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 of finding the velocity acquired by a mass of 1 kg carrying a charge of 2 C when accelerated through a potential of 1 V, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and electric potential energy
The kinetic energy (K.E.) acquired by a charged particle when it is accelerated through a potential difference (V) is given by the formula:
\[ K.E. = qV \]
where:
- \( q \) is the charge (in coulombs),
- \( V \) is the potential difference (in volts).

### Step 2: Write the expression for kinetic energy in terms of mass and velocity
The kinetic energy can also be expressed in terms of mass (m) and velocity (v) as:
\[ K.E. = \frac{1}{2} mv^2 \]

### Step 3: Set the two expressions for kinetic energy equal to each other
Since both expressions represent kinetic energy, we can set them equal to each other:
\[ qV = \frac{1}{2} mv^2 \]

### Step 4: Rearrange the equation to solve for velocity (v)
To find the velocity, we rearrange the equation:
\[ \frac{1}{2} mv^2 = qV \]
Multiplying both sides by 2 gives:
\[ mv^2 = 2qV \]
Now, divide both sides by m:
\[ v^2 = \frac{2qV}{m} \]
Taking the square root of both sides, we find:
\[ v = \sqrt{\frac{2qV}{m}} \]

### Step 5: Substitute the known values into the equation
Now we can substitute the known values into the equation:
- \( q = 2 \, \text{C} \)
- \( V = 1 \, \text{V} \)
- \( m = 1 \, \text{kg} \)

Substituting these values gives:
\[ v = \sqrt{\frac{2 \cdot 2 \cdot 1}{1}} = \sqrt{4} = 2 \, \text{m/s} \]

### Final Answer
The velocity acquired by the mass is:
\[ \boxed{2 \, \text{m/s}} \]