JEE Main 2023 — Mechanical Properties of Solids Question with Solution

The formula to calculate the Young's modulus of the material is given by

$\begin{array}{rcl}Y& =& \frac{\mathrm{stress}}{\mathrm{strain}}\\ & =& \frac{\mathrm{stress}}{{\displaystyle \frac{∆l}{L}}} ...\left(1\right)\end{array}$

The formula to calculate the potential energy $\left(U\right)$ stored into the wire is given by

$\begin{array}{rcl}U& =& \frac{1}{2}\times \mathrm{stress}\times \mathrm{strain}\times \mathrm{volume}\\ & =& \frac{1}{2}\times \mathrm{stress}\times \frac{∆l}{L}\times A\times L\\ & =& \frac{1}{2}\times \mathrm{stress}\times A\times ∆l ...\left(2\right)\end{array}$

From equation (1) and equation (2), it can be written that

$U=\frac{1}{2}\times Y\times \frac{{\left(∆l\right)}^2}{L}\times A ...\left(3\right)$

Substitute the values of the known parameters into equation (3) and solve to calculate the required cross-sectional area of the wire.

$\begin{array}{rcl}80 \mathrm{J}& =& \frac{1}{2}\times 2.0\times {10}^{11} {\mathrm{Nm}}^{-2}\times \frac{{\left(0.02 \mathrm{m}\right)}^2}{20 \mathrm{m}}\times A\\ & \Rightarrow & A= \frac{2\times 80 \mathrm{J}\times 20}{2.0\times {10}^{11}\times 4\times {10}^{-4} {\mathrm{Nm}}^{-1}}\\ & =& 4\times {10}^{-5} {\mathrm{m}}^2\times \frac{{10}^6 {\mathrm{mm}}^2}{1 {\mathrm{m}}^2}\\ & =& 40 {\mathrm{mm}}^2\end{array}$