A) 1 : 2
B) \[\sqrt{5}:\sqrt{6}\]
C) 2 : 3
D) 2 : 1
Correct Answer: B
Solution :
\[{{I}_{ring}}=\frac{m{{r}^{2}}}{2}+m{{r}^{2}}=\frac{3}{2}m{{r}^{2}}\] \[{{I}_{disc}}=\frac{m{{r}^{2}}}{4}+m{{r}^{2}}=\frac{5}{4}m{{r}^{2}}\] \[\frac{{{I}_{disc}}}{{{I}_{ring}}}=\frac{\frac{5\,m{{r}^{2}}}{4}}{\frac{3}{2}m{{r}^{2}}}=\frac{5}{6}\] \[\frac{mK_{disc}^{2}}{mK_{ring}^{2}}=\frac{5}{6}\] \[\Rightarrow \] \[\frac{K_{disc}^{2}}{K_{ring}^{2}}=\sqrt{\frac{5}{6}}\]
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