In a binary star system one star has thrice the mass of other. The stars rotate about their common centre of mass then :

Options

A. Both stars have same angular momentum about common centre of mass
B. Both stars have angular momentum of same magnitude about common centre of mass
C. Both stars have same angular speeds
D. Both stars have same linear speeds

Explanation

To solve the problem regarding the binary star system where one star has thrice the mass of the other, we need to analyze the situation step by step:

### Step 1: Understanding the Masses
Let the mass of the smaller star be \( m \) and the mass of the larger star be \( 3m \).

### Step 2: Center of Mass Calculation
The center of mass (CM) of a two-body system can be found using the formula:
\[ r_{CM} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2} \]
where \( r_1 \) and \( r_2 \) are the distances of the masses from the center of mass.

Since the total mass is \( m + 3m = 4m \), we can set the distances from the center of mass as \( r_1 \) for mass \( m \) and \( r_2 \) for mass \( 3m \). The center of mass will be closer to the more massive star.

Using the ratio of distances:
\[ \frac{r_1}{r_2} = \frac{m_2}{m_1} = \frac{3m}{m} = 3 \]
Let \( r_1 = x \) and \( r_2 = 3x \). The total distance \( r \) between the two stars is:
\[ r = r_1 + r_2 = x + 3x = 4x \]
Thus, we can express \( r_1 \) and \( r_2 \) in terms of \( r \):
\[ r_1 = \frac{r}{4}, \quad r_2 = \frac{3r}{4} \]

### Step 3: Angular Speed and Linear Speed
In a binary star system, both stars rotate about their common center of mass with the same angular speed \( \omega \).

The linear speed \( v \) of each star is given by:
\[ v = \omega r \]
For the smaller star:
\[ v_1 = \omega r_1 = \omega \frac{r}{4} \]
For the larger star:
\[ v_2 = \omega r_2 = \omega \frac{3r}{4} \]
Since \( v_1 \neq v_2 \), the linear speeds are not the same.

### Step 4: Angular Momentum Calculation
The angular momentum \( L \) of each star about the center of mass is given by:
\[ L = I \omega \]
where \( I \) is the moment of inertia.

The moment of inertia \( I \) for each star is:
\[ I_1 = m r_1^2 = m \left(\frac{r}{4}\right)^2 = \frac{mr^2}{16} \]
\[ I_2 = 3m r_2^2 = 3m \left(\frac{3r}{4}\right)^2 = 3m \cdot \frac{9r^2}{16} = \frac{27mr^2}{16} \]

Now, calculating the angular momentum for both stars:
\[ L_1 = I_1 \omega = \frac{mr^2}{16} \omega \]
\[ L_2 = I_2 \omega = \frac{27mr^2}{16} \omega \]

Since \( L_1 \neq L_2 \), the angular momenta are not the same.

### Conclusion
From the analysis:
1. Both stars do not have the same linear speed.
2. Both stars do not have the same angular momentum.
3. Both stars have the same angular speed.

Thus, the only correct option is that both stars have the same angular speed.