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\[I=M{{K}^{2}}\]. 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 \[I=M{{K}^{2}}\] \[\Rightarrow \] \[K=\sqrt{\frac{I}{M}}\] \[\therefore \] \[\frac{{{K}_{d}}}{{{K}_{r}}}=\sqrt{\frac{{{I}_{d}}}{{{I}_{r}}}\times \frac{{{M}_{r}}}{{{M}_{d}}}}\] 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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