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 :
\[{{T}_{\max }}=mg+\frac{m{{v}^{2}}}{l}\] \[=mg+\frac{2mg}{l}l(1-\cos {{\theta }_{0}})\] \[=mg+2mg\left[ 1-1+\frac{\theta _{0}^{2}}{2} \right]\] \[=mg(1+\theta _{0}^{2})\]You need to login to perform this action.
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