Two particles of masses m and 2m and charges 2q and q are placed in a uniform electric field E and allowed to move for the same time. The ratio of kinetic energies will be -

Options

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

Explanation

To solve the problem of finding the ratio of kinetic energies of two particles with different masses and charges placed in a uniform electric field, we can follow these steps:

### Step 1: Identify the forces acting on each particle
- For the first particle (mass = m, charge = 2q):
\[ F_1 = 2q \cdot E \]
- For the second particle (mass = 2m, charge = q):
\[ F_2 = q \cdot E \]

### Step 2: Calculate the accelerations of each particle
Using Newton's second law \( F = ma \):
- For the first particle:
\[ a_1 = \frac{F_1}{m} = \frac{2qE}{m} \]
- For the second particle:
\[ a_2 = \frac{F_2}{2m} = \frac{qE}{2m} \]

### Step 3: Determine the velocities after time \( T \)
Assuming both particles start from rest (initial velocity \( u = 0 \)):
- For the first particle:
\[ v_1 = u_1 + a_1 \cdot T = 0 + \left(\frac{2qE}{m}\right) \cdot T = \frac{2qET}{m} \]
- For the second particle:
\[ v_2 = u_2 + a_2 \cdot T = 0 + \left(\frac{qE}{2m}\right) \cdot T = \frac{qET}{2m} \]

### Step 4: Calculate the kinetic energies of each particle
The kinetic energy \( KE \) is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- For the first particle:
\[ KE_1 = \frac{1}{2} m \left(v_1\right)^2 = \frac{1}{2} m \left(\frac{2qET}{m}\right)^2 = \frac{1}{2} m \cdot \frac{4q^2E^2T^2}{m^2} = \frac{2q^2E^2T^2}{m} \]
- For the second particle:
\[ KE_2 = \frac{1}{2} (2m) \left(v_2\right)^2 = \frac{1}{2} (2m) \left(\frac{qET}{2m}\right)^2 = \frac{1}{2} (2m) \cdot \frac{q^2E^2T^2}{4m^2} = \frac{q^2E^2T^2}{4m} \]

### Step 5: Find the ratio of the kinetic energies
Now we can find the ratio \( \frac{KE_1}{KE_2} \):
\[ \frac{KE_1}{KE_2} = \frac{\frac{2q^2E^2T^2}{m}}{\frac{q^2E^2T^2}{4m}} = \frac{2q^2E^2T^2}{m} \cdot \frac{4m}{q^2E^2T^2} = \frac{8}{1} \]

### Final Answer
The ratio of the kinetic energies \( KE_1 : KE_2 = 8 : 1 \).

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