A) \[mg(1-{{\theta }_{0}})\]
B) \[mg(1+{{\theta }_{0}})\]
C) \[mg(1-\theta \,_{0}^{2})\]
D) \[mg(1+\theta \,_{0}^{2})\]
Correct Answer: D
Solution :
\[{{T}_{\max }}=mg+\frac{m{{v}^{2}}}{l}\]You need to login to perform this action.
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