A particle of mass m is moving in a circle of radius r. The centripetal acceleration `(a_(c))` of the particle varies with the time according to the relation, `a_(c)=Kt^(2)`, where K is a positive constant and t is the time. The magnitude of the time rate of change of angular momentum of the particle about the centre of the circle is

Options

A. mKr
B. `sqrt(m^(2)Kr^(3))`
C. `sqrt(mKr)`
D. `mKr^(2)`

Explanation

To solve the problem, we need to find the magnitude of the time rate of change of angular momentum of a particle moving in a circle with varying centripetal acceleration.

### Step-by-Step Solution:

1. **Understanding Centripetal Acceleration**:
The centripetal acceleration \( a_c \) is given by the formula:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the linear velocity of the particle and \( r \) is the radius of the circle.

2. **Given Relation**:
We have the relation for centripetal acceleration as:
\[ a_c = Kt^2 \]
where \( K \) is a positive constant and \( t \) is time.

3. **Equating the Two Expressions for Centripetal Acceleration**:
From the two expressions for centripetal acceleration, we can equate them:
\[ \frac{v^2}{r} = Kt^2 \]
Rearranging gives us:
\[ v^2 = Kr t^2 \]
Hence, the velocity \( v \) can be expressed as:
\[ v = \sqrt{K r} t \]

4. **Finding Angular Momentum**:
The angular momentum \( L \) of a particle about the center of the circle is given by:
\[ L = mvr \]
Substituting the expression for \( v \):
\[ L = m(\sqrt{K r} t) r = m \sqrt{K r} t r = m r \sqrt{K r} t \]

5. **Differentiating Angular Momentum with Respect to Time**:
To find the rate of change of angular momentum, we differentiate \( L \) with respect to time \( t \):
\[ \frac{dL}{dt} = \frac{d}{dt}(m r \sqrt{K r} t) = m r \sqrt{K r} \frac{d}{dt}(t) = m r \sqrt{K r} \]

6. **Final Result**:
Thus, the magnitude of the time rate of change of angular momentum of the particle about the center of the circle is:
\[ \frac{dL}{dt} = m r \sqrt{K r} \]

### Conclusion:
The magnitude of the time rate of change of angular momentum of the particle about the center of the circle is \( m r \sqrt{K r} \).

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