Question

The work done on a wire of volume \( 2 \) cm\(^3 \) is \( 16 \times 10^2 \) J. If the Young's modulus of the material of the wire is \( 4 \times 10^{12} \) Nm\(^{-2} \), then the strain produced in the wire is

• A. \( 0.03 \) m
• B. \( 0.04 \) m
• C. \( 0.01 \) m
• D. \( 0.02 \) m

Hint:

Remember the formula for work done in terms of Young's modulus, strain, and volume.

Answer 1

The Correct Option is D

Let \( V \) be the volume of the wire, \( W \) be the work done, \( Y \) be the Young's modulus, and \( \epsilon \) be the strain. Given: \( V = 2 \text{ cm}^3 = 2 \times 10^{
-6} \text{ m}^3 \) \( W = 16 \times 10^2 \text{ J} \) \( Y = 4 \times 10^{12} \text{ Nm}^{
-2} \) The work done on the wire is given by \[ W = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume} \] We know that Young's modulus \( Y = \frac{\text{stress}}{\text{strain}} \), so \(\text{stress} = Y \times \text{strain} = Y \epsilon \). Then \[ W = \frac{1}{2} \times Y \epsilon \times \epsilon \times V = \frac{1}{2} Y \epsilon^2 V \] \[ \epsilon^2 = \frac{2W}{YV} \] \[ \epsilon = \sqrt{\frac{2W}{YV}} \] Substituting the given values: \[ \epsilon = \sqrt{\frac{2 \times 16 \times 10^2}{4 \times 10^{12} \times 2 \times 10^{-6}}} \] \[ \epsilon = \sqrt{\frac{32 \times 10^2}{8 \times 10^6}} \] \[ \epsilon = \sqrt{4 \times 10^{-4}} \] \[ \epsilon = 2 \times 10^{-2} = 0.02 \] The strain produced in the wire is \( 0.02 \).