A spherical drop of capacitance `1 muF` is broken into eight drop of equal radius. Then, the capacitance of each small drop is

Options

A. `1/2 muF`
B. `1/4 muF`
C. `1/8 muF`
D. `8 muF`

Answer

To solve the problem, we need to determine the capacitance of each small drop after a spherical drop of capacitance \(1 \mu F\) is broken into eight smaller drops of equal radius.

### Step-by-Step Solution:

1. **Understanding Capacitance of a Sphere**:
The capacitance \(C\) of a spherical conductor is given by the formula:
\[ C = 4\pi \epsilon_0 R \]
where \(R\) is the radius of the sphere and \(\epsilon_0\) is the permittivity of free space.

2. **Initial Conditions**:
Let the radius of the original spherical drop be \(R\) and its capacitance be \(C_1 = 1 \mu F\). Therefore, we can write:
\[ C_1 = 4\pi \epsilon_0 R \]

3. **Volume Conservation**:
When the drop is broken into 8 smaller drops, the volume of the original drop is equal to the total volume of the smaller drops. The volume \(V\) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, for the original drop:
\[ V_1 = \frac{4}{3} \pi R^3 \]
For the 8 smaller drops, if each has radius \(r\):
\[ V_2 = 8 \left(\frac{4}{3} \pi r^3\right) \]
Setting \(V_1 = V_2\):
\[ \frac{4}{3} \pi R^3 = 8 \left(\frac{4}{3} \pi r^3\right) \]
Simplifying gives:
\[ R^3 = 8r^3 \]

4. **Finding the Radius of Smaller Drops**:
From the equation \(R^3 = 8r^3\), we can find \(r\):
\[ r = \frac{R}{2} \]

5. **Capacitance of Smaller Drops**:
Now we can find the capacitance \(C_2\) of each smaller drop using the radius \(r\):
\[ C_2 = 4\pi \epsilon_0 r = 4\pi \epsilon_0 \left(\frac{R}{2}\right) = \frac{4\pi \epsilon_0 R}{2} = \frac{1}{2} C_1 \]
Since \(C_1 = 1 \mu F\):
\[ C_2 = \frac{1}{2} \times 1 \mu F = 0.5 \mu F \]

### Final Answer:
The capacitance of each small drop is \(0.5 \mu F\).

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