JEE Physics Mechanical Properties Of Solids MCQs Set 10

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

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Multiple Choice Questions

Question 1: A steel wire of length L at 40°C is suspended from the ceiling and then a mass m is hung from its free end. The wire is cooled down from 40°C to 30°C to regain its original length L. The coefficient of linear thermal expansion of steel is 10⁻⁵ /°C, Young's modulus of steel is 10¹¹ N/m² and the radius of the wire is 1 mm. Assume L ≫ diameter of the wire. The value of m in kg is nearly (g = 10 m/s²)
(a) 1
(b) 2
(c) 3
(d) 5
Show Answer & Explanation

Answer: (c) 3

Explanation:
1. The thermal contraction LαΔT equals the extension from the load mgL/(AY).
2. So mg = AYαΔT = (π × 10⁻⁶)(10¹¹)(10⁻⁵)(10) = 31.4 N.
3. m ≈ 3.1 kg ≈ 3 kg.

Teacher's Note:
1. Thermal strain cancels the elastic strain.

Question 2: For the same cross-sectional area and for a given load, the ratio of depressions for a beam of square cross-section and circular cross-section is
(a) \( 3:\pi \)
(b) \( \pi:3 \)
(c) \( 1:\pi \)
(d) \( \pi:1 \)
Show Answer & Explanation

Answer: (a) \( 3:\pi \)

Explanation:
1. Depression δ ∝ 1/(bd³) ∝ 1/I for the same span and load.
2. Square of side a: I = a⁴/12. Circle with πr² = a²: I = πr⁴/4 = a⁴/4π.
3. δ_sq/δ_circ = (a⁴/4π)/(a⁴/12) = 3/π.

Teacher's Note:
1. Compare the second moments of area.

Question 3: A steel ring of radius r and cross-section area A is fitted onto a wooden disc of radius R (R > r). If Young's modulus of steel is E, then the force with which the steel ring is expanded is
(a) \( AE\frac Rr \)
(b) \( AE\left(\frac{R-r}{r}\right) \)
(c) \( \frac EA\left(\frac{R-r}{R}\right) \)
(d) \( \frac{Er}{AR} \)
Show Answer & Explanation

Answer: (b) \( AE\left(\frac{R-r}{r}\right) \)

Explanation:
1. Strain = (2πR − 2πr)/2πr = (R − r)/r.
2. Tension = stress × area = E × strain × A = AE(R − r)/r.

Teacher's Note:
1. Force = Y × strain × area.

Question 4: An elevator cable is to have a maximum stress of 7 × 10⁷ N/m² to allow for appropriate safety factors. If its maximum acceleration is 1.5 m/s² and 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) 3.29 cm²
(b) 2.38 cm²
(c) 0.329 cm²
(d) 8.23 cm²
Show Answer & Explanation

Answer: (a) 3.29 cm²

Explanation:
1. Maximum tension T = m(g + a) = 2000 × 11.5 = 23000 N.
2. A = T/stress = 23000/(7 × 10⁷) = 3.29 × 10⁻⁴ m² = 3.29 cm².

Teacher's Note:
1. Include the acceleration in the tension.

Question 5: A beam of metal supported at the two edges is loaded at the centre. The depression at the centre is proportional to
(a) \( Y^2 \)
(b) Y
(c) \( 1/Y \)
(d) \( 1/Y^2 \)
Show Answer & Explanation

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

Explanation:
1. The depression δ = Wl³/(4bd³Y).
2. So δ ∝ 1/Y.

Teacher's Note:
1. Stiffer material bends less.

Question 6: Which of the following statements is/are false? I. Normal stress = force/area. II. Young's modulus = (force × change in length)/(area × original length).
(a) I only
(b) II only
(c) Both I and II
(d) none of these
Show Answer & Explanation

Answer: (b) II only

Explanation:
1. Young's modulus = (force × original length)/(area × change in length).
2. Statement II has the lengths interchanged, so it is false.

Teacher's Note:
1. Y = (F/A)/(Δl/l).

Question 7: To break a wire, a force of 10⁶ N/m² is required. If the density of the material is 3 × 10³ kg/m³, then 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. The weight per unit area of a wire of length L = ρLg.
2. ρLg = 10⁶ gives L = 10⁶/(3 × 10³ × 9.8) = 34 m.

Teacher's Note:
1. Breaking stress equals ρgL.

Question 8: If a rubber ball is taken to a depth of 200 m in a pool, its volume decreases by 0.1%. If the density of water is 1 × 10³ kg/m³ and g = 10 m/s², then the volume elasticity in N/m² will be
(a) \( 10^8 \)
(b) \( 2\times10^8 \)
(c) \( 10^9 \)
(d) \( 2\times10^9 \)
Show Answer & Explanation

Answer: (d) \( 2\times10^9 \)

Explanation:
1. Excess pressure = ρgh = 10³ × 10 × 200 = 2 × 10⁶ N/m².
2. B = P/(ΔV/V) = 2 × 10⁶/10⁻³ = 2 × 10⁹ N/m².

Teacher's Note:
1. Hydrostatic pressure gives the stress.

Question 9: The modulus of rigidity of a liquid is
(a) constant
(b) infinite
(c) zero
(d) cannot be predicted
Show Answer & Explanation

Answer: (c) zero

Explanation:
1. A liquid cannot support a shear stress in equilibrium.
2. So its modulus of rigidity is zero.

Teacher's Note:
1. Liquids flow under shear.

Question 10: A and B are two wires. The radius of A is twice that of B. They are stretched by the same load. Then the stress on B is
(a) equal to that on A
(b) four times that on A
(c) twice that on A
(d) half that on A
Show Answer & Explanation

Answer: (b) four times that on A

Explanation:
1. Stress = F/πr².
2. With radius ratio 2 : 1, the area of A is 4 times that of B, so B's stress is 4 times A's.

Teacher's Note:
1. Stress is inversely proportional to r².

Question 11: Match Column I with Column II.
Column IColumn II
(A) Young's modulus of elasticity(1) Rubber
(B) Hooke's law(2) Solids
(C) Hydraulic stress(3) Straight line
(D) Elastomers(4) Solids, liquids and gases

(a) A-2, B-4, C-3, D-1
(b) A-3, B-2, C-4, D-1
(c) A-2, B-3, C-4, D-1
(d) A-2, B-1, C-4, D-3
Show Answer & Explanation

Answer: (c) A-2, B-3, C-4, D-1

Explanation:
1. Young's modulus applies to solids (A-2).
2. Hooke's law gives a straight line on a stress-strain graph (B-3).
3. Hydraulic stress and the bulk modulus apply to solids, liquids and gases (C-4).
4. Rubber is an elastomer (D-1).

Teacher's Note:
1. Match each term with its context.

Question 12: Consider four steel wires of dimensions given below (d is the diameter and l the length): (A) l = 1 m, d = 1 mm; (B) l = 2 m, d = 1 mm; (C) l = 2 m, d = 2 mm; (D) l = 1 m, d = 2 mm. If the same force is applied to all the wires, the elastic potential energy stored will be maximum in wire
(a) A
(b) B
(c) C
(d) D
Show Answer & Explanation

Answer: (b) B

Explanation:
1. U = F²l/(2AY) ∝ l/d².
2. A: 1, B: 2, C: 0.5, D: 0.25, so B stores the most.

Teacher's Note:
1. U ∝ l/d² at a fixed force.

Question 13: Which of the following affects the elasticity of a substance?
(a) Hammering and annealing
(b) Change in temperature
(c) Impurity in substance
(d) All of the above
Show Answer & Explanation

Answer: (d) All of the above

Explanation:
1. Hammering and annealing change the crystal structure, temperature changes the inter-atomic forces, and impurities alter the lattice.
2. All of them change the elastic properties.

Teacher's Note:
1. Many factors affect elasticity.

Question 14: In magnitude, hydraulic stress is equal to
(a) hydraulic force
(b) hydraulic pressure
(c) hydraulic strain
(d) restoring force
Show Answer & Explanation

Answer: (b) hydraulic pressure

Explanation:
1. Hydraulic stress is the normal force per unit area applied uniformly on the whole surface, which is the pressure.
2. So its magnitude equals the hydraulic pressure.

Teacher's Note:
1. Hydraulic stress is pressure.

Question 15:
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. W = ½kx² and k ∝ Y.
2. Steel has the larger Y, so more work is needed for the same stretch.

Teacher's Note:
1. Work is proportional to the force constant.

Question 16: Shearing strain is expressed by
(a) angle of shear
(b) angle of twist
(c) decrease in volume
(d) increase in volume
Show Answer & Explanation

Answer: (a) angle of shear

Explanation:
1. Shearing strain = Δx/L = tan θ ≈ θ.
2. It is the angle of shear.

Teacher's Note:
1. Shear strain is an angle.

Question 17:
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. A perfectly plastic body has no restoring stress, so stress/strain is zero.
2. The reason explains the assertion.

Teacher's Note:
1. Zero restoring force means Y = 0.

Question 18: Copper of fixed volume V is drawn into a wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is Δl. Which of the following graphs is a straight line?
(a) \( \Delta l\ \text{versus}\ \frac1l \)
(b) \( \Delta l\ \text{versus}\ l^2 \)
(c) \( \Delta l\ \text{versus}\ \frac1{l^2} \)
(d) \( \Delta l\ \text{versus}\ l \)
Show Answer & Explanation

Answer: (b) \( \Delta l\ \text{versus}\ l^2 \)

Explanation:
1. A = V/l, so Δl = Fl/(AY) = Fl²/(VY).
2. Δl is proportional to l², so Δl versus l² is a straight line.

Teacher's Note:
1. Area depends on length at constant volume.

Question 19: The length of a metal is ℓ₁ when the tension in it is T₁ and is ℓ₂ when the tension is T₂. The original length of the wire is
(a) \( \frac{\ell_1+\ell_2}{2} \)
(b) \( \frac{\ell_1T_2+\ell_2T_1}{T_1+T_2} \)
(c) \( \frac{\ell_1T_2-\ell_2T_1}{T_2-T_1} \)
(d) \( \sqrt{T_1T_2\ell_1\ell_2} \)
Show Answer & Explanation

Answer: (c) \( \frac{\ell_1T_2-\ell_2T_1}{T_2-T_1} \)

Explanation:
1. With natural length l: ℓ₁ = l + kT₁ and ℓ₂ = l + kT₂.
2. Eliminating k: l = (ℓ₁T₂ − ℓ₂T₁)/(T₂ − T₁).

Teacher's Note:
1. Extension is proportional to tension.

Question 20: The expression for the force constant of a wire (spring) following Hooke's law is given by (Y is Young's modulus, a the area of cross-section, ℓ the length)
(a) \( k=\frac{Ya}{\ell} \)
(b) \( k=\frac{Ya}{\Delta \ell} \)
(c) \( k=\frac{Ya\,\Delta\ell}{\ell} \)
(d) \( k=\frac{Ya\,\ell}{\Delta\ell} \)
Show Answer & Explanation

Answer: (a) \( k=\frac{Ya}{\ell} \)

Explanation:
1. F = (YaΔℓ)/ℓ = kΔℓ.
2. So k = Ya/ℓ.

Teacher's Note:
1. F = kΔℓ.

Question 21: For an equal stretching force F, the Young's modulus of steel (Yₛ) and rubber (Yᵣ) are related as
(a) \( Y_s=Y_r \)
(b) \( Y_s \lt Y_r \)
(c) \( Y_s \gt Y_r \)
(d) \( Y_s\geq Y_r \)
Show Answer & Explanation

Answer: (c) \( Y_s \gt Y_r \)

Explanation:
1. For the same stress, steel shows much less strain than rubber.
2. So the Young's modulus of steel is larger.

Teacher's Note:
1. Smaller strain means larger Y.

Question 22: Which of the following statements are correct? I. Elastic fatigue is the property by virtue of which behaviour becomes less elastic under the action of repeated alternating deforming forces. II. Elasticity is the property due to which a body regains its original configuration when deforming forces are removed. III. Plasticity is the property due to which the regain of original shape is delayed after the removal of deforming forces.
(a) I and II
(b) II and III
(c) I and III
(d) I, II and III
Show Answer & Explanation

Answer: (a) I and II

Explanation:
1. I and II are the correct definitions.
2. III describes the elastic after effect, not plasticity.

Teacher's Note:
1. Plasticity is permanent deformation.

Question 23: The ratio of shearing stress to the corresponding shearing strain is called
(a) bulk modulus
(b) Young's modulus
(c) modulus of rigidity
(d) none of these
Show Answer & Explanation

Answer: (c) modulus of rigidity

Explanation:
1. By definition, G = shearing stress/shearing strain.
2. It is also called the shear modulus.

Teacher's Note:
1. Definition of rigidity modulus.

Question 24: Which of the following is relevant only for solids?
(a) Young's modulus
(b) Shear modulus
(c) Bulk modulus
(d) Both (a) and (b)
Show Answer & Explanation

Answer: (d) Both (a) and (b)

Explanation:
1. Liquids and gases cannot sustain longitudinal or shear stress in equilibrium.
2. Only the bulk modulus applies to them, so Young's and shear moduli are relevant only to solids.

Teacher's Note:
1. Fluids have only a bulk modulus.

Question 25: The phenomenon of temporary delay in regaining the original configuration by an elastic body, after the removal of the deforming force, is called
(a) elastic fatigue
(b) elasticity
(c) plasticity
(d) elastic after effect
Show Answer & Explanation

Answer: (d) elastic after effect

Explanation:
1. The lag in recovery is called the elastic after effect.
2. Elastic fatigue is the loss of elasticity after repeated cycles.

Teacher's Note:
1. Delayed recovery means after effect.

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