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Mechanical Properties of Solids — NEET & JEE MCQs with answers
This chapter studies how solid bodies respond to applied forces through the concepts of stress, strain, and elastic moduli (Young's modulus, shear modulus, bulk modulus). It covers Hooke's law, the stress–strain curve, Poisson's ratio, elastic potential energy, and practical applications of the elastic behaviour of materials.
Topics in this chapter
- 8.1 Introduction
- 8.2 Stress and Strain
- 8.3 Hooke's Law and Elastic Moduli Overview
- 8.4 Stress–Strain Curve
- 8.5 Elastic Moduli
- 8.6 Applications of Elastic Behaviour of Materials
Free practice questions
A copper wire and a steel wire of equal length and equal cross-section are subjected to the same tension. Given $Y_{steel}>Y_{copper}$, which stretches more?
- A. the copper wire
- B. the steel wire
- C. both stretch equally
- D. neither stretches
Show answer
Answer: A — For equal $F$, $L$ and $A$, $\Delta L\propto1/Y$. Since copper has the smaller $Y$, the copper wire stretches more.
NCERT Class 11 Physics · Ch 8 · 8.5 Elastic Moduli
According to NCERT, the shear component of stress at the base of a mountain of height $h$ is approximately:
- A. $h\rho g/2$
- B. $2h\rho g$
- C. $h\rho g$
- D. zero
Show answer
Answer: C — NCERT states there is a shear component approximately equal to $h\rho g$ itself at the base of the mountain.
NCERT Class 11 Physics · Ch 8 · 8.6 Applications of Elastic Behaviour of Materials
Which one of the following sequences correctly describes what happens to a helical spring as per NCERT 8.1 when it is gently pulled at its ends and then released?
- A. Length decreases → on release it stays compressed
- B. Length increases slightly → on release it stays permanently elongated
- C. Length increases slightly → on release it regains original size and shape
- D. No change in length → on release it breaks
Show answer
Answer: C — NCERT 8.1 describes: on pulling, the length of the spring increases slightly; on releasing the ends, it regains its original size and shape — demonstrating elasticity.
NCERT Class 11 Physics · Ch 8 · 8.1 Introduction
Two materials P and Q both obey Hooke's law. For the same strain, P develops twice the stress of Q. The modulus of elasticity of P compared to Q is:
- A. twice that of Q
- B. half that of Q
- C. four times that of Q
- D. equal to that of Q
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Answer: A — Modulus of elasticity $k = \text{stress}/\text{strain}$. For the same strain, if P's stress is twice Q's, then P's modulus of elasticity is twice that of Q.
NCERT Class 11 Physics · Ch 8 · 8.3 Hooke's Law and Elastic Moduli Overview
The ratio of change in length $\Delta L$ to the original length $L$ of a body is known as:
- A. shearing strain
- B. volume strain
- C. longitudinal stress
- D. longitudinal strain
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Answer: D — Section 8.2, Eq. (8.2): Longitudinal strain $= \Delta L / L$, the ratio of the change in length to the original length.
NCERT Class 11 Physics · Ch 8 · 8.2 Stress and Strain
Which of the following is an example of an elastomer given in NCERT?
- A. Structural steel
- B. Concrete
- C. Glass
- D. The elastic tissue of aorta
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Answer: D — NCERT cites the tissue of aorta and rubber as elastomers that can be stretched to large strains.
NCERT Class 11 Physics · Ch 8 · 8.4 Stress-Strain Curve
For a given material, which is more effective in reducing the bending of a beam under a given load?
- A. Decreasing the depth $d$
- B. Increasing the length $l$
- C. Increasing the breadth $b$ rather than the depth $d$
- D. Increasing the depth $d$ rather than the breadth $b$
Show answer
Answer: D — Since $\delta \propto d^{-3}$ but only $\propto b^{-1}$, increasing the depth $d$ rather than the breadth $b$ is more effective in reducing bending.
NCERT Class 11 Physics · Ch 8 · 8.6 Applications of Elastic Behaviour of Materials
A wire of diameter $0.5\,\text{mm}$ is loaded with a normal force such that the stress is $4 \times 10^{8}\,\text{N m}^{-2}$. Taking $\pi = 3.14$, the applied force is approximately:
- A. $39.3\,\text{N}$
- B. $785\,\text{N}$
- C. $157\,\text{N}$
- D. $78.5\,\text{N}$
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Answer: D — Radius $r = 0.25\,\text{mm} = 2.5 \times 10^{-4}\,\text{m}$; area $A = \pi r^2 = 3.14 \times (2.5 \times 10^{-4})^2 = 3.14 \times 6.25 \times 10^{-8} = 1.96 \times 10^{-7}\,\text{m}^2$. Force $F = \text{stress} \times A = (4 \times 10^{8}) \times (1.96 \times 10^{-7}) \approx 78.5\,\text{N}$.
NCERT Class 11 Physics · Ch 8 · 8.2 Stress and Strain
The shear modulus $G$ relates the shearing stress $\sigma_s$ and angular displacement $\theta$ by $\sigma_s=G\theta$. If $G=2.5\times10^{10}\text{ Pa}$ and $\theta=4\times10^{-5}\text{ rad}$, then $\sigma_s$ equals
- A. $1\times10^{5}\text{ N m}^{-2}$
- B. $1\times10^{6}\text{ N m}^{-2}$
- C. $4\times10^{6}\text{ N m}^{-2}$
- D. $2.5\times10^{6}\text{ N m}^{-2}$
Show answer
Answer: B — $\sigma_s=G\theta=2.5\times10^{10}\times4\times10^{-5}=1\times10^{6}\text{ N m}^{-2}$.
NCERT Class 11 Physics · Ch 8 · 8.5 Elastic Moduli
Knowledge of the elastic properties of materials like steel and concrete is described in NCERT 8.1 as essential primarily in
- A. Engineering design such as buildings
- B. Cooking food
- C. Chemical reactions
- D. Measuring temperature
Show answer
Answer: A — NCERT 8.1 states that the elastic behaviour of materials plays an important role in engineering design; for example, while designing a building, knowledge of elastic properties of materials like steel, concrete, etc. is essential.
NCERT Class 11 Physics · Ch 8 · 8.1 Introduction
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