Thank you for reporting, we will resolve it shortly
Q.
The rotational kinetic energy of a body is $E$. In the absence of external torque, if mass of the body is halved and radius of gyration doubled, then its rotational kinetic energy will be :-
Solution:
$E =\frac{1}{2} I \omega^{2} \because L = I \omega$
$E=I \omega \Rightarrow E=\frac{L^{2}}{2 I}$
Now, M.O.I. is given by
$I = mk ^{2}$
$L_{i}=L_{f}$
Angular momentum must remain same always now the mass is half and radius of gyration is double.
The M.O.I is given by
$E'=\frac{L^{2}}{2(2 m k)^{2}}$
As we compare it with initial energy
$E'=\frac{E}{2}$