A) \[\frac{L}{4}\]
B) 2L
C) 4L
D) \[\frac{L}{2}\]
Correct Answer: A
Solution :
| [a] 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 |
| \[L'=\frac{2\left( \frac{K}{2} \right)}{2\omega }\] \[\left( \because K'=\frac{K}{2}and\omega '=2\omega \right)\] |
| \[\Rightarrow \] \[L'=\frac{L}{4}\] |
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