A) 4 U
B) 8 U
C) 16 U
D) U/4
Correct Answer: C
Solution :
| Let extension produced in a spring be \[x\] initially. In stretched condition spring will have potential energy |
| \[U=\frac{1}{2}k{{x}^{2}}\] |
| where k is spring constant or force constant. |
| \[\therefore \] \[\frac{{{U}_{1}}}{{{U}_{2}}}=\frac{x_{1}^{2}}{x_{2}^{2}}\] (i) |
| Given, \[{{U}_{1}}=U,\,{{x}_{1}}=2\,cm,\,{{x}_{2}}=8\,cm\] |
| putting these values in Eq. (i), we have |
| \[\frac{U}{{{U}_{2}}}=\frac{{{(2)}^{2}}}{{{(8)}^{2}}}=\frac{4}{64}=\frac{1}{16}\] |
| \[\therefore \] \[{{U}_{2}}=16U\] |
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