JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2024
- Hard
- Numerical
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: \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) )

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Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
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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