MECHANICAL PROPERTIES OF FLUIDS MCQs for NEET UG PHYSICS — Medium, Page 5

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Analyzing +4 | -1

\mathbf{A}

: An iron needle placed carefully on the surface of water may float, whereas a solid ball of same material will always sink

\mathbf{R}

: The buoyancy on an object depends both on the material and shape of object.

View answer and explanation

Correct option: C

(A) is true, (R) is false

Option C is correct because while the assertion is true due to surface tension, the reason is false. Buoyancy depends on the volume of fluid displaced, not the material or shape directly.

Solution approach

Correct Answer + Core Reason: (A) is true, (R) is false. The needle floats due to surface tension, but buoyancy depends on volume displaced, not material.

Step-by-Step Solution:

Step 1: Analyze the Assertion (A): An iron needle can float if placed carefully on the water surface due to surface tension. This is a true statement.

Step 2: Analyze the Reason (R): The buoyancy on an object depends both on the material and shape of the object. This statement is false. Buoyancy depends on the volume of fluid displaced by the object, as stated by Archimedes' principle. The material's density affects whether the object floats or sinks, but buoyancy itself is determined by the displaced fluid's weight.

Step 3: Determine the relationship between Assertion and Reason: The assertion is true, but the reason is false. Therefore, the correct answer is that (A) is true, and (R) is false.

Step 4: Final Answer: (A) is true, (R) is false

Why other options are wrong:
- Option A: The reason is not a correct explanation of the assertion, and the reason itself is false.
- Option B: The reason is false.
- Option D: The assertion is true.

Key Concepts:
- Surface Tension
- Buoyancy (Archimedes' Principle)
- Density

Memory Aids:
- Archimedes' Principle: Buoyant force equals the weight of the fluid displaced.

Quick Tips:
- Distinguish between surface tension and buoyancy.
- Remember that buoyancy depends on the volume of fluid displaced.

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Ice pieces are floating in beaker A containing water also in a beaker B containing miscible liquid of specific gravity 1.2. When ice melts, the level of

View answer and explanation

Correct option: D

liquid in B increases

Option D (liquid in B decreases) is correct because when ice melts in a liquid with a specific gravity greater than 1, the volume of the melted ice (water) is less than the volume of the liquid it displaced when it was ice, causing the liquid level to decrease.

Solution approach

Correct Answer + Core Reason: liquid in B increases. When ice melts in a liquid with specific gravity > 1, the level decreases.

Step-by-Step Solution:

Step 1: Analyze beaker A (water).\nWhen ice melts in water, the volume of water produced is equal to the volume of water displaced by the ice. Therefore, the level of water in beaker A remains the same.

Step 2: Analyze beaker B (miscible liquid with specific gravity 1.2).\nThe specific gravity of the liquid is 1.2, which means its density is 1.2 times the density of water. When ice floats, it displaces a volume of liquid such that the weight of the displaced liquid equals the weight of the ice.\nLet $V_i$ be the volume of ice and $\rho_i$ be the density of ice. Let $V_l$ be the volume of liquid displaced by the ice and $\rho_l$ be the density of the liquid.\nThen, $V_l \rho_l g = V_i \rho_i g$ , so $V_l = V_i \frac{\rho_i}{\rho_l}$ .

Step 3: Consider the melting process.\nWhen the ice melts, it turns into water. Let $V_w$ be the volume of water produced by melting the ice. The mass of the ice is equal to the mass of the water produced, so $V_i \rho_i = V_w \rho_w$ , which means $V_w = V_i \frac{\rho_i}{\rho_w}$ .

Step 4: Compare the volumes.\nWe have $V_l = V_i \frac{\rho_i}{\rho_l}$ and $V_w = V_i \frac{\rho_i}{\rho_w}$ . Since $\rho_l = 1.2 \rho_w$ , we have $V_l = V_i \frac{\rho_i}{1.2 \rho_w}$ .\nComparing $V_w$ and $V_l$ , we have $V_w = 1.2 V_l$ . This means that the volume of water produced is greater than the volume of liquid displaced. Therefore, the level of liquid in beaker B increases.

Why other options are wrong:
- Option A (water increases in A): The water level remains the same in A.
- Option B (water decreases in A): The water level remains the same in A.
- Option C (liquid in B decreases): The liquid level increases in B.

Key Concepts:
- Buoyancy
- Specific gravity
- Archimedes' principle

Memory Aids:
- Weight of displaced fluid equals the weight of the floating object.

Quick Tips:
- Consider the densities of the ice, water, and liquid when analyzing the change in levels.

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An application of Bernoulli’s equation for fluid flow is found in

View answer and explanation

Correct option: A

Dynamic lift of an aeroplane

Option A is correct because the dynamic lift of an aeroplane is a direct application of Bernoulli's principle, relating fluid speed to pressure.

Solution approach

Correct Answer + Core Reason: Dynamic lift of an aeroplane. Bernoulli's equation describes the relationship between fluid speed and pressure, directly explaining lift.

Step-by-Step Solution:

Step 1: Bernoulli's equation states that for an inviscid, incompressible fluid, an increase in fluid speed occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy.

Step 2: The shape of an airplane wing is designed such that the air flows faster over the top surface than the bottom surface.

Step 3: According to Bernoulli's principle, the pressure on the top surface is lower than the pressure on the bottom surface.

Step 4: This pressure difference creates an upward force, known as dynamic lift, which allows the airplane to fly.

Why other options are wrong:
- Option B: Viscosity meter measures the viscosity of a fluid, which is related to its resistance to flow, not directly to Bernoulli's equation.
- Option C: Capillary rise is due to surface tension and adhesive forces between the liquid and the tube, not Bernoulli's equation.
- Option D: Hydraulic press works on Pascal's principle, which states that pressure applied to a confined fluid is transmitted equally in all directions.

Key Concepts:
- Bernoulli's Equation
- Fluid Dynamics
- Pressure and Velocity Relationship
- Dynamic Lift

Memory Aids:
- Bernoulli's principle: Faster flow, lower pressure.

Quick Tips:
- Remember that Bernoulli's equation applies to ideal fluids (inviscid and incompressible).

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The rain drops are in spherical shape due to

View answer and explanation

Correct option: C

surface tension

Option C (surface tension) is correct because surface tension minimizes the surface area of the liquid, causing it to form a spherical shape.

Solution approach

Correct Answer + Core Reason: Surface tension. Surface tension causes liquids to minimize their surface area, resulting in a spherical shape for raindrops.

Step-by-Step Solution:

Step 1: Understand the concept of surface tension. Surface tension is the property of a liquid that causes it to behave as if its surface were covered with a stretched elastic membrane.

Step 2: Relate surface tension to the shape of raindrops. Surface tension tends to minimize the surface area of a liquid. For a given volume, the shape with the minimum surface area is a sphere.

Step 3: Conclude that surface tension is responsible for the spherical shape of raindrops.

Why other options are wrong:
- Option A: Residual pressure refers to the pressure remaining in a system after a process, and it does not directly cause the spherical shape of raindrops.
- Option B: Thrust is a force, and while it can affect the motion of a raindrop, it does not directly determine its shape.
- Option D: Viscosity is the resistance of a fluid to flow. While viscosity affects the speed at which a raindrop falls, it does not primarily determine its spherical shape.

Key Concepts:
- Surface Tension
- Minimization of Surface Area

Memory Aids:
- Remember that surface tension always tries to minimize the surface area of a liquid.

Quick Tips:
- Think about how liquids behave in small quantities. They tend to form spherical shapes due to surface tension.

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A man is sitting in a boat which is floating in a pond. If the man drink some water from the pond, what will happen to the level of water in the pond?

View answer and explanation

Correct option: C

Remains same

Option C (Remains same) is correct because the volume of water the man drinks is replaced by an equal volume of water from the pond, maintaining the water level.

Solution approach

Correct Answer + Core Reason: Remains same. The water level remains the same because the volume of water the man drinks is taken from the pond itself.

Step-by-Step Solution:

Step 1: Consider the situation: A man is sitting in a boat floating in a pond. The man drinks some water from the pond.

Step 2: When the man drinks water from the pond, he is essentially removing a certain volume of water from the pond.

Step 3: However, the water he drinks was already part of the pond. Therefore, the total volume of water in the pond decreases by the amount he drinks.

Step 4: Since the boat is floating, it displaces a weight of water equal to the weight of the boat plus the man. When the man drinks water, his weight doesn't change significantly (the weight of the water he drinks is negligible compared to his total weight). Therefore, the boat continues to displace the same weight of water.

Step 5: The volume of water displaced by the boat remains the same, and the water the man drank was already part of the pond. Therefore, the water level in the pond remains the same.

Step 6: Final Answer: Remains same

Why other options are wrong:
- Option A: Increases: The water level does not increase because the man is removing water from the pond.
- Option B: Decreases: The water level does not decrease because the volume removed is negligible compared to the total volume of the pond and the displacement by the boat remains the same.
- Option D: Depends on weight of boat: While the weight of the boat affects the overall water level, the act of the man drinking water doesn't change the water level.

Key Concepts:
- Buoyancy
- Displacement of water
- Archimedes' principle

Memory Aids:
- Floating objects displace their weight in water.

Quick Tips:
- Consider the source of the water being consumed.
- Think about the displacement of water by the boat.

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Why the dam of water reservoir is thick at the bottom

View answer and explanation

Correct option: C

Pressure of water increases with depth

Option C (Pressure of water increases with depth) is correct because the pressure exerted by a fluid increases linearly with depth, requiring a stronger dam at the bottom.

Solution approach

Correct Answer + Core Reason: Pressure of water increases with depth. Pressure increases linearly with depth.

Step-by-Step Solution:

Step 1: Recall the relationship between pressure and depth in a fluid.

Step 2: The pressure at a depth $h$ in a fluid is given by $P = P_0 + \rho g h$ , where $P_0$ is the atmospheric pressure, $\rho$ is the density of the fluid, and $g$ is the acceleration due to gravity.

Step 3: This equation shows that pressure increases linearly with depth.

Step 4: Therefore, the dam must be thicker at the bottom to withstand the higher pressure.

Why other options are wrong:
- Option A: The quantity of water increases with depth, but this is not the direct reason for the increased thickness.
- Option B: The density of water is approximately constant with depth (for typical dam depths).
- Option D: The temperature of water may vary with depth, but this is not the primary reason for the increased thickness.

Key Concepts:
- Fluid Pressure
- Pressure and Depth Relationship

Memory Aids:
- Pressure increases as you go deeper in a fluid.

Quick Tips:
- Remember the formula $P = \rho g h$ .

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Application +4 | -1

The radius of a soap bubble is increased from

\frac{1}{{\sqrt \pi }}\,cm\,\,to\,\,\frac{2}{{\sqrt \pi }}\,\,cm

. If the surface tension of water is 30 dynes per cm, then the work done will be

View answer and explanation

Correct option: C

720 ergs

Option C (720 ergs) is correct because it is the work done calculated from the change in surface area of the soap bubble.

Solution approach

Correct Answer + Core Reason: 720 ergs. The work done is equal to the surface tension multiplied by the change in surface area of the soap bubble (considering both inner and outer surfaces).

Step-by-Step Solution:

Step 1: Identify the given values: $r_1 = \frac{1}{\sqrt{\pi}}$ cm, $r_2 = \frac{2}{\sqrt{\pi}}$ cm, $T = 30$ dynes/cm.

Step 2: Calculate the initial surface area: $A_1 = 2 \times 4\pi r_1^2 = 8\pi (\frac{1}{\sqrt{\pi}})^2 = 8\pi \times \frac{1}{\pi} = 8 \text{{ cm}}^2$ (multiply by 2 because a soap bubble has two surfaces).

Step 3: Calculate the final surface area: $A_2 = 2 \times 4\pi r_2^2 = 8\pi (\frac{2}{\sqrt{\pi}})^2 = 8\pi \times \frac{4}{\pi} = 32 \text{{ cm}}^2$

Step 4: Calculate the change in surface area: $\Delta A = A_2 - A_1 = 32 - 8 = 24 \text{{ cm}}^2$

Step 5: Calculate the work done: $W = T \Delta A = 30 \text{{ dynes/cm}} \times 24 \text{{ cm}}^2 = 720 \text{{ ergs}}$

Why other options are wrong:
- Option A (180 ergs): This is incorrect, possibly due to not considering the two surfaces of the soap bubble or an incorrect calculation.
- Option B (360 ergs): This is incorrect, possibly due to not considering the two surfaces of the soap bubble or an incorrect calculation.
- Option D (960 ergs): This is incorrect, possibly due to an incorrect calculation.

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

Memory Aids:
- Soap bubble has two surfaces.

Quick Tips:
- Remember to multiply the surface area by 2 for soap bubbles.

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Application +4 | -1

The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface. Assuming atmospheric pressure to be

75cm

of

Hg

and the density of water to be 1/10 of the density of mercury, the depth of the lake is

View answer and explanation

Correct option: C

15m

The depth of the lake is 15 m. Using Boyle's law and the hydrostatic pressure equation, the depth can be calculated.

Solution approach

Correct Answer + Core Reason: $15m$ . The depth of the lake can be found using Boyle's law and the hydrostatic pressure equation.

Step-by-Step Solution:

Step 1: Let $P_1$ be the pressure at the bottom of the lake and $V_1$ be the volume of the air bubble at the bottom.

Step 2: Let $P_2$ be the pressure at the surface of the lake and $V_2$ be the volume of the air bubble at the surface. Given $V_2 = 3V_1$ .

Step 3: According to Boyle's law, $P_1V_1 = P_2V_2$ , so $P_1V_1 = P_2(3V_1)$ , which means $P_1 = 3P_2$ .

Step 4: $P_2$ is the atmospheric pressure, which is $75 cm$ of $Hg$ .

Step 5: $P_1 = P_2 + \rho_{water}gh$ , where $h$ is the depth of the lake.

Step 6: $3P_2 = P_2 + \rho_{water}gh$ , so $2P_2 = \rho_{water}gh$ .

Step 7: $2(75 cm \ Hg) = \rho_{water}gh$ . Since $\rho_{water} = 1/10 \rho_{Hg}$ , $2(75 cm \ Hg) = (1/10) \rho_{Hg}gh$ .

Step 8: $150 cm \ Hg = (1/10) \rho_{Hg}gh$ , so $1500 cm \ Hg = \rho_{Hg}gh$ .

Step 9: $h = 1500 cm = 15 m$ .

Why other options are wrong:
- Option A: $5m$ : Incorrect calculation.
- Option B: $10m$ : Incorrect calculation.
- Option D: $20m$ : Incorrect calculation.

Key Concepts:
- Boyle's Law
- Hydrostatic Pressure

Memory Aids:
- Remember Boyle's law: $P_1V_1 = P_2V_2$ .

Quick Tips:
- Use Boyle's law when the temperature is constant.

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Viscosity is closely related to -

View answer and explanation

Correct option: A

Friction

Option A (Friction) is correct because viscosity is essentially internal friction within a fluid, resisting its flow.

Solution approach

Correct Answer + Core Reason: Friction. Viscosity is a form of internal friction within a fluid.

Step-by-Step Solution:

Step 1: Define viscosity.

Step 2: Viscosity is the measure of a fluid's resistance to flow.

Step 3: This resistance arises from the internal friction between the fluid layers as they move relative to each other.

Step 4: Therefore, viscosity is closely related to friction.

Why other options are wrong:
- Option B: Adhesive molecular force - Adhesive forces are between different substances, not within the fluid itself.
- Option C: Cohesive molecular force - Cohesive forces contribute to viscosity, but viscosity is the manifestation of internal friction.
- Option D: Barnoulli's theorem - Bernoulli's theorem relates pressure, velocity, and height in a fluid flow, but it's not directly related to viscosity.

Key Concepts:
- Viscosity
- Friction
- Fluid Dynamics

Memory Aids:
- Viscosity = internal friction.

Quick Tips:
- Remember that viscosity is a type of friction within a fluid.

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Application +4 | -1

A water film is formed between two parallel wires of 10 cm length. The distance of 0.5 cm between the wires is increased by 1 mm. Which will be the work done? (Surface tension of water = 

72\mathrm{N}{\mathrm{m}^{ - 1}})

View answer and explanation

Correct option: B

144 erg

Option B (144 erg) 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: 144 erg. 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 $l = 10 \text{{ cm}}$ , Initial distance $d_1 = 0.5 \text{{ cm}}$ , Change in distance $\Delta d = 1 \text{{ mm}} = 0.1 \text{{ cm}}$ , Surface tension $T = 72 \text{{ dyne/cm}}$ (Note: the given surface tension is in N/m, but the answer is in erg, so we use dyne/cm).

Step 2: Calculate the change in area: $\Delta A = l \times \Delta d = 10 \text{{ cm}} \times 0.1 \text{{ cm}} = 1 \text{{ cm}}^2$

Step 3: Since the water film has two surfaces, the total change in area is $2\Delta A = 2 \times 1 = 2 \text{{ cm}}^2$

Step 4: Calculate the work done: $W = T \times 2\Delta A = 72 \text{{ dyne/cm}} \times 2 \text{{ cm}}^2 = 144 \text{{ ergs}}$

Why other options are wrong:
- Option A (288 erg): This is incorrect, possibly due to not dividing by 2 or an incorrect calculation.
- Option C (72 erg): This is incorrect, possibly due to not multiplying by 2 or an incorrect calculation.
- Option D (36 erg): This is incorrect, possibly due to dividing by 4 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.

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In the following fig. is shown the flow of liquid through a horizontal pipe. Three tubes

A,B

and

C

are connected to the pipe. The radii of the tubes

A,B

and

C

at the junction are respectively

2cm,1cm

and

2cm

. It can be said that the

View answer and explanation

Correct option: D

Height of the liquid in the tubes

A

and

C

is the same

Option D (Height of the liquid in the tubes

A

and

C

is the same) is correct because according to Pascal's law, the pressure at the same horizontal level is the same, and since tubes A and C have the same radius, the liquid height will be the same.

Solution approach

Correct Answer + Core Reason: Height of the liquid in the tubes $A$ and $C$ is the same. According to Pascal's law, pressure at the same horizontal level is the same.

Step-by-Step Solution:

Step 1: Apply Pascal's Law: In a horizontal pipe with connected vertical tubes, the height of the liquid in each tube indicates the pressure at that point in the pipe.

Step 2: Consider the Radii: The radii of the tubes affect the volume but not the pressure at the junction. Since tubes A and C have the same radius, the pressure at their junctions will be the same.

Step 3: Determine the Heights: Since the pressure is the same at the junctions of tubes A and C, the height of the liquid in those tubes will be the same.

Step 4: Conclusion: The height of the liquid in tubes A and C is the same.

Why other options are wrong:
- **Option A (Height of the liquid in the tube $A$ is maximum): This is incorrect because the height depends on the pressure, which is the same for A and C.
- Option B (Height of the liquid in the tubes $A$ and $B$ is the same): Tube B has a different radius, so the pressure might be different due to flow dynamics (though not directly related to radius in this static pressure scenario).
- Option C (Height of the liquid in all the three tubes is the same): This is incorrect because tube B has a different radius than A and C.

Key Concepts:
- Pascal's Law
- Fluid Pressure

Memory Aids:
- Pressure at the same level in a connected fluid is the same.

Quick Tips:**
- Understand how pressure relates to fluid height in connected tubes.

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Remembering +4 | -1

The working of venturimeter is based on

View answer and explanation

Correct option: C

Bernoulli’s theorem

Option C (Bernoulli’s theorem) is correct because the venturimeter works by measuring pressure differences related to fluid velocity changes, as described by Bernoulli's theorem.

Solution approach

Correct Answer + Core Reason: Bernoulli’s theorem. The venturimeter is a device used to measure the flow rate of a fluid in a pipe, and its operation is based on Bernoulli's theorem.

Step-by-Step Solution:

Step 1: Describe the venturimeter: A venturimeter consists of a converging section, a throat (narrow section), and a diverging section.

Step 2: Explain how it works: As the fluid flows through the converging section, its velocity increases, and its pressure decreases (according to Bernoulli's theorem).

Step 3: Measure the pressure difference: The pressure difference between the wider section and the throat is measured. This pressure difference is related to the flow rate of the fluid.

Step 4: Apply Bernoulli's theorem: Bernoulli's theorem relates the pressure, velocity, and height of a fluid in steady flow. In a venturimeter, the height is constant, so the equation simplifies to: $P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$ , where $P_1$ and $v_1$ are the pressure and velocity in the wider section, and $P_2$ and $v_2$ are the pressure and velocity in the throat.

Why other options are wrong:
- Option A: Torricelli's law describes the speed of efflux from an opening in a container.
- Option B: Pascal's law states that pressure applied to a fluid in a closed container is transmitted equally to every point of the fluid.
- Option D: Archimedes' principle describes the buoyant force on an object submerged in a fluid.

Key Concepts:
- Bernoulli's theorem
- Fluid dynamics
- Pressure
- Velocity

Memory Aids:
- Venturimeter uses Bernoulli's principle to measure flow rate.

Quick Tips:
- Remember that velocity increases and pressure decreases in the narrow section of the venturimeter.
- Understand the relationship between pressure difference and flow rate.

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Understanding +4 | -1

A piece of wood is floating in water kept in bottle. The bottle is connected to an air pump. Neglect the compressibility of water. When more air is pushed into the bottle from the pump, the piece of wood will float with

View answer and explanation

Correct option: D

Same part in water

The correct answer is D (Same part in water) because the density of water and the wood remain constant, so the fraction of the wood submerged remains the same regardless of the air pressure in the bottle.

Solution approach

Correct Answer + Core Reason: Same part in water. The buoyant force must equal the weight of the wood, and since the density of water and the weight of the wood are constant, the volume submerged remains constant.

Step-by-Step Solution:

Step 1: Understand the principle of floatation: For an object to float, the buoyant force must equal the weight of the object.

Step 2: Buoyant force is given by $F_B = V_{submerged} \times \rho_{water} \times g$ , where $V_{submerged}$ is the volume of the object submerged in water, $\rho_{water}$ is the density of water, and $g$ is the acceleration due to gravity.

Step 3: The weight of the wood is constant. Since the water is incompressible, its density remains constant.

Step 4: Since the buoyant force equals the weight of the wood, and the density of water is constant, the volume of the wood submerged must also remain constant.

Step 5: Pushing more air into the bottle increases the pressure, but it doesn't change the density of the water or the weight of the wood. Therefore, the submerged volume remains the same.

Why other options are wrong:
- Option A: Larger part in water: This would imply the buoyant force increased, which is not the case.
- Option B: Lesser part in water: This would imply the buoyant force decreased, which is not the case.
- Option C: It will sink: This would imply the weight of the wood is greater than the maximum buoyant force, which is not the case since it's floating.
- Option D: Same part in water: This is correct because the buoyant force equals the weight of the wood, and neither changes.

Key Concepts:
- Buoyancy
- Archimedes' Principle
- Floatation

Memory Aids:
- Remember that for an object to float, the buoyant force must equal its weight.

Quick Tips:
- Consider the factors affecting buoyant force and weight when analyzing floatation problems.

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Analyzing +4 | -1

The terminal velocity of a ball in air is

v

, where acceleration due to gravity is

g

. Now the same ball is taken in a gravity free space where all other conditions are same. The ball is now pushed at a speed

v v

, then

View answer and explanation

Correct option: D

The ball will finally stop (Given that density of the ball

\rho=2

times the density of air

\sigma

)

Option D is correct. In a gravity-free space, the only force acting on the ball is the viscous force due to air. Since the density of the ball is greater than the density of air, the ball will eventually stop due to this viscous drag.

Solution approach

Correct Answer + Core Reason: The ball will finally stop (Given that density of the ball $\rho=2$ times the density of air $\sigma$ ). In gravity-free space, viscous drag will eventually halt the ball.

Step-by-Step Solution:

Step 1: In a gravity-free space, the weight of the ball is zero, and there is no buoyant force.

Step 2: The only force acting on the ball is the viscous force, which opposes the motion of the ball. This force is given by Stokes' Law (approximately), $F_v = 6\pi \eta r v$ , where $\eta$ is the viscosity of air, $r$ is the radius of the ball, and $v$ is the velocity of the ball.

Step 3: Since there is a net force (viscous force) acting on the ball opposite to its velocity, the ball will decelerate.

Step 4: As the ball decelerates, its velocity decreases, and the viscous force also decreases. However, as long as there is a velocity, there will be a viscous force opposing the motion.

Step 5: Eventually, the viscous force will bring the ball to a complete stop.

Step 6: Final Answer: The ball will finally stop

Why other options are wrong:
- Option A: The terminal velocity concept applies when there is gravity and buoyancy. In gravity-free space, there is no terminal velocity in the traditional sense.
- Option B: The ball will not move with a constant velocity because the viscous force is always acting to decelerate it.
- Option C: The initial acceleration is not 2g. The acceleration is due to the viscous force and depends on the velocity and viscosity of the air.

Key Concepts:
- Viscous Force
- Stokes' Law (approximation)
- Newton's First Law (Inertia)

Memory Aids:
- Viscous force always opposes motion.

Quick Tips:
- In gravity-free space, consider only the forces that are present (e.g., viscous force).
- Remember that viscous force depends on velocity.

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Application +4 | -1

If the velocity head of a stream of water is equal to 10 cm, then its speed of flow is (

\mathrm{g} = 10\mathrm{m}{\mathrm{s}^{ - 2}}

)

View answer and explanation

Correct option: C

1.4\mathrm{m}{\mathrm{s}^{ - 1}}

Option C (

1.4\mathrm{m}{\mathrm{s}^{ - 1}}

) is correct because it is the result of calculating the speed of flow using the given velocity head and the formula relating velocity head to speed.

Solution approach

Correct Answer + Core Reason: $1.4\mathrm{m}{\mathrm{s}^{ - 1}}$ . The speed of flow is calculated using the formula for velocity head.

Step-by-Step Solution:

Step 1: Identify the given information: Velocity head = 10 cm = 0.1 m, g = $10\mathrm{m}{\mathrm{s}^{ - 2}}$ .

Step 2: Recall the formula for velocity head: Velocity head = $\frac{{{v^2}}}{{2g}}$ , where v is the speed of flow.

Step 3: Rearrange the formula to solve for v: $v = \sqrt {2g \times \text{Velocity head}}$

Step 4: Substitute the given values: $v = \sqrt {2 \times 10\mathrm{m}{\mathrm{s}^{ - 2}} \times 0.1\mathrm{m}}$

Step 5: Calculate the speed: $v = \sqrt {2} = 1.414\mathrm{m}{\mathrm{s}^{ - 1}}$

Step 6: Round to one decimal place: $v \approx 1.4\mathrm{m}{\mathrm{s}^{ - 1}}$

Final Answer: $1.4\mathrm{m}{\mathrm{s}^{ - 1}}$

Why other options are wrong:
- Option A: $10\mathrm{m}{\mathrm{s}^{ - 1}}$ : This is incorrect because it doesn't use the correct formula and substitution.
- Option B: $140\mathrm{m}{\mathrm{s}^{ - 1}}$ : This is incorrect due to a miscalculation or misunderstanding of the formula.
- Option D: $0.1\mathrm{m}{\mathrm{s}^{ - 1}}$ : This is incorrect because it likely involves dividing instead of multiplying and taking the square root.

Key Concepts:
- Velocity head
- Fluid dynamics

Memory Aids:
- Velocity head formula: $\frac{{{v^2}}}{{2g}}$

Quick Tips:
- Ensure consistent units before substituting values into the formula.

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Application +4 | -1

From a steel wire of density ρ is suspended a brass block of density

{\rho _B}

. The extension of steel wire comes to

l

. If the brass block is now fully immersed in a liquid of density

{\rho _L}

, the extension becomes

l'

. The ratio

l/l'

will be

View answer and explanation

Correct option: D

\frac{{{\rho _B}}}{{{\rho _B} - {\rho _L}}}

Option D (

\frac{{{\rho _B}}}{{{\rho _B} - {\rho _L}}}

) is correct because the extension of the wire is proportional to the tension, which changes when the brass block is immersed in the liquid due to the buoyant force.

Solution approach

Correct Answer + Core Reason: $\frac{{{\rho _B}}}{{{\rho _B} - {\rho _L}}}$ . The ratio of extensions depends on the buoyant force and the densities.

Step-by-Step Solution:

Step 1: The initial extension $l$ is proportional to the weight of the brass block: $l \propto W = V_B \rho_B g$ , where $V_B$ is the volume of the brass block.

Step 2: When the brass block is immersed in the liquid, the effective weight is reduced by the buoyant force: $W' = W - F_B = V_B \rho_B g - V_B \rho_L g = V_B (\rho_B - \rho_L)g$ .

Step 3: The new extension $l'$ is proportional to the effective weight: $l' \propto W' = V_B (\rho_B - \rho_L)g$ .

Step 4: The ratio of the extensions is $\frac{l}{l'} = \frac{V_B \rho_B g}{V_B (\rho_B - \rho_L)g} = \frac{\rho_B}{\rho_B - \rho_L}$ .

Why other options are wrong:
- Option A: $\frac{{{\rho _B} - \rho }}{{{\rho _L} - \rho }}$ . This is incorrect.
- Option B: $\frac{{{\rho _L}}}{{{\rho _B} - {\rho _L}}}$ . This is incorrect.
- Option C: $\frac{{{\rho _B} - {\rho _L}}}{{{\rho _B}}}$ . This is the inverse of the correct answer.

Key Concepts:
- Buoyancy
- Young's Modulus
- Tension and Extension

Memory Aids:
- Buoyant force reduces the effective weight.

Quick Tips:
- Consider the buoyant force when an object is immersed in a liquid.

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Understanding +4 | -1

A steel ball is dropped in oil then,

View answer and explanation

Correct option: A

the ball attains constant velocity after some time

Option A (the ball attains constant velocity after some time) is correct because the viscous drag increases with speed until it balances the gravitational force, resulting in terminal velocity.

Solution approach

Correct Answer + Core Reason: the ball attains constant velocity after some time. This is due to the balance between gravitational force and viscous drag.

Step-by-Step Solution:

Step 1: Consider the forces acting on the steel ball: gravity (downwards) and viscous drag (upwards).

Step 2: Initially, the gravitational force is greater than the viscous drag, so the ball accelerates downwards.

Step 3: As the ball's speed increases, the viscous drag force also increases (according to Stokes' Law: $F = 6\pi \eta r v$ ).

Step 4: Eventually, the viscous drag force becomes equal to the gravitational force. At this point, the net force on the ball is zero, and it stops accelerating.

Step 5: The ball then falls with a constant velocity, called the terminal velocity.

Why other options are wrong:
- Option B: the ball stops - This is incorrect because the ball reaches terminal velocity, not stopping.
- Option C: the speed of ball will keep on increasing - This is incorrect because the viscous drag force eventually balances the gravitational force.
- Option D: None of the above - This is incorrect because option A is correct.

Key Concepts:
- Viscous Drag
- Terminal Velocity
- Stokes' Law
- Gravitational Force

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

Quick Tips:
- Remember that viscous drag increases with speed.

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Understanding +4 | -1

After terminal velocity is reached, the acceleration of a body falling through a fluid is:

View answer and explanation

Correct option: B

Zero

Option B is correct because after terminal velocity is reached, the net force is zero, resulting in zero acceleration.

Solution approach

Correct Answer + Core Reason: Zero. At terminal velocity, the net force is zero, hence no acceleration.

Step-by-Step Solution:

Step 1: When a body falls through a fluid, it experiences two main forces: the force of gravity (weight) acting downwards and the drag force (viscous force) acting upwards.

Step 2: Initially, the weight is greater than the drag force, so the body accelerates downwards.

Step 3: As the velocity of the body increases, the drag force also increases. The drag force is dependent on the velocity of the body.

Step 4: Eventually, the drag force becomes equal in magnitude to the weight of the body. At this point, the net force on the body is zero.

Step 5: According to Newton's first law of motion, if the net force on an object is zero, the object will either remain at rest or continue to move at a constant velocity. Since the body is already moving, it will continue to move at a constant velocity, which is the terminal velocity.

Step 6: Since the velocity is constant, the acceleration is zero.

Why other options are wrong:
- Option A: If the acceleration were equal to g, the body would be in free fall, which is not the case when it reaches terminal velocity.
- Option C: If the acceleration were less than g, it would mean that there is a net force acting on the body, but it has already reached terminal velocity where the net force is zero.
- Option D: If the acceleration were greater than g, it would mean that there is a net downward force greater than gravity, which is not possible.

Key Concepts:
- Terminal Velocity
- Drag Force
- Newton's First Law of Motion
- Net Force
- Acceleration

Memory Aids:
- Terminal Velocity = Constant Velocity = Zero Acceleration

Quick Tips:
- Remember that at terminal velocity, the forces are balanced.

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Remembering +4 | -1

Dimensions of coefficient of viscosity are

View answer and explanation

Correct option: C

\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]

Option C (

\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]

) is correct because it represents the correct dimensional formula for the coefficient of viscosity.

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]$ . This is the correct dimensional formula for viscosity.

Step-by-Step Solution:

Step 1: Recall the formula for viscous force: $F = \eta A \frac{{dv}}{{dx}}$ , where $\eta$ is the coefficient of viscosity, $A$ is the area, and $\frac{{dv}}{{dx}}$ is the velocity gradient.

Step 2: Rearrange the formula to solve for $\eta$ : $\eta = \frac{F}{{A \frac{{dv}}{{dx}}}}$

Step 3: Determine the dimensions of each term:
- Force (F): $\left[\mathrm{MLT}^{-2}\right]$
- Area (A): $\left[\mathrm{L}^2\right]$
- Velocity gradient ( $\frac{{dv}}{{dx}}$ ): $\left[\mathrm{T}^{-1}\right]$

Step 4: Substitute the dimensions into the formula for $\eta$ : $\eta = \frac{{\left[\mathrm{MLT}^{-2}\right]}}{{\left[\mathrm{L}^2\right] \left[\mathrm{T}^{-1}\right]}}$

Step 5: Simplify the expression: $\eta = \left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]$

Why other options are wrong:
- Option A: $\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$ - Incorrect dimensions.
- Option B: $\left[\mathrm{MLT}^{-1}\right]$ - Incorrect dimensions.
- Option D: $\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right]$ - Incorrect dimensions.

Key Concepts:
- Viscosity
- Dimensional Analysis
- Viscous Force

Memory Aids:
- Remember the formula for viscous force and the dimensions of force, area, and velocity gradient.

Quick Tips:
- Use the formula for viscous force to derive the dimensions of viscosity.

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Application +4 | -1

The water flows from a tap of diameter

1.25cm

with a rate of

5 \times {10^{ - 5}}{m^3}{s^{ - 1}}

. The density and coefficient of viscosity of water are

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

and

{10^{ - 3}}

Pas, respectively. The flow of water is

View answer and explanation

Correct option: B

Turbulent with Reynolds number 5100

Option B (Turbulent with Reynolds number 5100) is correct because the calculated Reynolds number is greater than 4000, indicating turbulent flow.

Solution approach

Correct Answer + Core Reason: Turbulent with Reynolds number 5100. The Reynolds number is calculated and compared to the critical value to determine the flow type.

Step-by-Step Solution:

Step 1: Calculate the diameter of the tap in meters:\nd = 1.25 cm = $1.25 \times 10^{-2}$ m

Step 2: Calculate the area of the tap:\nA = $\pi (d/2)^2 = \pi (1.25 \times 10^{-2} / 2)^2 = \pi (0.625 \times 10^{-2})^2 \approx 1.227 \times 10^{-4} m^2$

Step 3: Calculate the velocity of the water flow:\nv = Q/A = $(5 \times 10^{-5} m^3/s) / (1.227 \times 10^{-4} m^2) \approx 0.407 m/s$

Step 4: Calculate the Reynolds number:\nRe = $\frac{\rho v d}{\eta} = \frac{(10^3 kg/m^3)(0.407 m/s)(1.25 \times 10^{-2} m)}{10^{-3} Pas} = \frac{5.0875}{10^{-3}} \approx 5087.5$

Step 5: Round the Reynolds number to 5100.

Step 6: Determine the flow type. Since Re > 4000, the flow is turbulent.

Why other options are wrong:
- Option A: Steady with Reynolds number 5100: This is incorrect because a Reynolds number of 5100 indicates turbulent flow, not steady flow.
- Option C: Steady with Reynolds number 3900: This is incorrect because the calculated Reynolds number is approximately 5100, not 3900, and 3900 would still likely be considered turbulent.
- Option D: Turbulent with Reynolds number 3900: This is incorrect because the calculated Reynolds number is approximately 5100, not 3900.

Key Concepts:
- Reynolds number
- Fluid flow
- Turbulent flow
- Steady flow

Memory Aids:
- Remember the formula for Reynolds number: $Re = \frac{\rho v d}{\eta}$
- Re < 2000: Laminar flow
- 2000 < Re < 4000: Transitional flow
- Re > 4000: Turbulent flow

Quick Tips:
- Ensure consistent units throughout the calculation.
- Remember the critical Reynolds number values for different flow regimes.

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Understanding +4 | -1

Statement – 1: It is difficult to stop bleeding from a cut in the body at high altitude
Statement – 2: The atmospheric pressure at high altitude is lesser than the blood pressure.

View answer and explanation

Correct option: A

Statement – 1 is true, Statement – 2 is true; Statement – 2 is a correct explanation for Statement – 1

Option A (Statement – 1 is true, Statement – 2 is true; Statement – 2 is a correct explanation for Statement – 1) is correct because bleeding is harder to stop at high altitudes due to lower atmospheric pressure.

Solution approach

Correct Answer + Core Reason: Statement – 1 is true, Statement – 2 is true; Statement – 2 is a correct explanation for Statement – 1. Lower atmospheric pressure makes it harder to stop bleeding.

Step-by-Step Solution:

Step 1: Evaluate Statement 1: It is difficult to stop bleeding at high altitudes. This is true because the lower atmospheric pressure provides less external pressure to counteract the blood pressure.

Step 2: Evaluate Statement 2: The atmospheric pressure at high altitude is less than blood pressure. This is also true.

Step 3: Determine if Statement 2 explains Statement 1: The lower atmospheric pressure (Statement 2) directly explains why it's harder to stop bleeding (Statement 1).

Why other options are wrong:
- Option B: Statement 2 is a correct explanation for Statement 1.
- Option C: Statement 2 is true.
- Option D: Statement 1 is true.

Key Concepts:
- Atmospheric Pressure
- Blood Pressure

Memory Aids:
- High Altitude = Low Pressure = Harder to Stop Bleeding

Quick Tips:
- Remember the relationship between atmospheric pressure and bleeding.

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Application +4 | -1

The work done in increasing the size of a rectangular soap film with dimensions

8\mathrm{cm} \times 3.75

cm to

10\mathrm{cm} \times 6\mathrm{cm}

is

2 \times {10^{ - 4}}

J. The surface tension of the film in

N{m^{- 1}}

is

View answer and explanation

Correct option: B

3.3 \times {10^{ - 2}}

Option B (

3.3 \times {10^{ - 2}}

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

Solution approach

Correct Answer + Core Reason: $3.3 \times {10^{ - 2}}$ . The surface tension is calculated by dividing the work done by twice the change in area of the film.

Step-by-Step Solution:

Step 1: Identify the given values: Initial area $A_1 = 8 \text{{ cm}} \times 3.75 \text{{ cm}} = 30 \text{{ cm}}^2$ , Final area $A_2 = 10 \text{{ cm}} \times 6 \text{{ cm}} = 60 \text{{ cm}}^2$ , Work done $W = 2 \times 10^{-4} \text{{ J}} = 2 \times 10^{-4} \times 10^7 \text{{ dyne cm}} = 2000 \text{{ dyne cm}}$

Step 2: Calculate the change in area: $\Delta A = A_2 - A_1 = 60 - 30 = 30 \text{{ cm}}^2$

Step 3: Since the soap film has two surfaces, the total change in area is $2\Delta A = 2 \times 30 = 60 \text{{ cm}}^2$

Step 4: Calculate the surface tension: $T = \frac{W}{2\Delta A} = \frac{2000 \text{{ dyne cm}}}{60 \text{{ cm}}^2} = \frac{100}{3} \text{{ dyne/cm}} \approx 33.33 \text{{ dyne/cm}}$

Step 5: Convert dyne/cm to N/m: $T = 33.33 \text{{ dyne/cm}} = 33.33 \times 10^{-3} \text{{ N/m}} = 3.33 \times 10^{-2} \text{{ N/m}} \approx 3.3 \times 10^{-2} \text{{ N/m}}$

Why other options are wrong:
- **Option A ( $1.65 \times {10^{ - 2}}$ ): This is incorrect, possibly due to not considering the two surfaces of the soap film.
- Option C ( $6.6 \times {10^{ - 2}}$ ): This is incorrect, possibly due to an incorrect calculation.
- Option D ( $8.25 \times {10^{ - 2}}$ ): This is incorrect, possibly due to an incorrect calculation.

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

Memory Aids:
- Soap film has two surfaces.

Quick Tips:**
- Remember to multiply the change in area by 2 for soap films.

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Analyzing +4 | -1

A block of ice in which a piece of stone is embedded is floating on water contained in a beaker. When all the ice melts, the level of water in the beaker

View answer and explanation

Correct option: B

falls

Option B (falls) is correct. The water level falls because the stone displaces its weight in water when embedded in ice, but only its volume when submerged after the ice melts. Since stone's density is greater than water, its volume is less than the volume of water of equal weight.

Solution approach

Correct Answer + Core Reason: falls. The water level falls because the stone displaces its weight in water as ice, but its volume in water when the ice melts.

Step-by-Step Solution:

Step 1: Initial situation: Ice with stone is floating.\nThe ice displaces a volume of water equal to the weight of the ice and stone combined.

Step 2: Final situation: Ice melts and the stone sinks.\nThe melted ice contributes a negligible volume change. The stone now displaces a volume of water equal to its own volume.

Step 3: Compare the volumes of water displaced.\nInitially, the volume of water displaced is equivalent to the weight of the ice and stone. After melting, the volume of water displaced is equivalent to the volume of the stone.\nSince the density of the stone is greater than the density of water, the volume of the stone is less than the volume of water with the same weight.

Step 4: Conclusion:\nThe water level falls because the stone displaces less volume after the ice melts compared to when it was embedded in the ice.

Why other options are wrong:
- Option A: The water level rises if the stone displaced more volume after the ice melts, which is not the case.
- Option C: The water level remains unchanged only if the volume displaced remains the same, which is not the case.
- Option D: One of the other options is correct.

Key Concepts:
- Archimedes' principle
- Density and volume relationship
- Buoyancy

Memory Aids:
- Density = Mass/Volume

Quick Tips:
- Consider what happens to the volume displaced by the stone before and after the ice melts.

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Understanding +4 | -1

Which of the following will vanish for a fluid in a space where there is no gravity?

View answer and explanation

Correct option: D

Both (1) and (2)

Option D (Both (1) and (2)) is correct because in the absence of gravity, there is no weight and hence no upward thrust.

Solution approach

Correct Answer + Core Reason: Both (1) and (2). In the absence of gravity, the weight of the fluid and the upward thrust (buoyant force) will vanish.

Step-by-Step Solution:

Step 1: Define weight: Weight is the force exerted on an object due to gravity: $W = mg$ , where $m$ is the mass and $g$ is the acceleration due to gravity.

Step 2: Define upward thrust (buoyant force): The buoyant force is the upward force exerted by a fluid on an object submerged in it. It is equal to the weight of the fluid displaced by the object: $F_B = V \rho g$ , where $V$ is the volume of the fluid displaced, $\rho$ is the density of the fluid, and $g$ is the acceleration due to gravity.

Step 3: Analyze the effect of no gravity: If there is no gravity ( $g = 0$ ), then:
- Weight of fluid: $W = m \times 0 = 0$
- Upward thrust: $F_B = V \rho \times 0 = 0$

Step 4: Surface tension: Surface tension is a property of the liquid due to cohesive forces between molecules and is independent of gravity.

Why other options are wrong:
- Option A: The weight of the fluid vanishes, but surface tension does not.
- Option B: The upward thrust vanishes, but surface tension does not.
- Option C: Surface tension is independent of gravity and does not vanish.

Key Concepts:
- Weight
- Buoyancy
- Surface tension
- Gravity

Memory Aids:
- Weight and buoyant force depend on gravity.
- Surface tension is a property of the liquid itself.

Quick Tips:
- Remember that weight and buoyant force are directly proportional to gravity.
- Surface tension is independent of gravity.

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Analyzing +4 | -1

Assertion(A): If angle of contact is

90^{\circ}

, then liquid will neither rise nor fall in a capillary.

Reason(R): When angle of contact is

90^{\circ}

, surface is neither convex nor concave inside the capillary. It is flat.

View answer and explanation

Correct option: B

If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.

Option B is correct. Both the assertion and the reason are true, but the reason does not fully explain why the liquid neither rises nor falls. The angle of contact determines the shape of the meniscus, which affects the pressure difference and thus the rise or fall.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion. The angle of contact determines the meniscus shape and capillary action.

Step-by-Step Solution:

Step 1: Understand the Assertion: An angle of contact of 90 degrees implies no net capillary action (neither rise nor fall).

Step 2: Understand the Reason: A 90-degree angle of contact results in a flat meniscus inside the capillary.

Step 3: Relate Assertion and Reason: While a flat meniscus (90-degree contact angle) is true, the *reason* for no capillary action is more directly related to the balance of adhesive and cohesive forces, leading to zero net force for the liquid to rise or fall.

Step 4: Consider the underlying physics: Capillary action arises from the interplay of adhesive forces (between liquid and tube) and cohesive forces (within the liquid). When the contact angle is 90 degrees, these forces are balanced, resulting in no net capillary rise or fall.

Step 5: Conclusion: Both Assertion and Reason are correct, but the Reason doesn't fully explain the Assertion because it doesn't explicitly mention the balance of adhesive and cohesive forces.

Why other options are wrong:
- Option A: Incorrect because the Reason is not the *correct* explanation.
- Option C: Incorrect because the Reason *is* correct.
- Option D: Incorrect because the Assertion *is* correct.

Key Concepts:
- Capillary action
- Angle of contact
- Adhesive and cohesive forces
- Meniscus formation

Memory Aids:
- "Contact angle dictates the curve, balance of forces the rise it serves."

Quick Tips:
- Remember that capillary action is a result of the balance between adhesive and cohesive forces, which is reflected in the contact angle.

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Application +4 | -1

A raindrop reaching the ground with terminal velocity has momentum p. Another drop of twice the radius, also reaching the ground with terminal velocity, will have momentum

View answer and explanation

Correct option: D

32p

Option D (32p) is correct because the momentum depends on the mass and terminal velocity, both of which are affected by the radius.

Solution approach

Correct Answer + Core Reason: 32p. The momentum depends on mass and terminal velocity, both related to the radius.

Step-by-Step Solution:

Step 1: The momentum $p = mv$ , where $m$ is the mass and $v$ is the terminal velocity.

Step 2: The mass of the raindrop is proportional to its volume, which is proportional to $r^3$ , where $r$ is the radius. So, $m \propto r^3$ .

Step 3: The terminal velocity $v \propto r^2$ . (Since viscous force = weight, $6\pi \eta r v = \frac{4}{3} \pi r^3 \rho g$ , so $v = \frac{{2r^2 \rho g}}{{9\eta}}$ )

Step 4: Therefore, the momentum $p \propto r^3 \times r^2 = r^5$ .

Step 5: If the radius is doubled, $r' = 2r$ , then the new momentum $p' \propto (2r)^5 = 32r^5$ .

Step 6: Since the original momentum $p \propto r^5$ , the new momentum $p' = 32p$ .

Final Answer: 32p

Why other options are wrong:
- Option A (4p): This is incorrect because it doesn't account for the effect of the radius on both mass and terminal velocity.
- Option B (8p): This is incorrect because it doesn't account for the effect of the radius on both mass and terminal velocity.
- Option C (16p): This is incorrect because it doesn't account for the effect of the radius on both mass and terminal velocity.

Key Concepts:
- Momentum
- Terminal velocity
- Relationship between radius, volume, and mass

Memory Aids:
- Momentum: $p = mv$
- Volume of a sphere: $V = \frac{4}{3}\pi r^3$

Quick Tips:
- Remember the relationships between radius, volume, mass, and terminal velocity for spherical objects falling through a fluid.

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Understanding +4 | -1

The value of g at a place decreases by 2%. The barometric
height of mercury:

View answer and explanation

Correct option: C

remain unchanged

Option C (remain unchanged) is correct because the barometric height depends on atmospheric pressure, which is independent of the local value of g.

Solution approach

Correct Answer + Core Reason: remain unchanged. Barometric height depends on atmospheric pressure, not 'g'.

Step-by-Step Solution:

Step 1: Recall the formula for barometric height: $h = \frac{P}{\rho g}$ , where $P$ is the atmospheric pressure, $\rho$ is the density of mercury, and $g$ is the acceleration due to gravity.

Step 2: The barometric height measures atmospheric pressure. A change in 'g' locally does not change the atmospheric pressure.

Step 3: Therefore, the barometric height remains unchanged.

Why other options are wrong:
- Option A: The barometric height does not increase because atmospheric pressure is not directly affected by the local value of g.
- Option B: The barometric height does not decrease because atmospheric pressure is not directly affected by the local value of g.
- Option D: The barometric height remains constant as it measures atmospheric pressure, which is independent of the local value of g.

Key Concepts:
- Barometric Height
- Atmospheric Pressure

Memory Aids:
- Barometer measures pressure, not gravity.

Quick Tips:
- Focus on what a barometer actually measures: atmospheric pressure.

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Application +4 | -1

A piece of wax weighs 18.03 g in air. A piece of metal is found to weigh 17.03 g in water. It is tied to the wax and both together weigh 15.23 g in water. Then, the specific gravity of wax is

View answer and explanation

Correct option: C

\frac{{18.03}}{{19.83}}

Option C (

\frac{{18.03}}{{19.83}}

) is correct because the specific gravity is calculated by dividing the weight of the wax in air by the loss of weight of the wax in water.

Solution approach

Correct Answer + Core Reason: $\frac{{18.03}}{{19.83}}$ . The specific gravity is calculated by dividing the weight of the wax in air by the loss of weight of the wax in water.

Step-by-Step Solution:

Step 1: Identify the given information: Weight of wax in air $W_{wax,air} = 18.03 \text{{ g}}$ , weight of metal in water $W_{metal,water} = 17.03 \text{{ g}}$ , weight of wax and metal in water $W_{total,water} = 15.23 \text{{ g}}$ .

Step 2: The weight of the wax in water is $W_{wax,water} = W_{total,water} - W_{metal,water} = 15.23 - 17.03 = -1.8 \text{{ g}}$ .

Step 3: The loss of weight of the wax in water is $W_{wax,air} - W_{wax,water} = 18.03 - (-1.8) = 18.03 + 1.8 = 19.83 \text{{ g}}$ .

Step 4: The specific gravity of the wax is given by $SG = \frac{W_{wax,air}}{W_{wax,air} - W_{wax,water}} = \frac{18.03}{19.83}$ .

Step 5: Final Answer: $\frac{{18.03}}{{19.83}}$

Why other options are wrong:
- **Option A ( $\frac{{18.03}}{{17.03}}$ ): This is incorrect. It uses the weight of the metal in water instead of the loss of weight of the wax in water.
- Option B ( $\frac{{17.03}}{{18.03}}$ ): This is incorrect. It inverts the specific gravity calculation.
- Option D ( $\frac{{15.03}}{{17.03}}$ ): This is incorrect and doesn't follow from any logical application of the specific gravity formula.

Key Concepts:
- Specific Gravity
- Archimedes' Principle

Memory Aids:
- Remember that specific gravity is the ratio of the density of a substance to the density of water.

Quick Tips:**
- Be careful with the signs when calculating the loss of weight in water.

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Understanding +4 | -1

If a drop of water is broken in to smaller drops the surface energy

View answer and explanation

Correct option: A

Increases

Option A is correct because when a drop of water is broken into smaller drops, the total surface area increases, leading to an increase in surface energy.

Solution approach

Correct Answer + Core Reason: Increases. Breaking a drop into smaller drops increases the total surface area, thus increasing surface energy.

Step-by-Step Solution:

Step 1: Understand the concept of surface energy. Surface energy is the energy required to create a new surface. It is proportional to the surface area.

Step 2: Consider a single drop of water with surface area $A_1$ . When this drop is broken into $n$ smaller drops, each with surface area $A_2$ , the total surface area becomes $n A_2$ .

Step 3: The total volume of water remains constant. Let $R$ be the radius of the original drop and $r$ be the radius of each smaller drop. Then, $\frac{4}{3}\pi R^3 = n \frac{4}{3}\pi r^3$ , which simplifies to $R^3 = n r^3$ .

Step 4: The initial surface area is $A_1 = 4\pi R^2$ , and the total final surface area is $A_f = n (4\pi r^2) = 4\pi n r^2$ .

Step 5: Since $R^3 = n r^3$ , we have $r = \frac{R}{n^{1/3}}$ . Substituting this into the expression for $A_f$ , we get $A_f = 4\pi n (\frac{R}{n^{1/3}})^2 = 4\pi n \frac{R^2}{n^{2/3}} = 4\pi R^2 n^{1/3} = A_1 n^{1/3}$ .

Step 6: Since $n > 1$ , $n^{1/3} > 1$ , which means $A_f > A_1$ . Therefore, the total surface area increases.

Step 7: Since surface energy is proportional to surface area, the surface energy also increases.

Step 8: Final Answer: Increases

Why other options are wrong:
- Option B: The surface energy decreases only if the surface area decreases, which is not the case here.
- Option C: The surface energy remains unchanged only if the surface area remains unchanged, which is not the case here.
- Option D: The surface energy always increases when a single drop is broken into smaller drops because the surface area increases.

Key Concepts:
- Surface Energy
- Surface Area
- Volume Conservation

Memory Aids:
- Breaking a larger drop into smaller ones always increases the total surface area.

Quick Tips:
- Remember that surface energy is directly proportional to surface area.

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Application +4 | -1

Blood is flowing at the rate of

200c{m^3}{s^{ - 1}}

in a capillary of cross sectional area

0.5{m^2}

. The velocity of flow, in

mm{s^{ - 1}}

, is

View answer and explanation

Correct option: D

0.4

Option D (0.4) is correct because velocity is calculated by dividing the flow rate by the cross-sectional area, with proper unit conversions.

Solution approach

Correct Answer + Core Reason: 0.4 $mm{s^{ - 1}}$ . The velocity is calculated by dividing the flow rate by the area.

Step-by-Step Solution:

Step 1: Identify given information: Flow rate $Q = 200 cm^3/s$ , Area $A = 0.5 m^2$ . We need to find the velocity $v$ in $mm/s$ .

Step 2: Formula: $Q = Av$ , so $v = Q/A$ .

Step 3: Convert units: $Q = 200 cm^3/s = 200 \times (10^{-2} m)^3/s = 200 \times 10^{-6} m^3/s = 2 \times 10^{-4} m^3/s$ .

Step 4: Substitute values: $v = (2 \times 10^{-4} m^3/s) / (0.5 m^2) = 4 \times 10^{-4} m/s$ .

Step 5: Convert to mm/s: $v = 4 \times 10^{-4} m/s \times (1000 mm/m) = 0.4 mm/s$ .

Step 6: Final Answer: $v = 0.4 mm/s$ .

Why other options are wrong:
- Option A (0.1): This is incorrect due to errors in unit conversion or calculation.
- Option B (0.2): This is incorrect due to errors in unit conversion or calculation.
- Option C (0.3): This is incorrect due to errors in unit conversion or calculation.
- Option D (0.4): This is the correct calculation as shown above.

Key Concepts:
- Fluid flow rate
- Area and velocity relationship
- Unit conversion

Memory Aids:
- Q = Av (Flow rate = Area x Velocity)

Quick Tips:
- Always convert to SI units before calculating.
- Pay attention to unit conversions (cm to m, etc.).

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