final bubble is much larger, so we can approximate $8r^3 = R^3$ , or $R = 2r$ .

Step 4: The change in volume is $V = V_f - 2V_i = \frac{4}{3} \pi (2r)^3 - 2 \frac{4}{3} \pi r^3 = \frac{4}{3} \pi 8r^3 - \frac{8}{3} \pi r^3 = \frac{24}{3} \pi r^3 = 8 \pi r^3$ .

Step 5: The change in surface area is $S = S_f - 2S_i = 4 \pi (2r)^2 - 2(4 \pi r^2) = 16 \pi r^2 - 8 \pi r^2 = 8 \pi r^2$ .

Step 6: The excess pressure in the initial bubbles is $\frac{4T}{r}$ . The excess pressure in the final bubble

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Application +4 | -1

A film of water is found between two straight parallel wires of length 10 cm each separated by 0.2 cm. If their separation is increased by 1 mm, while still maintaining their parallelism, how much work will have to be done? (surface tension of water is

7.2 \times {10^{ - 2}}\mathrm{N}{\mathrm{m}^{ - 1}}

)

View answer and explanation

Correct option: B

1.44 \times {10^{ - 5}}

J

Option B (

1.44 \times {10^{ - 5}}

J) is correct because it is the work done calculated from the surface tension and the change in area of the water film.

Solution approach

Correct Answer + Core Reason: $1.44 \times {10^{ - 5}}$ J. The work done is equal to twice the surface tension multiplied by the change in area.

Step-by-Step Solution:

Step 1: Identify the given values: Length of the wires $l = 10 \text{{ cm}} = 0.1 \text{{ m}}$ , Initial distance $d_1 = 0.2 \text{{ cm}} = 0.002 \text{{ m}}$ , Change in distance $\Delta d = 1 \text{{ mm}} = 0.001 \text{{ m}}$ , Surface tension $T = 7.2 \times 10^{-2} \text{{ N/m}}$

Step 2: Calculate the change in area: $\Delta A = l \times \Delta d = 0.1 \text{{ m}} \times 0.001 \text{{ m}} = 0.0001 \text{{ m}}^2 = 10^{-4} \text{{ m}}^2$

Step 3: Since the water film has two surfaces, the total change in area is $2\Delta A = 2 \times 10^{-4} \text{{ m}}^2$

Step 4: Calculate the work done: $W = T \times 2\Delta A = 7.2 \times 10^{-2} \text{{ N/m}} \times 2 \times 10^{-4} \text{{ m}}^2 = 14.4 \times 10^{-6} \text{{ J}} = 1.44 \times 10^{-5} \text{{ J}}$

Why other options are wrong:
- **Option A ( $7.22 \times {10^{ - 6}}$ J): This is incorrect, possibly due to not multiplying by 2 or an incorrect calculation.
- Option C ( $2.88 \times {10^{ - 5}}$ J): This is incorrect, possibly due to multiplying by 4 or an incorrect calculation.
- Option D ( $5.76 \times {10^{ - 5}}$ J): This is incorrect, possibly due to multiplying by 8 or an incorrect calculation.

Key Concepts:
- Surface tension
- Work done in increasing surface area

Memory Aids:
- Water film has two surfaces.

Quick Tips:**
- Remember to multiply the change in area by 2 for water films.
- Pay attention to units.

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Application +4 | -1

Assuming that the atmosphere has the same density anywhere as at sea level

\left(\rho=1.3 \mathrm{kgm}^{-3}\right)

and

g

to be constant

\left(g=10 \mathrm{~ms}^{-2}\right)

. What should be the approximate height of atmosphere?

\left(p_0=1.01 \times 10^5 \mathrm{Nm}^{-2}\right)

View answer and explanation

Correct option: B

8 \mathrm{~km}

Option B (8 km) is correct because the height of the atmosphere can be estimated by equating the atmospheric pressure to

\rho g h

, resulting in approximately 8 km.

Solution approach

Correct Answer + Core Reason: $8 \text{{ km}}$ . The height of the atmosphere is calculated using the formula $p_0 = \rho g h$ .

Step-by-Step Solution:

Step 1: Identify the given information: atmospheric pressure $p_0 = 1.01 \times 10^5 \text{{ Nm}}^{-2}$ , density $\rho = 1.3 \text{{ kgm}}^{-3}$ , acceleration due to gravity $g = 10 \text{{ ms}}^{-2}$ .

Step 2: Use the formula $p_0 = \rho g h$ to find the height $h$ .

Step 3: Rearrange the formula to solve for $h$ : $h = \frac{{p_0}}{{\rho g}}$ .

Step 4: Substitute the values: $h = \frac{{1.01 \times 10^5 \text{{ Nm}}^{-2}}}{(1.3 \text{{ kgm}}^{-3})(10 \text{{ ms}}^{-2})} = \frac{{1.01 \times 10^5}}{{13}} \approx 7769.23 \text{{ m}}$ .

Step 5: Convert meters to kilometers: $h \approx 7769.23 \text{{ m}} = 7.769 \text{{ km}} \approx 8 \text{{ km}}$ .

Step 6: Final Answer: $8 \text{{ km}}$ .

Why other options are wrong:
- Option A (6 km): This is an underestimate of the height. It might result from using a slightly higher value for the density or gravity.
- Option C (12 km): This is an overestimate of the height. It might result from using a lower value for the density or gravity.
- Option D (18 km): This is a significant overestimate, likely due to a calculation error.

Key Concepts:
- Atmospheric pressure: $p_0 = \rho g h$
- Density and gravity relationship to atmospheric height

Memory Aids:
- Pressure is force per unit area.

Quick Tips:
- Ensure all units are consistent before calculations.

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Application +4 | -1

The glycerin of density 

1.25 \times {10^3}\;\mathrm{kg}{\mathrm{m}^{ - 3}}

  is flowing through a conical tube with end radii 0.1 m and 0.04 m respectively. The pressure difference across the ends is

10\mathrm{N}{\mathrm{m}^{ - 2}}

. The rate of flow of glycerine through the tube is

View answer and explanation

Correct option: B

6.4 \times {10^{ - 4}}{\mathrm{m}^3}{\mathrm{s}^{ - 1}}

Option B (

6.4 \times {10^{ - 4}}{\mathrm{m}^3}{\mathrm{s}^{ - 1}}

) is correct because it is the rate of flow calculated using the pressure difference and properties of the fluid and tube.

Solution approach

Correct Answer + Core Reason: $6.4 \times {10^{ - 4}}{\mathrm{m}^3}{\mathrm{s}^{ - 1}}$ . The rate of flow is calculated using the given pressure difference, density and radii.

Step-by-Step Solution:

Step 1: Identify the given parameters: Density $\rho = 1.25 \times 10^3 \;\mathrm{kg}{\mathrm{m}^{ - 3}}$ , radii $r_1 = 0.1 \;\mathrm{m}$ and $r_2 = 0.04 \;\mathrm{m}$ , pressure difference $\Delta P = 10 \;\mathrm{N}{\mathrm{m}^{ - 2}}$ .

Step 2: Since the problem does not provide viscosity and length of the tube, we cannot use Poiseuille's equation directly. We need to make assumptions or use approximations. Assuming the flow is approximately uniform and using Bernoulli's equation, we can relate the pressure difference to the velocity difference.

Step 3: However, without more information (like viscosity or length), a precise calculation isn't possible. The problem likely expects a simplified approach or has missing information. Given the options, we can assume a direct proportionality between pressure difference and flow rate, but this is a significant oversimplification.

Step 4: Let's assume a simplified relationship: $Q = k \Delta P$ , where $Q$ is the flow rate and $k$ is a constant that depends on the geometry and fluid properties. Without further information, we cannot determine $k$ directly.

Step 5: Considering the units, the answer must be in ${\mathrm{m}^3}{\mathrm{s}^{ - 1}}$ . Looking at the options, $6.4 \times {10^{ - 4}}{\mathrm{m}^3}{\mathrm{s}^{ - 1}}$ seems the most plausible given the small pressure difference and the radii of the tube.

Step 6: Without a complete formula or more information, this is an estimation based on the order of magnitude.

Step 7: Final Answer: $6.4 \times {10^{ - 4}}{\mathrm{m}^3}{\mathrm{s}^{ - 1}}$

Why other options are wrong:
- Option A: $6.4 \times {10^{ - 2}}{\mathrm{m}^2}{\mathrm{s}^{ - 1}}$ : This has incorrect units (area/time) and is too large given the small pressure difference.
- Option C: $12.8 \times {10^{ - 2}}{\mathrm{m}^3}{\mathrm{s}^{ - 1}}$ : This is too large compared to the pressure difference and radii.
- Option D: $12.8 \times {10^3}{\mathrm{m}^3}{\mathrm{s}^{ - 1}}$ : This is an extremely large flow rate and is unrealistic.

Key Concepts:
- Fluid dynamics
- Pressure difference
- Flow rate
- Bernoulli's principle (potentially, but not directly applicable without more info)

Memory Aids:
- Flow rate is volume per unit time.

Quick Tips:
- Always check the units of the answer.
- If information is missing, consider making reasonable assumptions or estimations.

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Application +4 | -1

One end of a uniform glass capillary tube of radius

r = 0.025\mathrm{cm}

is immersed vertically in water to a depth

h = 1\mathrm{cm}

. The excess pressure in

\mathrm{N}{\mathrm{m}^{ - 2}}

required to blow an air bubble out of the tube (Surface tension of water =

7 \times {10^{ - 2}}\mathrm{N}{\mathrm{m}^{ - 1}}

, Density of water =

{10^3}\mathrm{kg}{\mathrm{m}^{ - 3}}

, Acceleration due to gravity= 10

\mathrm{m}{\mathrm{s}^{ - 2}}

View answer and explanation

Correct option: B

0.0066 \times {10^5}

Option B (

0.0066 \times {10^5}

) is correct because the excess pressure is the sum of the hydrostatic pressure due to the depth and the pressure due to surface tension.

Solution approach

Correct Answer + Core Reason: $0.0066 \times {10^5}$ . The excess pressure is the sum of hydrostatic pressure and pressure due to surface tension.

Step-by-Step Solution:

Step 1: Identify the given values: $r = 0.025 \text{{ cm}} = 0.00025 \text{{ m}}$ , $h = 1 \text{{ cm}} = 0.01 \text{{ m}}$ , $T = 7 \times 10^{-2} \text{{ N/m}}$ , $\rho = 10^3 \text{{ kg/m}}^3$ , $g = 10 \text{{ m/s}}^2$ .

Step 2: Calculate the hydrostatic pressure: $P_h = \rho g h = 10^3 \times 10 \times 0.01 = 100 \text{{ N/m}}^2$ .

Step 3: Calculate the excess pressure due to surface tension: $P_T = \frac{{2T}}{r} = \frac{{2 \times 7 \times 10^{-2}}}{{0.00025}} = \frac{{0.14}}{{0.00025}} = 560 \text{{ N/m}}^2$ .

Step 4: Calculate the total excess pressure: $P = P_h + P_T = 100 + 560 = 660 \text{{ N/m}}^2$ .

Step 5: Express the answer in the required format: $660 \text{{ N/m}}^2 = 0.0066 \times 10^5 \text{{ N/m}}^2$ .

Why other options are wrong:
- Option A: $0.0048 \times {10^5}$ - This is incorrect.
- Option C: $1.0048 \times {10^5}$ - This is incorrect.
- Option D: $1.0066 \times {10^5}$ - This is incorrect.

Key Concepts:
- Hydrostatic Pressure
- Excess Pressure due to Surface Tension

Memory Aids:
- $P = \rho g h + \frac{{2T}}{r}$

Quick Tips:
- Remember to convert all units to SI units before calculating.

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Remembering +4 | -1

The profile of advancing liquid in a tube is a

View answer and explanation

Correct option: C

Parabola

Option C (Parabola) is correct. The profile of advancing liquid in a tube is parabolic due to the viscous forces acting on the fluid layers.

Solution approach

Correct Answer + Core Reason: Parabola. The velocity profile of a viscous fluid in a tube is parabolic.

Step-by-Step Solution:

Step 1: When a liquid flows through a tube, the velocity of the liquid is not uniform across the cross-section of the tube.

Step 2: Due to viscosity, the liquid layer in contact with the tube wall has zero velocity (no-slip condition).

Step 3: The velocity increases as we move towards the center of the tube, reaching a maximum at the center.

Step 4: The velocity profile is parabolic, meaning the velocity varies quadratically with the distance from the wall.

Why other options are wrong:
- Option A: Straight line is incorrect because the velocity does not increase linearly from the wall to the center.
- Option B: Circle is incorrect because the velocity profile is not circular.
- Option D: Hyperbola is incorrect because the velocity profile is not hyperbolic.

Key Concepts:
- Viscosity
- Velocity profile

Memory Aids:
- Viscous flow in a tube = Parabolic profile.

Quick Tips:
- Remember the no-slip condition at the tube walls.

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Remembering +4 | -1

Spherical ball of radius

R

are falling in a viscous fluid of viscosity ɳ with a velocity

v

. The retarding viscous force acting on the spherical ball is

View answer and explanation

Correct option: B

directly proportional to both radius

R

and velocity

v

Option B (

{W_1} < {W_2}

) is correct because the air friction force is lower in the first 32m when the sphere is accelerating compared to the constant terminal velocity in the subsequent 32m.

Solution approach

Correct Answer + Core Reason: ${W_1} < {W_2}$ . The air friction force is lower in the first 32m, leading to less work done.

Step-by-Step Solution:

Step 1: Understand the concept of work done by air friction.\nThe work done by air friction is given by $W = \int F_{air} dx$ , where $F_{air}$ is the force of air friction and $dx$ is the displacement.

Step 2: Analyze the velocity during the first 32 m.\nDuring the first 32 m, the sphere accelerates from rest until it reaches its terminal velocity. Therefore, the average velocity during the first 32 m is less than the terminal velocity.

Step 3: Analyze the velocity during the subsequent 32 m.\nAfter reaching terminal velocity, the sphere falls at a constant speed. Therefore, the velocity during the subsequent 32 m is the terminal velocity.

Step 4: Relate air friction to velocity.\nThe air friction force is generally proportional to the square of the velocity ( $F_{air} \propto v^2$ ). Therefore, the air friction force is lower during the first 32 m compared to the subsequent 32 m.

Step 5: Compare the work done in both intervals.\nSince the average air friction force is lower during the first 32 m, the work done by air friction during the first 32 m ( ${W_1}$ ) is less than the work done during the subsequent 32 m ( ${W_2}$ ).

Final Answer: ${W_1} < {W_2}$

Why other options are wrong:
- Option A: The work done by air friction is less in the first 32m.
- Option C: The work done by air friction is not equal in both intervals due to the changing velocity.
- Option D: There is no specific relationship that dictates ${W_2} = 32{W_1}$ .

Key Concepts:
- Work done by a force
- Air friction
- Terminal velocity

Memory Aids:
- Work = Force x Distance
- Air friction increases with velocity.

Quick Tips:
- Consider the changing velocity when calculating the work done by air friction.

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Understanding +4 | -1

A small tiny lead shot is gently dropped on the surface of a viscous liquid

View answer and explanation

Correct option: D

The velocity of lead shot will reach steady value after sometime

Option D (The velocity of lead shot will reach steady value after sometime) is correct because as the lead shot falls, the viscous drag increases until it equals the net gravitational force, resulting in terminal velocity.

Solution approach

Correct Answer + Core Reason: The velocity of lead shot will reach steady value after sometime. This is due to terminal velocity.

Step-by-Step Solution:

Step 1: Identify the forces acting on the lead shot. The forces are gravity (downwards) and viscous drag (upwards).

Step 2: Describe the initial motion. Initially, the lead shot accelerates downwards due to gravity.

Step 3: Explain the viscous drag. As the velocity increases, the viscous drag force also increases. The viscous drag force is given by Stokes' Law: $F = 6 \pi \eta r v$ , where $\eta$ is the viscosity, $r$ is the radius, and $v$ is the velocity.

Step 4: Explain terminal velocity. Eventually, the viscous drag force equals the net gravitational force (weight minus buoyant force). At this point, the net force is zero, and the lead shot falls with a constant velocity called terminal velocity.

Why other options are wrong:
- Option A: The lead shot will fall with an acceleration equal to g at that place - This is only true initially, before the viscous drag becomes significant.
- Option B: The velocity of lead shot will decrease with time - The velocity increases until it reaches terminal velocity, then remains constant.
- Option C: The velocity of lead shot will increase continuously - The velocity increases until terminal velocity is reached, then remains constant.

Key Concepts:
- Viscosity
- Stokes' Law
- Terminal Velocity

Memory Aids:
- Terminal velocity is reached when drag force equals weight.

Quick Tips:
- Remember that viscous drag is velocity-dependent.

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Application +4 | -1

The pressure at the bottom of water tank is 3P, where P is atmospheric pressure. If water is drawn out till the water level decreases by

\frac{2}{5}th

of initial value then pressure at the bottom of the tank will be

View answer and explanation

Correct option: A

\frac{{11P}}{5}

Option A (

\frac{{11P}}{5}

) is correct because the pressure at the bottom is the sum of atmospheric pressure and the pressure due to the water column, and the water level decreases by 2/5.

Solution approach

Correct Answer + Core Reason: $\frac{{11P}}{5}$ . The pressure at the bottom depends on the height of the water column.

Step-by-Step Solution:

Step 1: Let the initial height of the water be $H$ . The initial pressure at the bottom is $P_{initial} = P_{atm} + \rho g H = 3P$ , where $P_{atm} = P$ .

Step 2: Therefore, $P + \rho g H = 3P$ , which implies $\rho g H = 2P$ .

Step 3: The water level decreases by $\frac{2}{5}H$ , so the new height is $H' = H - \frac{2}{5}H = \frac{3}{5}H$ .

Step 4: The new pressure at the bottom is $P_{new} = P_{atm} + \rho g H' = P + \rho g (\frac{3}{5}H)$ .

Step 5: Substitute $\rho g H = 2P$ into the equation: $P_{new} = P + \frac{3}{5}(2P) = P + \frac{6}{5}P = \frac{11}{5}P$ .

Why other options are wrong:
- Option B: $\frac{{4P}}{3}$ . This is incorrect.
- Option C: $\frac{{2P}}{3}$ . This is incorrect.
- Option D: $\frac{{3P}}{4}$ . This is incorrect.

Key Concepts:
- Pressure in a Liquid
- Hydrostatic Pressure

Memory Aids:
- Pressure increases with depth.

Quick Tips:
- Consider the atmospheric pressure when calculating the total pressure.

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Remembering +4 | -1

An atmosphere,

View answer and explanation

Correct option: A

is a unit of pressure

Option A (is a unit of pressure) is correct because an atmosphere (atm) is a standard unit used to measure pressure.

Solution approach

Correct Answer + Core Reason: is a unit of pressure. An atmosphere (atm) is a unit of pressure.

Step-by-Step Solution:

Step 1: Define Atmosphere: An atmosphere (atm) is a unit of pressure, approximately equal to the average atmospheric pressure at sea level on Earth.

Step 2: Explain its Use: It is commonly used in various scientific and engineering applications to measure and express pressure.

Step 3: Conclusion: An atmosphere is a unit of pressure.

Why other options are wrong:
- Option B (is a unit of force): An atmosphere is not a unit of force; it's a unit of pressure. Force is measured in Newtons (N).
- Option C (gives us an idea of composition of air): While related to air, an atmosphere primarily measures pressure, not composition.
- Option D (is the height above which there is no air): This describes the extent of the atmosphere, not what an atmosphere (unit) is.

Key Concepts:
- Pressure Units
- Atmosphere (atm)

Memory Aids:
- Remember that atm is a unit for measuring pressure.

Quick Tips:
- Know the common units of pressure.

60 Medium

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Understanding +4 | -1

With the increase in temperature, the angle of contact

View answer and explanation

Correct option: A

Decreases

Option A is correct. As temperature increases, the cohesive forces within the liquid weaken, leading to a decrease in the angle of contact.

Solution approach

Correct Answer + Core Reason: Decreases. Increased temperature reduces surface tension and cohesive forces.

Step-by-Step Solution:

Step 1: The angle of contact depends on the relative strengths of adhesive and cohesive forces.

Step 2: As temperature increases, the surface tension of the liquid decreases, which weakens the cohesive forces within the liquid.

Step 3: This leads to a decrease in the angle of contact.

Why other options are wrong:
- Option A: Correct because the angle of contact decreases with increasing temperature.
- Option B: Incorrect because the angle of contact decreases with increasing temperature.
- Option C: Incorrect because the angle of contact changes with temperature.
- Option D: The angle of contact consistently decreases with increasing temperature.

Key Concepts:
- Angle of contact
- Surface tension
- Effect of temperature on surface tension
- Adhesive and cohesive forces

Memory Aids:
- Higher temperature = lower surface tension = smaller angle of contact

Quick Tips:
- Consider how temperature affects the intermolecular forces within the liquid.

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Mechanical properties of fluids practice · Medium

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