Mechanical Properties of Solids Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A copper 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 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: \( \Delta L = \frac{250 \times 2.0}{2.5 \times 10^{-6} \times 1.1 \times 10^{11}} = \frac{500}{2.75 \times 10^5} \approx 1.82 \times 10^{-3} \, \text{m} = 1.82 \, \text{mm} \).

\( 1.5 \, \text{mm} \)
\( 2 \, \text{mm} \)
\( 1.8 \, \text{mm} \)
\( 1.82 \, \text{mm} \)
4

A copper wire of length \( 2.3 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) is stretched by a force of \( 200 \, \text{N} \). If the Young's modulus of copper is \( 1.1 \times 10^{11} \, \text{N/m}^2 \), what is the strain?

Stress: \( \text{Stress} = \frac{F}{A} = \frac{200}{2 \times 10^{-6}} = 1 \times 10^8 \, \text{N/m}^2 \).

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

Strain: \( \text{Strain} = \frac{\text{Stress}}{Y} = \frac{1 \times 10^8}{1.1 \times 10^{11}} \approx 9.09 \times 10^{-4} \).

\( 8 \times 10^{-4} \)
\( 1 \times 10^{-3} \)
\( 7 \times 10^{-4} \)
\( 9.09 \times 10^{-4} \)
4

A copper wire of length \( 2.4 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) is stretched by a force producing a strain of \( 1.5 \times 10^{-4} \). If the Young's modulus of copper is \( 1.1 \times 10^{11} \, \text{N/m}^2 \), what is the stress?

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

Stress: \( \text{Stress} = Y \times \text{Strain} = 1.1 \times 10^{11} \times 1.5 \times 10^{-4} = 1.65 \times 10^7 \, \text{N/m}^2 \).

\( 1.5 \times 10^7 \, \text{N/m}^2 \)
\( 1.65 \times 10^7 \, \text{N/m}^2 \)
\( 1.8 \times 10^7 \, \text{N/m}^2 \)
\( 1.2 \times 10^7 \, \text{N/m}^2 \)
2

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

A water sample of volume \( 1 \, \text{litre} \) is compressed by a pressure of \( 1 \times 10^6 \, \text{N/m}^2 \). If the bulk modulus of water is \( 2.2 \times 10^9 \, \text{N/m}^2 \), what is the fractional change in volume?

Bulk modulus: \( B = -\frac{p}{\frac{\Delta V}{V}} \).

Rearrange: \( \frac{\Delta V}{V} = -\frac{p}{B} = -\frac{1 \times 10^6}{2.2 \times 10^9} \approx -4.55 \times 10^{-4} \).

Magnitude: \( 4.55 \times 10^{-4} \).

\( 4 \times 10^{-4} \)
\( 5 \times 10^{-4} \)
\( 3 \times 10^{-4} \)
\( 4.55 \times 10^{-4} \)
4

A steel cable of radius 0.015 m is used to support a load. If the maximum stress it can withstand is 2 × 108 N/m2, what is the maximum load it can support? (Take π ≈ 3.14)

Stress: Stress = F / A.

Area: A = π r2 = 3.14 × (0.015)2 = 3.14 × 0.000225 = 7.065 × 10-4 m2.

Maximum force: F = Stress × A = 2 × 108 × 7.065 × 10-4 = 1.413 × 105 N.

1.413 × 105 N
1 × 105 N
2 × 105 N
1.5 × 105 N
1

A steel wire of length \( 2.7 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) is stretched by \( 0.54 \, \text{mm} \). If the Young's modulus of steel is \( 2 \times 10^{11} \, \text{N/m}^2 \), what is the stress?

Strain: \( \text{Strain} = \frac{\Delta L}{L} = \frac{0.54 \times 10^{-3}}{2.7} = 2 \times 10^{-4} \).

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

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

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

A brass block of dimensions \( 0.6 \, \text{m} \times 0.4 \, \text{m} \times 0.2 \, \text{m} \) is subjected to a shearing force of \( 9 \times 10^4 \, \text{N} \). 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: \( \text{Shear stress} = \frac{F}{A} \), \( A = 0.6 \times 0.4 = 0.24 \, \text{m}^2 \).

Shear stress: \( \frac{9 \times 10^4}{0.24} = 3.75 \times 10^5 \, \text{N/m}^2 \).

Shear strain: \( \text{Shear strain} = \frac{\text{Shear stress}}{G} = \frac{3.75 \times 10^5}{3.6 \times 10^{10}} \approx 1.04 \times 10^{-5} \).

\( 1.04 \times 10^{-5} \)
\( 1.5 \times 10^{-5} \)
\( 2 \times 10^{-5} \)
\( 1 \times 10^{-5} \)
1

A rectangular slab of length 0.4 m, width 0.2 m, and thickness 0.1 m is subjected to a shearing force of 8 × 104 N. If the shear modulus is 4 × 109 N/m2, what is the displacement of the upper face relative to the lower face?

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

Rearrange: Δx = (F L) / (A G).

Area A = 0.4 × 0.2 = 0.08 m2, L = 0.1 m.

Substitute: Δx = (8 × 104 × 0.1) / (0.08 × 4 × 109) = 8000 / (3.2 × 108) = 2.5 × 10-5 m.

2.5 × 10-5 m
5 × 10-5 m
1 × 10-5 m
3 × 10-5 m
1

An aluminium block of dimensions \( 0.6 \, \text{m} \times 0.4 \, \text{m} \times 0.2 \, \text{m} \) is subjected to a shearing force of \( 8 \times 10^4 \, \text{N} \). 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: \( A = 0.6 \times 0.4 = 0.24 \, \text{m}^2 \), \( L = 0.2 \, \text{m} \).

Substitute: \( \Delta x = \frac{8 \times 10^4 \times 0.2}{0.24 \times 2.5 \times 10^{10}} = \frac{16000}{6 \times 10^9} \approx 2.67 \times 10^{-6} \, \text{m} \).

\( 2.5 \times 10^{-6} \, \text{m} \)
\( 2.67 \times 10^{-6} \, \text{m} \)
\( 3 \times 10^{-6} \, \text{m} \)
\( 2 \times 10^{-6} \, \text{m} \)
2

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