A particle is moving with uniform speed along the circumference of a circle of radius R under the action of a central fictitious force F which is inversely proportional to `R^3`. its time period of revolution will be given by :

Options

A. `TpropR^2`
B. `TpropR^(3/2)`
C. `TpropR^(5/2)`
D. `TpropR^(4/3)`

Answer

To solve the problem, we need to find the time period of a particle moving in a circle of radius \( R \) under the influence of a central fictitious force \( F \) that is inversely proportional to \( R^3 \).

### Step-by-step Solution:

1. **Understanding the Forces**:
The particle is moving in a circle of radius \( R \) with uniform speed \( V \). The centripetal force \( F_c \) required to keep the particle in circular motion is given by:
\[ F_c = \frac{mV^2}{R} \]
where \( m \) is the mass of the particle.

2. **Fictitious Force**:
We are given that the fictitious force \( F \) is inversely proportional to \( R^3 \). We can express this as:
\[ F = \frac{k}{R^3} \]
where \( k \) is a constant.

3. **Equating Forces**:
For the particle to move in a circle, the fictitious force must equal the centripetal force:
\[ \frac{k}{R^3} = \frac{mV^2}{R} \]

4. **Solving for Velocity**:
Rearranging the equation gives:
\[ k = mV^2 R^2 \]
From this, we can express \( V^2 \):
\[ V^2 = \frac{k}{m} \cdot \frac{1}{R^2} \]
Taking the square root gives us the velocity:
\[ V = \sqrt{\frac{k}{m}} \cdot \frac{1}{R} \]

5. **Finding the Time Period**:
The time period \( T \) of revolution is given by the circumference of the circle divided by the velocity:
\[ T = \frac{2\pi R}{V} \]
Substituting for \( V \):
\[ T = \frac{2\pi R}{\sqrt{\frac{k}{m}} \cdot \frac{1}{R}} = \frac{2\pi R^2}{\sqrt{\frac{k}{m}}} \]

6. **Identifying the Proportionality**:
We can express the time period \( T \) as:
\[ T \propto R^2 \]
This indicates that the time period is directly proportional to the square of the radius.

### Final Answer:
The time period of revolution \( T \) is directly proportional to \( R^2 \).