When body rolls dawn on inclined plane with velocity V0 at bottom then body has both rotational and translational kinetic energy. Therefore, by law of conservation of energy, P.E.=K.Etrans +K.Erotational =21mV02+21Iω2=21mV02+21mk2R02V02… (i) [∵I=mk2,ω=R0V] When body is sliding down then body has only translatory motion. ∴P.E.=K.Etrans =21m(45v0)2 Dividing (i) by (ii) we get P.E.P.E.=21×1625×mV0221mv02[1+R02K2]=1625=1+R02K2⇒R02K2=169 or, K=43R0.