Mechanical Properties of Solids Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A steel wire of length \( 2.5 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) is stretched by a force producing a strain of \( 3 \times 10^{-4} \). If the Young's modulus of steel is \( 2 \times 10^{11} \, \text{N/m}^2 \), what is the force applied?

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

Stress: \( \text{Stress} = Y \times \text{Strain} = 2 \times 10^{11} \times 3 \times 10^{-4} = 6 \times 10^7 \, \text{N/m}^2 \).

Force: \( F = \text{Stress} \times A = 6 \times 10^7 \times 2 \times 10^{-6} = 120 \, \text{N} \).

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

A copper rod of radius \( 0.02 \, \text{m} \) and length \( 1.0 \, \text{m} \) is subjected to a tensile force producing a stress of \( 2 \times 10^7 \, \text{N/m}^2 \). What is the force applied? (Take \( \pi \approx 3.14 \))

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

Area: \( A = \pi r^2 = 3.14 \times (0.02)^2 = 3.14 \times 4 \times 10^{-4} = 1.256 \times 10^{-3} \, \text{m}^2 \).

Force: \( F = \text{Stress} \times A = 2 \times 10^7 \times 1.256 \times 10^{-3} = 2.512 \times 10^4 \, \text{N} \).

\( 2.512 \times 10^4 \, \text{N} \)
\( 3 \times 10^4 \, \text{N} \)
\( 2 \times 10^4 \, \text{N} \)
\( 1.5 \times 10^4 \, \text{N} \)
1

A steel wire of length 2.5 m and cross-sectional area 3 × 10-6 m2 is stretched by a force producing a strain of 2 × 10-4. If the Young's modulus of steel is 2 × 1011 N/m2, what is the force applied?

Young's modulus: Y = Stress / Strain.

Stress: Stress = Y × Strain = 2 × 1011 × 2 × 10-4 = 4 × 107 N/m2.

Force: F = Stress × A = 4 × 107 × 3 × 10-6 = 120 N.

100 N
120 N
150 N
80 N
2

What does the steepness of the initial linear portion of a stress-strain curve indicate?

The steepness of the initial linear portion indicates a higher Young’s modulus, meaning the material is stiffer and requires more stress to produce a given strain.

Lower stiffness
Greater ductility
Higher shear modulus
Higher Young’s modulus
4

A copper wire of length 1 m and cross-sectional area 1 × 10-6 m2 is stretched by a force of 50 N. If the Young's modulus of copper is 1.1 × 1011 N/m2, what is the elongation?

Young's modulus: Y = (F L) / (A ΔL).

Rearrange: ΔL = (F L) / (A Y).

Substitute: ΔL = (50 × 1) / (1 × 10-6 × 1.1 × 1011) = 50 / (1.1 × 105) ≈ 4.55 × 10-4 m = 0.455 mm.

0.4 mm
0.455 mm
0.5 mm
0.6 mm
2

Which type of stress leads to a change in length of a body without changing its shape?

Longitudinal stress (tensile or compressive) causes a change in length along the direction of the applied force without altering the shape of the body.

Longitudinal stress
Shearing stress
Hydraulic stress
Volumetric stress
1

In the stress-strain behavior of a material, what occurs after the maximum stress point when the material is ductile?

For a ductile material, after the maximum stress point (ultimate tensile strength), the stress decreases with increasing strain as the material necks, leading to fracture.

The material returns to its original shape
The stress decreases leading to fracture
The material becomes more elastic
The material obeys Hooke’s law
2

A glass slab of volume \( 0.02 \, \text{m}^3 \) is subjected to a hydraulic pressure of \( 3 \times 10^6 \, \text{N/m}^2 \). 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: \( \frac{\Delta V}{V} = -\frac{p}{B} = -\frac{3 \times 10^6}{3.7 \times 10^{10}} \approx -8.11 \times 10^{-5} \).

Magnitude: \( 8.11 \times 10^{-5} \).

\( 7 \times 10^{-5} \)
\( 9 \times 10^{-5} \)
\( 6 \times 10^{-5} \)
\( 8.11 \times 10^{-5} \)
4

Why are I-shaped beams commonly used in construction for load-bearing applications?

I-shaped beams provide a large load-bearing surface and sufficient depth to reduce bending, minimizing buckling while reducing weight compared to solid beams.

They have a higher shear modulus
They provide a large load-bearing surface with reduced bending
They have a lower density
They exhibit greater ductility
2

A copper wire of length \( 1.6 \, \text{m} \) 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 elongation?

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

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

Substitute: \( \Delta L = \frac{180 \times 1.6}{1.8 \times 10^{-6} \times 1.1 \times 10^{11}} = \frac{288}{1.98 \times 10^5} \approx 1.45 \times 10^{-3} \, \text{m} = 1.45 \, \text{mm} \).

\( 1.45 \, \text{mm} \)
\( 1.5 \, \text{mm} \)
\( 1.2 \, \text{mm} \)
\( 1.8 \, \text{mm} \)
1

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