A) \[\frac{1}{4}\]
B) \[\frac{4}{5}\]
C) \[\frac{2}{3}\]
D) \[\frac{1}{2}\]
Correct Answer: B
Solution :
\[{{T}_{disc}}=2\pi \sqrt{\frac{1}{mgd}}=2\pi \sqrt{\frac{5R}{4g}}\] \[{{T}_{pendulum}}=2\pi \sqrt{\frac{l}{g}}\] \[\therefore \] \[\frac{R}{l}=\frac{4}{5}\]You need to login to perform this action.
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