A) \[\alpha {{r}^{3}}\]
B) \[\frac{1}{2}\alpha {{r}^{3}}\]
C) \[\frac{4}{3}\alpha {{r}^{3}}\]
D) \[\frac{5}{6}\alpha {{r}^{3}}\]
Correct Answer: D
Solution :
\[\frac{m{{V}^{2}}}{r}=\alpha {{r}^{2}}\] \[\therefore \]\[K.E.=\frac{\alpha {{r}^{3}}}{2}\] \[\Delta P.E=\int\limits_{0}^{r}{\alpha {{r}^{2}}.dr}\] \[P.E=\frac{\alpha {{r}^{3}}}{3}\] \[T.E=\frac{\alpha {{r}^{3}}}{2}+\frac{\alpha {{r}^{3}}}{3}\] \[\] \[T.E=\frac{5}{6}\alpha {{r}^{3}}\]You need to login to perform this action.
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