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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