In vertical direction 2 Tsinθ=20 using small angle approximation sinθ=θ $\begin{aligned}
& \theta=\frac{1}{100} \
& \therefore \mathrm{T}=\frac{10}{\theta} \
& \mathrm{T}=1000 \mathrm{N}
\end{aligned}\begin{array}{ll}
\text {Change in length } \Delta \mathrm{L} =2 \sqrt{\mathrm{x}^2+\mathrm{L}^2}-2 \mathrm{L} \
=2 \mathrm{L}\left[1+\frac{\mathrm{x}^2}{2 \mathrm{L}^2}-1\right] \
\Delta \mathrm{L} =\frac{\mathrm{x}^2}{\mathrm{L}}
\end{array}\thereforeModulusofelasticity=\frac{\text { stress }}{\text { strain }}\begin{aligned}
& 2 \times 10^{11}=\frac{10^3}{\mathrm{A} \times \frac{\mathrm{x}^2}{\mathrm{L}}} \times 2 \mathrm{L} \
\therefore & \mathrm{A}=1 \times 10^{-4} \mathrm{~m}^2
\end{aligned}$
JEE Main 2024 — Physics Mechanics
2024integermedium
A wire of cross sectional area A, modulus of elasticity 2×1011Nm−2 and length 2 m is stretched between two vertical rigid supports. When a mass of 2 kg is suspended at the middle it sags lower from its original position making angle θ=1001 radian on the points of support. The value of A is _______ ×10−4 m2 (consider x<<L ). (given : g=10 m/s2 )
Official previous-year question
Held on 6 Apr 2024 · Verified 6 Jul 2026.
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