Sixty four spherical drops each of radius `2 cm` and carrying `5 C` charge combine to form a bigger drop. Its capacity is.

Options

A. `(8)/(9) xx 10^(-11)F`
B. `90 xx 10^(-11) F`
C. `1.1 xx 10^(-11) F`
D. `9 xx 10^(11) F`

Explanation

To solve the problem, we need to find the capacitance of the larger drop formed by the combination of 64 smaller spherical drops, each with a radius of 2 cm and carrying a charge of 5 C.

### Step-by-Step Solution:

1. **Calculate the Total Charge:**
Each of the 64 drops carries a charge of 5 C. Therefore, the total charge \( Q \) on the larger drop is given by:
\[ Q = n \times q = 64 \times 5 \, \text{C} = 320 \, \text{C} \]

2. **Calculate the Radius of the Larger Drop:**
When the smaller drops combine to form a larger drop, the volume of the larger drop is equal to the total volume of the smaller drops. The volume \( V \) of a single spherical drop is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
For one drop with radius \( r = 2 \, \text{cm} = 0.02 \, \text{m} \):
\[ V_{\text{small}} = \frac{4}{3} \pi (0.02)^3 = \frac{4}{3} \pi (8 \times 10^{-6}) = \frac{32\pi}{3} \times 10^{-6} \, \text{m}^3 \]
The total volume of 64 drops:
\[ V_{\text{total}} = 64 \times V_{\text{small}} = 64 \times \frac{32\pi}{3} \times 10^{-6} = \frac{2048\pi}{3} \times 10^{-6} \, \text{m}^3 \]

Let \( R \) be the radius of the larger drop. The volume of the larger drop is:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
Setting the volumes equal:
\[ \frac{4}{3} \pi R^3 = \frac{2048\pi}{3} \times 10^{-6} \]
Dividing both sides by \( \frac{4}{3} \pi \):
\[ R^3 = 512 \times 10^{-6} \]
Taking the cube root:
\[ R = (512 \times 10^{-6})^{1/3} = 8 \times 10^{-2} \, \text{m} = 8 \, \text{cm} \]

3. **Calculate the Capacitance of the Larger Drop:**
The capacitance \( C \) of a spherical conductor is given by:
\[ C = 4 \pi \epsilon_0 R \]
Where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{F/m} \). Substituting \( R = 0.08 \, \text{m} \):
\[ C = 4 \pi (8.85 \times 10^{-12}) (0.08) \]
\[ C \approx 4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.08 \]
\[ C \approx 8.85 \times 10^{-12} \times 1.0056 \approx 8.91 \times 10^{-12} \, \text{F} \]
\[ C \approx 0.89 \times 10^{-11} \, \text{F} = 8.91 \, \text{pF} \]

### Final Answer:
The capacitance of the bigger drop is approximately \( 8.91 \, \text{pF} \).