A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius R `(R lt lt L)` A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the galaxy and passing through its centre. If the time period of star is T and its distance from the galaxy's axis is r, then-

Options

A. `Tpropsqrt(r)`
B. `Tpropr`
C. `Tpropr^(2)`
D. `T^(2)propr^(3)`

Explanation

To solve the problem, we need to find the relationship between the time period \( T \) of a star orbiting a long cylindrical galaxy and its distance \( r \) from the axis of the galaxy. We will use the principles of gravitation and centripetal motion.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a long cylindrical galaxy of radius \( R \) and length \( L \) with uniformly distributed mass. A star is orbiting around this galaxy at a distance \( r \) from the axis.

2. **Using Gauss's Law**:
- To find the gravitational field \( g \) at the distance \( r \) from the axis of the cylinder, we can apply Gauss's law for gravitation. The gravitational field \( g \) is directed towards the center of the cylinder.

3. **Setting Up the Gaussian Surface**:
- Consider a cylindrical Gaussian surface of radius \( r \) and length \( L \). The mass enclosed by this surface can be expressed in terms of the mass per unit length \( \lambda \) of the galaxy.
- The mass enclosed by the Gaussian surface is \( M_{\text{enc}} = \lambda L \).

4. **Calculating the Gravitational Field**:
- According to Gauss's law:
\[ \oint g \cdot dA = -M_{\text{enc}} \]
- The area \( A \) of the curved surface of the cylinder is \( 2 \pi r L \). Thus, we have:
\[ g \cdot (2 \pi r L) = -\lambda L \]
- Rearranging gives:
\[ g = -\frac{\lambda}{2 \pi r} \]
- Since \( g \) is directed inward, we can write:
\[ g = \frac{2 G \lambda}{r} \]
- Here, \( G \) is the gravitational constant.

5. **Relating Gravitational Force to Centripetal Force**:
- The gravitational force acting on the star provides the necessary centripetal force for circular motion:
\[ F_{\text{gravity}} = m g \]
- This force is equal to the centripetal force \( F_{\text{centripetal}} = m \frac{v^2}{r} \), where \( v \) is the orbital velocity of the star.

6. **Expressing Velocity in Terms of Time Period**:
- The velocity \( v \) can be expressed in terms of the time period \( T \):
\[ v = \frac{2 \pi r}{T} \]
- Substituting this into the centripetal force equation gives:
\[ m g = m \frac{(2 \pi r)^2}{T^2 r} \]
- Simplifying this leads to:
\[ g = \frac{4 \pi^2 r}{T^2} \]

7. **Equating the Two Expressions for \( g \)**:
- We have two expressions for \( g \):
\[ \frac{2 G \lambda}{r} = \frac{4 \pi^2 r}{T^2} \]
- Rearranging gives:
\[ T^2 = \frac{2 \pi^2 r^2}{G \lambda} \]

8. **Conclusion**:
- From the above relationship, we can see that \( T^2 \) is proportional to \( r^2 \):
\[ T^2 \propto r^2 \quad \Rightarrow \quad T \propto r \]

### Final Answer:
The time period \( T \) of the star is directly proportional to its distance \( r \) from the axis of the galaxy.