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`

Answer

To solve the problem, we need to find the ratio of the speeds of two charged particles, A and B, after they fall through the same electric potential difference.

### Step-by-Step Solution:

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

2. **Using Energy Conservation**:
- The change in kinetic energy (\( \Delta KE \)) of each particle is equal to the change in potential energy (\( \Delta PE \)).
- The equation for kinetic energy is given by:
\[ \Delta KE = \frac{1}{2} mv^2 \]
- The potential energy change when a charge \( q \) moves through a potential difference \( V \) is:
\[ \Delta PE = qV \]

3. **Setting Up the Equations**:
- For particle A:
\[ \frac{1}{2} m v_A^2 = qV \quad \text{(1)} \]
- For particle B:
\[ \frac{1}{2} m v_B^2 = 4qV \quad \text{(2)} \]

4. **Dividing the Equations**:
- Divide equation (1) by equation (2):
\[ \frac{\frac{1}{2} m v_A^2}{\frac{1}{2} m v_B^2} = \frac{qV}{4qV} \]
- Simplifying this gives:
\[ \frac{v_A^2}{v_B^2} = \frac{1}{4} \]

5. **Finding the Ratio of Speeds**:
- Taking the square root of both sides:
\[ \frac{v_A}{v_B} = \frac{1}{2} \]

6. **Conclusion**:
- The ratio of the speeds of particles A and B is:
\[ \frac{v_A}{v_B} = \frac{1}{2} \]

### Final Answer:
The ratio of their speeds \( \frac{v_A}{v_B} \) is \( \frac{1}{2} \).

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