JEE Main 2023 — Physics Mechanics

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

$$\begin{matrix}Y & = & \frac{\mathrm{stress}}{\mathrm{strain}} \\ & = & \frac{\mathrm{stress}}{\frac{\Delta l}{L}}...(1)\end{matrix}$$Y​==​strainstress​LΔl​stress​...(1)​

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

$$\begin{matrix}U & = & \frac{1}{2}\times \mathrm{stress}\times \mathrm{strain}\times \mathrm{volume} \\ & = & \frac{1}{2}\times \mathrm{stress}\times \frac{\Delta l}{L}\times A\times L \\ & = & \frac{1}{2}\times \mathrm{stress}\times A\times \Delta l...(2)\end{matrix}$$U​===​21​×stress×strain×volume21​×stress×LΔl​×A×L21​×stress×A×Δl...(2)​

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

$$U=\frac{1}{2}\times Y\times \frac{{(\Delta l)}^{2}}{L}\times A...(3)$$U=21​×Y×L(Δl)2​×A...(3)

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

$$\begin{matrix}80J & = & \frac{1}{2}\times 2.0\times {10}^{11}{\mathrm{Nm}}^{-2}\times \frac{{(0.02m)}^{2}}{20m}\times A \\ & \Rightarrow & A=\frac{2\times 80J\times 20}{2.0\times {10}^{11}\times 4\times {10}^{-4}{\mathrm{Nm}}^{-1}} \\ & = & 4\times {10}^{-5}{m}^{2}\times \frac{{10}^{6}{\mathrm{mm}}^{2}}{1{m}^{2}} \\ & = & 40{\mathrm{mm}}^{2}\end{matrix}$$80J​=⇒==​21​×2.0×1011Nm−2×20m(0.02m)2​×AA=2.0×1011×4×10−4Nm−12×80J×20​4×10−5m2×1m2106mm2​40mm2​