A wire of cross sectional area A, modulus… - JEE Main 2024 Physics PYQ

JEE Main 2024 — Physics Mechanics

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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 $\mathrm{A}$ is _______ $\times 10^{-4} \mathrm{~m}^2$ (consider $x< < \mathrm{L}$ ). (given : $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ ) JEE Main 2024 Physics, Mechanical Properties of Solids — question figure

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Held on 6 Apr 2024 · Verified 6 Jul 2026.

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Solution

JEE Main 2024 Physics, Mechanical Properties of Solids — solution figure In vertical direction $2 \mathrm{~T} \sin \theta=20$ using small angle approximation $\sin \theta=\theta$ $\begin{aligned}
& \theta=\frac{1}{100} \
& \therefore \mathrm{T}=\frac{10}{\theta} \
& \mathrm{T}=1000 \mathrm{N}
\end{aligned}\begin{array}{ll}
\text {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}}
\end{array}\therefore $Modulus of elasticity$ =\frac{\text { stress }}{\text { strain }}\begin{aligned}
& 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
\end{aligned}$

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