Question

A $1m$ long steel wire of cross-sectional area $1m{m}^2$ is extended by $1mm$. If $Y=2\times 10^{11}N/m^2$ , then the work done is :

• A. 0.1 J
• B. 0.2 J
• C. 0.3 J
• D. 0.4 J

Answer 1

Answer: (A)

When a wire is stretched, work is done against the interatomic forces. This work is stored in the wire in the form of elastic potential energy.
If length of wire is $L$, and area of cross-section is $A$, suppose on applying a force $F$ along length of wire, the length increases by
$I$.
Then Youngs modulus $Y=\frac{F/A}{L/L}$
$\Rightarrow F=\frac{YA}{L}l$
Work done $dW=F\times dl=\underset{0}{\overset{l}{\int }}\frac{YA}{L}ldl$
$=\frac{1}{2}YA\frac{l^2}{L}$
Given, $A=1m{m}^2=10^{-6}$,
$l=1mm=10^{-3}m,$
$Y=2\times 10^{11}N/m^2,L=1m$
$\therefore W=\frac{1}{2}\times 2\times 10^{11}\times 10^{-6}\times {\left(10^{-3}\right)}^{2}W$
$=0.1J$