In Millikan's oil drop experiment an oil drop of radius `r` and change `Q` is held in equilibrium between the plates of a charged parallel plate capacitor when the potential change is `V`. To keep a drop radius `2r` and with a change `2Q` is equilibriu between the plates the potential difference `V'` required is:

Options

A. `V`
B. `2V`
C. `4V`
D. `8V`

Explanation

To solve the problem, we need to analyze the forces acting on the oil drop in the Millikan's oil drop experiment and how they relate to the potential difference required to maintain equilibrium.

### Step-by-Step Solution:

1. **Understanding the Forces**:
In the Millikan oil drop experiment, the oil drop experiences two main forces: the gravitational force acting downwards and the electric force acting upwards. The gravitational force can be expressed as:
\[ F_g = mg = \rho V g \]
where \( \rho \) is the density of the oil, \( V \) is the volume of the drop, and \( g \) is the acceleration due to gravity.

2. **Volume of the Oil Drop**:
The volume \( V \) of a spherical drop of radius \( r \) is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, the gravitational force becomes:
\[ F_g = \rho \left(\frac{4}{3} \pi r^3\right) g \]

3. **Electric Force**:
The electric force acting on the oil drop is given by:
\[ F_e = Q E \]
where \( Q \) is the charge on the drop and \( E \) is the electric field between the plates. The electric field \( E \) can be related to the potential difference \( V \) and the distance \( d \) between the plates:
\[ E = \frac{V}{d} \]
Thus, the electric force can be rewritten as:
\[ F_e = Q \frac{V}{d} \]

4. **Equilibrium Condition**:
For the drop to be in equilibrium, the gravitational force must equal the electric force:
\[ F_g = F_e \]
Substituting the expressions for \( F_g \) and \( F_e \):
\[ \rho \left(\frac{4}{3} \pi r^3\right) g = Q \frac{V}{d} \]

5. **New Conditions**:
Now, we consider the new drop with radius \( 2r \) and charge \( 2Q \). The volume of the new drop is:
\[ V' = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = \frac{32}{3} \pi r^3 \]
The new gravitational force becomes:
\[ F_g' = \rho \left(\frac{32}{3} \pi r^3\right) g \]

6. **Setting Up the New Equilibrium**:
The new equilibrium condition is:
\[ F_g' = F_e' \]
Thus:
\[ \rho \left(\frac{32}{3} \pi r^3\right) g = (2Q) \frac{V'}{d} \]

7. **Relating the Two Conditions**:
From the original equilibrium condition, we have:
\[ \rho \left(\frac{4}{3} \pi r^3\right) g = Q \frac{V}{d} \]
We can express \( V' \) in terms of \( V \):
\[ \frac{32}{3} \pi r^3 \cdot \rho g = 2Q \frac{V'}{d} \]
Dividing the new condition by the original condition gives:
\[ \frac{32/3 \cdot \rho g}{4/3 \cdot \rho g} = \frac{2Q \cdot V'}{Q \cdot V} \]
Simplifying this:
\[ 8 = 2 \frac{V'}{V} \]
Therefore:
\[ \frac{V'}{V} = 4 \]
Hence:
\[ V' = 4V \]

### Final Answer:
The potential difference \( V' \) required to keep the drop of radius \( 2r \) and charge \( 2Q \) in equilibrium is:
\[ V' = 4V \]