A) 5 : 2
B) 2 : 5
C) 1 : 3
D) 3 : 1
Correct Answer: B
Solution :
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 \[\frac{{{W}_{1}}}{{{W}_{2}}}={{\left( \frac{\Delta {{l}_{1}}}{\Delta {{l}_{2}}} \right)}^{2}}.\frac{{{L}_{2}}}{{{L}_{1}}}\] \[={{\left( \frac{2}{5} \right)}^{2}}.\left( \frac{5}{2} \right)\] \[=\frac{2}{5}\] \[=2.5\]
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