A wire of cross sectional area $A$, modulus of elasticity $2 \times 10^{11} \mathrm{Nm}^{-2}$ and length $2 \mathrm{~m}$ is stretched between two vertical rigid supports. When a mass of $2 \mathrm{~kg}$ is suspended at the middle it sags lower from its original position making angle $\theta=\frac{1}{100}$ radian on the points of support. The value of $A$ is. . . . . . $\times 10^{-4} \mathrm{~m}^2$ (consider $\mathrm{x}<\mathrm{L}$ ).
(given: $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
JEE MAIN 2024, Diffcult
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In vertical derection $2 \mathrm{~T} \sin \theta=20$
using small angle approximation $\sin \theta=\theta$
$\theta=\frac{1}{100}$
$\therefore \quad \mathrm{T}=\frac{10}{\theta}$
$T=1000 \mathrm{~N}$
Change in length $\Delta \mathrm{L} \quad=2 \sqrt{\mathrm{x}^2+\mathrm{L}^2}-2 \mathrm{~L}$
$=2 \mathrm{~L}\left[1+\frac{\mathrm{x}^2}{2 \mathrm{~L}^2}-1\right]$
$\Delta \mathrm{L} =\frac{\mathrm{x}^2}{\mathrm{~L}}$
$\therefore$ Modulus of elasticity $=\frac{\text { stress }}{\text { strain }}$
$2 \times 10^{11}=\frac{10^3}{\mathrm{~A} \times \frac{\mathrm{x}^2}{\mathrm{~L}}} \times 2 \mathrm{~L}$
$\therefore \quad \mathrm{A}=1 \times 10^{-4} \mathrm{~m}^2$
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