A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the `n^(th)` power of R. If the period of rotation of the particle is T, then :

Options

A. `T prop R^(n//2)`
B. `T prop R^(3//2)` for any n.
C. `T prop R^((n)/(2)+1)`
D. `T prop R^((n+1)//2)`

Answer

To solve the problem, we need to analyze the motion of a particle moving in a circular orbit under the influence of a central force that is inversely proportional to the \( n^{th} \) power of the radius \( R \).

### Step-by-Step Solution:

1. **Understanding the Forces**:
The centripetal force required to keep a particle moving in a circular path is given by:
\[ F_c = \frac{mv^2}{R} \]
where \( m \) is the mass of the particle, \( v \) is its speed, and \( R \) is the radius of the circular path.

2. **Expressing the Central Force**:
According to the problem, the central force \( F \) is inversely proportional to \( R^n \):
\[ F = \frac{k}{R^n} \]
where \( k \) is a constant of proportionality.

3. **Setting the Forces Equal**:
Since the centripetal force is provided by the central force, we can set them equal:
\[ \frac{mv^2}{R} = \frac{k}{R^n} \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ mv^2 = \frac{k}{R^{n-1}} \]
From this, we can express \( v^2 \):
\[ v^2 = \frac{k}{m} \cdot \frac{1}{R^{n-1}} \]

5. **Finding the Velocity**:
Taking the square root, we find the velocity \( v \):
\[ v = \sqrt{\frac{k}{m}} \cdot \frac{1}{R^{(n-1)/2}} \]

6. **Finding the Time Period \( T \)**:
The time period \( T \) for one complete revolution is given by:
\[ T = \frac{2\pi R}{v} \]
Substituting the expression for \( v \):
\[ T = \frac{2\pi R}{\sqrt{\frac{k}{m}} \cdot \frac{1}{R^{(n-1)/2}}} \]

7. **Simplifying the Expression**:
This simplifies to:
\[ T = 2\pi R \cdot \frac{R^{(n-1)/2}}{\sqrt{\frac{k}{m}}} \]
\[ T = 2\pi \cdot \frac{R^{(n+1)/2}}{\sqrt{\frac{k}{m}}} \]

8. **Final Result**:
Thus, we find that the time period \( T \) is proportional to \( R^{(n+1)/2} \):
\[ T \propto R^{(n+1)/2} \]

### Conclusion:
The correct answer is that the period of rotation \( T \) is proportional to \( R^{(n+1)/2} \).