A) \[\sqrt{3}Mg\]
B) \[\sqrt{2}Mg\]
C) \[\frac{Mg}{\sqrt{3}}\]
D) \[\frac{Mg}{2}\]
Correct Answer: A
Solution :
[a] \[T\,sin\,{{30}^{{}^\circ }}=Mg\] \[T\,\cos \,{{30}^{{}^\circ }}={{T}_{1}}\] \[\tan \,{{30}^{{}^\circ }}=\,\frac{Mg}{{{T}_{1}}}\] \[{{T}_{1}}=\sqrt{3}Mg\]You need to login to perform this action.
You will be redirected in
3 sec