Question

A wire of length $L$ and cross-sectional area $A$ is made of a material of Youngs modulus $Y$. The work done in stretching the wire by an amount $x$ is given by

• A. $\frac{YA{x}^2}{L}$
• B. $\frac{YA{x}^2}{2L}$
• C. $\frac{YA{L}^2}{x}$
• D. $\frac{YA{L}^2}{2x}$

Answer 1

Answer: (B)

In stretching a wire work is done against internal restoring forces.
This work is stored in the wire as elastic potential energy or strain energy.
If a force $F$ acts along the length $L$ of the wire of cross-section $A$ and stretches it by $x$, then
$Y=\frac{\text{ stress }}{\text{ strain }}=\frac{F/A}{x/L}=\frac{FL}{Ax}$
$\Rightarrow F=\frac{YAx}{L}$
So, the work done for an additional small increase $dx$ in length,
$dW=Fdx=\frac{YA}{L}xdx$
Hence, the total work done in increasing the length by $l$,
$W=\underset{0}{\overset{x}{\int }}dW=\underset{0}{\overset{x}{\int }}Fdx=\underset{0}{\overset{x}{\int }}\frac{YA}{L}xdx=\frac{1}{2}\frac{YA}{L}{x}^2$
This work done is stored in the wire.
$\therefore$ Energy stored in wire
$U=\frac{1}{2}\frac{YA{x}^2}{L}$