Two particle of masses in the ration 1:2 with charges in the ratio 1:1, are palced at rest in a uniform electric field. They are relesaded and allowed to move for the same time. The ratio of their kinetic energies will be finally

Options

A. `2:1`
B. `8:1`
C. `4:1`
D. `1:4`

Explanation

To solve the problem, we need to analyze the motion of the two particles under the influence of the electric field and derive the ratio of their kinetic energies after they have been released and allowed to move for the same time.

### Step-by-Step Solution:

1. **Identify the Masses and Charges**:
- Let the mass of the first particle be \( m_1 \) and the mass of the second particle be \( m_2 \).
- Given the mass ratio \( \frac{m_1}{m_2} = \frac{1}{2} \), we can express \( m_2 \) as \( m_2 = 2m_1 \).
- Let the charge of both particles be \( q_1 \) and \( q_2 \). Given the charge ratio \( \frac{q_1}{q_2} = \frac{1}{1} \), we have \( q_1 = q_2 \).

2. **Determine the Force on Each Particle**:
- The force \( F \) acting on a charged particle in an electric field \( E \) is given by \( F = qE \).
- Since both particles have the same charge and are in the same electric field, the force on both particles is the same:
\[ F_1 = q_1 E = qE \]
\[ F_2 = q_2 E = qE \]

3. **Calculate the Acceleration**:
- Using Newton's second law, the acceleration \( a \) of each particle can be calculated as:
\[ a_1 = \frac{F_1}{m_1} = \frac{qE}{m_1} \]
\[ a_2 = \frac{F_2}{m_2} = \frac{qE}{m_2} = \frac{qE}{2m_1} \]

4. **Find the Velocity After Time \( t \)**:
- The velocity \( v \) of an object under constant acceleration after time \( t \) is given by \( v = at \).
- For the first particle:
\[ v_1 = a_1 t = \frac{qE}{m_1} t \]
- For the second particle:
\[ v_2 = a_2 t = \frac{qE}{2m_1} t \]

5. **Calculate the Kinetic Energy of Each Particle**:
- The kinetic energy \( KE \) of an object is given by \( KE = \frac{1}{2} mv^2 \).
- For the first particle:
\[ KE_1 = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_1 \left(\frac{qE}{m_1} t\right)^2 = \frac{1}{2} m_1 \frac{q^2 E^2 t^2}{m_1^2} = \frac{q^2 E^2 t^2}{2 m_1} \]
- For the second particle:
\[ KE_2 = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} (2m_1) \left(\frac{qE}{2m_1} t\right)^2 = \frac{1}{2} (2m_1) \frac{q^2 E^2 t^2}{4 m_1^2} = \frac{q^2 E^2 t^2}{4 m_1} \]

6. **Find the Ratio of Kinetic Energies**:
- Now, we can find the ratio of the kinetic energies:
\[ \frac{KE_1}{KE_2} = \frac{\frac{q^2 E^2 t^2}{2 m_1}}{\frac{q^2 E^2 t^2}{4 m_1}} = \frac{4}{2} = 2 \]

### Final Answer:
The ratio of their kinetic energies \( KE_1 : KE_2 \) is \( 2 : 1 \).