JEE Mains · Physics · STD 11- 8. mechanical properties of solids
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: \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) )

- A \(4\)
- B \(5\)
- C \(1\)
- D \(3\)
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…
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