Question

A block of weight $100$ N is suspended by copper and steel wires of same cross sectional area $0.5c{m}^2$ and length $\sqrt{3}m$ and $1m$ respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are ${30}^o$ and ${60}^o$ respectively. If elongation in copper wire is $\Delta l_C$ and elongation in steel wire is $\Delta l_S$, then the ratio $\frac{\Delta l_C}{\Delta l_S}$ is

(Young's modulus for copper and steel are $1\times {10}^{11}N/m^2$ and $2\times {10}^{11}N/m^2$ respectively)

Answer 1

Let tension in the steel wire be $T_1$ and that in the copper wire be $T_2$.
Assuming the whole system is in mechanical equilibrium, balancing the horizontal components of the tension in both the wires.
$T_{1}sin30=T_{2}sin60$
$\frac{T_1}{2}=\frac{T_2\sqrt{3}}{2}$
$\begin{array}{}{T}_1=\sqrt{3}{T}_2& \text{(1)}\end{array}$


$\frac{\Delta l_2}{\Delta l_1}=\frac{T_2{l}_2}{A_2{Y}_2}\times \frac{A_1{Y}_1}{T_1{l}_1}$
Substituting the values and using equation $\left(1\right)$
$\frac{\Delta l_2}{\Delta l_1}=\frac{\sqrt{3}{T}_2}{T_1}\times \frac{2}{1}=2$