The answer to each of the following questions is a single-digit integer ranging from 0 to 9. Darken the correct digit.
There is one particle A having charge q and mass m. There is another particle B of charge eight times and mass two times to that of A. Both the particles are accelerated through the same potential difference. Let speeds acquired by the particles A and B be `v_(A)` and `v_(B)` respectively, then calculate `v_(B)//v_(A)`.

Explanation

To solve the problem, we need to find the ratio of the speeds acquired by particles A and B when they are accelerated through the same potential difference. Let's denote the charge and mass of particle A as \( q \) and \( m \), respectively. For particle B, the charge is \( 8q \) and the mass is \( 2m \).

### Step-by-Step Solution:

1. **Understanding the Kinetic Energy and Potential Energy Relationship**:
When a charged particle is accelerated through a potential difference \( V \), the kinetic energy (KE) gained by the particle is equal to the work done on it by the electric field, which is given by:
\[ KE = q \cdot V \]

2. **Kinetic Energy of Particle A**:
For particle A, the kinetic energy can be expressed as:
\[ KE_A = \frac{1}{2} m v_A^2 = q \cdot V \]
Rearranging this gives:
\[ v_A^2 = \frac{2qV}{m} \]

3. **Kinetic Energy of Particle B**:
For particle B, the kinetic energy is:
\[ KE_B = \frac{1}{2} (2m) v_B^2 = (8q) \cdot V \]
Simplifying this gives:
\[ v_B^2 = \frac{8qV}{m} \]

4. **Finding the Ratio \( \frac{v_B}{v_A} \)**:
Now, we can find the ratio of the speeds \( v_B \) and \( v_A \):
\[ \frac{v_B^2}{v_A^2} = \frac{\frac{8qV}{m}}{\frac{2qV}{m}} = \frac{8qV}{2qV} = \frac{8}{2} = 4 \]
Taking the square root of both sides gives:
\[ \frac{v_B}{v_A} = \sqrt{4} = 2 \]

### Final Answer:
The ratio of the speeds acquired by particles B and A is:
\[ \frac{v_B}{v_A} = 2 \]