A particle performing uniform circular motion has angular frequency is doubled & its kinetic energy halved, then the new angular momentum is-

Solution :
$$\eqalign{ & K.E.\, = \frac{1}{2}I{\omega ^2},\,\,{\text{but}}\,L = I\omega \Rightarrow I = \frac{L}{\omega } \cr & \therefore K.E. = \frac{1}{2}\frac{L}{\omega } \times {\omega ^2} = \frac{1}{2}L\omega \cr & \therefore \frac{{K.E.}}{{K.E.}} = \frac{{L \times \omega }}{{L' \times \omega '}} \cr & \Rightarrow \frac{{K.E.}}{{\frac{{K.E.}}{2}}} = \frac{{L \times \omega }}{{L' \times 2\omega }} \cr & \therefore L' = \frac{L}{4} \cr} $$