A) \[\frac{L}{4}\]
B) 2L
C) 4L
D) \[\frac{L}{2}\]
Correct Answer: A
Solution :
Angular momentum \[L=\operatorname{l}\omega \] ... (i) Kinetic energy \[K=\frac{1}{2}l{{\omega }^{2}}=\frac{1}{2}L\omega \] [from Eq. (i)] \[\therefore \] \[L=\frac{2K}{\omega }\] Now, the new angular momentum \[L'=\frac{2\left( \frac{K}{2} \right)}{2\omega }\] \[\left( \because K'=\frac{K}{2}and\omega '=2\omega \right)\] \[\Rightarrow \] \[L'=\frac{L}{4}\]
You need to login to perform this action.
You will be redirected in
3 sec
