A) \[{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}\]
B) \[\left( \frac{r}{\sqrt[3]{2}} \right)\]
C) \[\left( \frac{2r}{\sqrt{3}} \right)\]
D) \[\left( \frac{2r}{3} \right)\]
Correct Answer: B
Solution :
As \[Fe=mg\tan \theta \] We have\[Fe=\tan {{\theta }_{1}}\]and\[F_{e}^{'}=\tan \theta \] \[\therefore \] \[\frac{F_{e}^{'}}{{{F}_{e}}}=\frac{\tan {{\theta }_{1}}}{\tan {{\theta }_{1}}}\]You need to login to perform this action.
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