A particle moving in a circular path has an angular momentum of L. If the frequency of rotation is halved, then its angular momentum becomes

A particle moving in a circular path has an angular momentum of L. If

1.	A particle moving in a circular path has an angular momentum of L. If the frequency of rotation is halved, then its angular momentum becomes
Answer» `(L)/(2)` L `(L)/(3)` `(L)/(4)` Solution :We KNOE that `L=I omega=2pi mr^(2)f` Now, `omega'=omega//2` Hence, `L'=Iomega'=mr^(2)=(omega)/(2)=PIMR^(2)f RARR (L)/(L')=(2M pi r^(2)f)/(pi mr^(2)f)rArr L'=(L)/(2)`.