A) \[2\times {{10}^{-2}}\,J\]
B) \[4\times {{10}^{-2}}\,J\]
C) \[8\times {{10}^{-2}}\,J\]
D) \[16\times {{10}^{-2}}\,J\]
Correct Answer: A
Solution :
Max. K.E. of the system = Max. P.E. of the system \[\frac{1}{2}k{{x}^{2}}\]= \[=\frac{1}{2}\times (16)\times {{(5\times {{10}^{-2}})}^{2}}=2\times {{10}^{-2}}\,J\]
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