Question

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 $x \ll l$).  
(Given: $g = 10 \, \text{m/s}^2$)

Answer 1

Correct Answer: 1

In the vertical direction:
\[2T \sin \theta = 20\]
Using the small angle approximation \( \sin \theta \approx \theta \):
\[\theta = \frac{1}{100} \implies T = \frac{10}{\theta} = 1000 \, \text{N}\]
The change in length \( \Delta L \) is given by:
\[\Delta L = 2\sqrt{x^2 + L^2} - 2L\]
\[\Delta L \approx 2L \left( \frac{x^2}{2L^2} \right) = \frac{x^2}{L}\]
Modulus of elasticity \( E \) is defined as:
\[E = \frac{\text{stress}}{\text{strain}}\]
Substitute \( E = 2 \times 10^{11} \, \text{Nm}^{-2} \):
\[2 \times 10^{11} = \frac{10^3}{A} \times \frac{x^2}{L} \times 2L\]
Solve for \( A \):
\[A = 1 \times 10^{-4} \, \text{m}^2\]