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  4. Mechanical Properties Of Solids Chapter-Wise Test 3

Mechanical Properties Of Solids Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

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

A glass slab of volume \( 0.03 \, \text{m}^3 \) is subjected to a hydraulic pressure of \( 5 \times 10^6 \, \text{N/m}^2 \). 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: \( \frac{\Delta V}{V} = -\frac{p}{B} = -\frac{5 \times 10^6}{3.7 \times 10^{10}} \approx -1.35 \times 10^{-4} \).

Change in volume: \( \Delta V = \frac{\Delta V}{V} \times V = -1.35 \times 10^{-4} \times 0.03 \approx -4.05 \times 10^{-6} \, \text{m}^3 \).

\( 4 \times 10^{-6} \, \text{m}^3 \)
\( 4.05 \times 10^{-6} \, \text{m}^3 \)
\( 4.5 \times 10^{-6} \, \text{m}^3 \)
\( 3.5 \times 10^{-6} \, \text{m}^3 \)
2

A steel rod of length 1 m and cross-sectional area 1 × 10-5 m2 is subjected to a tensile force of 1000 N. If the Young's modulus of steel is 2 × 1011 N/m2, what is the strain produced?

Stress: Stress = F / A = 1000 / (1 × 10-5) = 1 × 108 N/m2.

Young's modulus: Y = Stress / Strain.

Strain: Strain = Stress / Y = (1 × 108) / (2 × 1011) = 5 × 10-4.

5 × 10-4
1 × 10-3
2.5 × 10-4
7.5 × 10-4
1

A steel wire of length \( 2.0 \, \text{m} \) and cross-sectional area \( 3 \times 10^{-6} \, \text{m}^2 \) is stretched by \( 0.4 \, \text{mm} \). 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: \( \Delta L = 0.4 \times 10^{-3} \, \text{m} \).

\( F = \frac{2 \times 10^{11} \times 3 \times 10^{-6} \times 0.4 \times 10^{-3}}{2} = \frac{240}{2} = 120 \, \text{N} \).

\( 100 \, \text{N} \)
\( 150 \, \text{N} \)
\( 120 \, \text{N} \)
\( 130 \, \text{N} \)
3

A glass slab of volume \( 0.02 \, \text{m}^3 \) is subjected to a hydraulic pressure of \( 5 \times 10^6 \, \text{N/m}^2 \). 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: \( \frac{\Delta V}{V} = -\frac{p}{B} = -\frac{5 \times 10^6}{3.7 \times 10^{10}} \approx -1.35 \times 10^{-4} \).

Magnitude: \( 1.35 \times 10^{-4} \).

\( 1.35 \times 10^{-4} \)
\( 2 \times 10^{-4} \)
\( 1 \times 10^{-4} \)
\( 1.5 \times 10^{-4} \)
1

A steel wire of length \( 2.8 \, \text{m} \) and cross-sectional area \( 4 \times 10^{-6} \, \text{m}^2 \) is stretched by a force producing a strain of \( 1.5 \times 10^{-4} \). If the Young's modulus of steel is \( 2 \times 10^{11} \, \text{N/m}^2 \), what is the elongation?

Strain: \( \text{Strain} = \frac{\Delta L}{L} \).

Rearrange: \( \Delta L = \text{Strain} \times L = 1.5 \times 10^{-4} \times 2.8 = 4.2 \times 10^{-4} \, \text{m} = 0.42 \, \text{mm} \).

\( 0.4 \, \text{mm} \)
\( 0.42 \, \text{mm} \)
\( 0.5 \, \text{mm} \)
\( 0.3 \, \text{mm} \)
2

A copper wire of length \( 1.8 \, \text{m} \) and cross-sectional area \( 3 \times 10^{-6} \, \text{m}^2 \) is stretched by a force producing a strain of \( 3 \times 10^{-4} \). 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: \( \text{Stress} = Y \times \text{Strain} = 1.1 \times 10^{11} \times 3 \times 10^{-4} = 3.3 \times 10^7 \, \text{N/m}^2 \).

\( 3 \times 10^7 \, \text{N/m}^2 \)
\( 3.3 \times 10^7 \, \text{N/m}^2 \)
\( 3.5 \times 10^7 \, \text{N/m}^2 \)
\( 2.5 \times 10^7 \, \text{N/m}^2 \)
2

A water sample of volume 2 litres is compressed by a pressure of 2 × 106 N/m2. If the bulk modulus of water is 2.2 × 109 N/m2, what is the fractional change in volume?

Bulk modulus: B = -p / (ΔV / V).

Rearrange: ΔV / V = -p / B = -(2 × 106) / (2.2 × 109) ≈ -9.09 × 10-4.

Magnitude: 9.09 × 10-4.

8 × 10-4
1 × 10-3
9.09 × 10-4
7 × 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 change in volume?

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

Rearrange: \( \frac{\Delta V}{V} = -\frac{p}{B} = -\frac{3 \times 10^6}{2.2 \times 10^9} \approx -1.36 \times 10^{-3} \).

Volume: \( V = 1.5 \, \text{litres} = 1.5 \times 10^{-3} \, \text{m}^3 \).

Change in volume: \( \Delta V = \frac{\Delta V}{V} \times V = -1.36 \times 10^{-3} \times 1.5 \times 10^{-3} \approx -2.04 \times 10^{-6} \, \text{m}^3 \).

\( 2 \times 10^{-6} \, \text{m}^3 \)
\( 2.04 \times 10^{-6} \, \text{m}^3 \)
\( 2.5 \times 10^{-6} \, \text{m}^3 \)
\( 1.5 \times 10^{-6} \, \text{m}^3 \)
2

What is the key difference between elastic and plastic deformation in a material?

Elastic deformation is reversible, meaning the material returns to its original shape after the load is removed, while plastic deformation results in permanent changes.

Elastic deformation occurs only in solids
Plastic deformation occurs at lower stress
Elastic deformation involves higher energy
Elastic deformation is reversible, plastic is not
4

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