Mechanical Properties of Solids Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A copper wire of length 1.5 m and diameter 1 mm is subjected to a tensile force of 100 N. If the Young's modulus of copper is 1.1 × 1011 N/m2, what is the stress in the wire?

Stress is given by: Stress = F / A.

Area A = π r2, where r = (1/2) × 10-3 m = 5 × 10-4 m.

So, A = 3.14 × (5 × 10-4)2 = 3.14 × 25 × 10-8 = 7.85 × 10-7 m2.

Stress: 100 / (7.85 × 10-7) ≈ 1.27 × 108 N/m2.

1.27 × 108 N/m2
1.5 × 108 N/m2
1.0 × 108 N/m2
2.0 × 108 N/m2
1

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 solid aluminium sphere is subjected to a hydraulic pressure of 1 × 107 N/m2. If the bulk modulus of aluminium is 7.2 × 1010 N/m2, what is the fractional change in volume?

Bulk modulus: B = -p / (ΔV / V).

Rearrange: ΔV / V = -p / B.

Substitute: ΔV / V = -(1 × 107) / (7.2 × 1010) ≈ -1.39 × 10-4.

Magnitude: 1.39 × 10-4.

1.39 × 10-4
2 × 10-4
1 × 10-4
3 × 10-4
1

An aluminium wire of length \( 2.0 \, \text{m} \) and cross-sectional area \( 1.5 \times 10^{-6} \, \text{m}^2 \) is stretched by a force of \( 150 \, \text{N} \). If the Young's modulus of aluminium is \( 7 \times 10^{10} \, \text{N/m}^2 \), what is the elongation?

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

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

Substitute: \( \Delta L = \frac{150 \times 2.0}{1.5 \times 10^{-6} \times 7 \times 10^{10}} = \frac{300}{1.05 \times 10^5} \approx 2.86 \times 10^{-3} \, \text{m} = 2.86 \, \text{mm} \).

\( 2.5 \, \text{mm} \)
\( 2.86 \, \text{mm} \)
\( 3 \, \text{mm} \)
\( 2 \, \text{mm} \)
2

Why do materials like steel require a significantly larger force to produce small deformations compared to materials like rubber?

Steel has a much larger Young’s modulus compared to rubber, meaning it has greater stiffness and requires more force to produce the same strain.

Due to lower shear modulus
Due to higher density
Due to larger Young’s modulus
Due to greater ductility
3

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 glass slab of volume \( 0.03 \, \text{m}^3 \) is subjected to a hydraulic pressure of \( 6 \times 10^6 \, \text{N/m}^2 \). 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: \( \frac{\Delta V}{V} = -\frac{p}{B} = -\frac{6 \times 10^6}{3.7 \times 10^{10}} \approx -1.62 \times 10^{-4} \).

Magnitude: \( 1.62 \times 10^{-4} \).

\( 1.62 \times 10^{-4} \)
\( 2 \times 10^{-4} \)
\( 1.5 \times 10^{-4} \)
\( 1 \times 10^{-4} \)
1

A copper rod of length \( 2.0 \, \text{m} \) and cross-sectional area \( 2.5 \times 10^{-6} \, \text{m}^2 \) is subjected to a tensile force of \( 500 \, \text{N} \). If the Young's modulus of copper is \( 1.1 \times 10^{11} \, \text{N/m}^2 \), what is the strain produced?

Stress: \( \text{Stress} = \frac{F}{A} = \frac{500}{2.5 \times 10^{-6}} = 2 \times 10^8 \, \text{N/m}^2 \).

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

Strain: \( \text{Strain} = \frac{\text{Stress}}{Y} = \frac{2 \times 10^8}{1.1 \times 10^{11}} \approx 1.82 \times 10^{-3} \).

\( 1.82 \times 10^{-3} \)
\( 2 \times 10^{-3} \)
\( 1.5 \times 10^{-3} \)
\( 1 \times 10^{-3} \)
1

A glass slab of volume \( 0.05 \, \text{m}^3 \) is subjected to a hydraulic pressure of \( 8 \times 10^6 \, \text{N/m}^2 \). 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: \( \frac{\Delta V}{V} = -\frac{p}{B} = -\frac{8 \times 10^6}{3.7 \times 10^{10}} \approx -2.16 \times 10^{-4} \).

Magnitude: \( 2.16 \times 10^{-4} \).

\( 2.16 \times 10^{-4} \)
\( 2.5 \times 10^{-4} \)
\( 2 \times 10^{-4} \)
\( 1.5 \times 10^{-4} \)
1

A copper block of dimensions \( 0.3 \, \text{m} \times 0.2 \, \text{m} \times 0.05 \, \text{m} \) is subjected to a shearing force of \( 3 \times 10^4 \, \text{N} \). If the shear modulus of copper is \( 4.2 \times 10^{10} \, \text{N/m}^2 \), what is the shear strain?

Shear modulus: \( G = \frac{\text{Shear stress}}{\text{Shear strain}} \).

Shear stress: \( \text{Shear stress} = \frac{F}{A} \), \( A = 0.3 \times 0.2 = 0.06 \, \text{m}^2 \).

Shear stress: \( \frac{3 \times 10^4}{0.06} = 5 \times 10^5 \, \text{N/m}^2 \).

Shear strain: \( \text{Shear strain} = \frac{\text{Shear stress}}{G} = \frac{5 \times 10^5}{4.2 \times 10^{10}} \approx 1.19 \times 10^{-5} \).

\( 1 \times 10^{-5} \)
\( 1.5 \times 10^{-5} \)
\( 1.19 \times 10^{-5} \)
\( 2 \times 10^{-5} \)
3

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