Question

A copper wire of cross-sectional area 0.01 cm\(^2\) is subjected to a tension of 22 N. If Young’s modulus and Poisson’s ratio of copper are \(1.1 \times 10^{11}\) Nm\(^2\) and 0.32 respectively, then the change in the cross-sectional area of the wire is:

• A. \( 1.28 \times 10^{-6} \, \text{cm}^2 \)
• B. \( 1.16 \times 10^{-6} \, \text{cm}^2 \)
• C. \( 0.64 \times 10^{-6} \, \text{cm}^2 \)
• D. \( 0.58 \times 10^{-6} \, \text{cm}^2 \)

Hint:

When calculating changes in dimensions due to mechanical forces, use Poisson’s ratio to relate longitudinal and lateral strains.

Answer 1

The Correct Option is A

- Stress: \[ \sigma = \frac{F}{A} = \frac{22}{1 \times 10^{-6}} = 2.2 \times 10^{7} \,\text{N/m}^2 \]
- Longitudinal strain: \[ \epsilon = \frac{\sigma}{E} = \frac{2.2 \times 10^{7}}{1.1 \times 10^{11}} = 2 \times 10^{-4} \]
- Lateral strain: \[ \epsilon_{\text{lateral}} = -\nu \epsilon = -0.32 \times 2 \times 10^{-4} = -6.4 \times 10^{-5} \]
- Fractional change in area (approx): \[ \frac{\Delta A}{A} = 2 \epsilon_{\text{lateral}} = 2(-6.4 \times 10^{-5}) = -1.28 \times 10^{-4} \]
- Change in area: \[ \Delta A = A \times \frac{\Delta A}{A} = 1 \times 10^{-6} \times (-1.28 \times 10^{-4}) = -1.28 \times 10^{-10} \,\text{m}^2 \]
- Converting to \(\text{cm}^2\): \[ \Delta A = -1.28 \times 10^{-6} \,\text{cm}^2 \]