Question

A wire of length \(100\text{ cm}\) is made of a material of Young’s modulus \(1.6\times10^{11}\text{ Nm}^{-2}\). If work done in stretching this wire by \(0.1\text{ cm}\) is \(2\text{ J}\), then the area of cross-section of the wire (in \(10^{-5}\text{ m}^2\)) is

• A. \(5.0\)
• B. \(1.25\)
• C. \(1.5\)
• D. \(2.5\)

Hint:

Memorize elastic potential energy formula: \[ W=\frac12\frac{YA(\Delta L)^2}{L} \] This directly solves most stretching problems.

Answer 1

The Correct Option is D

Concept: Work done in stretching elastic wire \[ W=\frac12 F\Delta L \] From Young modulus \[ Y=\frac{FL}{A\Delta L} \] Hence force \[ F=\frac{YA\Delta L}{L} \] Substituting in work formula \[ W=\frac12\frac{YA(\Delta L)^2}{L} \]

Step 1: Write known values Length \[ L=100cm=1m \] Extension \[ \Delta L=0.1cm=10^{-3}m \] Young modulus \[ Y=1.6\times10^{11} \] Work done \[ W=2J \]

Step 2: Substitute \[ 2=\frac12\frac{(1.6\times10^{11})A(10^{-3})^2}{1} \] \[ 2=\frac12(1.6\times10^5)A \] \[ 2=0.8\times10^5A \] \[ A=\frac{2}{8\times10^4} \] \[ A=2.5\times10^{-5} \] Hence answer is \[ \boxed{2.5} \]