The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is

Solution :
Moment of inertia of a disc and circular ring about a tangential axis in their planes are respectively.
Momentum inertia of disc about tangential axis $${I_d} = \frac{5}{4}{M_d}{R^2}$$
Moment of inertia of ring about a tangential axis $${I_r} = \frac{3}{2}{M_r}{R^2}$$
$$\eqalign{ & {\text{but}}\,\,I = M{k^2} \Rightarrow k = \sqrt {\frac{I}{M}} \cr & \therefore \frac{{{k_d}}}{{{k_r}}} = \sqrt {\frac{{{I_d}}}{{{I_r}}} \times \frac{{{M_r}}}{{{M_d}}}} \cr & {\text{or}}\,\,\frac{{{k_d}}}{{{k_r}}} = \sqrt {\frac{{\left( {\frac{5}{4}} \right){M_d}{R^2}}}{{\left( {\frac{3}{2}} \right){M_r}{R^2}}} \times \frac{{{M_r}}}{{{M_d}}}} = \sqrt {\frac{5}{6}} \cr & \therefore {k_d}:{k_r} = \sqrt 5 :\sqrt 6 \cr} $$