A) \[\frac{mg\,\cos \,\theta }{\pi iR}\]
B) \[\frac{mg\,}{\pi iR}\]
C) \[\frac{mg\,\tan \,\theta }{\pi iR}\]
D) \[\frac{mg\,\sin \,\theta }{\pi iR}\]
Correct Answer: B
Solution :
[b] For rotational equilibrium of the sphere about point P. \[(mg\,\sin \,\theta )R=M(B\,\sin \,\theta )\] or, \[mgR\,\sin \,\theta =i\pi {{R}^{2}}\,B\sin \,\theta \] \[B=mg/\pi iR\]You need to login to perform this action.
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