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

Explanation

To solve the problem, we need to determine the potential difference required to accelerate a bullet of mass 2 grams and charge 2 microcoulombs to a speed of 10 m/s. Here’s the step-by-step solution:

### Step 1: Convert the mass and charge to SI units
- Mass of the bullet, \( m = 2 \, \text{grams} = 2 \times 10^{-3} \, \text{kg} \)
- Charge of the bullet, \( Q = 2 \, \mu C = 2 \times 10^{-6} \, C \)

### Step 2: Calculate the kinetic energy (KE) of the bullet
The kinetic energy of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times (2 \times 10^{-3} \, \text{kg}) \times (10 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \times (2 \times 10^{-3}) \times 100 \]
\[ KE = \frac{1}{2} \times 0.2 = 0.1 \, \text{J} \]

### Step 3: Relate the kinetic energy to potential energy
The potential energy (PE) gained by the bullet when it is accelerated through a potential difference \( V \) is given by:
\[ PE = Q \cdot V \]
Since the bullet starts from rest, the change in potential energy will equal the change in kinetic energy:
\[ Q \cdot V = KE \]
Substituting the values:
\[ (2 \times 10^{-6} \, C) \cdot V = 0.1 \, J \]

### Step 4: Solve for the potential difference \( V \)
Rearranging the equation to solve for \( V \):
\[ V = \frac{KE}{Q} \]
\[ V = \frac{0.1 \, J}{2 \times 10^{-6} \, C} \]
\[ V = \frac{0.1}{2 \times 10^{-6}} = 5 \times 10^{4} \, V \]
\[ V = 50,000 \, V = 50 \, kV \]

### Conclusion
The potential difference required to accelerate the bullet to the desired speed is **50 kV**.

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