A) \[\frac{2\,{{m}^{2}}}{3}g\]
B) \[\frac{4\,{{m}^{2}}}{3}g\]
C) \[\frac{\,{{m}^{2}}}{\sqrt{2}}g\]
D) \[5\Omega \]
Correct Answer: B
Solution :
The radius of the circle followed by the masses is \[r=l\sin \alpha \]As, angular momentum, \[\xrightarrow{\text{ROOR (peroxide)}}\underset{Major}{\mathop{(X)}}\,+\underset{Minor}{\mathop{(Y)}}\,\] \[BrC{{H}_{2}}-C{{H}_{2}}-CH=C{{H}_{2}}\] \[{{C}_{2}}{{H}_{5}}-CHBr-C{{H}_{3}}\] \[|L|=l\sin \theta (m\omega lsin\theta )\] On differentiating, we get \[\frac{d|L|}{dt}=m\omega {{l}^{2}}2\sin \theta \frac{d\theta }{dt}\] \[\Rightarrow \]\[\left| \frac{dL}{dt} \right|=2m{{l}^{2}}{{\omega }^{2}}\sin \theta .cos\theta \] \[=m{{l}^{2}}{{\omega }^{2}}\sin 2\theta \]
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