A particle of mass M is moving in a circle of fixed radius R in such a way that its centripetal acceleration at time t is given by n2 R t2 where n is a constant. The power delivered to the particle by the force acting on it, is :

Work Power and Energy PYQ — JEE Main 2016 · 10 Apr · Shift 1 · Q61 with Solution

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Correct answer: A

For motion in a circle of radius $R$ , the centripetal acceleration is

$a_c = \frac{v^2}{R}$

Given:

$a_c = n^2 R t^2$

So,

$\frac{v^2}{R} = n^2 R t^2$

$v^2 = n^2 R^2 t^2$

Taking positive speed,

$v = nRt$

Tangential acceleration is

$a_t = \frac{dv}{dt} = \frac{d}{dt}(nRt) = nR$

Instantaneous power is

$P = \vec F \cdot \vec v$

Only the tangential component of force contributes to power, since centripetal force is perpendicular to velocity.

Thus,

$P = (Ma_t) v$

Substitute $a_t = nR$ and $v = nRt$ :

$P = M(nR)(nRt)$

$P = Mn^2R^2 t$

The correct expression is

$P = Mn^2R^2 t$

So the correct option is A.

Stored correct answer: A

Our derived answer: A

They match.