A) \[\frac{Mgd}{4}\]
B) \[\frac{-Mgd}{4}\]
C) \[\frac{3Mgd}{4}\]
D) \[\frac{-3Mgd}{4}\]
Correct Answer: C
Solution :
Tension in the chord is \[\frac{1}{2}\,m{{v}^{2}}=16\,\,J\]in upward direction, since Chord is displaced in the downward direction, so \[v=4\,m{{s}^{-1}}\] \[Mg\,\,\sin \,\,\theta \] \[Mg\,\,\sin \theta \times \frac{h}{2}Mg\,\cos \,\theta \,\times r\] Here, negative sign shows that T and dare in the opposite directions.You need to login to perform this action.
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