Question:
A copper wire of length \(2.4\,\text{m}\) and an aluminum wire of length \(0.7\,\text{m}\), both having diameter \(2\,\text{mm}\), are connected end to end. When stretched by a load, the obtained elongation is found to be \(0.6\,\text{mm}\). The applied load is \((Y_{\text{Cu}}=1.2\times10^{11}\,\text{N/m}^2,\ Y_{\text{Al}}=0.7\times10^{11}\,\text{N/m}^2)\):
A copper wire of length \(2.4\,\text{m}\) and an aluminum wire of length \(0.7\,\text{m}\), both having diameter \(2\,\text{mm}\), are connected end to end. When stretched by a load, the obtained elongation is found to be \(0.6\,\text{mm}\). The applied load is \((Y_{\text{Cu}}=1.2\times10^{11}\,\text{N/m}^2,\ Y_{\text{Al}}=0.7\times10^{11}\,\text{N/m}^2)\):
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For wires connected in series, the same force acts through each wire and total elongation is the sum of individual elongations.
Updated On: Jun 24, 2026
- \(12\pi\,\text{N}\)
- \(24\pi\,\text{N}\)
- \(20\pi\,\text{N}\)
- \(80\pi\,\text{N}\)
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The Correct Option is C
Solution and Explanation
Step 1: Use the formula for elongation.
For a wire, \[ \Delta l=\frac{FL}{AY} \] Since the two wires are connected in series, total elongation is \[ \Delta l=\Delta l_1+\Delta l_2 \] Thus, \[ \Delta l= \frac{FL_1}{AY_1} + \frac{FL_2}{AY_2} \]
Step 2: Substitute the given values.
Given, \[ L_1=2.4\,\text{m} \] for copper and \[ Y_1=1.2\times10^{11}\,\text{N/m}^2 \] Also, \[ L_2=0.7\,\text{m} \] for aluminum and \[ Y_2=0.7\times10^{11}\,\text{N/m}^2 \] Diameter: \[ d=2\,\text{mm}=2\times10^{-3}\,\text{m} \] Radius: \[ r=1\times10^{-3}\,\text{m} \] Area: \[ A=\pi r^2 \] \[ A=\pi(10^{-3})^2 \] \[ A=\pi\times10^{-6}\,\text{m}^2 \] Total elongation: \[ \Delta l=0.6\,\text{mm}=0.6\times10^{-3}\,\text{m} \]
Step 3: Form the equation.
\[ 0.6\times10^{-3} = \frac{F(2.4)}{\pi\times10^{-6}\times1.2\times10^{11}} + \frac{F(0.7)}{\pi\times10^{-6}\times0.7\times10^{11}} \] Simplify first term: \[ \frac{2.4}{1.2\times10^{11}} = 2\times10^{-11} \] Second term: \[ \frac{0.7}{0.7\times10^{11}} = 10^{-11} \] Thus, \[ 0.6\times10^{-3} = \frac{F}{\pi\times10^{-6}} \left(2\times10^{-11}+10^{-11}\right) \] \[ 0.6\times10^{-3} = \frac{F}{\pi\times10^{-6}} (3\times10^{-11}) \]
Step 4: Solve for \(F\).
\[ 0.6\times10^{-3} = \frac{3F\times10^{-11}}{\pi\times10^{-6}} \] \[ 0.6\times10^{-3} = \frac{3F\times10^{-5}}{\pi} \] \[ F= \frac{0.6\times10^{-3}\times\pi}{3\times10^{-5}} \] \[ F= 20\pi\,\text{N} \]
Step 5: Final conclusion.
Hence, the applied load is \[ \boxed{20\pi\,\text{N}} \]
For a wire, \[ \Delta l=\frac{FL}{AY} \] Since the two wires are connected in series, total elongation is \[ \Delta l=\Delta l_1+\Delta l_2 \] Thus, \[ \Delta l= \frac{FL_1}{AY_1} + \frac{FL_2}{AY_2} \]
Step 2: Substitute the given values.
Given, \[ L_1=2.4\,\text{m} \] for copper and \[ Y_1=1.2\times10^{11}\,\text{N/m}^2 \] Also, \[ L_2=0.7\,\text{m} \] for aluminum and \[ Y_2=0.7\times10^{11}\,\text{N/m}^2 \] Diameter: \[ d=2\,\text{mm}=2\times10^{-3}\,\text{m} \] Radius: \[ r=1\times10^{-3}\,\text{m} \] Area: \[ A=\pi r^2 \] \[ A=\pi(10^{-3})^2 \] \[ A=\pi\times10^{-6}\,\text{m}^2 \] Total elongation: \[ \Delta l=0.6\,\text{mm}=0.6\times10^{-3}\,\text{m} \]
Step 3: Form the equation.
\[ 0.6\times10^{-3} = \frac{F(2.4)}{\pi\times10^{-6}\times1.2\times10^{11}} + \frac{F(0.7)}{\pi\times10^{-6}\times0.7\times10^{11}} \] Simplify first term: \[ \frac{2.4}{1.2\times10^{11}} = 2\times10^{-11} \] Second term: \[ \frac{0.7}{0.7\times10^{11}} = 10^{-11} \] Thus, \[ 0.6\times10^{-3} = \frac{F}{\pi\times10^{-6}} \left(2\times10^{-11}+10^{-11}\right) \] \[ 0.6\times10^{-3} = \frac{F}{\pi\times10^{-6}} (3\times10^{-11}) \]
Step 4: Solve for \(F\).
\[ 0.6\times10^{-3} = \frac{3F\times10^{-11}}{\pi\times10^{-6}} \] \[ 0.6\times10^{-3} = \frac{3F\times10^{-5}}{\pi} \] \[ F= \frac{0.6\times10^{-3}\times\pi}{3\times10^{-5}} \] \[ F= 20\pi\,\text{N} \]
Step 5: Final conclusion.
Hence, the applied load is \[ \boxed{20\pi\,\text{N}} \]
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