Mechanical Properties of Solids Chapter-Wise Test 8

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A glass slab of volume \( 0.01 \, \text{m}^3 \) is subjected to a hydraulic pressure of \( 2 \times 10^6 \, \text{N/m}^2 \). If the bulk modulus of glass is \( 3.7 \times 10^{10} \, \text{N/m}^2 \), what is the change in volume?

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

Rearrange: \( \frac{\Delta V}{V} = -\frac{p}{B} = -\frac{2 \times 10^6}{3.7 \times 10^{10}} \approx -5.41 \times 10^{-5} \).

Change in volume: \( \Delta V = \frac{\Delta V}{V} \times V = -5.41 \times 10^{-5} \times 0.01 \approx -5.41 \times 10^{-7} \, \text{m}^3 \).

\( 5 \times 10^{-7} \, \text{m}^3 \)
\( 6 \times 10^{-7} \, \text{m}^3 \)
\( 5.41 \times 10^{-7} \, \text{m}^3 \)
\( 4 \times 10^{-7} \, \text{m}^3 \)
3

A brass wire of length \( 1.6 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) is stretched by a force producing a stress of \( 4 \times 10^7 \, \text{N/m}^2 \). If the Young's modulus of brass is \( 9 \times 10^{10} \, \text{N/m}^2 \), what is the strain?

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

Strain: \( \text{Strain} = \frac{\text{Stress}}{Y} = \frac{4 \times 10^7}{9 \times 10^{10}} \approx 4.44 \times 10^{-4} \).

\( 4 \times 10^{-4} \)
\( 5 \times 10^{-4} \)
\( 3 \times 10^{-4} \)
\( 4.44 \times 10^{-4} \)
4

A brass wire of length \( 2.4 \, \text{m} \) and cross-sectional area \( 3 \times 10^{-6} \, \text{m}^2 \) is stretched by a force producing a stress of \( 5 \times 10^7 \, \text{N/m}^2 \). If the Young's modulus of brass is \( 9 \times 10^{10} \, \text{N/m}^2 \), what is the strain?

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

Strain: \( \text{Strain} = \frac{\text{Stress}}{Y} = \frac{5 \times 10^7}{9 \times 10^{10}} \approx 5.56 \times 10^{-4} \).

\( 5 \times 10^{-4} \)
\( 5.56 \times 10^{-4} \)
\( 6 \times 10^{-4} \)
\( 4 \times 10^{-4} \)
2

A steel wire of length \( 2.6 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) is stretched by a force of \( 400 \, \text{N} \). If the Young's modulus of steel is \( 2 \times 10^{11} \, \text{N/m}^2 \), what is the elongation?

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

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

Substitute: \( \Delta L = \frac{400 \times 2.6}{2 \times 10^{-6} \times 2 \times 10^{11}} = \frac{1040}{4 \times 10^5} = 2.6 \times 10^{-3} \, \text{m} = 2.6 \, \text{mm} \).

\( 2 \, \text{mm} \)
\( 2.6 \, \text{mm} \)
\( 3 \, \text{mm} \)
\( 1.5 \, \text{mm} \)
2

A brass rod of radius \( 0.008 \, \text{m} \) and length \( 1.0 \, \text{m} \) is subjected to a tensile force producing a stress of \( 5 \times 10^7 \, \text{N/m}^2 \). What is the force applied? (Take \( \pi \approx 3.14 \))

Stress: \( \text{Stress} = \frac{F}{A} \).

Area: \( A = \pi r^2 = 3.14 \times (0.008)^2 = 3.14 \times 6.4 \times 10^{-5} \approx 2.01 \times 10^{-4} \, \text{m}^2 \).

Force: \( F = \text{Stress} \times A = 5 \times 10^7 \times 2.01 \times 10^{-4} \approx 1.005 \times 10^4 \, \text{N} \).

\( 1.005 \times 10^4 \, \text{N} \)
\( 1.5 \times 10^4 \, \text{N} \)
\( 1 \times 10^4 \, \text{N} \)
\( 2 \times 10^4 \, \text{N} \)
1

A steel rod of radius \( 0.01 \, \text{m} \) and length \( 1.5 \, \text{m} \) is subjected to a tensile force producing a stress of \( 4 \times 10^7 \, \text{N/m}^2 \). What is the force applied? (Take \( \pi \approx 3.14 \))

Stress: \( \text{Stress} = \frac{F}{A} \).

Area: \( A = \pi r^2 = 3.14 \times (0.01)^2 = 3.14 \times 10^{-4} \, \text{m}^2 \).

Force: \( F = \text{Stress} \times A = 4 \times 10^7 \times 3.14 \times 10^{-4} = 1.256 \times 10^4 \, \text{N} \).

\( 1.256 \times 10^4 \, \text{N} \)
\( 1.5 \times 10^4 \, \text{N} \)
\( 1 \times 10^4 \, \text{N} \)
\( 2 \times 10^4 \, \text{N} \)
1

What property of a material primarily determines its resistance to uniform compression?

The bulk modulus determines a material’s resistance to uniform compression by measuring how much it resists volume change under pressure applied in all directions.

Shear modulus
Young’s modulus
Bulk modulus
Yield strength
3

A copper rod of length 1 m and diameter 2 mm is subjected to a tensile force. If the stress is 4 × 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 = 4 × 107 × 3.14 × 10-6 = 125.6 N.

100 N
150 N
120 N
125.6 N
4

A copper block of dimensions 0.4 m × 0.3 m × 0.1 m is subjected to a shearing force of 3 × 104 N. If the shear modulus of copper is 4.2 × 1010 N/m2, what is the displacement of the top face?

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

Rearrange: Δx = (F L) / (A G).

Area: A = 0.4 × 0.3 = 0.12 m2, L = 0.1 m.

Substitute: Δx = (3 × 104 × 0.1) / (0.12 × 4.2 × 1010) = 3000 / (5.04 × 109) ≈ 5.95 × 10-7 m.

5 × 10-7 m
6.5 × 10-7 m
5.95 × 10-7 m
7 × 10-7 m
3

A brass wire of length \( 1.8 \, \text{m} \) and cross-sectional area \( 2.5 \times 10^{-6} \, \text{m}^2 \) is stretched by a force producing a strain of \( 2 \times 10^{-4} \). If the Young's modulus of brass is \( 9 \times 10^{10} \, \text{N/m}^2 \), what is the stress?

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

Stress: \( \text{Stress} = Y \times \text{Strain} = 9 \times 10^{10} \times 2 \times 10^{-4} = 1.8 \times 10^7 \, \text{N/m}^2 \).

\( 8 \times 10^6 \, \text{N/m}^2 \)
\( 1.8 \times 10^7 \, \text{N/m}^2 \)
\( 2 \times 10^7 \, \text{N/m}^2 \)
\( 1.5 \times 10^7 \, \text{N/m}^2 \)
2

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