Correct Answer - Option 3 :
\(\sqrt 5 \;:\;\sqrt 6\)
MI of a circular disc about a tangential axis in the plane of the disc,
\({I_1}\; = \;MK_1^2\; = \;\frac{5}{4}M{R^2}\)
\(\Rightarrow {K_1}\; = \;\sqrt {\frac{5}{4}} \;R\)
MI of a circular ring of the same radius about tangential axis in the plane of ring
\({I_2}\; = \;MK_2^2\; = \;\frac{3}{2}M{R^2}\)
\({K_2}\; = \;\sqrt {\frac{3}{2}} R\)
\(\frac{{{k_1}}}{{{k_2}}}\; = \;\frac{{\sqrt {\frac{5}{4}} R}}{{\sqrt {\frac{3}{2}} R}}\; = \;\frac{{\sqrt 5 }}{{\sqrt 6 }}\; = \;\sqrt 5 \;:\;\sqrt 6\)