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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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