1.

A particle performing uniform circular motion has angular momentum L. If its angular frequency is doubled and its kinetic halved, then the new angular momentum is :

Answer»

`(L)/(2)`
`(L)/(4)`
`2L`
`4L`

Solution :Here `E=(1)/(2)Iomega^(2)=(1)/(2)Iomega.omega`
`E=(1)/(2)xxL.omega`
or `L=(2E)/(omega)orLprop(E)/(omega)`
Similarly `L.=(2E.)/(omega.)=(2xx(E)/(2))/(2omega)=(2E.)/(4OMEGA.)=(E)/(2omega)`
`therefore (L.)/(L)=((E)/(2omega))/((2E)/(omega))=(1)/(4)`
or `L.=(L)/(4)`


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