64 small drops of mercury each of radius r and charge q coalesce to form a big drop. The ratio of the surface density of charge of each small drop withh that of big drop is

Options

A. `1:4`
B. `4:1`
C. `1:64`
D. `64:1`

Explanation

To solve the problem, we need to find the ratio of the surface charge density of each small drop to that of the big drop formed by the coalescence of 64 small drops.

### Step-by-Step Solution:

1. **Understanding Surface Charge Density**:
The surface charge density (σ) is defined as the charge (Q) per unit area (A) of the surface. Mathematically, it is given by:
\[ \sigma = \frac{Q}{A} \]

2. **Surface Area of Small Drops**:
Each small drop has a radius \( r \). The surface area \( A_s \) of one small drop is given by:
\[ A_s = 4\pi r^2 \]

3. **Charge on Small Drops**:
Each small drop has a charge \( q \). Therefore, the surface charge density \( \sigma_s \) of one small drop is:
\[ \sigma_s = \frac{q}{4\pi r^2} \]

4. **Finding the Radius of the Big Drop**:
When 64 small drops coalesce, the total charge \( Q \) of the big drop is:
\[ Q = 64q \]
The volume of the small drop is:
\[ V_s = \frac{4}{3}\pi r^3 \]
The total volume of 64 small drops is:
\[ V_{total} = 64 \times V_s = 64 \times \frac{4}{3}\pi r^3 = \frac{256}{3}\pi r^3 \]
The volume of the big drop \( V_b \) is:
\[ V_b = \frac{4}{3}\pi R^3 \]
Setting the volumes equal gives:
\[ \frac{4}{3}\pi R^3 = \frac{256}{3}\pi r^3 \]
Simplifying, we find:
\[ R^3 = 64r^3 \implies R = 4r \]

5. **Surface Area of the Big Drop**:
The surface area \( A_b \) of the big drop is:
\[ A_b = 4\pi R^2 = 4\pi (4r)^2 = 64\pi r^2 \]

6. **Surface Charge Density of the Big Drop**:
The surface charge density \( \sigma_b \) of the big drop is:
\[ \sigma_b = \frac{Q}{A_b} = \frac{64q}{64\pi r^2} = \frac{q}{\pi r^2} \]

7. **Finding the Ratio of Surface Charge Densities**:
Now, we can find the ratio of the surface charge density of the small drop to that of the big drop:
\[ \text{Ratio} = \frac{\sigma_s}{\sigma_b} = \frac{\frac{q}{4\pi r^2}}{\frac{q}{\pi r^2}} = \frac{1/4}{1} = \frac{1}{4} \]

### Final Answer:
The ratio of the surface density of charge of each small drop to that of the big drop is \( \frac{1}{4} \).