A binary star has a time period `3` years (time period of earth is one year) while distance between the two stars is 9 times distance between earth and the sun. Mass of one star is equal to mass of the sun and mass of other is `20n` times mass of the sun then calculate `n`.

Explanation

To solve the problem, we will use the concepts of gravitational force, centripetal force, and the relationship between the time period and angular speed of the binary star system.

### Step-by-Step Solution:

1. **Understanding the Given Data:**
- Time period of the binary star system, \( T_2 = 3 \) years.
- Time period of the Earth around the Sun, \( T_1 = 1 \) year.
- Distance between the two stars, \( d = 9R \) (where \( R \) is the distance between the Earth and the Sun).
- Mass of one star, \( M_1 = M_{sun} \).
- Mass of the other star, \( M_2 = 20n \times M_{sun} \).

2. **Relating Time Periods to Angular Speeds:**
\[ \frac{T_1}{T_2} = \frac{\omega_2}{\omega_1} \]
Substituting the values:
\[ \frac{1}{3} = \frac{\omega_2}{\omega_1} \implies \omega_2 = \frac{1}{3} \omega_1 \]

3. **Using the Gravitational Force as Centripetal Force:**
For the Earth-Sun system:
\[ \frac{G M_{sun} M_{earth}}{R^2} = M_{earth} \omega_1^2 R \]
Simplifying, we get:
\[ \omega_1^2 = \frac{G M_{sun}}{R^3} \]

4. **For the Binary Star System:**
The stars rotate around their center of mass. The distance between them is \( 9R \). Let \( r_1 \) and \( r_2 \) be the distances of the stars from the center of mass.
\[ r_1 + r_2 = 9R \]
Using the center of mass formula:
\[ \frac{M_1 r_1}{M_2 r_2} = 1 \implies r_1 = \frac{M_2}{M_1 + M_2} \times 9R \]
Substituting \( M_1 \) and \( M_2 \):
\[ r_1 = \frac{20n \times M_{sun}}{M_{sun} + 20n \times M_{sun}} \times 9R = \frac{20n}{1 + 20n} \times 9R \]
Thus, \( r_2 = 9R - r_1 = \frac{9R}{1 + 20n} \).

5. **Centripetal Force for the Binary System:**
The gravitational force between the two stars provides the centripetal force:
\[ \frac{G M_1 M_2}{(9R)^2} = M_2 \omega_1^2 r_1 \]
Substituting the values:
\[ \frac{G M_{sun} (20n M_{sun})}{81R^2} = (20n M_{sun}) \omega_1^2 \left(\frac{20n}{1 + 20n} \times 9R\right) \]
Simplifying, we find:
\[ \frac{20n G M_{sun}^2}{81R^2} = \frac{20n^2 \omega_1^2 M_{sun} \times 9R}{1 + 20n} \]

6. **Relating Angular Speeds:**
From the earlier steps, we have:
\[ \omega_2^2 = \frac{G M_{sun}}{(9R)^3} \implies \omega_1^2 = \frac{G M_{sun}}{R^3} \]
Therefore:
\[ \left(\frac{1}{3} \omega_1\right)^2 = \frac{G M_{sun}}{729R^3} \]

7. **Equating the Two Expressions:**
Now, we can equate the expressions for \( \omega_1^2 \) and solve for \( n \):
\[ \frac{20n G M_{sun}^2}{81R^2} = \frac{20n^2 G M_{sun}}{729R^3} \cdot 9R \]
Simplifying gives:
\[ 20n = 80n^2 \implies n = 4 \]

### Final Answer:
Thus, the value of \( n \) is \( 4 \).