Question

The work done in increasing the length of a $1\text{m}$ long wire of cross-sectional area $1{\text{mm}}^2$ through $1\text{mm}$ will be
[Given, $Y=2\times {10}^{11}{\text{N/m}}^2$]

• A. $250\text{J}$
• B. $10\text{J}$
• C. $5\text{J}$
• D. $0.1\text{J}$

Answer 1

The correct option is D $0.1\text{J}$

Work done by external force on the wire in increasing its length,
$W=\frac{1}{2}F\times \Delta L$
$\Rightarrow W=\frac{1}{2}\times AY\frac{\Delta L}{L}\times \Delta L$
$\Rightarrow W=\frac{1}{2}\left(AL\right)\times Y\times {\left(\frac{\Delta L}{L}\right)}^2$
Or, $W=\frac{AY\Delta L^2}{2L}$
$\Rightarrow W=\frac{1\times {10}^{-6}\times 2\times {10}^{11}\times {10}^{-6}}{2\times 1}={10}^{-1}$
$\therefore W=0.1\text{J}$