A wire of cross sectional area A, modulus of elasticity 2 × 10¹¹ Nm⁻²… | JEE Main 2024 (6 Apr, Shift 2)

Q8JEE Main 2024 · 6 APR · SHIFT 2Difficult

Wire Combinations & Special Cases

A wire of cross sectional area A, modulus of elasticity $2 \times 10^{11}\text{ Nm}^{-2}$ and length $2\text{ m}$ is stretched between two vertical rigid supports. When a mass of $2\text{ kg}$ 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 ________ $\times 10^{-4}\text{ m}^2$ (consider $x << \text{L}$ ).

(given: $\text{g} = 10\text{ m/s}^2$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 1

At the middle point, the vertical parts of the two tensions hold the weight: $2T\sin\theta = mg$ .
$T = \dfrac{mg}{2\theta} = \dfrac{20}{2 \times 0.01} = 1000\text{ N}$
Each half of length $L$ becomes $L\sec\theta$ , so strain $= \sec\theta - 1 \approx \dfrac{\theta^2}{2} = 5 \times 10^{-5}$ .
$A = \dfrac{T}{Y \times \text{strain}} = \dfrac{1000}{2 \times 10^{11} \times 5 \times 10^{-5}} = 1 \times 10^{-4}\text{ m}^2$
So the answer is 1.

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