Mechanical Properties of Solids Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A brass wire of length \( 2.0 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) is stretched by a force of \( 300 \, \text{N} \). If the Young's modulus of brass is \( 9 \times 10^{10} \, \text{N/m}^2 \), what is the strain produced?

Stress: \( \text{Stress} = \frac{F}{A} = \frac{300}{2 \times 10^{-6}} = 1.5 \times 10^8 \, \text{N/m}^2 \).

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

Strain: \( \text{Strain} = \frac{\text{Stress}}{Y} = \frac{1.5 \times 10^8}{9 \times 10^{10}} \approx 1.67 \times 10^{-3} \).

\( 1.67 \times 10^{-3} \)
\( 2 \times 10^{-3} \)
\( 1.5 \times 10^{-3} \)
\( 1 \times 10^{-3} \)
1

A copper wire of length 2 m and cross-sectional area 2 × 10-6 m2 is stretched by a force of 200 N. If the elongation is 0.727 mm, what is the Young's modulus of copper?

Young's modulus: Y = (F L) / (A ΔL).

Substitute: ΔL = 0.727 × 10-3 m.

Y = (200 × 2) / (2 × 10-6 × 0.727 × 10-3) = 400 / (1.454 × 10-9) ≈ 2.75 × 1011 N/m2.

2 × 1011 N/m2
3 × 1011 N/m2
2.5 × 1011 N/m2
2.75 × 1011 N/m2
4

A water sample of volume \( 2.5 \, \text{litres} \) is compressed by a pressure of \( 2 \times 10^6 \, \text{N/m}^2 \). 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: \( \frac{\Delta V}{V} = -\frac{p}{B} = -\frac{2 \times 10^6}{2.2 \times 10^9} \approx -9.09 \times 10^{-4} \).

Magnitude: \( 9.09 \times 10^{-4} \).

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

A glass slab of volume \( 0.02 \, \text{m}^3 \) is subjected to a hydraulic pressure of \( 4 \times 10^6 \, \text{N/m}^2 \). If the bulk modulus of glass is \( 3.7 \times 10^{10} \, \text{N/m}^2 \), what is the fractional change in volume?

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

Rearrange: \( \frac{\Delta V}{V} = -\frac{p}{B} = -\frac{4 \times 10^6}{3.7 \times 10^{10}} \approx -1.08 \times 10^{-4} \).

Magnitude: \( 1.08 \times 10^{-4} \).

\( 1 \times 10^{-4} \)
\( 1.2 \times 10^{-4} \)
\( 5 \times 10^{-5} \)
\( 1.08 \times 10^{-4} \)
4

An aluminium wire of length \( 2.3 \, \text{m} \) and cross-sectional area \( 1.5 \times 10^{-6} \, \text{m}^2 \) is stretched by a force producing a stress of \( 2 \times 10^7 \, \text{N/m}^2 \). If the Young's modulus of aluminium is \( 7 \times 10^{10} \, \text{N/m}^2 \), what is the strain?

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

Strain: \( \text{Strain} = \frac{\text{Stress}}{Y} = \frac{2 \times 10^7}{7 \times 10^{10}} \approx 2.86 \times 10^{-4} \).

\( 2.5 \times 10^{-4} \)
\( 2.86 \times 10^{-4} \)
\( 3 \times 10^{-4} \)
\( 2 \times 10^{-4} \)
2

A glass block of volume 0.01 m3 is subjected to a hydraulic pressure of 4 × 106 N/m2. If the bulk modulus of glass is 3.7 × 1010 N/m2, what is the change in volume?

Bulk modulus: B = -p / (ΔV / V).

Rearrange: ΔV / V = -p / B = -(4 × 106) / (3.7 × 1010) ≈ -1.08 × 10-4.

Change in volume: ΔV = (ΔV / V) × V = -1.08 × 10-4 × 0.01 = -1.08 × 10-6 m3.

1 × 10-6 m3
1.08 × 10-6 m3
1.2 × 10-6 m3
1.5 × 10-6 m3
2

A copper cube of edge length 0.1 m is subjected to a hydraulic pressure of 5 × 106 N/m2. If the bulk modulus of copper is 1.4 × 1011 N/m2, what is the change in volume?

Bulk modulus: B = -p / (ΔV / V).

Volume: V = (0.1)3 = 0.001 m3.

Fractional change: ΔV / V = -p / B = -(5 × 106) / (1.4 × 1011) ≈ -3.57 × 10-5.

Change in volume: ΔV = (ΔV / V) × V = -3.57 × 10-5 × 0.001 ≈ -3.57 × 10-8 m3.

3 × 10-8 m3
3.57 × 10-8 m3
4 × 10-8 m3
5 × 10-8 m3
2

What is the primary reason solids are far less compressible than gases?

The tight coupling between neighboring atoms in solids results in a much higher resistance to volume change compared to gases, where molecules are weakly coupled.

Tight coupling between neighboring atoms
High temperature of solids
Low density of solids
High molecular weight of solids
1

A steel wire of length \( 1.5 \, \text{m} \) and cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) is stretched by a force of \( 100 \, \text{N} \). 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: \( \Delta L = \frac{100 \times 1.5}{1 \times 10^{-6} \times 2 \times 10^{11}} = \frac{150}{2 \times 10^5} = 7.5 \times 10^{-4} \, \text{m} = 0.75 \, \text{mm} \).

\( 0.75 \, \text{mm} \)
\( 1 \, \text{mm} \)
\( 0.5 \, \text{mm} \)
\( 1.5 \, \text{mm} \)
1

A brass wire of length 1.5 m and cross-sectional area 2.5 × 10-6 m2 is stretched by a force of 150 N. If the elongation is 0.25 mm, what is the Young's modulus of brass?

Young's modulus: Y = (F L) / (A ΔL).

Substitute: ΔL = 0.25 × 10-3 m.

Y = (150 × 1.5) / (2.5 × 10-6 × 0.25 × 10-3) = 225 / (6.25 × 10-10) = 3.6 × 1011 N/m2.

2 × 1011 N/m2
4 × 1011 N/m2
3 × 1011 N/m2
3.6 × 1011 N/m2
4

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