A) \[\frac{mg}{2\pi }\]
B) \[\frac{mg}{2}\]
C) \[\frac{\sqrt{3}mg}{\pi }\]
D) \[\frac{\sqrt{3}mg}{2\pi }\]
Correct Answer: D
Solution :
[d] For equilibrium of an elemental part 'dm? \[2T\left( \frac{d\theta }{2} \right)\sin \alpha =dmg\cos \alpha \] \[Td\theta =\left( \frac{m}{2\pi }d\theta g \right)\cot \alpha \] \[T=\frac{mg}{2\pi }\left( \frac{\sqrt{3}R}{2(R/2)} \right)\] \[T=\frac{\sqrt{3}mg}{2\pi }\]You need to login to perform this action.
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