Question:
Two copper wires of length $L$ and $2L$ have radii $r$ and $2r$ respectively. If they are subjected to the same tension force, the ratio of their extension ($\Delta L_1 / \Delta L_2$) is:
Two copper wires of length $L$ and $2L$ have radii $r$ and $2r$ respectively. If they are subjected to the same tension force, the ratio of their extension ($\Delta L_1 / \Delta L_2$) is:
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Use proportionality directly: $\Delta L \propto \frac{L}{r^2}$. Doubling the length doubles the extension, but doubling the radius quarter-sizes it. Net change is $2 \times \frac{1}{4} = \frac{1}{2}$, meaning the second wire stretches half as much.
Updated On: Jun 3, 2026
- $2 : 1$
- $1 : 2$
- $1 : 1$
- $4 : 1$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
According to Hooke's Law, Young's modulus is $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L}$. Rearranging, the elongation is given by $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.
Step 2: Meaning
Since both wires are made of the same material (copper), they have the same Young's Modulus $Y$. The applied tension force $F$ is also identical.
Step 3: Analysis
Elongation is proportional to: \[ \Delta L \propto \frac{L}{r^2} \] For the first wire: \[ \Delta L_1 \propto \frac{L}{r^2} \] For the second wire: \[ \Delta L_2 \propto \frac{2L}{(2r)^2} = \frac{2L}{4r^2} = \frac{1}{2} \frac{L}{r^2} \] Taking the ratio: \[ \frac{\Delta L_1}{\Delta L_2} = \frac{1}{\frac{1}{2}} = 2 \implies \Delta L_1 : \Delta L_2 = 2 : 1 \]
Step 4: Conclusion
The ratio of their extension is $2 : 1$.
Final Answer: (A)
According to Hooke's Law, Young's modulus is $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L}$. Rearranging, the elongation is given by $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.
Step 2: Meaning
Since both wires are made of the same material (copper), they have the same Young's Modulus $Y$. The applied tension force $F$ is also identical.
Step 3: Analysis
Elongation is proportional to: \[ \Delta L \propto \frac{L}{r^2} \] For the first wire: \[ \Delta L_1 \propto \frac{L}{r^2} \] For the second wire: \[ \Delta L_2 \propto \frac{2L}{(2r)^2} = \frac{2L}{4r^2} = \frac{1}{2} \frac{L}{r^2} \] Taking the ratio: \[ \frac{\Delta L_1}{\Delta L_2} = \frac{1}{\frac{1}{2}} = 2 \implies \Delta L_1 : \Delta L_2 = 2 : 1 \]
Step 4: Conclusion
The ratio of their extension is $2 : 1$.
Final Answer: (A)
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