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

Options

A. same as that for smaller droplet
B. `1//2` of that for smaller droplet
C. `1//4` of that for smaller droplet
D. twice of that for smaller droplet

Answer

To solve the problem, we need to determine the pressure difference between the inner and outer sides of a larger water drop when it is divided into eight equal smaller droplets. We will use the concept of excess pressure in a droplet, which is given by the formula:

\[ \Delta P = \frac{2\sigma}{R} \]

where:
- \(\Delta P\) is the pressure difference,
- \(\sigma\) is the surface tension of the liquid,
- \(R\) is the radius of the droplet.

### Step-by-Step Solution

1. **Define the Radius of the Drops**:
- Let the radius of the larger drop be \(R\).
- When the larger drop is divided into 8 smaller droplets, let the radius of each smaller droplet be \(r\).

2. **Volume Conservation**:
- The volume of the larger drop is given by:
\[ V_{\text{big}} = \frac{4}{3} \pi R^3 \]
- The volume of one smaller droplet is:
\[ V_{\text{small}} = \frac{4}{3} \pi r^3 \]
- Since there are 8 smaller droplets, the total volume of the smaller droplets is:
\[ V_{\text{total small}} = 8 \times V_{\text{small}} = 8 \times \frac{4}{3} \pi r^3 = \frac{32}{3} \pi r^3 \]
- Setting the volumes equal (since volume is conserved):
\[ \frac{4}{3} \pi R^3 = \frac{32}{3} \pi r^3 \]
- Canceling \(\frac{4}{3} \pi\) from both sides:
\[ R^3 = 8r^3 \]
- Taking the cube root:
\[ R = 2r \]

3. **Calculate Pressure Difference for the Larger Drop**:
- The pressure difference for the larger drop is:
\[ \Delta P_{\text{big}} = \frac{2\sigma}{R} = \frac{2\sigma}{2r} = \frac{\sigma}{r} \]

4. **Calculate Pressure Difference for the Smaller Drop**:
- The pressure difference for one smaller droplet is:
\[ \Delta P_{\text{small}} = \frac{2\sigma}{r} \]

5. **Relate the Pressure Differences**:
- From the previous calculations, we see:
\[ \Delta P_{\text{big}} = \frac{1}{2} \Delta P_{\text{small}} \]

6. **Conclusion**:
- Therefore, the pressure difference between the inner and outer sides of the larger drop is half that of the smaller drops.

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