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