An oil drop carrying charge `Q` is held in equilibrium by a potential difference of `600V` between the horizontal plates.In order to hold another drop of radius in equilibrium a potential drop of `1600V` had to be maintained .The charge on the second drop is

Options

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

Explanation

To solve the problem, we need to analyze the relationship between the charge on the oil drops, their radii, and the potential differences required to hold them in equilibrium.

### Step-by-Step Solution:

1. **Understanding the Forces on the Oil Drop:**
The oil drop is held in equilibrium by the electric force due to the electric field between the plates, which balances the gravitational force acting on the drop. The electric force can be expressed as:
\[ F_e = Q \cdot E \]
where \( Q \) is the charge on the drop and \( E \) is the electric field.

2. **Electric Field Calculation:**
The electric field \( E \) between two parallel plates with a potential difference \( V \) and separation \( D \) is given by:
\[ E = \frac{V}{D} \]
For the first drop, with a potential difference of \( V_1 = 600 \, \text{V} \):
\[ E_1 = \frac{600}{D} \]

3. **Force Balance for the First Drop:**
The gravitational force acting on the drop is:
\[ F_g = m \cdot g \]
where \( m \) is the mass of the drop. Setting the electric force equal to the gravitational force gives:
\[ Q_1 \cdot E_1 = m \cdot g \]
Substituting for \( E_1 \):
\[ Q_1 \cdot \frac{600}{D} = m \cdot g \]

4. **For the Second Drop:**
The second drop has a radius \( R_2 = 2R \) and requires a potential difference of \( V_2 = 1600 \, \text{V} \) to maintain equilibrium. The electric field for the second drop is:
\[ E_2 = \frac{1600}{D} \]

5. **Force Balance for the Second Drop:**
Similarly, for the second drop:
\[ Q_2 \cdot E_2 = m_2 \cdot g \]
where \( m_2 \) is the mass of the second drop. The mass of the second drop can be expressed in terms of its volume and density:
\[ m_2 = \frac{4}{3} \pi (2R)^3 \rho = \frac{4}{3} \pi (8R^3) \rho = 8 \cdot \frac{4}{3} \pi R^3 \rho = 8m_1 \]
(where \( m_1 \) is the mass of the first drop).

6. **Setting Up the Equation for the Second Drop:**
Thus, we have:
\[ Q_2 \cdot \frac{1600}{D} = 8m \cdot g \]

7. **Relating the Two Drops:**
From the equations for both drops, we can set up the ratio:
\[ \frac{Q_1 \cdot \frac{600}{D}}{Q_2 \cdot \frac{1600}{D}} = \frac{m \cdot g}{8m \cdot g} \]
Simplifying gives:
\[ \frac{Q_1 \cdot 600}{Q_2 \cdot 1600} = \frac{1}{8} \]
Rearranging gives:
\[ Q_2 = Q_1 \cdot \frac{1600}{600} \cdot 8 = Q_1 \cdot \frac{8}{3} \]

8. **Final Calculation:**
Therefore, if we denote the charge on the first drop as \( Q \):
\[ Q_2 = Q \cdot \frac{8}{3} \]

### Conclusion:
The charge on the second drop is \( \frac{8}{3} Q \).