JEE Main 2022 — Mathematics in Physics Question with Solution

Given here, 

Length, $L=1 \mathrm{m}$ and extension in length,  $\Delta L=0.4\times {10}^{-3} \mathrm{m}$

Mass of load, $m=1 \mathrm{kg}$ and diameter of wire, $d=0.4\times {10}^{-3} \mathrm{m}$

Young's modulus is given by

$Y=\frac{FL}{A\Delta L}=\frac{\left(mg\right)\times \left(1\right)}{\left(\frac{\pi d^2}{4}\right)0.4\times {10}^{-3}}$

$\Rightarrow Y=\frac{10\times 4}{\pi {\left(0.4\times {10}^{-3}\right)}^2\times 0.4\times {10}^{-3}}$

$\Rightarrow Y=\frac{40\times 7}{22\times 64\times {10}^{-3}\times {10}^{-9}}$   

$\Rightarrow Y=0.199\times {10}^{-12} \mathrm{N} {\mathrm{m}}^{-2}$

Error in measurement of Young's modulus,

$\frac{\Delta Y}{Y}=\frac{\Delta F}{F}+\frac{\Delta L}{L}+\frac{\Delta A}{A}+\frac{\Delta \left(\Delta L\right)}{\left(\Delta L\right)}$

$=\frac{0.02}{0.4}+2\frac{\Delta d}{d}=\frac{0.2}{4}+2\times \frac{0.01}{0.4}$

$=\frac{0.1}{2}+\frac{0.1}{2}=0.1$

$=0.199\times {10}^{11}\approx 2\times {10}^{10}$

Hence, the value of $x=2$.