A) \[{{v}_{B}}>{{v}_{D}}>{{v}_{A}}\]
B) tension in string at D = 3 mg
C) \[{{v}_{D}}=\sqrt{3gl}\]
D) All of the above
Correct Answer: D
Solution :
At A, \[{{v}_{A}}=\sqrt{gl}\] At B, \[{{v}_{B}}=\sqrt{5gl}\] and at D, \[{{v}_{D}}=\sqrt{3gl}\] Thus, \[{{v}_{B}}>{{v}_{D}}>{{v}_{A}}\] Also, \[T=3mg\,(1+\cos \theta )\] So, at D, \[\theta ={{90}^{o}}\] \[\therefore \] T = 3mg (1 + 0) = 3mg
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