An oil drop of 10 excess electron is held stationary under a consatnt electric field of `3.6xx10^(4)NC^(-1)` in Millikan's oil drop experiment. The density of oil is `1.26gcm^(-3)`. Radius of the oil drop is
(Take, `g=9.8ms^(-2), e=1.6xx10^(-19)C`)

Options

A. `1.04xx10^(-6)m`
B. `4.8xx10^(-5)m`
C. `4.8xx10^(-18)m`
D. `1.13xx10^(-18)m`

Explanation

To find the radius of the oil drop in Millikan's oil drop experiment, we can follow these steps:

### Step 1: Understand the Forces Acting on the Oil Drop
The oil drop is held stationary under the influence of an electric field. The forces acting on it are:
- The upward electric force (\(F_e\))
- The downward gravitational force (\(F_g\))

Since the drop is stationary, these two forces are equal in magnitude:
\[ F_e = F_g \]

### Step 2: Calculate the Electric Force
The electric force acting on the oil drop can be calculated using the formula:
\[ F_e = qE \]
where:
- \(q\) is the charge on the oil drop
- \(E\) is the electric field strength

Given that the oil drop has 10 excess electrons, the charge \(q\) can be calculated as:
\[ q = 10 \times e \]
where \(e = 1.6 \times 10^{-19} \, C\).

Thus,
\[ q = 10 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-18} \, C \]

Now, substituting the values into the electric force equation:
\[ F_e = (1.6 \times 10^{-18}) \times (3.6 \times 10^{4}) \]

### Step 3: Calculate the Gravitational Force
The gravitational force acting on the oil drop is given by:
\[ F_g = mg \]
where:
- \(m\) is the mass of the oil drop
- \(g\) is the acceleration due to gravity (\(g = 9.8 \, m/s^2\))

The mass \(m\) can be calculated using the volume and density of the oil drop:
\[ m = \text{Volume} \times \text{Density} \]
The volume \(V\) of the spherical oil drop is given by:
\[ V = \frac{4}{3} \pi r^3 \]
where \(r\) is the radius of the drop.

Thus,
\[ m = \frac{4}{3} \pi r^3 \times \rho \]
where \(\rho\) is the density of the oil. Given \(\rho = 1.26 \, g/cm^3 = 1.26 \times 10^{3} \, kg/m^3\).

### Step 4: Set the Forces Equal
Now we set the electric force equal to the gravitational force:
\[ qE = mg \]
Substituting the expressions for \(q\) and \(m\):
\[ (1.6 \times 10^{-18}) \times (3.6 \times 10^{4}) = \left(\frac{4}{3} \pi r^3 \times (1.26 \times 10^{3})\right) \times 9.8 \]

### Step 5: Solve for \(r^3\)
Rearranging the equation to solve for \(r^3\):
\[ r^3 = \frac{(1.6 \times 10^{-18}) \times (3.6 \times 10^{4})}{\left(\frac{4}{3} \pi (1.26 \times 10^{3}) \times 9.8\right)} \]

### Step 6: Calculate \(r^3\)
Calculating the right-hand side:
1. Calculate \(qE\):
\[ qE = (1.6 \times 10^{-18}) \times (3.6 \times 10^{4}) = 5.76 \times 10^{-14} \, N \]

2. Calculate the denominator:
\[ \frac{4}{3} \pi (1.26 \times 10^{3}) \times 9.8 \approx 4.94 \times 10^{4} \]

3. Now substituting back:
\[ r^3 = \frac{5.76 \times 10^{-14}}{4.94 \times 10^{4}} \approx 1.16 \times 10^{-18} \]

### Step 7: Calculate \(r\)
Taking the cube root to find \(r\):
\[ r = (1.16 \times 10^{-18})^{1/3} \approx 1.06 \times 10^{-6} \, m \]

### Final Answer
The radius of the oil drop is approximately:
\[ r \approx 1.06 \times 10^{-6} \, m \]