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

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