A ring of mass M and R is rotating about its axis with angular velocity `omega`. Two identical bodies each o mass m are now gently attached at the two ends of a diameter of the ring. Because of this the kinetic energy loss will be :

Options

A. `(m(M+2m))/(M)omega^(2)R^(2)`
B. `(Mm)/((M+m))omega^(2)R^(2)`
C. `(Mm)/((M+2m))omega^(2)R^(2)`
D. `((M+m)M)/((M+2m))omega^(2)R^(2)`

Answer

C