A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration a꜀ is varying with time t as a꜀= k²rt² , where k is a constant. The power delivered to the particle by the forces acting on it is

The correct answer is mk²r²t.

Let's derive the power delivered to the particle step-by-step.

Step 1: Understand the motion and the acceleration components
The particle of mass $m$ moves in a circular path of constant radius $r$. The centripetal acceleration $a_c$ is given as a function of time $t$ by:
$a_c=k^{2}r{t}^2$
where $k$ is a constant.

Step 2: Relate centripetal acceleration to tangential velocity
We know that centripetal acceleration is related to the tangential speed $v$ of the particle and the radius $r$ by the formula:
$a_c=\frac{v^2}{r}$

Equating the two expressions for centripetal acceleration, we get:
$\frac{v^2}{r}=k^{2}r{t}^2$
Solving for $v^2$:
$v^2=k^2{r}^2{t}^2$
Taking the square root, the tangential speed is:
$v=krt$

Step 3: Determine the tangential acceleration
The tangential acceleration $a_t$ is the rate of change of tangential speed with respect to time:
$a_t=\frac{dv}{dt}$
Differentiating $v=krt$ with respect to $t$:
$a_t=\frac{d}{dt}\left(krt\right)=kr$

Step 4: Find the tangential force and the power delivered
The centripetal force acts perpendicular to the direction of motion, so it does no work and delivers zero power. Only the tangential force $F_t$ delivers power to the particle.
The tangential force is given by Newton's second law:
$F_t=m{a}_t=mkr$

The power $P$ delivered by the forces is the rate at which work is done, which is the product of the tangential force and the tangential velocity:
$P=F_t\cdot v$
Substituting the expressions for $F_t$ and $v$:
$P=\left(mkr\right)\cdot \left(krt\right)$
$P=m{k}^2{r}^{2}t$

Thus, the power delivered to the particle by the forces acting on it is $m{k}^2{r}^{2}t$.