Question:
Two wires A and B of same material and of equal length with the radii in the ratio 1 : 2 are subjected to identical loads. If the length of A increases by 8 mm, then the increase in length of B is
Two wires A and B of same material and of equal length with the radii in the ratio 1 : 2 are subjected to identical loads. If the length of A increases by 8 mm, then the increase in length of B is
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Extension is inversely proportional to cross-sectional area \((\propto r^2)\).
Updated On: May 8, 2026
- 2 mm
- 4 mm
- 8 mm
- 16 mm
- 1 mm
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The Correct Option is A
Solution and Explanation
Concept:
Extension in a wire:
\[
\Delta L = \frac{FL}{AY}
\]
For same material, same length, and same force:
\[
\Delta L \propto \frac{1}{A} \propto \frac{1}{r^2}
\]
Step 1: Write ratio of extensions. \[ \frac{\Delta L_A}{\Delta L_B} = \frac{r_B^2}{r_A^2} \]
Step 2: Substitute radius ratio. \[ \frac{r_A}{r_B} = \frac{1}{2} \Rightarrow \frac{\Delta L_A}{\Delta L_B} = \frac{(2)^2}{(1)^2} = 4 \]
Step 3: Substitute value of A. \[ \frac{8}{\Delta L_B} = 4 \]
Step 4: Solve. \[ \Delta L_B = 2 \, \text{mm} \]
Step 5: Conclusion. \[ \boxed{2 \, \text{mm}} \]
Step 1: Write ratio of extensions. \[ \frac{\Delta L_A}{\Delta L_B} = \frac{r_B^2}{r_A^2} \]
Step 2: Substitute radius ratio. \[ \frac{r_A}{r_B} = \frac{1}{2} \Rightarrow \frac{\Delta L_A}{\Delta L_B} = \frac{(2)^2}{(1)^2} = 4 \]
Step 3: Substitute value of A. \[ \frac{8}{\Delta L_B} = 4 \]
Step 4: Solve. \[ \Delta L_B = 2 \, \text{mm} \]
Step 5: Conclusion. \[ \boxed{2 \, \text{mm}} \]
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