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 $\mathrm{A}$ is _______ $\times 10^{-4} \mathrm{~m}^2$ (consider $x< < \mathrm{L}$ ). (given : $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Solution

In vertical direction $2 \mathrm{~T} \sin \theta=20$ using small angle approximation $\sin \theta=\theta$ $\begin{aligned} & \theta=\frac{1}{100} \\ & \therefore \mathrm{T}=\frac{10}{\theta} \\ & \mathrm{T}=1000 \mathrm{~N} \end{aligned}$ $\begin{array}{ll} \text {Change in length } \Delta \mathrm{L} =2 \sqrt{\mathrm{x}^2+\mathrm{L}^2}-2 \mathrm{~L} \\ =2 \mathrm{~L}\left[1+\frac{\mathrm{x}^2}{2 \mathrm{~L}^2}-1\right] \\ \Delta \mathrm{L} =\frac{\mathrm{x}^2}{\mathrm{~L}} \end{array}$ $\therefore$ Modulus of elasticity $=\frac{\text { stress }}{\text { strain }}$ $\begin{aligned} & 2 \times 10^{11}=\frac{10^3}{\mathrm{~A} \times \frac{\mathrm{x}^2}{\mathrm{~L}}} \times 2 \mathrm{~L} \\ \therefore & \mathrm{A}=1 \times 10^{-4} \mathrm{~m}^2 \end{aligned}$

Hence, the correct answer is 1.