JEE Main 2023 — Mechanical Properties of Solids Question with Solution

The external force or tension $\left(F\right)$ acting on the wire is its own weight. It is given by

 $\begin{array}{rcl}F& =& mg\\ & =& 4 \mathrm{kg}\times \frac{10}{4} \mathrm{m} {\mathrm{s}}^{-2}\\ & =& 10 \mathrm{N}\\ & & \end{array}$

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

$\begin{array}{rcl}Y& =& \frac{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}}{{\displaystyle \raisebox{1ex}{$∆L$}\!\left/ \!\raisebox{-1ex}{$L$}\right.}}\\ & =& \frac{FL}{∆LA}..........................\left(1\right)\end{array}$

where, $A$ is the cross-sectional area, $L$ is the original length and $∆L$ is the elongation of the wire.

Substitute the given values of the known parameters into equation (1) and solve to calculate the required elongation of the wire.

$\begin{array}{rcl}2\times {10}^{11} \mathrm{N}/{\mathrm{m}}^2& =& \frac{10 \mathrm{N}\times 6 \mathrm{m}}{∆L\times 3 {\mathrm{mm}}^2\times {\displaystyle \frac{1 {\mathrm{m}}^2}{{10}^6 {\mathrm{mm}}^2}}}\\ ∆L& =& \frac{10 \mathrm{N}\times 6 \mathrm{m}}{2\times {10}^{11} \mathrm{N}/{\mathrm{m}}^2\times 3 {\mathrm{mm}}^2\times \frac{1 {\mathrm{m}}^2}{{10}^6 {\mathrm{mm}}^2}}\\ & =& 0.0001 \mathrm{m}\times \frac{{10}^3 \mathrm{mm}}{1 \mathrm{m}}\\ & =& 0.1 \mathrm{mm}\end{array}$