A) \[2mk{{r}^{2}}t\]
B) \[mk{{r}^{2}}{{t}^{2}}\]
C) \[m{{k}^{2}}{{r}^{2}}t\]
D) \[m{{k}^{2}}r{{t}^{2}}\]
Correct Answer: B
Solution :
Tension in the string \[T-mg\cos \theta =\frac{m{{v}^{2}}}{L}\] T is maximum, when \[\theta ={{0}^{o}}\] \[\therefore \] \[{{T}_{\max }}=mg+\frac{m{{v}^{2}}_{\max }}{L}=mg+\frac{m{{a}^{2}}{{\omega }^{2}}}{L}\] \[\because \] a is amplitude \[{{T}_{\max }}=mg+\frac{m{{a}^{2}}}{L}\times \frac{g}{L}\] \[=mg\left[ 1+{{\left( \frac{a}{L} \right)}^{2}} \right]\]You need to login to perform this action.
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