In Millikan's oil drop experiment an oil drop carrying a charge `Q` is held stationary by a potential difference `2400 V` between the plates. To keep a drop of half the radius stationary the potential difference had to be made `600 V`. What is the charge on the second drop ?

Options

A. `(Q)/(4)`
B. `(Q)/(2)`
C. `Q`
D. `(3Q)/(2)`

Explanation

To solve the problem, we need to analyze the relationship between the charge on the oil drops, their radii, and the potential difference required to keep them stationary in an electric field.

### Step-by-Step Solution:

1. **Understanding the Forces**: In Millikan's oil drop experiment, an oil drop carrying a charge \( Q \) is held stationary by an electric field created by a potential difference \( V \). The forces acting on the drop are the gravitational force \( mg \) (downwards) and the electric force \( QE \) (upwards), where \( E \) is the electric field strength.

2. **Electric Field Calculation**: The electric field \( E \) between two plates is given by:
\[ E = \frac{V}{d} \]
where \( V \) is the potential difference and \( d \) is the distance between the plates.

3. **Force Balance**: For the drop to be stationary, the electric force must balance the gravitational force:
\[ QE = mg \]

4. **Mass of the Oil Drop**: The mass \( m \) of the oil drop can be expressed in terms of its volume and density:
\[ m = \rho V = \rho \left(\frac{4}{3} \pi r^3\right) \]
where \( \rho \) is the density of the oil and \( r \) is the radius of the drop.

5. **Relating Charge, Radius, and Potential**: From the force balance equation, we can express the charge \( Q \) in terms of the radius \( r \) and the potential \( V \):
\[ Q = \frac{mg}{E} = \frac{\rho \left(\frac{4}{3} \pi r^3\right) g}{\frac{V}{d}} = \frac{4 \pi \rho g d r^3}{3V} \]
This shows that \( Q \) is directly proportional to \( r^3 \) and inversely proportional to \( V \).

6. **Setting Up Ratios**: Let \( Q_1 \) be the charge on the first drop (radius \( r \), potential \( 2400 \, \text{V} \)) and \( Q_2 \) be the charge on the second drop (radius \( \frac{r}{2} \), potential \( 600 \, \text{V} \)):
\[ Q_1 = k \frac{r^3}{2400} \quad \text{and} \quad Q_2 = k \frac{\left(\frac{r}{2}\right)^3}{600} \]
where \( k = \frac{4 \pi \rho g d}{3} \).

7. **Calculating \( Q_2 \)**:
\[ Q_2 = k \frac{\left(\frac{r}{2}\right)^3}{600} = k \frac{\frac{r^3}{8}}{600} = \frac{Q_1}{8} \cdot \frac{2400}{600} = \frac{Q_1}{8} \cdot 4 = \frac{Q_1}{2} \]

8. **Final Result**: Thus, the charge on the second drop is:
\[ Q_2 = \frac{Q_1}{2} \]

### Conclusion:
If the charge on the first drop is \( Q \), then the charge on the second drop is:
\[ Q_2 = \frac{Q}{2} \]