A drop is charged to 2V Now 512 drop & identical are combined to form a single drop there the voltage of bigger drop is?

Options

A. 148V
B. 128V
C. 125V
D. 127V

Answer

To solve the problem, we need to find the voltage of a larger drop formed by combining 512 smaller drops, each charged to a potential of 2 volts.

### Step-by-Step Solution:

1. **Understanding Voltage of a Single Drop**:
The voltage \( V \) of a charged drop is given by the formula:
\[ V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r} \]
where \( Q \) is the charge and \( r \) is the radius of the drop.

2. **Initial Conditions**:
For a single drop, we know:
\[ V = 2 \, \text{V} \]
Therefore, we can express the charge \( Q \) of a single drop as:
\[ 2 = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r} \implies Q = 2 \cdot 4 \pi \epsilon_0 \cdot r \]

3. **Combining Drops**:
When 512 identical drops are combined, the total charge \( Q' \) of the larger drop becomes:
\[ Q' = 512 \cdot Q = 512 \cdot (2 \cdot 4 \pi \epsilon_0 \cdot r) = 1024 \cdot 4 \pi \epsilon_0 \cdot r \]

4. **Volume Conservation**:
The volume of the smaller drops combined equals the volume of the larger drop. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, for 512 smaller drops:
\[ 512 \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
where \( R \) is the radius of the larger drop. Simplifying this gives:
\[ 512 r^3 = R^3 \implies R = 8r \]

5. **Finding Voltage of the Larger Drop**:
Now we can find the voltage \( V' \) of the larger drop:
\[ V' = \frac{1}{4 \pi \epsilon_0} \frac{Q'}{R} = \frac{1}{4 \pi \epsilon_0} \frac{1024 \cdot 4 \pi \epsilon_0 \cdot r}{8r} \]
Simplifying this:
\[ V' = \frac{1024}{8} \cdot 2 = 128 \, \text{V} \]

### Final Answer:
The voltage of the larger drop is \( 128 \, \text{V} \).