STATEMENT-1: A charged particle is accelerated by a potential difference of V volts. It then enters perpendicularly to a uniform magnetic field. It rotates in a circle. Its angular momentum about centre is say L. Now if V is doubled, L also becomes two times because
STATEMENT-2: If V is doubled, kinetic energy will become two times and therefore, L also becomes two times.

Options

A. Statement-1 is True , Statement-2 is True , Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True , Statement-2 is True , Statement-2 is NOT a correct explanation for Statement-1.
C. Statement-1 is True , Statement-2 is False
D. Statement-1 is False , Statement-2 is True

Answer

To solve the problem, we will analyze the statements given and derive the relationships between potential difference, kinetic energy, velocity, radius, and angular momentum for a charged particle moving in a magnetic field.

### Step-by-Step Solution:

1. **Understanding the Kinetic Energy**:
When a charged particle with charge \( Q \) is accelerated through a potential difference \( V \), its kinetic energy \( KE \) is given by:
\[ KE = QV \]

2. **Relating Kinetic Energy to Velocity**:
The kinetic energy can also be expressed in terms of velocity \( v \):
\[ KE = \frac{1}{2} mv^2 \]
Equating the two expressions for kinetic energy, we have:
\[ QV = \frac{1}{2} mv^2 \]
From this, we can solve for velocity \( v \):
\[ v = \sqrt{\frac{2QV}{m}} \]

3. **Effect of Doubling the Potential Difference**:
If the potential difference \( V \) is doubled (i.e., \( V' = 2V \)), the new velocity \( v' \) becomes:
\[ v' = \sqrt{\frac{2Q(2V)}{m}} = \sqrt{\frac{4QV}{m}} = 2\sqrt{\frac{QV}{m}} = \sqrt{2}v \]
Thus, the velocity increases by a factor of \( \sqrt{2} \).

4. **Radius of Circular Motion**:
The radius \( r \) of the circular path of the charged particle in a magnetic field \( B \) is given by:
\[ r = \frac{mv}{QB} \]
If the velocity is now \( v' = \sqrt{2}v \), the new radius \( r' \) becomes:
\[ r' = \frac{m(\sqrt{2}v)}{QB} = \sqrt{2} \cdot \frac{mv}{QB} = \sqrt{2}r \]

5. **Angular Momentum Calculation**:
The angular momentum \( L \) of the particle about the center is given by:
\[ L = mvr \]
Substituting the new values of \( v' \) and \( r' \):
\[ L' = m(\sqrt{2}v)(\sqrt{2}r) = 2mvr = 2L \]
Therefore, if the potential difference is doubled, the angular momentum \( L \) also becomes two times the original angular momentum.

### Conclusion:
- **Statement 1** is true: If the potential difference \( V \) is doubled, the angular momentum \( L \) becomes two times.
- **Statement 2** is also true: Doubling the potential difference results in doubling the kinetic energy, which leads to the increase in both velocity and radius, thereby doubling the angular momentum.