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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