A wire of cross-sectional area $A$, modulus of elasticity $2\times10^{11}\ \text{N m}^{-2}$ and length $2\ \text{m}$ is stretched between two vertical rigid supports. When a mass of $2\ \text{kg}$ is suspended at the middle, the wire sags lower from its original position, making an angle $\theta = \dfrac{1}{100}$ radian at the points of support. The value of $A$ is ______ $\times10^{-4}\ \text{m}^2$. (Consider the sag $x \ll L$, where $2L$ is the length of the wire; $g = 10\ \text{m/s}^2$)
Numerical value type. Enter your answer.
Answer: 1
Vertical balance at the midpoint: $2T\sin\theta = mg \Rightarrow T \approx \dfrac{20}{2\times0.01} = 1000\ \text{N}$.
Each half stretches from $L$ to $\dfrac{L}{\cos\theta} \approx L\left(1 + \dfrac{\theta^2}{2}\right)$, so the strain is $\dfrac{\theta^2}{2} = 5\times10^{-5}$.
$$A = \frac{T}{Y\times\text{strain}} = \frac{1000}{2\times10^{11}\times5\times10^{-5}} = 1\times10^{-4}\ \text{m}^2$$
Solution by Sreeraj P, M.Sc Physics