JEE Main 2024 — 6 April, Shift 2 — Question 57

In vertical derection

$2 \mathrm{~T} \sin \theta=20$

using small angle approximation $\sin \theta=\theta$

$\theta=\frac{1}{100}$ $\therefore \mathrm{T}=\frac{10}{\theta}$ $\mathrm{T}=1000 \mathrm{~N}$ Change in length

$\Delta \mathrm{L}=2 \sqrt{\mathrm{x}^{2}+\mathrm{L}^{2}}-2 \mathrm{~L}$

$$=2 \mathrm{~L}\left[1+\frac{\mathrm{x}^{2}}{2 \mathrm{~L}^{2}}-1\right]$$

$\Delta \mathrm{L}=\frac{\mathrm{x}^{2}}{\mathrm{~L}}$ $\therefore$

Modulus of elasticity $=\frac{\text { stress }}{\text { strain }}$

$$2 \times 10^{11}=\frac{10^{3}}{\mathrm{~A} \times \frac{\mathrm{x}^{2}}{\mathrm{~L}}} \times 2 \mathrm{~L}$$

$\therefore \mathrm{A}=1 \times 10^{-4} \mathrm{~m}^{2}$