A particle is moving in a circle of radius r under the action $F = α r^{2}$ of a force which is directed towards centre of the circle. Total mechanical energy (kinetic energy+ potential energy) of the particle is (take potential energy $y = 0$ for $r = 0$ );

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$\frac{1}{2} α r^{2}$

$\frac{4}{3} α r^{3}$

$\frac{5}{6} α r^{3}$

$α r^{3}$

answer is B.

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Detailed Solution

As we know, $d U = - F . d r, U = \int_{0}^{r} α r^{2} d r = \frac{α r^{3}}{3} ...... (i)$

As, $\frac{m v^{2}}{r} = α r^{2}, m^{2} v^{2} = m α r^{3}, o r, (K E) = \frac{1}{2} α r^{3} ........ (i i)$

Total energy potential energy + kinetic energy Now, from equation (i) and (ii)
Total energy $= K . E + P . E = \frac{α r^{3}}{3} + \frac{α r^{3}}{2} = \frac{5}{6} α r^{3}$

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