A) \[mg\]
B) \[3mg\]
C) \[5\,mg\]
D) \[6\,mg\]
Correct Answer: B
Solution :
\[T=mg+\frac{m{{v}^{2}}}{l}\] \[=mg+2mg=3mg\] where \[v=\sqrt{2gl}\] from \[\frac{1}{2}m{{v}^{2}}=mgl\]You need to login to perform this action.
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