Question

A weight of $100N$ is suspended by two wires made of steel and copper as shown in the figure. Length of steel wire is $1m$ and copper wire is $\sqrt{3}m$. Find the ratio of change in length of copper wire $\left(\Delta l_c\right)$ to change in length of steel wire $\left(\Delta l_s\right)$. Given Young's modulus: $Y_{steel}=2\times {10}^{11}N/m^2$, $Y_{copper}=1\times {10}^{11}N/m^2$

Answer 1

$\frac{\Delta l_c}{\Delta l_s}=?$
As, Stress $=\frac{F}{A}=\frac{Y\Delta l}{l}$
$\Delta l=\frac{F}{A}\times \frac{l}{Y}$
$\Delta l_c=\frac{F_c}{A_c}\times \frac{l_c}{Y_c}$--- (1)
$\Delta l_s=\frac{F_s}{A_s}\times \frac{l_s}{Y_s}$------(2)
As area is same so $A_s=A_c=A$
Dividing equation (1) by equation (2)
​$\frac{\Delta l_c}{\Delta l_s}=\frac{F_c\times l_c\times Y_s}{F_s\times l_s\times Y_c}$

In the above expression, we don't know the ratio of forces. So,

Equating horizontal component of the two tension forces, we get
$T_s\cos {60}^0=T_C\cos {30}^0$
$\frac{T_s}{2}=T_C\frac{\sqrt{3}}{2}$
$\frac{T_c}{T_s}=\frac{1}{\sqrt{3}}$
​$\frac{\Delta l_c}{\Delta l_s}=\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{1}\times \frac{2\times {10}^{11}}{1\times {10}^{11}}=2$