A body moves in a circular orbit of radius R under the action of a central force. Potential due to the central force is given by V(r) = kr (k is a positive constant). Period of revolution of the body is proportional to-

Options

A. `R^(1//2)`
B. `R^(-1//2)`
C. `R^(-3//2)`
D. `R^(-5//2)`

Explanation

To solve the problem, we need to determine how the period of revolution \( T \) of a body moving in a circular orbit under a central force relates to the radius \( R \) of the orbit, given that the potential \( V(r) = kr \), where \( k \) is a positive constant.

### Step-by-Step Solution:

1. **Identify the Potential Energy**:
The potential energy due to the central force is given by:
\[ V(r) = kr \]

2. **Calculate the Central Force**:
The force \( F \) associated with this potential is obtained by taking the negative gradient of the potential energy:
\[ F = -\frac{dV}{dr} = -\frac{d(kr)}{dr} = -k \]
Since we are dealing with a central force, we consider the magnitude:
\[ F = k \]

3. **Equate to Centripetal Force**:
For a body moving in a circular path of radius \( R \) with mass \( m \) and speed \( v \), the centripetal force required is given by:
\[ F_c = \frac{mv^2}{R} \]
Setting the central force equal to the centripetal force gives:
\[ k = \frac{mv^2}{R} \]

4. **Solve for Velocity**:
Rearranging the equation to solve for \( v^2 \):
\[ mv^2 = kR \implies v^2 = \frac{kR}{m} \]
Taking the square root gives:
\[ v = \sqrt{\frac{kR}{m}} \]

5. **Determine the Period of Revolution**:
The period \( T \) of revolution is given by the formula:
\[ T = \frac{2\pi R}{v} \]
Substituting the expression for \( v \):
\[ T = \frac{2\pi R}{\sqrt{\frac{kR}{m}}} \]
Simplifying this expression:
\[ T = 2\pi R \cdot \sqrt{\frac{m}{kR}} = 2\pi \sqrt{\frac{m}{k}} \cdot \sqrt{R} \]

6. **Final Expression**:
Therefore, we find that the period \( T \) is proportional to \( R^{1/2} \):
\[ T \propto R^{1/2} \]

### Conclusion:
The period of revolution \( T \) of the body is proportional to the square root of the radius \( R \) of its circular orbit.