A) \[\frac{1}{2}\alpha {{r}^{3}}\]
B) \[\frac{5}{6}\alpha {{r}^{3}}\]
C) \[\frac{4}{3}\alpha {{r}^{3}}\]
D) \[\alpha {{r}^{3}}\]
Correct Answer: B
Solution :
[b] As we know, dU = F.dr \[U=\int\limits_{0}^{r}{\alpha {{r}^{2}}dr=\frac{a{{r}^{3}}}{3}}\] ?(i) As, \[\frac{m{{v}^{2}}}{r}=\alpha {{r}^{2}}\] \[{{m}^{2}}{{v}^{2}}=m\alpha {{r}^{3}}\] or, \[2m(KE)=\frac{1}{2}\alpha {{r}^{3}}\] ...(ii) Total energy = Potential energy + kinetic energy Now, from eqn (i) and (ii) Total energy = K.E. + RE. \[=\frac{\alpha {{r}^{3}}}{3}+\frac{\alpha {{r}^{3}}}{2}=\frac{5}{6}\alpha {{r}^{3}}\]
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