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

Answer: (A)

$\frac{1}{2}(A.L)\left( \frac{Yx}{L} \right)\left( \frac{x}{L} \right)$

Answer 2

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