Mechanical Properties of Solids Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A steel wire of length \( 2.2 \, \text{m} \) 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?

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

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

A brass block of dimensions 0.3 m × 0.2 m × 0.1 m is subjected to a shearing force of 5 × 104 N. If the shear modulus of brass is 3.6 × 1010 N/m2, what is the displacement of the upper face?

Shear modulus: G = (F / A) / (Δx / L).

Rearrange: Δx = (F L) / (A G).

Area: A = 0.3 × 0.2 = 0.06 m2, L = 0.1 m.

Substitute: Δx = (5 × 104 × 0.1) / (0.06 × 3.6 × 1010) = 5000 / (2.16 × 109) ≈ 2.31 × 10-6 m.

2 × 10-6 m
3 × 10-6 m
2.31 × 10-6 m
2.5 × 10-6 m
3

An aluminium block of dimensions \( 0.5 \, \text{m} \times 0.3 \, \text{m} \times 0.1 \, \text{m} \) is subjected to a shearing force of \( 7 \times 10^4 \, \text{N} \). If the shear modulus of aluminium is \( 2.5 \times 10^{10} \, \text{N/m}^2 \), what is the displacement of the top face?

Shear modulus: \( G = \frac{F / A}{\Delta x / L} \).

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

Area: \( A = 0.5 \times 0.3 = 0.15 \, \text{m}^2 \), \( L = 0.1 \, \text{m} \).

Substitute: \( \Delta x = \frac{7 \times 10^4 \times 0.1}{0.15 \times 2.5 \times 10^{10}} = \frac{7000}{3.75 \times 10^9} \approx 1.87 \times 10^{-6} \, \text{m} \).

\( 1.5 \times 10^{-6} \, \text{m} \)
\( 2 \times 10^{-6} \, \text{m} \)
\( 1.8 \times 10^{-6} \, \text{m} \)
\( 1.87 \times 10^{-6} \, \text{m} \)
4

An aluminium rod of length \( 1.2 \, \text{m} \) and cross-sectional area \( 2.5 \times 10^{-6} \, \text{m}^2 \) is compressed by a force producing a strain of \( 2 \times 10^{-4} \). If the Young's modulus of aluminium is \( 7 \times 10^{10} \, \text{N/m}^2 \), what is the stress?

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

Stress: \( \text{Stress} = Y \times \text{Strain} = 7 \times 10^{10} \times 2 \times 10^{-4} = 1.4 \times 10^7 \, \text{N/m}^2 \).

\( 1 \times 10^7 \, \text{N/m}^2 \)
\( 2 \times 10^7 \, \text{N/m}^2 \)
\( 1.4 \times 10^7 \, \text{N/m}^2 \)
\( 1.5 \times 10^7 \, \text{N/m}^2 \)
3

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

A steel rod of length 0.5 m and cross-sectional area 1 × 10-5 m2 is compressed by a force of 500 N. If the strain is 5 × 10-4, what is the Young's modulus?

Young's modulus: Y = Stress / Strain.

Stress: Stress = F / A = 500 / (1 × 10-5) = 5 × 107 N/m2.

Strain: 5 × 10-4.

Substitute: Y = (5 × 107) / (5 × 10-4) = 1 × 1011 N/m2.

0.5 × 1011 N/m2
2 × 1011 N/m2
1.5 × 1011 N/m2
1 × 1011 N/m2
4

A brass wire of length \( 2.6 \, \text{m} \) and cross-sectional area \( 4 \times 10^{-6} \, \text{m}^2 \) is stretched by a force of \( 400 \, \text{N} \). If the Young's modulus of brass is \( 9 \times 10^{10} \, \text{N/m}^2 \), what is the elongation?

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

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

Substitute: \( \Delta L = \frac{400 \times 2.6}{4 \times 10^{-6} \times 9 \times 10^{10}} = \frac{1040}{3.6 \times 10^5} \approx 2.89 \times 10^{-3} \, \text{m} = 2.89 \, \text{mm} \).

\( 2.5 \, \text{mm} \)
\( 3 \, \text{mm} \)
\( 2 \, \text{mm} \)
\( 2.89 \, \text{mm} \)
4

Which modulus describes a material’s resistance to volume change under uniform pressure in all directions?

The bulk modulus measures a material’s resistance to volume change when subjected to uniform pressure (hydraulic stress) in all directions, affecting volume but not shape.

Young’s modulus
Shear modulus
Bulk modulus
Elastic modulus
3

A steel wire of length \( 2.4 \, \text{m} \) and cross-sectional area \( 4 \times 10^{-6} \, \text{m}^2 \) is stretched by \( 0.6 \, \text{mm} \). If the Young's modulus of steel is \( 2 \times 10^{11} \, \text{N/m}^2 \), what is the stress?

Strain: \( \text{Strain} = \frac{\Delta L}{L} = \frac{0.6 \times 10^{-3}}{2.4} = 2.5 \times 10^{-4} \).

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

Stress: \( \text{Stress} = Y \times \text{Strain} = 2 \times 10^{11} \times 2.5 \times 10^{-4} = 5 \times 10^7 \, \text{N/m}^2 \).

\( 4 \times 10^7 \, \text{N/m}^2 \)
\( 6 \times 10^7 \, \text{N/m}^2 \)
\( 5 \times 10^7 \, \text{N/m}^2 \)
\( 3 \times 10^7 \, \text{N/m}^2 \)
3

A copper block of dimensions 0.5 m × 0.3 m × 0.2 m is subjected to a shearing force of 7 × 104 N on its top face. If the shear modulus of copper is 4.2 × 1010 N/m2, what is the shear strain?

Shear modulus: G = (Shear stress) / (Shear strain).

Shear stress: Shear stress = F / A, A = 0.5 × 0.3 = 0.15 m2.

Shear stress = (7 × 104) / 0.15 = 4.67 × 105 N/m2.

Shear strain: Shear strain = Shear stress / G = (4.67 × 105) / (4.2 × 1010) ≈ 1.11 × 10-5.

1.11 × 10-5
2 × 10-5
1 × 10-5
1.5 × 10-5
1

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