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 \[I=M{{K}^{2}}\] |
| \[\Rightarrow \] \[K=\sqrt{\frac{1}{M}}\] |
| \[\frac{{{K}_{d}}}{{{K}_{r}}}=\sqrt{\frac{{{I}_{d}}}{{{I}_{r}}}\times \frac{{{M}_{r}}}{{{M}_{d}}}}\] |
| 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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