JEE Physics Mechanical Properties Of Solids MCQs Set 14

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Chapter-wise Objective Questions: Mechanical Properties Of Solids

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Multiple Choice Questions

Question 1: Wires A and B are made from the same material. A has twice the diameter and three times the length of B. If the elastic limits are not reached, when each is stretched by the same tension, the ratio of energy stored in A to that in B is
(a) 2 : 3
(b) 3 : 4
(c) 3 : 2
(d) 6 : 1
Show Answer & Explanation

Answer: (b) 3 : 4

Explanation:
1. Energy stored \( U = \frac{F^2L}{2AY} \propto \frac{L}{d^2} \) for the same F and Y.
2. \( \frac{U_A}{U_B} = \frac{3}{2^2} = \frac34 \).

Teacher's Note:
1. Express U in terms of F, L, A and Y.
2. A ∝ d².

Question 2: The breaking stress for a wire of radius r of a given material is F N/m². The breaking stress for a wire of the same material of radius 2r is
(a) F/4
(b) F/2
(c) 2F
(d) F
Show Answer & Explanation

Answer: (d) F

Explanation:
1. Breaking stress is a property of the material and does not depend on the dimensions of the wire.
2. So it remains F.

Teacher's Note:
1. Breaking force changes with area, breaking stress does not.
2. Stress is force per unit area.

Question 3: The bulk modulus of rubber is \( 9.1\times10^8 \) N/m². To what depth should a rubber ball be taken in a lake so that its volume is decreased by 0.1%? (g = 10 m/s², density of water 1000 kg/m³)
(a) 91 m
(b) 25 m
(c) 200 m
(d) 500 m
Show Answer & Explanation

Answer: (a) 91 m

Explanation:
1. \( \Delta P = B\frac{\Delta V}{V} = 9.1\times10^8\times10^{-3} = 9.1\times10^5 \) Pa.
2. \( h = \frac{\Delta P}{\rho g} = \frac{9.1\times10^5}{10^4} = 91 \) m.

Teacher's Note:
1. Excess pressure at depth h is ρgh.
2. 0.1% = 10⁻³.

Question 4: A solid sphere of radius R made of a material of bulk modulus B is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the surface of the liquid. When a mass M is placed on the piston to compress the liquid, the fractional change in the radius of the sphere is
(a) \( \frac{Mg}{BA} \)
(b) \( \frac{3Mg}{4BA} \)
(c) \( \frac{Mg}{3BA} \)
(d) \( \frac{Mg}{4BAR} \)
Show Answer & Explanation

Answer: (c) \( \frac{Mg}{3BA} \)

Explanation:
1. Pressure increase \( \Delta P = \frac{Mg}{A} \) and \( \frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{Mg}{AB} \).
2. For a sphere \( \frac{\Delta V}{V} = 3\frac{\Delta R}{R} \), so \( \frac{\Delta R}{R} = \frac{Mg}{3BA} \).

Teacher's Note:
1. V ∝ R³.
2. Pressure is transmitted equally.

Question 5: If the pressure of a fluid is increased from \( 1.01\times10^5 \) Pa to \( 1.165\times10^5 \) Pa and the volume decreases by 10% at constant temperature, the bulk modulus of the fluid is
(a) \( 15.5\times10^5 \) Pa
(b) \( 1.4\times10^5 \) Pa
(c) \( 1.55\times10^5 \) Pa
(d) \( 0.0155\times10^5 \) Pa
Show Answer & Explanation

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

Explanation:
1. \( \Delta P = 0.155\times10^5 \) Pa and \( \frac{\Delta V}{V} = 0.10 \).
2. \( B = \frac{\Delta P}{\Delta V/V} = 1.55\times10^5 \) Pa.

Teacher's Note:
1. Use the change in pressure, not the final pressure.
2. 10% = 0.10.

Question 6: The rubber cord of a catapult has cross-section area 1 mm² and total unstretched length 10 cm. It is stretched to 12 cm and then released to project a particle of mass 5 g. The velocity of projection is (Y for rubber = \( 5\times10^8 \) N/m²)
(a) 20 m/s
(b) 5 m/s
(c) 10 m/s
(d) 40 m/s
Show Answer & Explanation

Answer: (a) 20 m/s

Explanation:
1. Strain = 2/10 = 0.2; volume = \( 10^{-6}\times0.1 = 10^{-7} \) m³.
2. Stored energy \( = \tfrac12Y(\text{strain})^2V = \tfrac12\times5\times10^8\times0.04\times10^{-7} = 1 \) J.
3. \( \tfrac12(0.005)v^2 = 1 \Rightarrow v = 20 \) m/s.

Teacher's Note:
1. Elastic energy per unit volume is ½ × stress × strain.
2. All of it becomes kinetic energy.

Question 7: A wire of natural length L hangs from a fixed support. With mass M₁ hung from it, its length is l₁, and with both M₁ and M₂ hanging from it, its length is l₂. The natural length of the wire is
(a) \( \frac{M_2l_1-M_1l_2}{M_1+M_2} \)
(b) \( \frac{l_1+l_2}{2} \)
(c) \( \sqrt{l_1l_2} \)
(d) \( \frac{M_1}{M_2}(l_1-l_2)+l_1 \)
Show Answer & Explanation

Answer: (d) \( \frac{M_1}{M_2}(l_1-l_2)+l_1 \)

Explanation:
1. Extension is proportional to the load: \( l_1-L = kM_1 \) and \( l_2-L = k(M_1+M_2) \).
2. Subtracting, \( l_2-l_1 = kM_2 \), so \( l_1-L = \frac{M_1}{M_2}(l_2-l_1) \) and \( L = l_1+\frac{M_1}{M_2}(l_1-l_2) \).

Teacher's Note:
1. Hooke's law: extension ∝ load.
2. Eliminate k between the two equations.

Question 8: The Young's modulus of a wire of length L and radius r is Y N/m². If the length is reduced to L/2 and the radius to r/2, its Young's modulus will be
(a) Y/2
(b) Y
(c) 2Y
(d) 4Y
Show Answer & Explanation

Answer: (b) Y

Explanation:
1. Young's modulus is a property of the material only.
2. It does not depend on length or radius.

Teacher's Note:
1. Dimensions change the force needed, not Y.
2. Y = stress/strain.

Question 9: When a load of 5 kg is hung on a wire, an extension of 30 cm takes place. The work done in stretching the wire is (g = 10 m/s²)
(a) 15 J
(b) 7.5 J
(c) 0.5 J
(d) 1.5 J
Show Answer & Explanation

Answer: (b) 7.5 J

Explanation:
1. Force = 50 N. Work = \( \tfrac12Fx = \tfrac12\times50\times0.30 = 7.5 \) J.

Teacher's Note:
1. The force rises from 0 to F, so use the average.
2. Convert cm to m.

Question 10: The Young's moduli of brass and steel are \( 1\times10^{10} \) N/m² and \( 2\times10^{10} \) N/m² respectively. A brass wire and a steel wire of the same length are extended by 1 mm under the same force. The radii \( R_B \) and \( R_S \) satisfy
(a) \( R_S = \sqrt2R_B \)
(b) \( R_S = 4R_B \)
(c) \( R_S = \frac{R_B}{\sqrt2} \)
(d) \( R_S = \frac{R_B}{4} \)
Show Answer & Explanation

Answer: (c) \( R_S = \frac{R_B}{\sqrt2} \)

Explanation:
1. \( \Delta L = \frac{FL}{\pi R^2Y} \) is the same for both, so \( R^2Y \) is the same.
2. \( R_B^2(1) = R_S^2(2) \Rightarrow R_S = \frac{R_B}{\sqrt2} \).

Teacher's Note:
1. Equal F, L and ΔL imply equal R²Y.
2. A stiffer wire can be thinner.

Question 11: If \( \rho \) is the density of the material of a wire and \( \sigma \) is the breaking stress, the greatest length of the wire that can hang freely without breaking is
(a) \( \frac{\sigma}{\rho g} \)
(b) \( \frac{2\sigma}{\rho g} \)
(c) \( \frac{\rho}{\sigma g} \)
(d) \( \frac{\rho g}{2\sigma} \)
Show Answer & Explanation

Answer: (a) \( \frac{\sigma}{\rho g} \)

Explanation:
1. Stress at the top \( = \frac{\rho ALg}{A} = \rho Lg \).
2. Setting this equal to σ: \( L = \frac{\sigma}{\rho g} \).

Teacher's Note:
1. Top section bears the whole weight.
2. Area cancels.

Question 12: The Young's modulus of the material of a wire is \( 2\times10^{10} \) N/m². If the longitudinal strain is 1%, the energy stored in the wire per unit volume is
(a) \( 0.5\times10^6 \) J/m³
(b) \( 2\times10^6 \) J/m³
(c) \( 4\times10^6 \) J/m³
(d) \( 10^6 \) J/m³
Show Answer & Explanation

Answer: (d) \( 10^6 \) J/m³

Explanation:
1. \( u = \tfrac12Y\varepsilon^2 = \tfrac12\times2\times10^{10}\times(10^{-2})^2 = 10^6 \) J/m³.

Teacher's Note:
1. Energy density = ½ stress × strain.
2. 1% = 0.01.

Question 13: A cube of side 40 cm has its upper face displaced by 0.1 mm by a tangential force of 8 kN. The shearing modulus of the cube is
(a) \( 2\times10^9 \) N/m²
(b) \( 2\times10^8 \) N/m²
(c) \( 4\times10^9 \) N/m²
(d) \( 4\times10^8 \) N/m²
Show Answer & Explanation

Answer: (b) \( 2\times10^8 \) N/m²

Explanation:
1. Shear stress \( = \frac{8000}{0.4^2} = 5\times10^4 \) N/m²; shear strain \( = \frac{0.1\times10^{-3}}{0.4} = 2.5\times10^{-4} \).
2. \( G = \frac{5\times10^4}{2.5\times10^{-4}} = 2\times10^8 \) N/m².

Teacher's Note:
1. Shear strain = displacement / height.
2. Area of a face = side².

Question 14: The bulk modulus for an incompressible liquid is
(a) zero
(b) unity
(c) between 0 and 1
(d) infinity
Show Answer & Explanation

Answer: (d) infinity

Explanation:
1. \( B = -\frac{\Delta P}{\Delta V/V} \). An incompressible liquid has \( \Delta V = 0 \) for any pressure change, so B is infinite.

Teacher's Note:
1. Compressibility is 1/B.
2. No volume change means infinite B.

Question 15: A steel wire of cross-sectional area \( 3\times10^{-6} \) m² can withstand a maximum strain of \( 10^{-3} \). The Young's modulus of steel is \( 2\times10^{11} \) N/m². The maximum mass the wire can hold is (g = 10 m/s²)
(a) 60 kg
(b) 40 kg
(c) 80 kg
(d) 100 kg
Show Answer & Explanation

Answer: (a) 60 kg

Explanation:
1. \( F = YA\varepsilon = 2\times10^{11}\times3\times10^{-6}\times10^{-3} = 600 \) N.
2. \( m = \frac{600}{10} = 60 \) kg.

Teacher's Note:
1. F = stress × area.
2. Stress = Y × strain.

Question 16: The Young's modulus of the material of a wire is equal to the
(a) stress required to increase its length four times
(b) strain produced in it
(c) stress required to produce unit strain
(d) half the strain produced in it
Show Answer & Explanation

Answer: (c) stress required to produce unit strain

Explanation:
1. Y = stress / strain, which is the stress when the strain equals 1.
2. Stress to increase the length four times would correspond to strain 3, giving 3Y.

Teacher's Note:
1. Unit strain means ΔL = L.
2. Y is measured in N/m².

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