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  4. Mechanical Properties Of Solids Chapter-Wise Test 2

Mechanical Properties Of Solids Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A steel wire and a copper wire have the same length and cross-sectional area. Both are stretched by the same force. If Young's modulus of steel is 2 × 1011 N/m2 and that of copper is 1.1 × 1011 N/m2, what is the ratio of elongation of the steel wire to the copper wire?

Elongation: ΔL = (F L) / (A Y).

Ratio: (ΔLsteel)/(ΔLcopper) = Ycopper / Ysteel = (1.1 × 1011) / (2 × 1011) = 11/20 = 0.55.

0.45
0.55
0.65
0.75
2

A steel rod of length 1 m and diameter 2 mm is subjected to a tensile force producing a stress of 5 × 107 N/m2. What is the force applied? (Take π ≈ 3.14)

Stress: Stress = F / A.

Area: A = π r2, r = 1 × 10-3 m.

A = 3.14 × (1 × 10-3)2 = 3.14 × 10-6 m2.

Force: F = Stress × A = 5 × 107 × 3.14 × 10-6 = 157 N.

100 N
200 N
150 N
157 N
4

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

A steel wire of length \( 2.0 \, \text{m} \) and cross-sectional area \( 2.5 \times 10^{-6} \, \text{m}^2 \) is stretched by a force of \( 250 \, \text{N} \). If the Young's modulus of steel is \( 2 \times 10^{11} \, \text{N/m}^2 \), what is the strain produced?

Stress: \( \text{Stress} = \frac{F}{A} = \frac{250}{2.5 \times 10^{-6}} = 1 \times 10^8 \, \text{N/m}^2 \).

Young's modulus: \( Y = \frac{\text{Stress}}{\text{Strain}} \).

Strain: \( \text{Strain} = \frac{\text{Stress}}{Y} = \frac{1 \times 10^8}{2 \times 10^{11}} = 5 \times 10^{-4} \).

\( 5 \times 10^{-4} \)
\( 6 \times 10^{-4} \)
\( 4 \times 10^{-4} \)
\( 3 \times 10^{-4} \)
1

A copper wire of length \( 2.8 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) is stretched by a force of \( 160 \, \text{N} \). If the Young's modulus of copper is \( 1.1 \times 10^{11} \, \text{N/m}^2 \), what is the elongation?

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

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

Substitute: \( \Delta L = \frac{160 \times 2.8}{2 \times 10^{-6} \times 1.1 \times 10^{11}} = \frac{448}{2.2 \times 10^5} \approx 2.04 \times 10^{-3} \, \text{m} = 2.04 \, \text{mm} \).

\( 2 \, \text{mm} \)
\( 1.5 \, \text{mm} \)
\( 2.5 \, \text{mm} \)
\( 2.04 \, \text{mm} \)
4

In designing bridges, why is it critical to understand the elastic properties of materials?

Understanding elastic properties ensures that materials can withstand various loads (traffic, wind, weight) without undergoing permanent deformation, maintaining structural integrity.

To increase the weight of the bridge
To reduce the cost of materials
To enhance thermal conductivity
To ensure structural integrity
4

A brass block of dimensions \( 0.5 \, \text{m} \times 0.2 \, \text{m} \times 0.1 \, \text{m} \) is subjected to a shearing force of \( 5 \times 10^4 \, \text{N} \). If the shear modulus of brass is \( 3.6 \times 10^{10} \, \text{N/m}^2 \), what is the shear strain?

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

Shear stress: \( \text{Shear stress} = \frac{F}{A} \), \( A = 0.5 \times 0.2 = 0.1 \, \text{m}^2 \).

Shear stress: \( \frac{5 \times 10^4}{0.1} = 5 \times 10^5 \, \text{N/m}^2 \).

Shear strain: \( \text{Shear strain} = \frac{\text{Shear stress}}{G} = \frac{5 \times 10^5}{3.6 \times 10^{10}} \approx 1.39 \times 10^{-5} \).

\( 1.39 \times 10^{-5} \)
\( 1.5 \times 10^{-5} \)
\( 2 \times 10^{-5} \)
\( 1 \times 10^{-5} \)
1

What characterizes the stress-strain behavior of a material in the region where plastic deformation begins?

When plastic deformation begins, the material experiences permanent deformation, meaning it does not return to its original shape even after the load is removed.

The material obeys Hooke’s law
The material experiences permanent deformation
The material fractures immediately
The material undergoes purely elastic deformation
2

A steel wire of length \( 2.6 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) is stretched by \( 0.52 \, \text{mm} \). If the Young's modulus of steel is \( 2 \times 10^{11} \, \text{N/m}^2 \), what is the force applied?

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

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

Substitute: \( \Delta L = 0.52 \times 10^{-3} \, \text{m} \).

\( F = \frac{2 \times 10^{11} \times 2 \times 10^{-6} \times 0.52 \times 10^{-3}}{2.6} = \frac{208}{2.6} = 80 \, \text{N} \).

\( 70 \, \text{N} \)
\( 90 \, \text{N} \)
\( 60 \, \text{N} \)
\( 80 \, \text{N} \)
4

What does the steepness of the initial linear portion of a stress-strain curve indicate?

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

Lower stiffness
Greater ductility
Higher shear modulus
Higher Young’s modulus
4

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