A particle `A` has chrage `+q` and a particle `B` has charge `+4q` with each of them having the same mass `m`. When allowed to fall from rest through the same electric potential difference, the ratio of their speed `(v_(A))/(v_(B))` will become

Options

A. `2 : 1`
B. `1 : 2`
C. `1 : 4`
D. `4 : 1`

Explanation

To solve the problem, we need to find the ratio of the speeds of two charged particles (A and B) after they have fallen through the same electric potential difference. Let's denote the speed of particle A as \( v_A \) and the speed of particle B as \( v_B \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Particle A has charge \( +q \) and mass \( m \).
- Particle B has charge \( +4q \) and the same mass \( m \).
- Both particles are allowed to fall through the same electric potential difference \( V \).

2. **Using the Work-Energy Theorem**:
- The work done on a charged particle in an electric field is equal to the change in its kinetic energy.
- The work done \( W \) on each particle can be expressed as:
\[ W = qV \]
- For particle A, the work done is:
\[ W_A = qV \]
- For particle B, the work done is:
\[ W_B = 4qV \]

3. **Relating Work Done to Kinetic Energy**:
- The change in kinetic energy for each particle is given by:
\[ \Delta KE = \frac{1}{2} mv^2 \]
- For particle A:
\[ \frac{1}{2} mv_A^2 = qV \quad \text{(1)} \]
- For particle B:
\[ \frac{1}{2} mv_B^2 = 4qV \quad \text{(2)} \]

4. **Solving for Speeds**:
- From equation (1):
\[ mv_A^2 = 2qV \implies v_A^2 = \frac{2qV}{m} \]
- From equation (2):
\[ mv_B^2 = 8qV \implies v_B^2 = \frac{8qV}{m} \]

5. **Finding the Ratio of Speeds**:
- Now, we can find the ratio of the speeds \( \frac{v_A}{v_B} \):
\[ \frac{v_A^2}{v_B^2} = \frac{\frac{2qV}{m}}{\frac{8qV}{m}} = \frac{2}{8} = \frac{1}{4} \]
- Taking the square root to find the ratio of speeds:
\[ \frac{v_A}{v_B} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Final Answer:
The ratio of the speeds of particles A and B is:
\[ \frac{v_A}{v_B} = \frac{1}{2} \]