A) 2 : 3
B) 2 : 1
C) \[\sqrt{5}:\sqrt{6}\]
D) 1 : \[\sqrt{2}\]
Correct Answer: C
Solution :
Key Idea: If a body has mass M and radius of gyration is K, then \[\text{I}=\text{M}{{\text{K}}^{\text{2}}}.\] Moment of inertia of a disc and circular ring about a tangential axis in their planes are respectively. \[{{I}_{d}}=\frac{5}{4}{{M}_{d}}{{R}^{2}}\] \[{{I}_{r}}=\frac{3}{2}{{M}_{r}}{{R}^{2}}\] but\[I=M{{K}^{2}}\] \[\Rightarrow \]\[K=\sqrt{\frac{1}{M}}\] \[\frac{{{K}_{d}}}{{{K}_{r}}}=\sqrt{\frac{{{I}_{d}}}{{{I}_{r}}}\times \frac{{{M}_{r}}}{{{M}_{d}}}}\] or\[\frac{{{I}_{d}}}{{{I}_{r}}}\sqrt{\frac{(5/4){{M}_{d}}{{R}^{2}}}{(3/2){{M}_{r}}{{R}^{2}}}\times \frac{{{M}_{r}}}{{{M}_{d}}}}=\sqrt{\frac{5}{6}}\] \[\therefore \]\[{{I}_{d}}:{{I}_{r}}=\sqrt{5}:\sqrt{6}\]
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