A) 1 : 2
B) 2 : 1
C) 1 : 4
D) 4 : 1
Correct Answer: B
Solution :
Key Idea: when spring is pulled and left, it oscillation in SHM. The work done in pulling the string is stored as potential energy in the spring ?(i) \[U=\frac{1}{2}k{{x}^{2}}\] where k is spring constant and \[x\] is distance through which it is pulled. Also in SHM Force \[\propto \]displacement \[F=kx\] ?(ii) where, k is spring constant. Putting \[x=\frac{F}{k}\]in Eq. (i), we get \[U=\frac{1}{2}k{{\left( \frac{F}{R} \right)}^{2}}=\frac{{{F}^{2}}}{2k}\] \[\therefore \] \[\frac{{{U}_{1}}}{{{U}_{2}}}=\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{3000}{1500}=\frac{2}{1}\] \[\therefore \] \[{{U}_{1}}:{{U}_{2}}=2:1\]You need to login to perform this action.
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