A) \[2\sqrt{3}J\]
B) \[3J\]
C) \[\sqrt{3}J\]
D) \[\frac{3J}{2}\]
Correct Answer: B
Solution :
[b] \[W=MB\,[cos{{\theta }_{1}}-cos{{\theta }_{2}}]\] \[=MB\,\,[cos{{0}^{0}}-cos{{60}^{0}}]\,\therefore {{\theta }_{1}}=0,{{\theta }_{2}}=60\] \[W=MB\,\left[ \frac{1}{2} \right]\] \[MB=\,2W\] \[MB=\,\,=2\sqrt{3}\therefore W=\sqrt{3}\] \[\tau =MB\,\cos \theta \,(\theta ={{60}^{0}})\] \[\tau =MB\,\sin \,{{60}^{0}}\] \[\tau =\frac{\sqrt{3}}{2}MB\] \[\tau =\frac{\sqrt{3}}{2}\times 2\sqrt{3}\] \[\tau =3J\]You need to login to perform this action.
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