P is the excess pressure inside a water drop . If that drop is divided into 8 indentical droplets , excess pressure inside smaller droplet is

Options

A. P
B. P/2
C. 2P
D. P/8

Explanation

To solve the problem of finding the excess pressure inside smaller droplets formed by dividing a larger water drop, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Excess Pressure**: The excess pressure \( P \) inside a spherical droplet is given by the formula:
\[ P = \frac{2T}{R} \]
where \( T \) is the surface tension and \( R \) is the radius of the droplet.

2. **Initial Conditions**: Let the radius of the original water drop be \( R_1 \) and the excess pressure inside it be \( P_1 = P \).

3. **Dividing the Drop**: When the original drop is divided into 8 identical smaller droplets, let the radius of each smaller droplet be \( R_2 \).

4. **Volume Conservation**: The total volume before and after division must be equal. The volume of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the volume of the original drop is:
\[ V_1 = \frac{4}{3} \pi R_1^3 \]
The total volume of the 8 smaller droplets is:
\[ V_2 = 8 \times \frac{4}{3} \pi R_2^3 = \frac{32}{3} \pi R_2^3 \]

5. **Setting Volumes Equal**: Since the volumes are equal, we have:
\[ \frac{4}{3} \pi R_1^3 = \frac{32}{3} \pi R_2^3 \]
Simplifying this gives:
\[ R_1^3 = 8 R_2^3 \]

6. **Finding the Relationship Between Radii**: Taking the cube root of both sides, we find:
\[ R_1 = 2 R_2 \]

7. **Relating Excess Pressures**: Now we can relate the excess pressures using the formula for excess pressure:
\[ \frac{P_1}{P_2} = \frac{R_2}{R_1} \]
Substituting \( R_1 = 2 R_2 \) into the equation gives:
\[ \frac{P}{P_2} = \frac{R_2}{2 R_2} = \frac{1}{2} \]
Therefore, we can express \( P_2 \) as:
\[ P_2 = 2P \]

8. **Conclusion**: The excess pressure inside each of the smaller droplets is:
\[ P_2 = 2P \]

### Final Answer:
The excess pressure inside each smaller droplet is \( 2P \).

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