A bullet of mass `2 gm` is having a charge of `2 muc`. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of `10 m//s`

Options

A. 50 kV
B. 5V
C. 50 V
D. 5kV

Explanation

To find the potential difference through which a bullet of mass \(2 \, \text{g}\) and charge \(2 \, \mu\text{C}\) must be accelerated to acquire a speed of \(10 \, \text{m/s}\), we can follow these steps:

### Step-by-Step Solution:

1. **Convert Units**:
- Convert the mass from grams to kilograms:
\[ m = 2 \, \text{g} = 2 \times 10^{-3} \, \text{kg} \]
- Convert the charge from microcoulombs to coulombs:
\[ Q = 2 \, \mu\text{C} = 2 \times 10^{-6} \, \text{C} \]

2. **Calculate Initial and Final Kinetic Energy**:
- The initial kinetic energy (\(K_i\)) is zero since the bullet starts from rest:
\[ K_i = 0 \]
- The final kinetic energy (\(K_f\)) can be calculated using the formula:
\[ K_f = \frac{1}{2} m v^2 \]
- Substitute the values:
\[ K_f = \frac{1}{2} \times (2 \times 10^{-3}) \times (10)^2 = \frac{1}{2} \times (2 \times 10^{-3}) \times 100 = 0.1 \, \text{J} \]

3. **Apply Conservation of Energy**:
- According to the conservation of energy, the loss in potential energy equals the gain in kinetic energy:
\[ QV = K_f - K_i \]
- Since \(K_i = 0\):
\[ QV = K_f \]

4. **Solve for Potential Difference (V)**:
- Rearranging the equation gives:
\[ V = \frac{K_f}{Q} \]
- Substitute the values:
\[ V = \frac{0.1}{2 \times 10^{-6}} = \frac{0.1}{2 \times 10^{-6}} = 50,000 \, \text{V} = 50 \, \text{kV} \]

5. **Final Answer**:
- The potential difference required is \(50 \, \text{kV}\).

### Summary:
The potential difference through which the bullet must be accelerated to acquire a speed of \(10 \, \text{m/s}\) is \(50 \, \text{kV}\).