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 analyze the relationship between the electric potential difference, the charge of the particles, and their resulting speeds after falling through that potential difference.

### Step-by-Step Solution:

1. **Understand the Energy Conversion**:
When a charged particle falls through an electric potential difference \( V \), it converts electric potential energy into kinetic energy. The kinetic energy \( K \) gained by a particle is given by:
\[ K = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the particle and \( v \) is its final speed.

2. **Electric Potential Energy**:
The electric potential energy \( U \) gained by a charged particle when it falls through a potential difference \( V \) is given by:
\[ U = qV \]
where \( q \) is the charge of the particle.

3. **Setting Up the Equation**:
For both particles A and B, we can set up the equations for kinetic energy:
- For particle A:
\[ \frac{1}{2} mv_A^2 = qV \]
- For particle B:
\[ \frac{1}{2} mv_B^2 = 4qV \]

4. **Solving for Speeds**:
Rearranging both equations to solve for \( v_A \) and \( v_B \):
- For particle A:
\[ v_A^2 = \frac{2qV}{m} \implies v_A = \sqrt{\frac{2qV}{m}} \]
- For particle B:
\[ v_B^2 = \frac{8qV}{m} \implies v_B = \sqrt{\frac{8qV}{m}} = \sqrt{8} \sqrt{\frac{qV}{m}} = 2\sqrt{2} \sqrt{\frac{qV}{m}} \]

5. **Finding the Ratio of Speeds**:
Now, we can find the ratio of the speeds \( \frac{v_A}{v_B} \):
\[ \frac{v_A}{v_B} = \frac{\sqrt{\frac{2qV}{m}}}{2\sqrt{2} \sqrt{\frac{qV}{m}}} \]
Simplifying this:
\[ \frac{v_A}{v_B} = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2} \]

6. **Final Result**:
Thus, the ratio of their speeds is:
\[ \frac{v_A}{v_B} = \frac{1}{2} \]

### Conclusion:
The ratio of the speeds of particles A and B is \( \frac{1}{2} \).