Two particles move on a circular path (one just inside and the other just outside) with the angular velocities `omega` and `5omega` starting from the same point. Then

Options

A. they cross each other at regular intervals of time `(2pi)/(4omega)` when their angular velocities are oppositely directed.
B. they cross each other at points on the path subtending an angle of `60^(@)` at the centre if their angular velocities are oppositely directed.
C. they cross at intervals of time `(pi)/(3omega)` if their angular velocities are oppositely directed
D. they cross each other at points on the path subtending `90^(@)` at the centre if their angular velocities are in the same sense.

Answer

To solve the problem step by step, we will analyze the motion of the two particles moving in circular paths with different angular velocities.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- Let particle A move with angular velocity \( \omega \) and particle B move with angular velocity \( 5\omega \).
- Both particles start from the same point on their respective circular paths.

2. **Case 1: Opposite Directions**:
- Assume particle A moves in the positive direction and particle B moves in the negative direction.
- The angle covered by particle A after time \( t \) is given by:
\[ \theta_A = \omega t \]
- The angle covered by particle B after time \( t \) is:
\[ \theta_B = 2\pi - \theta_A = 2\pi - \omega t \]
- For particle B, we can also express the angle as:
\[ \theta_B = 5\omega t \]
- Setting the two expressions for \( \theta_B \) equal:
\[ 2\pi - \omega t = 5\omega t \]
- Rearranging gives:
\[ 2\pi = 6\omega t \implies t = \frac{2\pi}{6\omega} = \frac{\pi}{3\omega} \]

3. **Calculating the Angle Subtended**:
- The angle subtended at the center by particle A when they meet is:
\[ \theta_A = \omega t = \omega \left(\frac{\pi}{3\omega}\right) = \frac{\pi}{3} \text{ radians} = 60^\circ \]

4. **Conclusion for Case 1**:
- They cross each other at regular intervals of time \( \frac{\pi}{3\omega} \) and subtend an angle of \( 60^\circ \) at the center when moving in opposite directions.

5. **Case 2: Same Direction**:
- Now, assume both particles move in the same direction.
- The angle covered by particle A after time \( t \) is still:
\[ \theta_A = \omega t \]
- For particle B, the angle is:
\[ \theta_B = 5\omega t \]
- Setting the angles equal when they meet:
\[ \theta_B = \theta_A + 2\pi \implies 5\omega t = \omega t + 2\pi \]
- Rearranging gives:
\[ 4\omega t = 2\pi \implies t = \frac{2\pi}{4\omega} = \frac{\pi}{2\omega} \]

6. **Calculating the Angle Subtended in Same Direction**:
- The angle subtended at the center by particle A when they meet is:
\[ \theta_A = \omega t = \omega \left(\frac{\pi}{2\omega}\right) = \frac{\pi}{2} \text{ radians} = 90^\circ \]

7. **Conclusion for Case 2**:
- They cross each other at intervals of \( \frac{\pi}{2\omega} \) and subtend an angle of \( 90^\circ \) at the center when moving in the same direction.

### Final Results:
- The correct options based on the analysis are:
- They cross each other at points on the path subtending an angle of \( 60^\circ \) at the center when moving in opposite directions.
- They cross at intervals of \( \frac{\pi}{3\omega} \) if their angular velocities are oppositely directed.
- They cross each other at points subtending \( 90^\circ \) at the center when moving in the same direction.