A) \[L/4\]
B) \[L/2\]
C) \[L\]
D) \[2L\]
Correct Answer: D
Solution :
Angular momentum during uniform circular motion \[L=mvr\] Kinetic energy \[K=\frac{1}{2}\,m{{v}^{2}}\] \[=\frac{1}{2}m\times {{\left( \frac{L}{mr} \right)}^{2}}=\frac{{{L}^{2}}}{2m{{r}^{2}}}\] According to the question, \[K'=\frac{{{(L')}^{2}}}{2m{{r}^{2}}}=4K\Rightarrow \frac{{{(L')}^{2}}}{2m{{r}^{2}}}=\frac{4\times {{L}^{2}}}{2m{{r}^{2}}}\] \[\Rightarrow \] \[L'=2L\]You need to login to perform this action.
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