Question

Two wires of same material and same diameter have lengths in the ratio 2 : 5. They are stretched by same force. The ratio of work done in stretching them is:

• A. 5 : 2
• B. 2 : 5
• C. 1 : 3
• D. 3 : 1

Answer 1

Answer: (B)

Stretching force $ F=\frac{YA}{L}.\Delta l $ Here, both are of same material (i.e., same Y), same diameter (i.e., same A) and are stretched by same force. So, $ \frac{\Delta {{l}_{1}}}{{{L}_{1}}}=\frac{\Delta {{l}_{2}}}{{{L}_{2}}} $ $ \therefore $ $ \frac{\Delta {{l}_{1}}}{\Delta {{l}_{2}}}=\frac{{{L}_{2}}}{{{L}_{2}}} $ $ =\frac{2}{5} $ Work done $ W=\frac{1}{2}YA\frac{\Delta {{l}^{2}}}{L} $ Ratio of work done $ \frac{{{W}_{1}}}{{{W}_{2}}}={{\left( \frac{\Delta {{l}_{1}}}{\Delta {{l}_{2}}} \right)}^{2}}.\frac{{{L}_{2}}}{{{L}_{1}}} $ $ ={{\left( \frac{2}{5} \right)}^{2}}.\left( \frac{5}{2} \right) $ $ =\frac{2}{5} $ $ =2.5 $