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).`
Acceleration of the first particle was

Options

A. zero
B. `4 xx 10^(16)ms^(-2)`
C. `2 xx 10^(16)ms^(-2)`
D. `2.5 xx 10^(15)ms^(-2)`

Explanation

To find the acceleration of the first particle in the two-particle system, we can use the concept of centripetal acceleration. Since the distance between the two particles remains constant, we can treat their motion as circular motion around a common center of mass.

1. **Identify the given data**:
- Velocity of the first particle, \( v_1 = 10^3 \, \text{m/s} \)
- Velocity of the second particle, \( v_2 = 2 \times 10^3 \, \text{m/s} \)
- Mass of the first particle, \( m_1 = 2 \times 10^{-30} \, \text{kg} \)
- Mass of the second particle, \( m_2 = 10^{-30} \, \text{kg} \)
- Distance between the particles, \( R = 12 \, \text{Å} = 12 \times 10^{-10} \, \text{m} \)

2. **Use the formula for centripetal acceleration**:
The centripetal acceleration \( a \) of an object moving in a circular path is given by:
\[ a = \frac{v^2}{R} \]
where \( v \) is the velocity of the particle and \( R \) is the radius of the circular path.

3. **Calculate the acceleration of the first particle**:
Substituting the values for the first particle:
\[ a_1 = \frac{v_1^2}{R} = \frac{(10^3 \, \text{m/s})^2}{12 \times 10^{-10} \, \text{m}} \]
\[ a_1 = \frac{10^6 \, \text{m}^2/\text{s}^2}{12 \times 10^{-10} \, \text{m}} = \frac{10^6}{12 \times 10^{-10}} \, \text{m/s}^2 \]
\[ a_1 = \frac{10^6}{12} \times 10^{10} \, \text{m/s}^2 \]
\[ a_1 = \frac{10^{16}}{12} \, \text{m/s}^2 \approx 8.33 \times 10^{15} \, \text{m/s}^2 \]

4. **Final Result**:
Therefore, the acceleration of the first particle is approximately:
\[ a_1 \approx 8.33 \times 10^{15} \, \text{m/s}^2 \]