Mechanical Properties of Solids Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

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

Why do engineers prefer materials with a large Young’s modulus for designing columns and beams?

Materials with a large Young’s modulus are stiffer, requiring greater force to produce small deformations, which ensures better resistance to bending and stretching.

They are more ductile
They have lower density
They are more brittle
They provide greater stiffness
4

In designing bridges, why is it critical to understand the elastic properties of materials?

Understanding elastic properties ensures that materials can withstand various loads (traffic, wind, weight) without undergoing permanent deformation, maintaining structural integrity.

To increase the weight of the bridge
To reduce the cost of materials
To enhance thermal conductivity
To ensure structural integrity
4

An aluminium block of dimensions \( 0.3 \, \text{m} \times 0.2 \, \text{m} \times 0.05 \, \text{m} \) is subjected to a shearing force of \( 1.5 \times 10^4 \, \text{N} \). If the shear modulus of aluminium is \( 2.5 \times 10^{10} \, \text{N/m}^2 \), 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: \( \frac{1.5 \times 10^4}{0.06} = 2.5 \times 10^5 \, \text{N/m}^2 \).

Shear strain: \( \text{Shear strain} = \frac{\text{Shear stress}}{G} = \frac{2.5 \times 10^5}{2.5 \times 10^{10}} = 1 \times 10^{-5} \).

\( 5 \times 10^{-6} \)
\( 1.5 \times 10^{-5} \)
\( 2 \times 10^{-5} \)
\( 1 \times 10^{-5} \)
4

A steel rod of length 0.8 m and cross-sectional area 2 × 10-5 m2 is compressed by a force of 800 N. If the Young's modulus of steel is 2 × 1011 N/m2, what is the compression?

Young's modulus: Y = (F L) / (A ΔL).

Rearrange: ΔL = (F L) / (A Y).

Substitute: ΔL = (800 × 0.8) / (2 × 10-5 × 2 × 1011) = 640 / (4 × 106) = 1.6 × 10-4 m = 0.16 mm.

0.1 mm
0.16 mm
0.2 mm
0.25 mm
2

A cylindrical rod of length 0.5 m and radius 0.01 m is compressed by a force of 5000 N. If the compressive stress is 5 × 106 N/m2, what is the cross-sectional area of the rod?

Stress: Stress = F / A.

Rearrange: A = F / Stress.

Substitute: A = 5000 / (5 × 106) = 10-3 m2.

10-3 m2
10-2 m2
5 × 10-3 m2
2 × 10-3 m2
1

Why do engineers prefer materials with a large Young’s modulus for designing columns and beams?

Materials with a large Young’s modulus are stiffer, requiring greater force to produce small deformations, which ensures better resistance to bending and stretching.

They are more ductile
They have lower density
They are more brittle
They provide greater stiffness
4

An aluminium wire of length \( 2.5 \, \text{m} \) 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 stress?

Stress: \( \text{Stress} = \frac{F}{A} = \frac{150}{1.5 \times 10^{-6}} = 1 \times 10^8 \, \text{N/m}^2 \).

\( 1 \times 10^8 \, \text{N/m}^2 \)
\( 1.5 \times 10^8 \, \text{N/m}^2 \)
\( 2 \times 10^8 \, \text{N/m}^2 \)
\( 5 \times 10^7 \, \text{N/m}^2 \)
1

A glass slab of volume 0.005 m3 is compressed by a pressure of 3 × 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 = -(3 × 106) / (3.7 × 1010) ≈ -8.11 × 10-5.

Change in volume: ΔV = (ΔV / V) × V = -8.11 × 10-5 × 0.005 ≈ -4.05 × 10-7 m3.

3 × 10-7 m3
5 × 10-7 m3
4 × 10-7 m3
4.05 × 10-7 m3
4

A steel wire of length \( 2.5 \, \text{m} \) and cross-sectional area \( 3.5 \times 10^{-6} \, \text{m}^2 \) is stretched by a force of \( 350 \, \text{N} \). If the elongation is \( 0.2 \, \text{mm} \), what is the Young's modulus of steel?

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

Substitute: \( \Delta L = 0.2 \times 10^{-3} \, \text{m} \).

\( Y = \frac{350 \times 2.5}{3.5 \times 10^{-6} \times 0.2 \times 10^{-3}} = \frac{875}{7 \times 10^{-10}} \approx 1.25 \times 10^{12} \, \text{N/m}^2 \).

\( 1 \times 10^{12} \, \text{N/m}^2 \)
\( 1.5 \times 10^{12} \, \text{N/m}^2 \)
\( 1.2 \times 10^{12} \, \text{N/m}^2 \)
\( 1.25 \times 10^{12} \, \text{N/m}^2 \)
4

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