In Milikan's oil drop experiment, a charged drop falls with terminal velocity V. If an electric field E is applied in vertically upward direction then it starts moving in upward direction with terminal velocity 2V. If magnitude of electric field is decreased to `(E)/(2)`, then terminal velocity will become

Options

A. `V/2`
B. V
C. `3V/2`
D. 2V

Explanation

To solve the problem, we will analyze the forces acting on the charged oil drop in different scenarios and derive the terminal velocities step by step.

### Step 1: Understand the forces acting on the drop without electric field
When the charged drop is falling under gravity without any electric field, the forces acting on it are:
- Weight of the drop (downward): \( mg \)
- Viscous drag force (upward): \( F_b = 6 \pi \eta r v \)

At terminal velocity, these forces balance each other:
\[ mg = 6 \pi \eta r v \quad \text{(1)} \]

### Step 2: Analyze the situation with electric field E
When an electric field \( E \) is applied in the upward direction, the drop experiences an additional upward force due to the electric field:
- Electric force (upward): \( F_e = Q E \)

Now, the forces acting on the drop are:
- Weight (downward): \( mg \)
- Viscous drag force (upward): \( F_b = 6 \pi \eta r (2v) \) (since the terminal velocity is now \( 2v \))
- Electric force (upward): \( Q E \)

Setting up the balance of forces gives us:
\[ mg + Q E = 6 \pi \eta r (2v) \quad \text{(2)} \]

### Step 3: Substitute the expression for mg from equation (1)
From equation (1), we know that \( mg = 6 \pi \eta r v \). Substituting this into equation (2):
\[ 6 \pi \eta r v + Q E = 12 \pi \eta r v \]
Rearranging gives:
\[ Q E = 12 \pi \eta r v - 6 \pi \eta r v \]
\[ Q E = 6 \pi \eta r v \quad \text{(3)} \]

### Step 4: Analyze the situation with electric field \( \frac{E}{2} \)
Now, we decrease the electric field to \( \frac{E}{2} \). The forces acting on the drop now are:
- Weight (downward): \( mg \)
- Viscous drag force (upward): \( F_b = 6 \pi \eta r v' \) (where \( v' \) is the new terminal velocity)
- Electric force (upward): \( Q \left(\frac{E}{2}\right) \)

Setting up the balance of forces gives us:
\[ mg + Q \left(\frac{E}{2}\right) = 6 \pi \eta r v' \quad \text{(4)} \]

### Step 5: Substitute the expressions for mg and QE
From equation (1), we have \( mg = 6 \pi \eta r v \), and from equation (3), we have \( Q E = 6 \pi \eta r v \). Therefore, \( Q \left(\frac{E}{2}\right) = \frac{1}{2} Q E = \frac{1}{2} (6 \pi \eta r v) = 3 \pi \eta r v \).

Substituting these into equation (4):
\[ 6 \pi \eta r v + 3 \pi \eta r v = 6 \pi \eta r v' \]
\[ 9 \pi \eta r v = 6 \pi \eta r v' \]

### Step 6: Solve for the new terminal velocity \( v' \)
Dividing both sides by \( 6 \pi \eta r \):
\[ v' = \frac{9}{6} v = \frac{3}{2} v \]

Thus, the new terminal velocity \( v' \) when the electric field is decreased to \( \frac{E}{2} \) is:
\[ \boxed{\frac{3}{2} v} \]