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 energy (kinetic energy + potential energy) of the particle is (take potential energy $= 0 f o r r = 0$ ):

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$\frac{4}{3} α r^{3}$

$\frac{1}{2} α r^{3}$

$\frac{5}{6} α r^{3}$

$α r^{3}$

answer is B.

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Detailed Solution

As we know $d U = F . d r$

$U = \int_{0}^{r} α r^{2} d r = \frac{α r^{3}}{3} \dots \dots \dots \dots (i)$

As, $\frac{m v^{2}}{r} = α r^{2} \Rightarrow m^{2} v^{2} = m α r^{3}$

or $2 m (K E) = \frac{1}{2} α r^{3} \dots \dots \dots . (i i)$

Total energy = Potential energy + kinetic energy
Now, from eqn (i) and (ii)
The 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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