Question

A block of weight $100\text{N}$ is suspended by copper and steel wires of same cross-sectional area $0.5{\text{cm}}^2$ and lengths $\sqrt{3}\text{m}$ and $1\text{m}$, respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are ${30}^{\circ }$ and ${60}^{\circ }$, respectively. If elongation in copper wire is $\left(\Delta L_c\right)$ and elongation in steel wire is $\left(\Delta L_s\right)$, then the ratio $\frac{\Delta L_c}{\Delta L_s}$ is
[Young's modulus for copper and steel are $1\times {10}^{11}$ $\text{N}{\text{m}}^2$ and $2\times {10}^{11}\text{N}{\text{m}}^2$ respectively]

• A. 2.00
• B. 2
• C. 2.0

Answer 1

Answer: (A), (B), (C)

Let,
$T_S$ = tension in steel wire
$T_C$ = Tension in copper wire
At equilibrium :


Net force along $+X$ direction :
$\sum F_x=0$
$\Rightarrow T_c\cos {30}^{\circ }=T_s\cos {60}^{\circ }$
$\Rightarrow T_c\times \frac{\sqrt{3}}{2}=T_s\times \frac{1}{2}$

$\Rightarrow \sqrt{3}{T}_C=T_s\dots \dots \left(i\right)$
Net force along $+Y$ direction
$\sum F_y=0$
$\Rightarrow T_c\sin {30}^{\circ }+T_s\sin {60}^{\circ }-100=0$
$\Rightarrow \frac{T_c}{2}+\frac{T_s\sqrt{3}}{2}=100$
As, $T_s=\sqrt{3}{T}_C$ from equation $\left(i\right)$
$\Rightarrow T_c=50\text{N}$
$\therefore T_s=50\sqrt{3}N$
$\text{We know}$
$\Delta L=\frac{FL}{AY}$
$\Rightarrow \frac{\Delta L_c}{\Delta L_s}=\frac{T_c{L}_c}{A_c{Y}_c}\times \frac{A_s{Y}_s}{T_s{L}_s}$
$\text{On solving above equation}$
$\Rightarrow \frac{\Delta L_c}{\Delta L_s}=2$