A particle of mass in is moving in a circular with of constant radius `r` such that its contripetal accelenation `a_(c) ` is varying with time `t` as `a_(c) = K^(2) rt^(2)` where K` is a constant . The power delivered to the particles by the force action on it is

Options

A. (a) zero
B. (b) `mk^2r^2t^2`
C. (c) `mk^2r^2t`
D. (d) `mk^2rt`

Explanation

To solve the problem, we need to find the power delivered to a particle of mass \( m \) moving in a circular path of constant radius \( r \), where the centripetal acceleration \( a_c \) is given by \( a_c = K^2 r t^2 \).

### Step-by-Step Solution:

1. **Understanding Centripetal Acceleration**:
The centripetal acceleration \( a_c \) for an object moving in a circular path is given by the formula:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the tangential velocity and \( r \) is the radius of the circular path.

2. **Relating Centripetal Acceleration to Velocity**:
From the given equation \( a_c = K^2 r t^2 \), we can equate the two expressions for centripetal acceleration:
\[ \frac{v^2}{r} = K^2 r t^2 \]
Multiplying both sides by \( r \) gives:
\[ v^2 = K^2 r^2 t^2 \]

3. **Finding Tangential Velocity**:
Taking the square root of both sides, we find the tangential velocity \( v \):
\[ v = K r t \]

4. **Finding Tangential Acceleration**:
The tangential acceleration \( a_t \) is the rate of change of velocity. Since \( v = K r t \), we differentiate with respect to time \( t \):
\[ a_t = \frac{dv}{dt} = K r \]

5. **Calculating Tangential Force**:
The tangential force \( F_t \) acting on the particle can be calculated using Newton's second law:
\[ F_t = m a_t = m (K r) \]

6. **Calculating Power**:
Power \( P \) delivered to the particle by the tangential force is given by the formula:
\[ P = F_t \cdot v \]
Substituting the expressions for \( F_t \) and \( v \):
\[ P = (m K r) \cdot (K r t) = m K^2 r^2 t \]

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