Question

A weight of $100N$is suspended by two wires made by 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

Step 1. Given data

A weight of $100N$

Length of steel wire, $I_s$ is $1m$ and copper wire, $I_c$ is $\sqrt{3}m$.

Young’s modulus of steel, $Y_s=2\times {10}^{11}N{m}^{-2}$

Young's modulus of copper, $Y_c=1\times {10}^{11}N{m}^{-2}$

Step 2. Calculating the ratio of the change in length of steel and copper

From the given figure,

$T_c$ is the tension of copper

$T_s$ is the tension of steel

$∆I_c$ is the change in length of copper wire

$∆I_s$ is the change in length of steel wire

$Y$ is the young's modulus

Therefore, we have horizontal components,

$$\begin{gathered} T_c\cos 30°=T_s\cos 60° \\ {T}_c\frac{\sqrt{3}}{2}=T_s\frac{1}{2} \\ {T}_c\sqrt{3}=T_s \\ \Delta l\propto \frac{T\cdot l}{Y} \\ \frac{∆I_c}{∆I_s}=\frac{T_c}{T_s}\left(\frac{I_c}{I_s}\right)\left(\frac{Y_s}{Y_c}\right) \\ \frac{∆I_c}{∆I_s}=\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}\times 2\times {10}^{11}}{1\times {10}^{11}\times 1} \\ \frac{∆I_c}{∆I_s}=2 \end{gathered}$$ $$\begin{gathered} \left[\frac{T_c}{T_s}=\frac{1}{\sqrt{3}} \\ {I}_c=\sqrt{3}m \\ {I}_s=1m \\ {Y}_s=2\times {10}^{11}N{m}^{-2} \\ {Y}_c=1\times {10}^{11}N{m}^{-2}\right] \end{gathered}$$

Therefore, the ratio of change in length of copper wire is $2$.