Mechanical properties of solids examples for practice with answers

Mcq on mechanical properties of solids with answers

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1. A wire of length 2 m and cross-sectional area 1 mm² is stretched by a force of 200 N. The Young’s modulus of the wire is 2 × 10¹¹ Pa. What is the elongation of the wire?

A. 1.5 mm

B. 0.1 mm

C. 0.2 mm

D. 2 mm

The correct answer is C. 0.2 mm.

Explanation:
Using the formula: ΔL = (F × L) / (A × Y)
A = 1 mm2 = 1 × 10-6 m2
ΔL = (200 × 2) / (1 × 10-6 × 2 × 1011)
ΔL = 2 × 10-4 m = 0.2 mm
Hence, the elongation is 0.2 mm.

2. A steel wire of length 1.5 m and radius 0.5 mm is stretched by a force of 100 N. Calculate the stress on the wire.

A. 2.5 × 10⁷ Pa

B. 1.27 × 10⁸ Pa

C. 3.14 × 10⁸ Pa

D. 7.85 × 10⁶ Pa

The correct answer is B. 1.27 × 10⁸ Pa.

Explanation:
Stress is given by: σ = F / A
A = πr2 = π × (0.5 × 10-3)2 = 7.85 × 10-7 m2
σ = 100 / 7.85 × 10-7 = 1.27 × 108 Pa
Hence, the stress is 1.27 × 10⁸ Pa.

3. A copper rod of length 50 cm and diameter 2 mm is stretched by a load of 500 N. What is the strain in the rod? (Young’s modulus for copper = 1.1 × 10¹¹ Pa)

A. 1.44 × 10^-4

B. 2.86 × 10^-4

C. 7.15 × 10^-4

D. 1.15 × 10^-3

The correct answer is A. 1.44 × 10^-4.

Explanation:
Strain = Stress / Young’s modulus
Stress: σ = F / A
A = πr2 = π × (1 × 10-3)2 = 3.14 × 10-6 m2
σ = 500 / 3.14 × 10-6 = 1.59 × 108 Pa
Strain: ε = σ / Y = 1.59 × 108 / 1.1 × 1011 = 1.44 × 10-4

4. A cylindrical rod of steel of radius 2 cm and length 3 m is subjected to a tensile force of 5000 N. Calculate the tensile stress in the rod.

A. 3.98 × 10⁵ Pa

B. 3.98 × 10⁶ Pa

C. 1.99 × 10⁶ Pa

D. 9.95 × 10⁵ Pa

The correct answer is B. 3.98 × 10⁶ Pa.

Explanation:
Tensile stress: σ = F / A
A = πr2 = π × (2 × 10-2)2 = 1.26 × 10-3 m2
σ = 5000 / 1.26 × 10-3 = 3.98 × 106 Pa
Hence, the tensile stress is 3.98 × 10⁶ Pa.

5. A brass rod of length 1 m and cross-sectional area 2 cm² is stretched by a force of 1000 N. The Young’s modulus for brass is 1 × 10¹¹ Pa. Calculate the elongation of the rod.

A. 0.5 mm

B. 1.0 mm

C. 0.2 mm

D. 2.0 mm

The correct answer is A. 0.5 mm.

Explanation:
Using the formula: ΔL = (F × L) / (A × Y)
A = 2 cm2 = 2 × 10-4 m2
ΔL = (1000 × 1) / (2 × 10-4 × 1 × 1011)
ΔL = 0.5 × 10-3 m = 0.5 mm

6. A steel wire of length 2 m and cross-sectional area 1 mm² is stretched by a load of 10 kg. Calculate the elongation of the wire. (Young’s modulus for steel = 2 × 10¹¹ Pa)

A. 0.5 mm

B. 1.0 mm

C. 0.98 mm

D. 2.0 mm

The correct answer is C. 0.98 mm.

Explanation:
Elongation: ΔL = (F × L) / (A × Y)
F = mg = 10 × 9.8 = 98 N
A = 1 mm2 = 1 × 10-6 m2
ΔL = (98 × 2) / (1 × 10-6 × 2 × 1011)
ΔL = 0.00098 m = 0.98 mm

7. A spherical ball of radius 5 cm is subjected to a pressure of 200 kPa. The bulk modulus of the material is 2 × 10⁹ Pa. Calculate the change in the volume of the sphere.

A. 2.5 cm³

B. 0.52 cm³

C. 1.04 cm³

D. 5 cm³

The correct answer is B. 0.52 cm³.

Explanation:
Change in volume: ΔV = (P × V) / K
V = (4/3)πr3 = (4/3)π(5 × 10-2)3 = 5.24 × 10-4 m3
ΔV = (200 × 103 × 5.24 × 10-4) / 2 × 109
ΔV = 5.24 × 10-7 m3 = 0.52 cm3

8. A steel cable of length 10 m and diameter 1 cm is used to lift a load of 2000 kg. Find the extension of the cable. (Young’s modulus for steel = 2 × 10¹¹ Pa)

A. 0.1 cm

B. 1 cm

C. 0.01 cm

D. 10 cm

The correct answer is A. 0.1 cm.

Explanation:
Extension: ΔL = (F × L) / (A × Y)
F = mg = 2000 × 9.8 = 19600 N
A = πr2 = π × (0.005)2 = 7.85 × 10-5 m2
ΔL = (19600 × 10) / (7.85 × 10-5 × 2 × 1011)
ΔL = 0.001 m = 0.1 cm

9. A rod of length 1 m is subjected to a compressive force of 1000 N. The cross-sectional area of the rod is 5 cm². The bulk modulus is 1.2 × 10¹¹ Pa. Calculate the change in the length of the rod.

A. 0.83 × 10^-5 m

B. 8.3 × 10^-5 m

C. 1.2 × 10^-5 m

D. 5 × 10^-5 m

The correct answer is B. 8.3 × 10^-5 m.

Explanation:
Change in length: ΔL = (F × L) / (A × B)
A = 5 cm2 = 5 × 10-4 m2
ΔL = (1000 × 1) / (5 × 10-4 × 1.2 × 1011)
ΔL = 8.3 × 10-5 m

10. A hollow cylindrical pipe of inner radius 2 cm and outer radius 4 cm is subjected to a shearing stress τ of 50 × 10⁶ Pa. If the shear modulus is 8 × 10¹⁰ Pa, calculate the angular deformation.

A. 2 × 10^-4 rad

B. 1.25 × 10^-4 rad

C. 5 × 10^-5 rad

D. 1 × 10^-3 rad

The correct answer is B. 1.25 × 10^-4 rad.

Explanation:
Angular deformation: θ = τ / G
τ = 50 × 106 Pa, G = 8 × 1010 Pa
θ = 50 × 106 / 8 × 1010 = 1.25 × 10-4 rad

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Mechanical properties of solids examples for practice with answers

Numericals on Mechanical Properties of Solids Class 11