A) \[\sqrt{3}\,W\]
B) W
C) \[\frac{\sqrt{3}}{2}W\]
D) 2W
Correct Answer: A
Solution :
\[W=MB(\cos {{\theta }_{1}}-\cos {{\theta }_{2}})=MB(\cos {{0}^{o}}-\cos {{60}^{o}})\] \[=MB\,\left( 1-\frac{1}{2} \right)=\frac{MB}{2}\] and \[\tau =MB\sin \theta =MB\sin {{60}^{o}}=MB\frac{\sqrt{3}}{2}\] \[\therefore \ \tau =\left( \frac{MB}{2} \right)\sqrt{3}\Rightarrow \tau =\sqrt{3}\,W\]
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