A) \[2L\]
B) \[4L\]
C) \[\frac{L}{2}\]
D) \[\frac{L}{4}\]
Correct Answer: D
Solution :
\[\because \] \[L=m\upsilon r=m{{r}^{2}}\omega \] ?.(i) Also kinetic energy\[K=\frac{1}{2}m{{\upsilon }^{2}}\] Or \[K=\frac{1}{2}m{{(r\omega )}^{2}}=\frac{1}{2}m{{r}^{2}}{{\omega }^{2}}\] \[K=\frac{1}{2}\frac{L}{\omega }{{\omega }^{2}}=\frac{L\omega }{2}\] \[\Rightarrow \] \[L=\frac{2K}{\omega }\] Hence, \[\omega =2\omega \] \[K=\frac{1}{2}K\] \[\therefore \] \[L=\frac{2K}{\omega }=\frac{2\left( \frac{1}{2}K \right)}{2\omega }=\frac{L}{4}\]
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