Mechanical Properties of Solids Chapter-Wise Test 10

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which type of stress causes a change in the shape of a body without altering its volume?

Shearing stress causes a change in the shape of a body by producing relative displacement of opposite faces (pure shear) without changing its volume.

Tensile stress
Compressive stress
Shearing stress
Hydraulic stress
3

A copper block of dimensions \( 0.5 \, \text{m} \times 0.3 \, \text{m} \times 0.1 \, \text{m} \) is subjected to a shearing force of \( 6 \times 10^4 \, \text{N} \). If the shear modulus of copper is \( 4.2 \times 10^{10} \, \text{N/m}^2 \), 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: \( \Delta x = \frac{6 \times 10^4 \times 0.1}{0.15 \times 4.2 \times 10^{10}} = \frac{6000}{6.3 \times 10^9} \approx 9.52 \times 10^{-7} \, \text{m} \).

\( 9 \times 10^{-7} \, \text{m} \)
\( 9.52 \times 10^{-7} \, \text{m} \)
\( 1 \times 10^{-6} \, \text{m} \)
\( 8 \times 10^{-7} \, \text{m} \)
2

A steel wire of length \( 3.2 \, \text{m} \) and cross-sectional area \( 5 \times 10^{-6} \, \text{m}^2 \) is stretched by \( 0.8 \, \text{mm} \). 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: \( \Delta L = 0.8 \times 10^{-3} \, \text{m} \).

\( F = \frac{2 \times 10^{11} \times 5 \times 10^{-6} \times 0.8 \times 10^{-3}}{3.2} = \frac{800}{3.2} = 250 \, \text{N} \).

\( 200 \, \text{N} \)
\( 300 \, \text{N} \)
\( 250 \, \text{N} \)
\( 150 \, \text{N} \)
3

Why is steel often chosen over materials like copper for heavy-duty structural applications?

Steel has a higher Young’s modulus (\( \approx 200 \times 10^9 \, \text{N/m}^2 \)) compared to copper (\( \approx 110 \times 10^9 \, \text{N/m}^2 \)), requiring a larger force to produce the same deformation, making it more suitable for resisting bending and stretching.

Higher Young’s modulus
Lower density
Greater ductility
Higher thermal conductivity
1

A glass slab of volume \( 0.015 \, \text{m}^3 \) 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: \( \Delta V = \frac{\Delta V}{V} \times V = -5.41 \times 10^{-5} \times 0.015 \approx -8.11 \times 10^{-7} \, \text{m}^3 \).

\( 8 \times 10^{-7} \, \text{m}^3 \)
\( 9 \times 10^{-7} \, \text{m}^3 \)
\( 8.11 \times 10^{-7} \, \text{m}^3 \)
\( 7 \times 10^{-7} \, \text{m}^3 \)
3

A copper wire of length \( 2.2 \, \text{m} \) and cross-sectional area \( 1.8 \times 10^{-6} \, \text{m}^2 \) is stretched by a force of \( 180 \, \text{N} \). If the Young's modulus of copper is \( 1.1 \times 10^{11} \, \text{N/m}^2 \), what is the strain?

Stress: \( \text{Stress} = \frac{F}{A} = \frac{180}{1.8 \times 10^{-6}} = 1 \times 10^8 \, \text{N/m}^2 \).

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

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

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

A water sample of volume \( 1.5 \, \text{litres} \) is compressed by a pressure of \( 3 \times 10^6 \, \text{N/m}^2 \). 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: \( \frac{\Delta V}{V} = -\frac{p}{B} = -\frac{3 \times 10^6}{2.2 \times 10^9} \approx -1.36 \times 10^{-3} \).

Magnitude: \( 1.36 \times 10^{-3} \).

\( 1 \times 10^{-3} \)
\( 1.5 \times 10^{-3} \)
\( 2 \times 10^{-3} \)
\( 1.36 \times 10^{-3} \)
4

A copper wire of length 2 m and cross-sectional area 3 × 10-6 m2 is stretched by a force of 150 N. If the strain produced is 1 × 10-3, what is the Young's modulus of copper?

Young's modulus: Y = Stress / Strain.

Stress: Stress = F / A = 150 / (3 × 10-6) = 5 × 107 N/m2.

Strain: 1 × 10-3.

So, Y = (5 × 107) / (1 × 10-3) = 5 × 1010 N/m2.

4 × 1010 N/m2
6 × 1010 N/m2
5 × 1010 N/m2
7 × 1010 N/m2
3

A steel wire of length \( 3.0 \, \text{m} \) 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 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: \( \Delta L = \frac{200 \times 3.0}{2 \times 10^{-6} \times 2 \times 10^{11}} = \frac{600}{4 \times 10^5} = 1.5 \times 10^{-3} \, \text{m} = 1.5 \, \text{mm} \).

\( 1.5 \, \text{mm} \)
\( 2 \, \text{mm} \)
\( 1 \, \text{mm} \)
\( 3 \, \text{mm} \)
1

A glass slab of volume \( 0.025 \, \text{m}^3 \) 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 change in volume?

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

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

Change in volume: \( \Delta V = \frac{\Delta V}{V} \times V = -1.08 \times 10^{-4} \times 0.025 \approx -2.7 \times 10^{-6} \, \text{m}^3 \).

\( 2.5 \times 10^{-6} \, \text{m}^3 \)
\( 2.7 \times 10^{-6} \, \text{m}^3 \)
\( 3 \times 10^{-6} \, \text{m}^3 \)
\( 2 \times 10^{-6} \, \text{m}^3 \)
2

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