A) mg
B) mg \[\left[ 1+{{\left( \frac{a}{L} \right)}^{2}} \right]\]
C) \[mg{{\left[ 1+\left( \frac{a}{2L} \right) \right]}^{2}}\]
D) \[mg{{\left[ 1+\left( \frac{a}{L} \right) \right]}^{2}}\]
Correct Answer: B
Solution :
\[T-mg\cos \theta =\frac{m{{v}^{2}}}{L}\] \[\therefore \] \[{{T}_{\max }}=mg+\frac{mv_{\max }^{2}}{L}=mg+\frac{m{{a}^{2}}{{\omega }^{2}}}{L}\] \[=mg+\frac{m{{a}^{2}}}{L}\times \frac{g}{L}=mg\left[ 1+{{\left( \frac{a}{L} \right)}^{2}} \right]\]You need to login to perform this action.
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