AIIMS 2017 Physics Centripetal Force and Acceleration MCQ Question

To find the power delivered to the particle, we first determine the tangential acceleration $a_t$ using the relationship between centripetal acceleration $a_p$ and the radius $r$. Given $a_p = k^2 r t^2$, we can express the tangential velocity $v$ as $v = r \omega$ where $\omega$ is the angular velocity. The tangential acceleration $a_t$ can be derived from the change in $v$, leading to $a_t = \frac{dv}{dt} = k^2 r t$. The power $P$ is then calculated as $P = F_t v = m a_t v = m (k^2 r t) (r \omega) = mk^2 r^2 t$, confirming option B.

Options A and C are incorrect because they either miscalculate the relationship between force, acceleration, and velocity or introduce unnecessary variables, while option D is incorrect as there is indeed power being delivered due to the tangential acceleration.