A mass of 1kg carrying a charge of 2C is accelerated through a potential of 1V. The velocity acquired by it is

Options

A. `sqrt(2) ms^(-1)`
B. `2 ms^(-1)`
C. `1/(sqrt2) ms^(-1)`
D. `1/2 ms^(-1)`

Answer

To solve the problem, we need to determine the velocity acquired by a mass of 1 kg carrying a charge of 2 C when it is accelerated through a potential difference of 1 V. We can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between potential energy and kinetic energy**:
When a charged particle moves through a potential difference, it gains kinetic energy equal to the work done on it by the electric field. The change in potential energy (ΔPE) is given by the formula:
\[ \Delta PE = q \cdot V \]
where \( q \) is the charge and \( V \) is the potential difference.

2. **Calculate the change in potential energy**:
Here, the charge \( q = 2 \, \text{C} \) and the potential difference \( V = 1 \, \text{V} \). Thus, the change in potential energy is:
\[ \Delta PE = 2 \, \text{C} \cdot 1 \, \text{V} = 2 \, \text{J} \]

3. **Apply the conservation of energy principle**:
The kinetic energy (KE) gained by the mass is equal to the change in potential energy:
\[ KE = \Delta PE = 2 \, \text{J} \]

4. **Relate kinetic energy to velocity**:
The kinetic energy can also be expressed in terms of mass and velocity:
\[ KE = \frac{1}{2} m v^2 \]
where \( m \) is the mass and \( v \) is the velocity. Substituting the known values:
\[ 2 \, \text{J} = \frac{1}{2} \cdot 1 \, \text{kg} \cdot v^2 \]

5. **Solve for velocity**:
Rearranging the equation to solve for \( v^2 \):
\[ 2 = \frac{1}{2} v^2 \implies v^2 = 4 \]
Taking the square root gives:
\[ v = 2 \, \text{m/s} \]

### Final Answer:
The velocity acquired by the mass is \( 2 \, \text{m/s} \).

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