Question

A $1\,m$ long steel wire of cross-sectional area $ 1\,mm^{2}$ is extended by $1\, mm$. 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 $Y=\frac{F / A}{L / L}$
$\Rightarrow F=\frac{Y A}{L} l$
Work done $d W=F \times d l=\int\limits_{0}^{l} \frac{Y A}{L} l d l$
$=\frac{1}{2} Y A \frac{l^{2}}{L}$
Given, $A=1\, mm ^{2}=10^{-6}$,
$l=1\, m m=10^{-3} m,$
$Y=2 \times 10^{11} N / m^{2}, L=1\, m$
$\therefore W=\frac{1}{2} \times 2 \times 10^{11} \times 10^{-6} \times\left(10^{-3}\right)^{2} W$
$=0.1 \,J$