Question:
A particle is moving in a circle of radius $\mathrm{r}$ under the action of a force $\mathrm{F}=\alpha \mathrm{r}^{2}$ which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy $=0$ for $\mathrm{r}=0$ ) :
$\frac{1}{2} \alpha r^{3}$$
$\frac{5}{6} \alpha \mathrm{r}^{3}$
$\frac{4}{3} \alpha r^{3}$
$\alpha \mathrm{r}^{3}
Question of from chapter.
JEE Main Previous Year April 11, 2015
Correct Option: 2
Solution:
Related Questions
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