JEE Physics Mechanical Properties Of Fluids MCQs Set 04

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Multiple Choice Questions

Question 1: The lower surface of a cube is fixed. A force is applied on its upper surface at an angle of 30 degrees from the surface. The change will be in its:
(a) shape
(b) size
(c) volume
(d) both shape and size
Show Answer & Explanation

Answer: (d) both shape and size

Explanation:
1. The component along the surface produces shear, which changes shape
2. The component normal to the surface changes the size

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

Question 2: One end of a uniform wire of length L and weight W is attached rigidly to the roof and a weight W1 is hung from its lower end. If s is the cross-sectional area, the stress in the wire at a height L/4 from its lower end is:
(a) \( W_1/s \)
(b) \( (W_1+W/4)/s \)
(c) \( (W_1+3W/4)/s \)
(d) \( (W_1+W)/s \)
Show Answer & Explanation

Answer: (b) \( (W_1+W/4)/s \)

Explanation:
1. At height L/4 from the bottom, the part below it has weight W/4 plus the hanging weight W1
2. Stress = (W1 + W/4)/s

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

Question 3: For steel the breaking stress is \( 6\times10^6 \) N/m2 and the density is \( 8\times10^3 \) kg/m3. The maximum length of steel wire which can be suspended without breaking under its own weight is (g = 10 m/s2):
(a) 140 m
(b) 120 m
(c) 75 m
(d) 200 m
Show Answer & Explanation

Answer: (c) 75 m

Explanation:
1. Stress at the top = L rho g
2. L = stress/(rho g) = 6e6/(8e3 x 10) = 75 m

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

Question 4: Two wires A and B have the same dimensions but different materials. Their load-extension graphs are straight lines through the origin, with the line of A steeper than that of B. If YA and YB are their Young's moduli, then:
(a) \( Y_A>Y_B \)
(b) \( Y_A\lt Y_B \)
(c) \( Y_A=Y_B \)
(d) \( Y_B=2Y_A \)
Show Answer & Explanation

Answer: (a) \( Y_A>Y_B \)

Explanation:
1. Slope of the load-extension graph is YA/L
2. For equal dimensions a steeper line means a larger Young's modulus

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

Question 5: A fixed volume of iron is drawn into a wire of length l. The extension produced in this wire by a constant force F is proportional to:
(a) \( 1/l^2 \)
(b) \( 1/l \)
(c) \( l^2 \)
(d) \( l \)
Show Answer & Explanation

Answer: (c) \( l^2 \)

Explanation:
1. Volume V = A l is fixed, so A = V/l
2. Extension = F l/(A Y) = F l^2/(V Y), which is proportional to l^2

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

Question 6: A wire elongates by l mm when a load W is hung from it. If the wire goes over a pulley and two weights W each are hung at its two ends, the elongation of the wire will be (in mm):
(a) l
(b) 2l
(c) zero
(d) l/2
Show Answer & Explanation

Answer: (a) l

Explanation:
1. The tension in the wire is W in both cases
2. So the elongation is the same, l

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

Question 7: The Young's modulus of a rubber string 8 cm long and of density 1.5 kg/m3 is \( 5\times10^8 \) N/m2. It is suspended from the ceiling in a room. The increase in length due to its own weight will be:
(a) \( 9.6\times10^{-5} \) m
(b) \( 9.6\times10^{-11} \) m
(c) \( 9.6\times10^{-3} \) m
(d) 9.6 m
Show Answer & Explanation

Answer: (b) \( 9.6\times10^{-11} \) m

Explanation:
1. Elongation due to own weight = rho g L^2/(2Y)
2. = 1.5 x 10 x (0.08)^2/(2 x 5e8) = 9.6e-11 m

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

Question 8: A ball falling in a lake of depth 200 m shows a 0.1% decrease in its volume at the bottom. The bulk modulus of the material of the ball is:
(a) \( 19.6\times10^8 \) N/m2
(b) \( 19.6\times10^{-10} \) N/m2
(c) \( 19.6\times10^{10} \) N/m2
(d) \( 19.6\times10^{-8} \) N/m2
Show Answer & Explanation

Answer: (a) \( 19.6\times10^8 \) N/m2

Explanation:
1. Pressure rise = rho g h = 1000 x 9.8 x 200 = 1.96e6 Pa
2. B = 1.96e6/0.001 = 1.96e9 N/m2

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

Question 9:
Assertion (A): Work is required to stretch a wire, and this work is stored in the wire as elastic potential energy.
Reason (R): Work is required to be done against the intermolecular forces of attraction.
(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: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Stretching increases the separation of the atoms against their attractive forces
2. The work done is stored as elastic energy, so R explains A

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

Question 10: The pressure of a medium is changed from \( 1.01\times10^5 \) Pa to \( 1.165\times10^5 \) Pa and the change in volume is 10 percent, keeping the temperature constant. The bulk modulus of the medium is:
(a) \( 204.8\times10^5 \) Pa
(b) \( 102.4\times10^5 \) Pa
(c) \( 51.2\times10^5 \) Pa
(d) \( 1.55\times10^5 \) Pa
Show Answer & Explanation

Answer: (d) \( 1.55\times10^5 \) Pa

Explanation:
1. Change in pressure = 0.155e5 Pa
2. B = 0.155e5/0.1 = 1.55e5 Pa

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

Question 11: A weight is suspended from a long metal wire. If the wire suddenly breaks, its temperature:
(a) rises
(b) falls
(c) remains unchanged
(d) attains a value of 0 K
Show Answer & Explanation

Answer: (a) rises

Explanation:
1. The elastic energy stored in the wire is released as heat, so the temperature rises

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

Question 12: A mass of 0.5 kg is suspended from a wire and the length of the wire increases by 3 mm. The work done is (g = 9.8 m/s2):
(a) \( 4.5\times10^{-3} \) J
(b) \( 7.3\times10^{-3} \) J
(c) \( 9.3\times10^{-2} \) J
(d) \( 2.5\times10^{-2} \) J
Show Answer & Explanation

Answer: (b) \( 7.3\times10^{-3} \) J

Explanation:
1. Work done = (1/2) F x extension
2. = 0.5 x (0.5 x 9.8) x 3e-3 = 7.35e-3 J

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

Question 13: If the strain in a wire is not more than 1/1000 and Y = \( 2\times10^{11} \) N/m2, with a wire diameter of 1 mm, the maximum weight that can be hung from it is:
(a) 110 N
(b) 125 N
(c) 157 N
(d) 168 N
Show Answer & Explanation

Answer: (c) 157 N

Explanation:
1. Stress = Y x strain = 2e8 N/m2
2. Area = pi (0.5e-3)^2 = 7.85e-7 m2
3. F = 157 N

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

Question 14:
Assertion (A): Bridges are declared unsafe after a long use.
Reason (R): The elastic strength of bridges is lost with time.
(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: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Repeated stress causes elastic fatigue
2. Unsafe declaration follows from the loss of strength

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

Question 15: When a tension F is applied to a uniform wire of length l and radius r, the elongation is e. When a tension 2F is applied to another wire of the same material of length 2l and radius 2r, the elongation is:
(a) 0.5 e
(b) 1.0 e
(c) 1.5 e
(d) 2.0 e
Show Answer & Explanation

Answer: (b) 1.0 e

Explanation:
1. Elongation = F l/(pi r^2 Y)
2. The new value = (2F)(2l)/(pi 4r^2 Y) = F l/(pi r^2 Y) = e

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

Question 16: How much force is required to produce an increase of 0.2% in the length of a brass wire of diameter 0.6 mm? (Y = \( 0.9\times10^{11} \) N/m2)
(a) nearly 17 N
(b) nearly 34 N
(c) nearly 51 N
(d) nearly 68 N
Show Answer & Explanation

Answer: (c) nearly 51 N

Explanation:
1. F = Y A (strain) = 0.9e11 x pi(0.3e-3)^2 x 0.002
2. = 0.9e11 x 2.83e-7 x 0.002 = 51 N

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

Question 17: If the interatomic spacing in a steel wire is \( 2.8\times10^{-10} \) m and Y of steel = \( 2\times10^{11} \) N/m2, the force constant in N/m is:
(a) 5.6
(b) 56
(c) 0.56
(d) 560
Show Answer & Explanation

Answer: (b) 56

Explanation:
1. Y = k/r0, so k = Y r0
2. = 2e11 x 2.8e-10 = 56 N/m

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

Question 18: Load versus elongation graphs for four wires of the same material and the same length are drawn as straight lines OA, OB, OC and OD, where OA has the smallest slope and OD the largest. The thinnest wire is represented by:
(a) OA
(b) OB
(c) OC
(d) OD
Show Answer & Explanation

Answer: (a) OA

Explanation:
1. Slope = Y A/L, and Y and L are the same, so slope is proportional to area
2. The thinnest wire has the smallest slope, which is OA

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

Question 19:
Assertion (A): Young's modulus for a perfectly plastic body is zero.
Reason (R): For a perfectly plastic body the restoring force is zero.
(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: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Y = stress/strain
2. With no restoring force, the stress is zero for any strain, so Y = 0

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

Question 20: The mean distance between the atoms of iron is \( 3\times10^{-10} \) m and the interatomic force constant for iron is 7 N/m. The Young's modulus of iron is:
(a) \( 2.33\times10^5 \) N/m2
(b) \( 23.3\times10^{10} \) N/m2
(c) \( 233\times10^{10} \) N/m2
(d) \( 2.33\times10^{10} \) N/m2
Show Answer & Explanation

Answer: (d) \( 2.33\times10^{10} \) N/m2

Explanation:
1. Y = k/r0 = 7/(3e-10) = 2.33e10 N/m2

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

Question 21: The cross-section area of a steel wire (Y = \( 2.0\times10^{11} \) N/m2) is 0.1 cm2. The force required to make its length double is:
(a) \( 2\times10^{12} \) N
(b) \( 2\times10^{11} \) N
(c) \( 2\times10^{10} \) N
(d) \( 2\times10^{6} \) N
Show Answer & Explanation

Answer: (d) \( 2\times10^{6} \) N

Explanation:
1. Doubling the length means strain = 1
2. F = Y A strain = 2e11 x 0.1e-4 = 2e6 N

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

Question 22: For a given material, the Young's modulus is 2.4 times its modulus of rigidity. Its Poisson's ratio is:
(a) 2.4
(b) 1.2
(c) 0.4
(d) 0.2
Show Answer & Explanation

Answer: (d) 0.2

Explanation:
1. Y = 2 eta (1 + sigma)
2. 2.4 = 2(1 + sigma), so sigma = 0.2

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

Question 23: The diameter of a brass rod is 4 mm and Young's modulus of brass is \( 9\times10^{10} \) N/m2. The force required to stretch it by 0.1% of its length is:
(a) \( 360\pi \) N
(b) 36 N
(c) \( 144\pi\times10^3 \) N
(d) \( 36\pi\times10^5 \) N
Show Answer & Explanation

Answer: (a) \( 360\pi \) N

Explanation:
1. F = Y A strain = 9e10 x pi (2e-3)^2 x 1e-3
2. = 9e10 x 4e-6 pi x 1e-3 = 360 pi N

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

Question 24:
Assertion (A): Identical springs of steel and copper are equally stretched. More work will be done on the steel spring.
Reason (R): Steel is more elastic than copper.
(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: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Steel has the larger Young's modulus, and hence the larger force constant
2. Work = (1/2) k x^2 is greater for steel

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

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