A spherical drop of capacitane `2muF` is broken into eight drops of equal radius. Then the capacitance of each small drop is `("in "muF)`.

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript:

### Step 1: Understand the initial conditions
We have a spherical drop with a capacitance of \( C = 2 \, \mu F \).

### Step 2: Determine the volume of the original drop
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
where \( R \) is the radius of the original drop.

### Step 3: Calculate the volume of the smaller drops
When the original drop is broken into 8 smaller drops of equal radius, the total volume remains the same. Thus, the volume of each smaller drop \( V_s \) is:
\[ V_s = \frac{V}{8} = \frac{1}{8} \left( \frac{4}{3} \pi R^3 \right) = \frac{4}{24} \pi R^3 = \frac{1}{6} \pi R^3 \]

### Step 4: Relate the volume of the smaller drops to their radius
Let \( r \) be the radius of each smaller drop. The volume of one smaller drop can also be expressed as:
\[ V_s = \frac{4}{3} \pi r^3 \]
Setting the two expressions for \( V_s \) equal gives:
\[ \frac{4}{3} \pi r^3 = \frac{1}{6} \pi R^3 \]

### Step 5: Solve for the radius of the smaller drops
Cancelling \( \pi \) from both sides and rearranging gives:
\[ \frac{4}{3} r^3 = \frac{1}{6} R^3 \]
Multiplying both sides by \( 6 \):
\[ 8 r^3 = R^3 \]
Thus,
\[ r^3 = \frac{R^3}{8} \]
Taking the cube root:
\[ r = \frac{R}{2} \]

### Step 6: Calculate the capacitance of the smaller drops
The capacitance \( C_s \) of a spherical capacitor is given by:
\[ C = \frac{r}{k} \]
where \( k \) is a constant. For the smaller drops:
\[ C_s = \frac{r}{k} = \frac{\frac{R}{2}}{k} = \frac{1}{2} \cdot \frac{R}{k} \]
From the original drop, we know:
\[ C = \frac{R}{k} = 2 \, \mu F \]
Thus,
\[ C_s = \frac{1}{2} \cdot 2 \, \mu F = 1 \, \mu F \]

### Final Answer
The capacitance of each small drop is \( 1 \, \mu F \).

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