Two charged particles having masses in the ratio `2: 3` and charges in the ratio `1: 2` are released from rest in a uniform electric field. After a time 1 minute their K.E. will be in the ratio of

Options

A. `3: 8`
B. `3: 4`
C. `1: 3`
D. `2: 5`

Answer

To solve the problem, we need to find the ratio of the kinetic energies of two charged particles after they are released from rest in a uniform electric field. Here's a step-by-step solution:

### Step 1: Define the quantities
Let:
- Mass of particle 1, \( m_1 = 2x \)
- Mass of particle 2, \( m_2 = 3x \)
- Charge of particle 1, \( q_1 = y \)
- Charge of particle 2, \( q_2 = 2y \)

### Step 2: Calculate the acceleration of each particle
The force on each particle in a uniform electric field \( E \) is given by:
\[ F = qE \]

Using Newton's second law, we have:
\[ F = ma \]
Thus, for each particle:
1. For particle 1:
\[ F_1 = q_1 E = m_1 a_1 \]
\[ a_1 = \frac{q_1 E}{m_1} = \frac{yE}{2x} \]

2. For particle 2:
\[ F_2 = q_2 E = m_2 a_2 \]
\[ a_2 = \frac{q_2 E}{m_2} = \frac{2yE}{3x} \]

### Step 3: Calculate the final velocity of each particle after time \( t \)
Since they are released from rest, the initial velocity \( u = 0 \). Using the equation of motion:
\[ v = u + at \]
We have:
1. For particle 1:
\[ v_1 = 0 + a_1 t = \left(\frac{yE}{2x}\right) t \]

2. For particle 2:
\[ v_2 = 0 + a_2 t = \left(\frac{2yE}{3x}\right) t \]

### Step 4: Calculate the kinetic energy of each particle
The kinetic energy \( KE \) is given by:
\[ KE = \frac{1}{2} mv^2 \]

1. For particle 1:
\[ KE_1 = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} (2x) \left(\frac{yE t}{2x}\right)^2 = \frac{1}{2} (2x) \left(\frac{y^2 E^2 t^2}{4x^2}\right) = \frac{y^2 E^2 t^2}{4x} \]

2. For particle 2:
\[ KE_2 = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} (3x) \left(\frac{2yE t}{3x}\right)^2 = \frac{1}{2} (3x) \left(\frac{4y^2 E^2 t^2}{9x^2}\right) = \frac{2y^2 E^2 t^2}{3x} \]

### Step 5: Find the ratio of the kinetic energies
Now we can find the ratio of the kinetic energies:
\[ \frac{KE_1}{KE_2} = \frac{\frac{y^2 E^2 t^2}{4x}}{\frac{2y^2 E^2 t^2}{3x}} = \frac{y^2 E^2 t^2}{4x} \cdot \frac{3x}{2y^2 E^2 t^2} = \frac{3}{8} \]

### Step 6: Conclusion
Thus, the ratio of the kinetic energies of the two particles after 1 minute is:
\[ \frac{KE_1}{KE_2} = \frac{3}{8} \]

### Final Answer
The kinetic energy will be in the ratio of \( 3:8 \).

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