A) \[4\Omega \]
B) \[\frac{M{{R}^{2}}T}{2\pi }\]
C) \[\frac{M{{R}^{2}}T}{4\pi }\]
D) \[\frac{4\pi M{{R}^{2}}}{5T}\]
Correct Answer: A
Solution :
Since, \[\frac{1}{2}m{{V}^{2}}=mgh=mg(1-\cos \theta )\] \[\frac{1}{2}m\times 2gl=mgl-mgl\cos \theta \] But \[\cos \theta =0\,\,or\,\,\theta ={{90}^{o}}\]
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