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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