Question

A point mass $m$ is suspended using two wires of different material as shown in the figure. If cross-section area of wire$-1$ and wire$-2$ are $3m{m}^2$ and $\surd 3m{m}^2$ respectively, which of the following is correct?

• A. stress in wire $-1>$ stress in wire $-2$
• B. stress in wire $-1<$ stress in wire $-2$
• C. stress in wire $-1=$ stress in wire $-2$
• D. value of Young's modulus of both the wires is needed

Answer 1

The correct option is C stress in wire $-1=$ stress in wire $-2$

$\begin{array}{c}{T}_2\sin 30=T_1\sin 60\Rightarrow T_2=\sqrt{3}{T}_1--1\\ T_2\cos {30}^{\circ }+T_1\cos 6\theta =mg\\ \sqrt{3}{T}_2+T_1=2mg-\left(2\right)\\ \text{Using equation (1) and}\left(2\right)\text{we get}\end{array}$

$\begin{array}{cc}4T_1& =2mg\\ \Rightarrow T_1& =\frac{mg}{2}\\ \Rightarrow T_2& =\frac{\sqrt{3}mg}{2}mg\end{array}$

$\text{Now, area of cross - section of were}A=3\text{area of cross - section of were}B=\sqrt{3}m$

$\therefore \text{Stress in wire}B=\frac{T_2}{A_2}=\frac{mg}{2\sqrt{3}}\text{Stress in wire}B=\frac{T_2}{A_2}=\frac{mg}{2\sqrt{3}}$

$\therefore \text{Stress in}A=\text{Stress in}B$