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 an angle $$\theta = \frac{1}{100}$$ radian on the points of support. The value of A is ___________ $$\times 10^{-4} \text{ m}^2$$ (consider $$x << L$$). (given : $$g = 10 \text{ m/s}^2$$)

When the mass $$m$$ is suspended, the vertical components of the tension ($$T$$) in the two halves of the wire balance the weight: $$2T \sin \theta = mg$$

For small angles ($$\theta = 1/100$$ rad), we use the approximation $$\sin \theta \approx \theta$$

$$2T\theta = mg \implies T = \frac{mg}{2\theta}$$

The original length of half the wire is $$L$$. The new length $$L'$$ is $$L' = \frac{L}{\cos \theta}$$

$$\text{Strain} = \frac{L' - L}{L} = \frac{1}{\cos \theta} - 1$$

Using the small angle approximation $$\cos \theta \approx 1 - \frac{\theta^2}{2}$$:

$$\text{Strain} \approx \left(1 + \frac{\theta^2}{2}\right) - 1 = \frac{\theta^2}{2}$$

$$Y = \frac{T / A}{\theta^2 / 2} = \frac{2T}{A\theta^2}$$

$$Y = \frac{2(mg / 2\theta)}{A\theta^2} = \frac{mg}{A\theta^3}$$

$$A = \frac{mg}{Y\theta^3}$$

$$A = \frac{2 \times 10}{(2 \times 10^{11}) \times (10^{-2})^3}$$

$$A = \frac{20}{2 \times 10^{11} \times 10^{-6}} = \frac{10}{10^5} = 1 \times 10^{-4} \text{ m}^2$$

The answer is 1.