Physics · 2015
JEE Main · 11 April 2015, Shift 1 · Q4
A particle is moving in a circle of radius r under the action of a force F =α r^2 which is directed towards centre of the circle. Total mechanical…
A particle is moving in a circle of radius $\displaystyle r$ under the action of a force $\displaystyle \mathrm{F}=\alpha \mathrm{r}^2$ which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy $\displaystyle =0$ for $\displaystyle \mathrm{r}=0$ ) :
Official answer
(4)
$\displaystyle \frac{5}{6} \alpha r^3$
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JEE Main 2015 Physics question, with the answer from NTA’s final answer key. Where our answers come from.