JEE Physics Mechanical Properties Of Fluids MCQs Set 10

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Chapter-wise Objective Questions: Mechanical Properties Of Fluids

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Multiple Choice Questions

Question 1: The Young's modulus of steel is twice that of brass. Two wires of the same length and the same area of cross-section, one of steel and one of brass, are suspended from the same roof. If the lower ends are to be at the same level, the weights added to the steel and brass wires must be in the ratio:
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 4 : 1
Show Answer & Explanation

Answer: (c) 2 : 1

Explanation:
1. Equal extension means F/Y is the same
2. F_steel/F_brass = Y_steel/Y_brass = 2 : 1

Teacher's Note:
1. Apply the standard formula and check the units.

Question 2: Water rises to a height h in a capillary tube. If the length of the capillary tube above the surface of the water is made less than h, then:
(a) water does not rise at all
(b) water rises up to the tip of the tube and then starts overflowing like a fountain
(c) water rises up to the top of the tube and stays there without overflowing
(d) water rises up to a point a little below the top and stays there
Show Answer & Explanation

Answer: (c) water rises up to the top of the tube and stays there without overflowing

Explanation:
1. The radius of curvature of the meniscus adjusts so that h r = constant
2. The water stays at the top without overflowing

Teacher's Note:
1. Apply the standard formula and check the units.

Question 3: Two objects A and B of equal density and radii rA = 1 mm and rB = 2 mm move in the same medium. The ratio vB/vA of their terminal velocities is:
(a) 1/4
(b) 1/2
(c) 4
(d) 2
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. Terminal velocity is proportional to r^2
2. The ratio is (2/1)^2 = 4

Teacher's Note:
1. Apply the standard formula and check the units.

Question 4: The total area of the wings of an aeroplane is 10 m2. The speeds of air above and below the wings are 140 m/s and 110 m/s. The force on the aeroplane by air is (density of air = 1.28 kg/m3):
(a) 48750 N
(b) 48000 N
(c) 95000 N
(d) 50000 N
Show Answer & Explanation

Answer: (b) 48000 N

Explanation:
1. Pressure difference = (1/2) rho (140^2 - 110^2) = 0.64 x 7500 = 4800 Pa
2. Force = 4800 x 10 = 48000 N

Teacher's Note:
1. Apply the standard formula and check the units.

Question 5:
Assertion (A): Bernoulli's theorem holds true for ideal fluids in laminar flow.
Reason (R): In laminar flow, viscous force is not exerted.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. Bernoulli's theorem needs steady, non-viscous, incompressible flow, so A is true
2. Laminar flow can still have viscous force, so R is false

Teacher's Note:
1. Decide whether each statement is true, then whether R explains A.

Question 6: Two non-mixing liquids of densities rho and n rho (n > 1) are put in a container, each to a height h. A solid cylinder of length L and density d floats with its axis vertical and length pL (p < 1) in the denser liquid. The density d is equal to:
(a) \( \{1+(n+1)p\}\rho \)
(b) \( \{2+(n+1)p\}\rho \)
(c) \( \{2+(n-1)p\}\rho \)
(d) \( \{1+(n-1)p\}\rho \)
Show Answer & Explanation

Answer: (d) \( \{1+(n-1)p\}\rho \)

Explanation:
1. Weight = buoyancy: d L = n rho p L + rho (1 - p) L
2. d = rho {1 + (n - 1) p}

Teacher's Note:
1. Apply the standard formula and check the units.

Question 7: A rectangular film of liquid is extended from (4 cm x 2 cm) to (5 cm x 4 cm). If the work done is \( 3\times10^{-4} \) J, the surface tension of the liquid is:
(a) \( 0.2 \) N/m
(b) 8.0 N/m
(c) 0.250 N/m
(d) 0.125 N/m
Show Answer & Explanation

Answer: (d) 0.125 N/m

Explanation:
1. Increase in area = 20 - 8 = 12 cm2 = 1.2e-3 m2
2. The film has two surfaces, so W = 2 T dA
3. T = 3e-4/(2 x 1.2e-3) = 0.125 N/m

Teacher's Note:
1. Apply the standard formula and check the units.

Question 8: Three liquids of densities rho1 > rho2 > rho3, having the same surface tension T, rise to the same height in three identical capillaries. The angles of contact obey:
(a) \( \pi/2\lt\theta_1\lt\theta_2\lt\theta_3\lt\pi \)
(b) \( \pi>\theta_1>\theta_2>\theta_3>\pi/2 \)
(c) \( \pi/2>\theta_1>\theta_2>\theta_3\ge0 \)
(d) \( 0\le\theta_1\lt\theta_2\lt\theta_3\lt\pi/2 \)
Show Answer & Explanation

Answer: (d) \( 0\le\theta_1\lt\theta_2\lt\theta_3\lt\pi/2 \)

Explanation:
1. h = 2T cos(theta)/(r rho g) is the same for all
2. cos(theta) is proportional to rho, so cos(theta1) > cos(theta2) > cos(theta3)
3. Hence the angles increase from 1 to 3, all acute

Teacher's Note:
1. Apply the standard formula and check the units.

Question 9: Two arteries of the same length have the ratio of their radii 1 : 3. The pressure difference across the ends of both is the same. The ratio of the rates of flow is:
(a) \( 1/9 \)
(b) \( 1/27 \)
(c) \( 1/81 \)
(d) \( 1/243 \)
Show Answer & Explanation

Answer: (c) \( 1/81 \)

Explanation:
1. Poiseuille's law: rate of flow is proportional to r^4
2. Ratio = (1/3)^4 = 1/81

Teacher's Note:
1. Apply the standard formula and check the units.

Question 10:
Assertion (A): Viscous force opposes relative motion between fluid layers.
Reason (R): Some kinetic energy is lost in the form of heat.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both are true
2. Viscous force opposes relative motion by definition, and heat is a consequence rather than the explanation

Teacher's Note:
1. Decide whether each statement is true, then whether R explains A.

Question 11: An object is taken 1.0 km deep in the sea. The density of sea water is \( 1.025\times10^3 \) kg/m3 and the bulk modulus of the object is \( 1.6\times10^6 \) kPa. The percentage change in density of the object is:
(a) 0.36%
(b) 0.64%
(c) 0.40%
(d) 0.60%
Show Answer & Explanation

Answer: (b) 0.64%

Explanation:
1. Pressure = 1025 x 10 x 1000 = 1.025e7 Pa
2. Fractional change in density = P/B = 1.025e7/1.6e9 = 0.64%

Teacher's Note:
1. Apply the standard formula and check the units.

Question 12: A capillary tube is dipped into water. Given T = 0.072 N/m, r = 5 mm and g = 10 m/s2, the height up to which water will rise is:
(a) 2.8 cm
(b) 2.9 mm
(c) 28 cm
(d) 2.8 m
Show Answer & Explanation

Answer: (b) 2.9 mm

Explanation:
1. h = 2T/(r rho g) = 0.144/(5e-3 x 1000 x 10) = 2.88 mm

Teacher's Note:
1. Apply the standard formula and check the units.

Question 13: The bulk modulus of a spherical object is B. If it is subjected to uniform pressure p, the fractional decrease in radius is:
(a) \( B/3p \)
(b) \( 3p/B \)
(c) \( p/3B \)
(d) \( p/B \)
Show Answer & Explanation

Answer: (c) \( p/3B \)

Explanation:
1. Delta V/V = p/B and V is proportional to r^3, so delta r/r = p/(3B)

Teacher's Note:
1. Apply the standard formula and check the units.

Question 14: The maximum weight of a needle that can float on water of surface tension 0.073 N/m, if the length of the needle is 4 cm, is (g = 10 m/s2):
(a) 0.58 kg
(b) 0.584 g
(c) 1.3 g
(d) 1.3 kg
Show Answer & Explanation

Answer: (b) 0.584 g

Explanation:
1. Maximum weight = 2 T L = 2 x 0.073 x 0.04 = 5.84e-3 N, which is a mass of 0.584 g

Teacher's Note:
1. Apply the standard formula and check the units.

Question 15: A small sphere of radius r falls from rest in a viscous liquid. Heat is produced due to the viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to:
(a) \( r^3 \)
(b) \( r^2 \)
(c) \( r^5 \)
(d) \( r^4 \)
Show Answer & Explanation

Answer: (c) \( r^5 \)

Explanation:
1. Power = F v = 6 pi eta r v^2
2. v is proportional to r^2, so power is proportional to r x r^4 = r^5

Teacher's Note:
1. Apply the standard formula and check the units.

Question 16: Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second 3A. If the length of the first wire is increased by delta l on applying a force F, the force needed to stretch the second wire by the same amount is:
(a) 9F
(b) 6F
(c) 4F
(d) F
Show Answer & Explanation

Answer: (a) 9F

Explanation:
1. The second wire is one-third as long
2. Force for the same extension = Y A delta l/l, so it is 3 x 3 = 9 times F

Teacher's Note:
1. Apply the standard formula and check the units.

Question 17: A needle of length 5 cm floats on the surface of water of surface tension 0.073 N/m. The maximum weight of the needle is:
(a) \( 73\times10^{-4} \) N
(b) \( 36.5\times10^{-4} \) N
(c) \( 146\times10^{-4} \) N
(d) \( 292\times10^{-4} \) N
Show Answer & Explanation

Answer: (a) \( 73\times10^{-4} \) N

Explanation:
1. W = 2 T L = 2 x 0.073 x 0.05 = 7.3e-3 N

Teacher's Note:
1. Apply the standard formula and check the units.

Question 18: Water emerges as a fountain from a vertical tube at the rate of 3000 litres/min up to a height of 10 m. The area of cross-section of the tube is (g = 10 m/s2):
(a) 35 cm2
(b) 40 cm2
(c) 80 cm2
(d) 17 cm2
Show Answer & Explanation

Answer: (a) 35 cm2

Explanation:
1. Speed = sqrt(2 x 10 x 10) = 14.1 m/s
2. Flow = 0.05 m3/s, so A = 0.05/14.1 = 3.5e-3 m2 = 35 cm2

Teacher's Note:
1. Apply the standard formula and check the units.

Question 19: Water is filled in a cylindrical tube of diameter 1 mm. If the surface tension of water is 0.055 N/m, the pressure difference across the meniscus (in N/m2) is:
(a) 220
(b) 110
(c) 45
(d) 440
Show Answer & Explanation

Answer: (a) 220

Explanation:
1. Pressure difference = 2T/r = 2 x 0.055/(0.5e-3) = 220 N/m2

Teacher's Note:
1. Apply the standard formula and check the units.

Question 20: A tank is filled with water up to a height of 1 m. A hole is made 20 cm from the top. The horizontal distance from the base of the tank where the water strikes the ground is:
(a) 60 cm
(b) 80 cm
(c) 100 cm
(d) 40 cm
Show Answer & Explanation

Answer: (b) 80 cm

Explanation:
1. Range R = 2 sqrt(h (H - h)) with h = 0.2 m, H = 1 m
2. R = 2 sqrt(0.2 x 0.8) = 0.8 m

Teacher's Note:
1. Apply the standard formula and check the units.

Question 21: A ring of radius 2.5 cm and mass 10 g floats on a liquid surface of surface tension 0.06 N/m. The force required to just lift the ring from the liquid surface is (g = 10 m/s2):
(a) 0.06 N
(b) 0.12 N
(c) 0.24 N
(d) 0.03 N
Show Answer & Explanation

Answer: (b) 0.12 N

Explanation:
1. Force = weight + surface tension force on both inner and outer circumference
2. = 0.1 + 0.06 x 2 pi(0.025) x 2 = 0.1 + 0.019 = 0.12 N

Teacher's Note:
1. Apply the standard formula and check the units.

Question 22: A force is applied to a wire of initial length 1 m and the extension is recorded. In the straight-line graph, extension changes from 2 mm to 4 mm as the stress changes from 4000 kN/m2 to 8000 kN/m2. The Young's modulus of the wire is:
(a) \( 2\times10^9 \) N/m2
(b) \( 1\times10^9 \) N/m2
(c) \( 2\times10^{10} \) N/m2
(d) \( 1\times10^{10} \) N/m2
Show Answer & Explanation

Answer: (a) \( 2\times10^9 \) N/m2

Explanation:
1. Change in stress = 4e6 N/m2
2. Change in strain = 2e-3 m/1 m = 2e-3
3. Y = 4e6/2e-3 = 2e9 N/m2

Teacher's Note:
1. Apply the standard formula and check the units.

Question 23: To what height h should a cylindrical vessel of diameter d be filled with a liquid so that the force on its vertical surface equals the force on its bottom?
(a) h = d
(b) h = 2d
(c) h = 3d
(d) h = d/2
Show Answer & Explanation

Answer: (d) h = d/2

Explanation:
1. Mean pressure on the side is rho g h/2 acting over area pi d h
2. Force on the side = rho g h^2 pi d/2
3. Force on the bottom = rho g h pi d^2/4
4. Equating gives h = d/2

Teacher's Note:
1. Apply the standard formula and check the units.

Question 24: A solid uniform ball of volume V and density rho floats at the interface of two immiscible liquids of densities rho1 and rho2, with rho1 < rho < rho2. The fraction of the volume of the ball in the lower liquid is:
(a) \( \frac{\rho-\rho_2}{\rho_1-\rho_2} \)
(b) \( \frac{\rho_1}{\rho_1-\rho_2} \)
(c) \( \frac{\rho_1-\rho}{\rho_1-\rho_2} \)
(d) \( \frac{\rho_1-\rho_2}{\rho_2} \)
Show Answer & Explanation

Answer: (c) \( \frac{\rho_1-\rho}{\rho_1-\rho_2} \)

Explanation:
1. rho V = rho1 V1 + rho2 V2 with V1 + V2 = V
2. V2/V = (rho - rho1)/(rho2 - rho1) = (rho1 - rho)/(rho1 - rho2)

Teacher's Note:
1. Apply the standard formula and check the units.

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