Question

A wire of length $ L $ and area of cross-section $ A $ is stretched through a distance $ x $ metre by applying a force $ F $ along length, then the work done in this process is: (Y is Youngs modulus of the material)

• A. $ \frac{1}{2}(A.L)\left( \frac{Yx}{L} \right)\left( \frac{x}{L} \right) $
• B. $ (A.L)(YL)\left( \frac{x}{L} \right) $
• C. $ 2(A.L)(YL)\left( \frac{x}{L} \right) $
• D. $ 3(A.L)(YL)\left( \frac{x}{L} \right) $

Answer 1

The Correct Option is A

Work done $ =\frac{1}{2}\times $ stress $ \times $ strain $ \times $ volume
or $ W=\frac{1}{2}Y\frac{x}{L}\times \frac{x}{L}\times AL $
$ =\frac{1}{2}(AL)\,\left( Y\frac{x}{L} \right)\,\left( \frac{x}{L} \right) $