A) V/25
B) 5V
C) V/5
D) 25V
Correct Answer: D
Solution :
\[U=\frac{1}{2}\left( \frac{YA}{L} \right)\,{{l}^{2}}\] \ \[U\propto {{l}^{2}}\] \[\frac{{{U}_{2}}}{{{U}_{1}}}={{\left( \frac{{{l}_{2}}}{{{l}_{1}}} \right)}^{2}}={{\left( \frac{10}{2} \right)}^{2}}=25\]Þ \[{{U}_{2}}=25{{U}_{1}}\] i.e. potential energy of the spring will be 25 V
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