A) 2 : 1
B) 1 : 2
C) 4 : 1
D) 1 : 4
Correct Answer: A
Solution :
Force, \[F=kx\] and potential energy, \[U=\frac{1}{2}k{{x}^{2}}=\frac{1}{2}\frac{{{F}^{2}}}{k}\] \[\Rightarrow \] \[U\propto \frac{1}{k}\] \[\therefore \] \[\frac{{{U}_{1}}}{{{U}_{2}}}=\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{3000}{1500}=\frac{2}{1}\]You need to login to perform this action.
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