An oil drop has a charge `-9.6xx10^(-19) C` and mass `1.6xx10^(-15)` gm. When allowed to fall, due to air resistance force it attains a constant velocity. Then if a uniform electric field is to be applied vertically to make the oil drop ascend up with the same constant speed, which of the following are correct. `(g=10 ms^(-2))` (Assume that the magnitude of resistance force is same in both the cases)

Options

A. The electric field is directed upward
B. The electric field is directed downward
C. The intensity of electric field is `1/3xx10^(2) NC^(-1)`
D. The intensity of electric field is `1/6xx10^(5) NC^(-1)`

Answer

To solve the problem, we need to analyze the forces acting on the oil drop when it is in equilibrium, both when it is falling and when it is ascending due to an electric field.

### Step 1: Understand the forces acting on the oil drop
When the oil drop is falling, it experiences three forces:
1. Gravitational force (weight) acting downward, \( F_g = mg \).
2. Air resistance force acting upward, \( F_r \).
3. Electric force (when the electric field is applied) acting upward, \( F_e = QE \).

At terminal velocity (constant velocity), the net force acting on the drop is zero. Therefore, the gravitational force is balanced by the air resistance force:
\[ F_g = F_r \]

### Step 2: Calculate the gravitational force
Given:
- Mass of the oil drop, \( m = 1.6 \times 10^{-15} \) gm = \( 1.6 \times 10^{-18} \) kg (since \( 1 \text{ gm} = 10^{-3} \text{ kg} \))
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

The gravitational force is calculated as:
\[ F_g = mg = (1.6 \times 10^{-18} \, \text{kg})(10 \, \text{m/s}^2) = 1.6 \times 10^{-17} \, \text{N} \]

### Step 3: Set up the equation for electric force
When the electric field is applied, the oil drop ascends with the same constant speed. In this case, the electric force must balance the gravitational force:
\[ F_e = F_g \]
\[ QE = mg \]

### Step 4: Solve for the electric field \( E \)
Rearranging the equation gives:
\[ E = \frac{mg}{Q} \]

Where:
- Charge of the oil drop, \( Q = -9.6 \times 10^{-19} \, \text{C} \) (the negative sign indicates the charge is negative, but we will use the magnitude for calculations).

Substituting the values:
\[ E = \frac{(1.6 \times 10^{-18} \, \text{kg})(10 \, \text{m/s}^2)}{9.6 \times 10^{-19} \, \text{C}} \]

Calculating:
\[ E = \frac{1.6 \times 10^{-17}}{9.6 \times 10^{-19}} \]
\[ E = \frac{16}{96} \times 10^{2} \]
\[ E = \frac{1}{6} \times 10^{2} \]
\[ E \approx 1.67 \, \text{N/C} \]

### Conclusion
The electric field required to make the oil drop ascend with the same constant speed is approximately \( 1.67 \, \text{N/C} \).