In a certain experiments to measure the ratio of charge to mass of elementry particles, a surprising result was obtained in which two particle, a surprising result was obtained in which two particles moved in such a way that the distance between them always remained constant. It was also noticed that this two-particle system was isolated from all other particles and no force was acting on this system except the force between these two mases. After careful observation followed bu intensive calculation, it was deduced that velocity of these two particles was always opposite in direction and magnitude of velocity was `10^(3) ms^(-1) and 2 xx 10^(3) ms^(-1)` for first and second particle, respectively, and mass of these particles were `2 xx 10^(-30) kg and 10^(-30)kg`, respectively. Distance between them were 12Å(1Å = 10^(-`10)m).`
Paths of the two particles was

Options

A. intersecting straight lines
B. parabolic
C. circular
D. straight line with respect

Explanation

To solve the problem, we need to analyze the motion of the two particles based on the information provided. The key points to consider are:

1. The distance between the two particles remains constant.
2. The velocities of the two particles are equal in magnitude but opposite in direction.
3. The particles have different masses and are moving in such a way that they maintain a constant distance from each other.

### Step-by-Step Solution:

**Step 1: Understand the motion of the particles.**
- Since the distance between the two particles remains constant, they must be moving in such a way that their paths are circular. This is because only circular motion can maintain a constant distance between two objects while allowing them to move.

**Step 2: Define the parameters of the particles.**
- Let:
- Mass of particle 1, \( m_1 = 2 \times 10^{-30} \, \text{kg} \)
- Velocity of particle 1, \( v_1 = 10^3 \, \text{m/s} \)
- Mass of particle 2, \( m_2 = 10^{-30} \, \text{kg} \)
- Velocity of particle 2, \( v_2 = 2 \times 10^3 \, \text{m/s} \)
- Distance between the particles, \( d = 12 \, \text{Å} = 12 \times 10^{-10} \, \text{m} \)

**Step 3: Analyze the circular motion.**
- Since the particles are moving in circular paths, we can denote the radius of the circular path of particle 1 as \( r_1 \) and for particle 2 as \( r_2 \).
- The relationship between the radii and the distance \( d \) can be expressed as:
\[ d = r_1 + r_2 \]

**Step 4: Apply the concept of centripetal force.**
- The centripetal force required to keep each particle in circular motion is provided by the gravitational force between them. The gravitational force \( F \) can be expressed as:
\[ F = \frac{G m_1 m_2}{d^2} \]
where \( G \) is the gravitational constant.

**Step 5: Set up the equations for centripetal acceleration.**
- The centripetal acceleration for each particle can be expressed as:
\[ a_1 = \frac{v_1^2}{r_1} \quad \text{and} \quad a_2 = \frac{v_2^2}{r_2} \]
- The centripetal force acting on each particle can be equated to the gravitational force:
\[ m_1 a_1 = \frac{G m_1 m_2}{d^2} \quad \text{and} \quad m_2 a_2 = \frac{G m_1 m_2}{d^2} \]

**Step 6: Solve for the radii.**
- From the equations of centripetal acceleration, we can express \( r_1 \) and \( r_2 \):
\[ r_1 = \frac{m_1 v_1^2}{\frac{G m_1 m_2}{d^2}} \quad \text{and} \quad r_2 = \frac{m_2 v_2^2}{\frac{G m_1 m_2}{d^2}} \]

**Step 7: Conclude the paths of the particles.**
- Since both particles are moving in circular paths with a common center, the paths of the two particles are circular.

### Final Answer:
The paths of the two particles are circular.