1.

In the above the question the loss of the kinetic energy during the above process is :

Answer»

`(IOMEGA^(2))/(2)`
`(Iomega^(2))/(3)`
`(Iomega^(2))/(4)`
`(Iomega^(2))/(6)`

Solution :Here loss in the K.E. is :
`E_(k)=E_(i)-E_(f)`
`=[(1)/(2)xxIxx(2omega)^(2)+(1)/(2)xx2Ixxomega^(2)]-(1)/(2)(I+2I)xx((4omega)/(3))^(2)`
`=3Iomega^(2)-(8)/(3)Iomega^(2)=(1)/(2)Iomega^(2)`


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