A water drop is divided into eight equal droplets. The pressure difference between inner and outer sides of big drop

Options

A. will be the same as for smaller droplet
B. will be half of that for smaller droplet
C. will be one-fourth of that for smaller droplet
D. will be twice of that for smaller droplet

Explanation

To solve the problem of finding the pressure difference between the inner and outer sides of a big water drop that has been divided into eight equal smaller droplets, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Excess Pressure**:
The excess pressure (ΔP) inside a droplet is given by the formula:
\[ \Delta P = \frac{2\sigma}{r} \]
where σ is the surface tension of the liquid and r is the radius of the droplet.

2. **Define the Initial Conditions**:
Let the radius of the big drop be \( R \). The excess pressure in the big drop can be expressed as:
\[ \Delta P_{\text{big}} = \frac{2\sigma}{R} \]

3. **Volume Conservation**:
When the big drop is divided into 8 smaller droplets, the total volume must remain the same. The volume of the big drop is:
\[ V_{\text{big}} = \frac{4}{3} \pi R^3 \]
The volume of one smaller droplet with radius \( r \) is:
\[ V_{\text{small}} = \frac{4}{3} \pi r^3 \]
Since there are 8 smaller droplets, the total volume of the smaller droplets is:
\[ V_{\text{total small}} = 8 \times \frac{4}{3} \pi r^3 = \frac{32}{3} \pi r^3 \]

4. **Equate the Volumes**:
Setting the volume of the big drop equal to the total volume of the smaller droplets gives:
\[ \frac{4}{3} \pi R^3 = \frac{32}{3} \pi r^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides results in:
\[ R^3 = 8r^3 \]

5. **Solve for the Relationship Between Radii**:
Taking the cube root of both sides:
\[ R = 2r \]
This means the radius of the big drop is twice that of the smaller droplets.

6. **Calculate the Excess Pressure for Smaller Droplets**:
The excess pressure for one smaller droplet is:
\[ \Delta P_{\text{small}} = \frac{2\sigma}{r} \]

7. **Calculate the Excess Pressure for the Big Drop**:
Substituting \( R = 2r \) into the formula for the big drop:
\[ \Delta P_{\text{big}} = \frac{2\sigma}{R} = \frac{2\sigma}{2r} = \frac{\sigma}{r} \]

8. **Relate the Excess Pressures**:
Now we can relate the excess pressures:
\[ \Delta P_{\text{big}} = \frac{1}{2} \Delta P_{\text{small}} \]
This shows that the excess pressure in the big drop is half that of the smaller droplets.

9. **Conclusion**:
Therefore, the pressure difference between the inner and outer sides of the big drop is half that of the smaller droplets.

### Final Answer:
The pressure difference between the inner and outer sides of the big drop is half that for the smaller droplets.