Mechanical Properties of Solids Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A copper block of dimensions \( 0.4 \, \text{m} \times 0.2 \, \text{m} \times 0.05 \, \text{m} \) is subjected to a shearing force of \( 1.5 \times 10^4 \, \text{N} \). If the shear modulus of copper is \( 4.2 \times 10^{10} \, \text{N/m}^2 \), what is the shear strain?

Shear modulus: \( G = \frac{\text{Shear stress}}{\text{Shear strain}} \).

Shear stress: \( \text{Shear stress} = \frac{F}{A} \), \( A = 0.4 \times 0.2 = 0.08 \, \text{m}^2 \).

Shear stress: \( \frac{1.5 \times 10^4}{0.08} = 1.875 \times 10^5 \, \text{N/m}^2 \).

Shear strain: \( \text{Shear strain} = \frac{\text{Shear stress}}{G} = \frac{1.875 \times 10^5}{4.2 \times 10^{10}} \approx 4.46 \times 10^{-6} \).

\( 4 \times 10^{-6} \)
\( 5 \times 10^{-6} \)
\( 3 \times 10^{-6} \)
\( 4.46 \times 10^{-6} \)
4

What does the term "elastic limit" refer to in the context of a material’s stress-strain behavior?

The elastic limit is the maximum stress a material can withstand while still returning to its original shape upon removal of the load; beyond this, permanent deformation occurs.

The point where the material fractures
The maximum stress before plastic deformation begins
The maximum stress beyond which the material does not return to its original shape
The minimum stress required for permanent deformation
3

Which type of stress leads to a change in length of a body without changing its shape?

Longitudinal stress (tensile or compressive) causes a change in length along the direction of the applied force without altering the shape of the body.

Longitudinal stress
Shearing stress
Hydraulic stress
Volumetric stress
1

Which of the following best defines stress in the context of mechanical properties of solids?

Stress is defined as the restoring force per unit area developed in a body due to an external deforming force, given by \( \text{Stress} = \frac{F}{A} \).

Restoring force per unit area
Deforming force per unit volume
Change in length per unit area
Change in volume per unit force
1

What does the steepness of the linear portion of a stress-strain curve typically indicate about a material?

The steepness of the linear portion of the stress-strain curve indicates a higher Young’s modulus, meaning the material is stiffer and requires more stress to produce a given strain.

Lower stiffness
Higher Young’s modulus
Greater ductility
Higher toughness
2

An aluminium rod of length \( 1.6 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) is compressed by a force of \( 600 \, \text{N} \). If the Young's modulus of aluminium is \( 7 \times 10^{10} \, \text{N/m}^2 \), what is the compression?

Young's modulus: \( Y = \frac{F L}{A \Delta L} \).

Rearrange: \( \Delta L = \frac{F L}{A Y} \).

Substitute: \( \Delta L = \frac{600 \times 1.6}{2 \times 10^{-6} \times 7 \times 10^{10}} = \frac{960}{1.4 \times 10^5} \approx 6.86 \times 10^{-3} \, \text{m} = 6.86 \, \text{mm} \).

\( 6 \, \text{mm} \)
\( 6.86 \, \text{mm} \)
\( 7 \, \text{mm} \)
\( 5 \, \text{mm} \)
2

A water sample of volume 1 litre is compressed by 0.05%. If the bulk modulus of water is 2.2 × 109 N/m2, what is the pressure applied?

Bulk modulus: B = -p / (ΔV / V).

Rearrange: p = -B × (ΔV / V).

Given: ΔV / V = -0.05% = -0.0005.

Substitute: p = 2.2 × 109 × 0.0005 = 1.1 × 106 N/m2.

1 × 106 N/m2
1.5 × 106 N/m2
2 × 106 N/m2
1.1 × 106 N/m2
4

A steel wire of length 4 m and cross-sectional area 2 × 10-6 m2 is stretched by 0.4 mm. If the force applied is 200 N, what is the Young's modulus of the steel?

Young's modulus: Y = (F L) / (A ΔL).

Substitute: ΔL = 0.4 × 10-3 m.

Y = (200 × 4) / (2 × 10-6 × 0.4 × 10-3) = 800 / (8 × 10-10) = 1 × 1012 N/m2.

0.5 × 1012 N/m2
2 × 1012 N/m2
1.5 × 1012 N/m2
1 × 1012 N/m2
4

A steel wire of length 3 m and cross-sectional area 4 × 10-6 m2 is stretched by 0.3 mm. If the Young's modulus of steel is 2 × 1011 N/m2, what is the force applied?

Young's modulus: Y = (F L) / (A ΔL).

Rearrange: F = (Y A ΔL) / L.

Substitute: ΔL = 0.3 × 10-3 m, so F = (2 × 1011 × 4 × 10-6 × 0.3 × 10-3) / 3 = 80 N.

60 N
100 N
80 N
120 N
3

Why is steel often chosen over materials like copper for heavy-duty structural applications?

Steel has a higher Young’s modulus (\( \approx 200 \times 10^9 \, \text{N/m}^2 \)) compared to copper (\( \approx 110 \times 10^9 \, \text{N/m}^2 \)), requiring a larger force to produce the same deformation, making it more suitable for resisting bending and stretching.

Higher Young’s modulus
Lower density
Greater ductility
Higher thermal conductivity
1

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