A) \[2\sqrt{3}\,W\]
B) \[\sqrt{3}\,W\]
C) \[\frac{\sqrt{3}}{2}W\]
D) \[\frac{\sqrt{3}}{4}W\]
Correct Answer: B
Solution :
Given, work done \[=W\]and \[\theta ={{60}^{o}}\] We know that \[W=MB(1-cos\theta )\] \[W=MB(1-cos{{60}^{o}})\] \[W=\frac{MB}{2}\] Hence, \[|\tau |=MB\,sin{{60}^{o}}=\sqrt{3}\,W\] | t | = MB sin 60° = V3 WYou need to login to perform this action.
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