Mechanical Properties Of Solids Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A rectangular block of aluminium with dimensions 0.5 m × 0.2 m × 0.1 m is subjected to a shearing force of 6 × 104 N. If the shear modulus of aluminium is 2.5 × 1010 N/m2, what is the shear strain?

Shear modulus: G = (Shear stress) / (Shear strain).

Shear stress: Shear stress = F / A, A = 0.5 × 0.2 = 0.1 m2.

So, Shear stress = (6 × 104) / 0.1 = 6 × 105 N/m2.

Shear strain: Shear strain = Shear stress / G = (6 × 105) / (2.5 × 1010) = 2.4 × 10-5.

2 × 10-5
2.4 × 10-5
3 × 10-5
1.5 × 10-5
2

In the stress-strain behavior of a ductile material, what happens after the maximum stress the material can endure?

After the maximum stress (ultimate tensile strength), a ductile material experiences a decrease in stress with additional strain, leading to eventual fracture.

The material begins to fracture
The material returns to its original shape
The proportionality between stress and strain holds
The material undergoes elastic deformation only
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

An aluminium wire of length \( 1.8 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) is stretched by a force producing a strain of \( 2 \times 10^{-4} \). If the Young's modulus of aluminium is \( 7 \times 10^{10} \, \text{N/m}^2 \), what is the force applied?

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

Stress: \( \text{Stress} = Y \times \text{Strain} = 7 \times 10^{10} \times 2 \times 10^{-4} = 1.4 \times 10^7 \, \text{N/m}^2 \).

Force: \( F = \text{Stress} \times A = 1.4 \times 10^7 \times 2 \times 10^{-6} = 28 \, \text{N} \).

\( 20 \, \text{N} \)
\( 30 \, \text{N} \)
\( 28 \, \text{N} \)
\( 25 \, \text{N} \)
3

A brass block of dimensions \( 0.5 \, \text{m} \times 0.2 \, \text{m} \times 0.05 \, \text{m} \) is subjected to a shearing force of \( 2 \times 10^4 \, \text{N} \). If the shear modulus of brass is \( 3.6 \times 10^{10} \, \text{N/m}^2 \), what is the shear strain?

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

Shear stress: \( \text{Shear stress} = \frac{F}{A} \), \( A = 0.5 \times 0.2 = 0.1 \, \text{m}^2 \).

Shear stress: \( \frac{2 \times 10^4}{0.1} = 2 \times 10^5 \, \text{N/m}^2 \).

Shear strain: \( \text{Shear strain} = \frac{\text{Shear stress}}{G} = \frac{2 \times 10^5}{3.6 \times 10^{10}} \approx 5.56 \times 10^{-6} \).

\( 5 \times 10^{-6} \)
\( 6 \times 10^{-6} \)
\( 4 \times 10^{-6} \)
\( 5.56 \times 10^{-6} \)
4

A water sample of volume \( 3 \, \text{litres} \) is compressed by a pressure of \( 4 \times 10^6 \, \text{N/m}^2 \). 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: \( \frac{\Delta V}{V} = -\frac{p}{B} = -\frac{4 \times 10^6}{2.2 \times 10^9} \approx -1.82 \times 10^{-3} \).

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

Change in volume: \( \Delta V = \frac{\Delta V}{V} \times V = -1.82 \times 10^{-3} \times 3 \times 10^{-3} \approx -5.45 \times 10^{-6} \, \text{m}^3 \).

\( 5 \times 10^{-6} \, \text{m}^3 \)
\( 5.45 \times 10^{-6} \, \text{m}^3 \)
\( 6 \times 10^{-6} \, \text{m}^3 \)
\( 4 \times 10^{-6} \, \text{m}^3 \)
2

A steel block of dimensions 0.2 m × 0.1 m × 0.05 m is subjected to a shearing force of 1 × 104 N. If the shear modulus of steel is 8.4 × 1010 N/m2, what is the shear strain?

Shear modulus: G = (Shear stress) / (Shear strain).

Shear stress: Shear stress = F / A, A = 0.2 × 0.1 = 0.02 m2.

Shear stress = (1 × 104) / 0.02 = 5 × 105 N/m2.

Shear strain: Shear strain = Shear stress / G = (5 × 105) / (8.4 × 1010) ≈ 5.95 × 10-6.

5 × 10-6
6.5 × 10-6
4 × 10-6
5.95 × 10-6
4

What distinguishes a ductile material from a brittle material in terms of stress-strain behavior?

A ductile material undergoes significant plastic deformation before fracture, allowing it to withstand larger strains, while a brittle material fractures with minimal plastic deformation.

Ductile materials have a larger shear modulus
Ductile materials undergo significant plastic deformation before fracture
Ductile materials break at a lower stress
Ductile materials have a lower bulk modulus
2

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 brass block of dimensions \( 0.4 \, \text{m} \times 0.3 \, \text{m} \times 0.1 \, \text{m} \) is subjected to a shearing force of \( 3 \times 10^4 \, \text{N} \). If the shear modulus of brass is \( 3.6 \times 10^{10} \, \text{N/m}^2 \), 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: \( A = 0.4 \times 0.3 = 0.12 \, \text{m}^2 \), \( L = 0.1 \, \text{m} \).

Substitute: \( \Delta x = \frac{3 \times 10^4 \times 0.1}{0.12 \times 3.6 \times 10^{10}} = \frac{3000}{4.32 \times 10^9} \approx 6.94 \times 10^{-7} \, \text{m} \).

\( 6 \times 10^{-7} \, \text{m} \)
\( 7.5 \times 10^{-7} \, \text{m} \)
\( 6.94 \times 10^{-7} \, \text{m} \)
\( 5 \times 10^{-7} \, \text{m} \)
3

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