A circular ring of mass 6kg and radius a is placed such that its center of lies at the origin. Two particles of masses 2 kg each are placed at the intersecting points of the circle with positive X-axis and positive Y-axis. Then, the angle made by the position vector of center of mass of entire system with X-axis is

Options

A. `45^(@)`
B. `60^(@)`
C. `tan ^(-)(4/5)`
D. `30^(@)`

Explanation

To solve the problem of finding the angle made by the position vector of the center of mass of the entire system with the X-axis, we can follow these steps:

### Step 1: Identify the masses and their positions
- We have a circular ring of mass \( m_1 = 6 \, \text{kg} \) located at the origin (0, 0).
- There are two particles, each of mass \( m_2 = 2 \, \text{kg} \), placed at the points where the circle intersects the positive X-axis and positive Y-axis:
- Particle 2 is at \( (a, 0) \) (on the positive X-axis).
- Particle 3 is at \( (0, a) \) (on the positive Y-axis).

### Step 2: Write down the coordinates of the center of mass
The center of mass \( (X_{cm}, Y_{cm}) \) of the system can be calculated using the formula:
\[ X_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} \]
\[ Y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} \]

### Step 3: Substitute the values into the formulas
- For the X-coordinates:
- \( x_1 = 0 \) (ring at origin)
- \( x_2 = a \) (particle 2)
- \( x_3 = 0 \) (particle 3)

\[ X_{cm} = \frac{6 \cdot 0 + 2 \cdot a + 2 \cdot 0}{6 + 2 + 2} = \frac{2a}{10} = \frac{a}{5} \]

- For the Y-coordinates:
- \( y_1 = 0 \) (ring at origin)
- \( y_2 = 0 \) (particle 2)
- \( y_3 = a \) (particle 3)

\[ Y_{cm} = \frac{6 \cdot 0 + 2 \cdot 0 + 2 \cdot a}{6 + 2 + 2} = \frac{2a}{10} = \frac{a}{5} \]

### Step 4: Calculate the angle \( \theta \)
The angle \( \theta \) made by the position vector of the center of mass with the X-axis can be found using:
\[ \tan \theta = \frac{Y_{cm}}{X_{cm}} = \frac{\frac{a}{5}}{\frac{a}{5}} = 1 \]

### Step 5: Find \( \theta \)
Since \( \tan \theta = 1 \):
\[ \theta = \tan^{-1}(1) = 45^\circ \]

### Final Answer
The angle made by the position vector of the center of mass of the entire system with the X-axis is \( 45^\circ \).

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