A) \[1:\sqrt{2}\]
B) \[\sqrt{2}:1\]
C) \[1:2\]
D) \[2:1\]
Correct Answer: B
Solution :
\[{{t}_{1}}=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}\left( 1+\frac{{{K}^{2}}}{{{r}^{2}}} \right)}\] For ring \[=\frac{{{K}^{2}}}{{{r}^{2}}}=1\] so \[{{t}_{1}}=\frac{1}{\sin \theta }\sqrt{\frac{4h}{g}}\] In sliding \[\frac{{{K}^{2}}}{{{r}^{2}}}=0\] \[{{t}_{2}}=\frac{I}{\sin \theta }\sqrt{\frac{2h}{g}}\] so \[{{t}_{1}}:{{t}_{2}}=\sqrt{2}:1\]You need to login to perform this action.
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