A) \[2U\]
B) \[25U\]
C) \[U/5\]
D) \[5U\]
Correct Answer: A
Solution :
Potential energy in a stretched spring is given by \[U=\frac{1}{2}K{{x}^{2}}\] \[\frac{{{U}_{1}}}{{{U}_{2}}}={{\left( \frac{{{x}_{1}}}{{{x}_{2}}} \right)}^{2}}\] Given \[{{X}_{1}}=2cm=.02m,\text{ }{{X}_{2}}=10cm=0.1m\] Substituting the values \[\frac{{{U}_{1}}}{{{U}_{2}}}={{\left( \frac{0.02}{0.1} \right)}^{2}}={{\left( \frac{1}{5} \right)}^{2}}=\frac{1}{25}\] \[{{U}_{2}}=25{{U}_{1}}=2U\]
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