A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration `a_(c )` is varying with time t as `a_(c ) = k^(2) rt^(2)`, where k is a constant. The power delivered to the particle by the force acting on its is

Options

A. `2pi m k^(2)r^(2)t`
B. `mk^(2)r^(2)t `
C. `(mk^(4)r^(2)t^(5))//3`
D. zero

Explanation

To solve the problem step by step, we will derive the power delivered to the particle by the force acting on it.

### Step 1: Understand the given centripetal acceleration
We are given the centripetal acceleration of the particle as:
\[ a_c = k^2 r t^2 \]
where \( k \) is a constant, \( r \) is the radius, and \( t \) is time.

### Step 2: Relate centripetal acceleration to velocity
Centripetal acceleration can also be expressed in terms of velocity \( v \) and radius \( r \):
\[ a_c = \frac{v^2}{r} \]
Setting the two expressions for centripetal acceleration equal gives:
\[ \frac{v^2}{r} = k^2 r t^2 \]

### Step 3: Solve for velocity \( v \)
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = k^2 r^2 t^2 \]
Taking the square root of both sides gives:
\[ v = k r t \]

### Step 4: Find tangential acceleration \( a_t \)
Tangential acceleration \( a_t \) is the rate of change of velocity with respect to time:
\[ a_t = \frac{dv}{dt} \]
Substituting \( v = k r t \):
\[ a_t = \frac{d(k r t)}{dt} = k r \]

### Step 5: Calculate the tangential force \( F_t \)
The tangential force \( F_t \) acting on the particle is given by:
\[ F_t = m a_t \]
Substituting \( a_t = k r \):
\[ F_t = m k r \]

### Step 6: Calculate the power \( P \)
Power \( P \) delivered to the particle by the force is given by the product of force and velocity:
\[ P = F_t \cdot v \]
Substituting \( F_t = m k r \) and \( v = k r t \):
\[ P = (m k r)(k r t) = m k^2 r^2 t \]

### Final Answer
Thus, the power delivered to the particle by the force acting on it is:
\[ P = m k^2 r^2 t \]