JEE Physics Mechanical Properties Of Solids MCQs Set 15

Multiple Choice Questions (MCQs) for JEE Physics: Mechanical Properties Of Solids

Review structured MCQ sets for JEE Physics Mechanical Properties Of Solids. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Mechanical Properties Of Solids MCQs for JEE Physics

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Multiple Choice Questions

Question 1: Two rods of different materials having coefficients of linear expansion α₁, α₂ and Young's moduli Y₁, Y₂ are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rods. If α₁ : α₂ = 2 : 3, the thermal stresses developed in the two rods are equal provided Y₁ : Y₂ is equal to
(a) 2 : 3
(b) 1 : 1
(c) 3 : 2
(d) 4 : 9
Show Answer & Explanation

Answer: (c) 3 : 2

Explanation:
1. Thermal stress \( =Y\alpha\Delta T \), so equal stress needs \( Y_1\alpha_1=Y_2\alpha_2 \).
2. \( \frac{Y_1}{Y_2}=\frac{\alpha_2}{\alpha_1}=\frac32 \).

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Question 2: Two rods A and B of the same material and length have radii r₁ and r₂ respectively. When they are rigidly fixed at one end and twisted by the same couple applied at the other end, the ratio (angle of twist at the end of A)/(angle of twist at the end of B) is
(a) \( r_1^2/r_2^2 \)
(b) \( r_2^3/r_1^3 \)
(c) \( r_2^4/r_1^4 \)
(d) \( r_1^4/r_2^4 \)
Show Answer & Explanation

Answer: (c) \( r_2^4/r_1^4 \)

Explanation:
1. Angle of twist \( \theta=\frac{2\tau l}{\pi\eta r^4}\propto\frac1{r^4} \).
2. \( \frac{\theta_A}{\theta_B}=\frac{r_2^4}{r_1^4} \).

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Question 3: The ratio (stress/strain) remains constant for a small deformation of a material. When the deformation is made larger, this ratio
(a) increases
(b) decreases
(c) remains constant
(d) becomes zero
Show Answer & Explanation

Answer: (b) decreases

Explanation:
1. Hooke's law holds only up to the proportional limit.
2. Beyond it, the stress–strain curve bends and stress grows more slowly than strain, so the ratio decreases.

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

Answer: (c) \( \frac{W_1+3W/4}{s} \)

Explanation:
1. The part of the wire below that point has length 3L/4 and weight 3W/4.
2. Tension there = W₁ + 3W/4, so the stress is (W₁ + 3W/4)/s.

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Question 5: A wire of length L and cross-sectional area A is made of a material of Young's modulus Y. If the wire is stretched by an amount x, the work done is
(a) \( \frac{YAx^2}{2L} \)
(b) \( \frac{YAx^2}{L} \)
(c) \( \frac{YAx}{2L} \)
(d) \( YAx^2L \)
Show Answer & Explanation

Answer: (a) \( \frac{YAx^2}{2L} \)

Explanation:
1. Force \( F=\frac{YA}{L}x \) increases linearly with extension.
2. Work \( =\int_0^xFdx=\frac{YAx^2}{2L} \).

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Question 6: The pressure on a fluid is changed from 1.01 × 10⁵ Pa to 1.165 × 10⁵ Pa, and the volume changes by 10%. The bulk modulus of the fluid is
(a) 1.55 × 10⁵ Pa
(b) 0.015 × 10⁵ Pa
(c) 1.015 × 10⁵ Pa
(d) 1.55 × 10⁶ Pa
Show Answer & Explanation

Answer: (a) 1.55 × 10⁵ Pa

Explanation:
1. \( K=\frac{\Delta P}{\Delta V/V}=\frac{0.155\times10^5}{0.10} \).
2. K = 1.55 × 10⁵ Pa.

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Question 7: If s is the stress and Y is Young's modulus of the material of a wire, the energy stored in the wire per unit volume is
(a) \( \frac{s^2}{2Y} \)
(b) \( 2s^2Y \)
(c) \( \frac{2Y}{s^2} \)
(d) \( \frac12Ys^2 \)
Show Answer & Explanation

Answer: (a) \( \frac{s^2}{2Y} \)

Explanation:
1. Energy density \( =\frac12\times\text{stress}\times\text{strain}=\frac12s\cdot\frac sY \).
2. \( =\frac{s^2}{2Y} \).

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Question 8: An elevator cable is to have a maximum stress of 7 × 10⁷ N/m² to allow for appropriate safety factors. Its maximum upward acceleration is 1.5 m/s². If the cable has to support the total weight of 2000 kg of a loaded elevator, the area of cross-section of the cable should be (take g = 10 m/s²)
(a) 3.28 cm²
(b) 2.38 cm²
(c) 0.328 cm²
(d) 8.23 cm²
Show Answer & Explanation

Answer: (a) 3.28 cm²

Explanation:
1. Maximum tension \( T=m(g+a)=2000\times11.5=23000 \) N.
2. \( A=\frac{T}{\text{stress}}=\frac{23000}{7\times10^7}=3.29\times10^{-4} \) m² ≈ 3.28 cm².

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Question 9: The temperature of a wire of length 1 m and area of cross-section 1 cm² is increased from 0°C to 100°C. If the rod is not allowed to increase in length, the force required will be (α = 10⁻⁵/°C and Y = 10¹¹ N/m²)
(a) 10³ N
(b) 10⁴ N
(c) 10⁵ N
(d) 10⁹ N
Show Answer & Explanation

Answer: (b) 10⁴ N

Explanation:
1. \( F=YA\alpha\Delta T \).
2. \( =10^{11}\times10^{-4}\times10^{-5}\times100=10^4 \) N.

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Question 10: A rod is fixed between two points at 20°C. The coefficient of linear expansion of the material of the rod is 1.1 × 10⁻⁵/°C and Young's modulus is 1.2 × 10¹¹ N/m². The stress developed in the rod when the temperature falls to 10°C is
(a) 1.32 × 10⁷ N/m²
(b) 1.10 × 10¹⁵ N/m²
(c) 1.32 × 10⁸ N/m²
(d) 1.10 × 10⁶ N/m²
Show Answer & Explanation

Answer: (a) 1.32 × 10⁷ N/m²

Explanation:
1. Stress \( =Y\alpha\Delta T \).
2. \( =1.2\times10^{11}\times1.1\times10^{-5}\times10=1.32\times10^7 \) N/m².

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Question 11: When a force is applied on a wire of uniform cross-section 3 × 10⁻⁶ m² and length 4 m, the increase in length is 1 mm. The energy stored in it will be (Y = 2 × 10¹¹ N/m²)
(a) 6250 J
(b) 0.177 J
(c) 0.075 J
(d) 0.150 J
Show Answer & Explanation

Answer: (c) 0.075 J

Explanation:
1. Energy \( =\frac12Y(\text{strain})^2\times\text{volume} \), strain \( =\frac{10^{-3}}4=2.5\times10^{-4} \).
2. \( =\frac12\times2\times10^{11}\times6.25\times10^{-8}\times(3\times10^{-6}\times4)=0.075 \) J.

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Question 12: The value of Poisson's ratio lies between
(a) \( -1\ \text{to}\ \frac12 \)
(b) \( -\frac34\ \text{to}\ -\frac12 \)
(c) \( -\frac12\ \text{to}\ 1 \)
(d) 1 to 2
Show Answer & Explanation

Answer: (a) \( -1\ \text{to}\ \frac12 \)

Explanation:
1. Positive Young's and bulk moduli require −1 < σ < ½ theoretically.
2. Most real materials have values between 0 and 0.5.

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Question 13: To break a wire, a force of 10⁶ N/m² is required. If the density of the material is 3 × 10³ kg/m³, the length of the wire which will break by its own weight will be (g = 9.8 m/s²)
(a) 34 m
(b) 30 m
(c) 300 m
(d) 3 m
Show Answer & Explanation

Answer: (a) 34 m

Explanation:
1. Stress at the top \( =\rho Lg \).
2. \( L=\frac{10^6}{3\times10^3\times9.8}=34 \) m.

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Question 14: A solid sphere of radius R made of a material of bulk modulus K 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 δR/R is
(a) \( \frac{mg}{AK} \)
(b) \( \frac{mg}{3AK} \)
(c) \( \frac{mg}{A} \)
(d) \( \frac{3mg}{AK} \)
Show Answer & Explanation

Answer: (b) \( \frac{mg}{3AK} \)

Explanation:
1. Excess pressure \( =\frac{mg}{A} \), so \( \frac{\Delta V}V=\frac{mg}{AK} \).
2. \( V\propto R^3\Rightarrow\frac{\delta R}R=\frac13\frac{\Delta V}V=\frac{mg}{3AK} \).

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Question 15: The table gives the areas of the four surfaces and the magnitude of a force applied perpendicular to the surface and uniformly across it: A: area 0.25A, force F; B: area 0.50A, force 2F; C: area 2A, force 4F; D: area 3A, force 5F. Which surface(s) has the greatest stress?
(a) A
(b) A and B
(c) B and C
(d) D
Show Answer & Explanation

Answer: (b) A and B

Explanation:
1. Stress = force/area: A gives 4F/A, B gives 4F/A, C gives 2F/A, D gives 1.67F/A.
2. The greatest stress is shared by A and B.

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Question 16: A rod of length l and radius r is joined to a rod of length l/2 and radius r/2 of the same material. The free end of the small rod is fixed to a rigid base and the free end of the larger rod is given a twist of θ°. The twist angle at the joint will be
(a) \( \theta/4 \)
(b) \( \theta/2 \)
(c) \( 5\theta/6 \)
(d) \( 8\theta/9 \)
Show Answer & Explanation

Answer: (d) \( 8\theta/9 \)

Explanation:
1. Same torque in both rods and twist \( \phi\propto\frac{L}{r^4} \). Large rod: \( \frac{l}{r^4} \); small rod: \( \frac{l/2}{(r/2)^4}=\frac{8l}{r^4} \).
2. The small rod twists 8 times as much as the large one: \( \phi_s=\frac89\theta \), which is the angle at the joint.

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Question 17: A light rod of length 2 m is suspended from the ceiling horizontally by means of two vertical wires of equal length. A weight W is hung from the rod. One wire is steel with cross-sectional area A₁ = 0.1 cm² and the other is brass with A₂ = 0.2 cm². To have equal stress in both wires, T₁/T₂ equals
(a) \( \frac13 \)
(b) \( \frac14 \)
(c) \( \frac43 \)
(d) \( \frac12 \)
Show Answer & Explanation

Answer: (d) \( \frac12 \)

Explanation:
1. Equal stress: \( \frac{T_1}{A_1}=\frac{T_2}{A_2} \).
2. \( \frac{T_1}{T_2}=\frac{A_1}{A_2}=\frac{0.1}{0.2}=\frac12 \).

Teacher's Note:
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Question 18: A wire of length L and radius r is rigidly fixed at one end. On stretching the other end with a force F, the increase in its length is l. If another wire of the same material but of length 2L and radius 2r is stretched with a force 2F, the increase in its length will be
(a) l
(b) 2l
(c) \( \frac l2 \)
(d) \( \frac l4 \)
Show Answer & Explanation

Answer: (a) l

Explanation:
1. \( \Delta L=\frac{FL}{\pi r^2Y} \).
2. New: \( \frac{(2F)(2L)}{\pi(2r)^2Y}=\frac{4FL}{4\pi r^2Y}=l \).

Teacher's Note:
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Question 19: The length of an elastic string is a metre when the longitudinal tension is 4 N and b metre when the longitudinal tension is 5 N. The length of the string in metre when the longitudinal tension is 9 N is
(a) a − b
(b) 5b − 4a
(c) \( 2b-\frac14a \)
(d) 4a − 3b
Show Answer & Explanation

Answer: (b) 5b − 4a

Explanation:
1. Extension is proportional to the extra tension: each extra newton adds \( (b-a) \) m.
2. Length at 9 N = a + 5(b − a) = 5b − 4a.

Teacher's Note:
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Question 20: Four rods A, B, C and D of the same length and material but of different radii r, r√2, r√3 and 2r respectively are held between two rigid walls. The temperature of all rods is increased through the same range. If the rods do not bend, which of the following are correct? (1) the stress in the rods A, B, C and D is in the ratio 1 : 2 : 3 : 4 (2) the forces on them exerted by the walls are in the ratio 1 : 2 : 3 : 4 (3) the energy stored in the rods due to elasticity is in the ratio 1 : 2 : 3 : 4 (4) the strains produced in the rods are in the ratio 1 : 2 : 3 : 4
(a) 1 and 2
(b) 2 and 3
(c) 3 and 4
(d) 1 and 4
Show Answer & Explanation

Answer: (b) 2 and 3

Explanation:
1. Strain \( =\alpha\Delta T \) and stress \( =Y\alpha\Delta T \) are the same for all rods (1 and 4 false).
2. Force \( =\text{stress}\times\pi r^2 \) and stored energy \( =\frac12\text{stress}\times\text{strain}\times\text{volume} \) are proportional to \( r^2 \), that is 1 : 2 : 3 : 4 (2 and 3 true).

Teacher's Note:
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72. .

Question 21: A uniform bar of length 5l is fixed at both ends. A load W = 5000 N is applied axially at an intermediate section that divides the bar into an upper part of length 2l and a lower part of length 3l. Which of the following are correct? (1) the reaction at the top support is 3000 N (2) the reaction at the bottom support is 2000 N (3) the reaction at the top support is 1000 N (4) the reaction at the bottom support is 3000 N
(a) 1 and 2
(b) 3 and 4
(c) 1 and 4
(d) 2 and 3
Show Answer & Explanation

Answer: (a) 1 and 2

Explanation:
1. Top part is stretched/compressed by \( x \): its reaction \( R_1 \) and the bottom reaction \( R_2 \) satisfy \( R_1+R_2=W \) and equal deformation: \( \frac{R_1(2l)}{AY}=\frac{R_2(3l)}{AY} \).
2. \( R_1:R_2=3:2 \), so \( R_1=3000 \) N and \( R_2=2000 \) N.

Teacher's Note:
1. T
2. h
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Question 22: A metal wire of length L is suspended vertically from a rigid support. When a body of mass M is attached to the lower end of the wire, the elongation of the wire is l. Which are correct? (1) the loss in gravitational potential energy of the mass M is Mgl (2) the elastic potential energy stored in the wire is Mgl (3) the elastic potential energy stored in the wire is ½Mgl (4) the heat produced is ½Mgl
(a) 1, 2 and 3
(b) 1, 3 and 4
(c) 2 and 4
(d) 1 and 2
Show Answer & Explanation

Answer: (b) 1, 3 and 4

Explanation:
1. The mass falls through l, losing \( Mgl \) of PE (1 true).
2. The wire stores \( \frac12Mgl \) (3 true, 2 false); the remaining \( \frac12Mgl \) is dissipated as heat when the mass oscillates and settles (4 true).

Teacher's Note:
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96. .

Question 23: When a wire is stretched to double its length, which statements are correct? (1) strain is unity (2) stress is equal to Young's modulus of elasticity (3) its radius is halved (4) Young's modulus is equal to twice the elastic potential energy per unit volume
(a) 1, 2 and 3
(b) 1, 2 and 4
(c) 2 and 3
(d) 3 and 4
Show Answer & Explanation

Answer: (b) 1, 2 and 4

Explanation:
1. Strain = Δl/l = 1 (1 true), so stress = Y × 1 = Y (2 true).
2. Energy density \( =\frac12Y\times1^2 \), so Y = 2u (4 true). The radius falls by a factor √2 (volume conserved), not 2 (3 false).

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Question 24: The axle of a pulley of mass 1 kg is attached to the end of an elastic string of length 1 m, cross-sectional area 10⁻³ m² and Young's modulus 2 × 10⁵ N/m² whose other end is fixed to the ceiling (take g = 10 m/s²). The elongation of the string before any force is applied to the rope is
(a) 0.05 cm
(b) 0.5 cm
(c) 5 cm
(d) 50 cm
Show Answer & Explanation

Answer: (c) 5 cm

Explanation:
1. Force constant \( k=\frac{YA}{L}=\frac{2\times10^5\times10^{-3}}{1}=200 \) N/m.
2. Extension \( =\frac{mg}{k}=\frac{10}{200}=0.05 \) m = 5 cm.

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Question 25: The viscous liquid flows through a fixed tube of length l and radius R under a pressure difference P between its ends. The velocity profile is v = v₀(1 − r²/R²) with v₀ = PR²/(4ηl), where η is the coefficient of viscosity. The force acting on the tube due to the liquid is
(a) \( \pi\eta lv_0 \)
(b) \( 2\pi\eta lv_0 \)
(c) \( 4\pi\eta lv_0 \)
(d) \( 6\pi\eta lv_0 \)
Show Answer & Explanation

Answer: (c) \( 4\pi\eta lv_0 \)

Explanation:
1. At the wall \( \left|\frac{dv}{dr}\right|=\frac{2v_0}{R} \).
2. Force \( =\eta(2\pi Rl)\frac{2v_0}R=4\pi\eta lv_0 \).

Teacher's Note:
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Question 26: The viscous liquid flows through a fixed tube of length l and radius R under a pressure difference P between its ends. The velocity profile is v = v₀(1 − r²/R²) with v₀ = PR²/(4ηl), where η is the coefficient of viscosity. The momentum of the liquid confined in the tube is
(a) \( \rho\pi R^2lv_0 \)
(b) \( \rho\pi R^2lv_0/2 \)
(c) \( 2\rho\pi R^2lv_0 \)
(d) \( \rho\pi R^2lv_0/4 \)
Show Answer & Explanation

Answer: (b) \( \rho\pi R^2lv_0/2 \)

Explanation:
1. Momentum \( =\int\rho v\,dV=\rho v_0\int_0^R\left(1-\frac{r^2}{R^2}\right)2\pi rl\,dr \).
2. \( =\rho v_0\,2\pi l\left(\frac{R^2}2-\frac{R^2}4\right)=\frac{\rho\pi R^2lv_0}{2} \).

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Question 27:
Assertion (A): Steel is more elastic than rubber.
Reason (R): Under a given deforming force, steel is deformed less than rubber.
(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. Elasticity is the ability to resist deformation and regain shape; steel has the much larger Young's modulus: assertion true.
2. Statement 2 is a true observation, but 'deformed less' only restates the larger modulus rather than explaining why elasticity is greater.

Teacher's Note:
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Question 28:
Assertion (A): Bridges are declared unsafe after a long use.
Reason (R): The elastic strength of bridges decreases 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 loading causes elastic fatigue, so the elastic strength falls with time.
2. Hence old bridges are declared unsafe.

Teacher's Note:
1. E
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Question 29:
Assertion (A): Wet clay does not regain its shape.
Reason (R): Young's modulus for a perfectly plastic body 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. Clay is nearly perfectly plastic and does not recover after deformation.
2. A perfectly plastic body has no restoring force, so its Young's modulus is zero.

Teacher's Note:
1. A
2.
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43. .

Practice MCQs for JEE Physics Mechanical Properties Of Solids

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