Question

A big water drop is divided into 8 equal droplets. $\Delta P_S$ and $\Delta P_B$ be the excess pressure inside a smaller and bigger drop respectively. The relation between $\Delta P_S$ and $\Delta P_B$ is

• A. $\Delta P_B = \Delta P_S$
• B. $\Delta P_B = \frac{1}{2} \Delta P_S$
• C. $\Delta P_B = \frac{1}{4} \Delta P_S$
• D. $\Delta P_B = 2 \Delta P_S$

Hint:

Since volume scales with the cube of the radius ($V \propto r^3$), splitting an object into 8 pieces means each new piece has a radius that is $\sqrt[3]{8} = 2$ times smaller. Because excess pressure is inversely proportional to radius ($\Delta P \propto 1/r$), halving the radius doubles the pressure, meaning $\Delta P_S = 2 \Delta P_B$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A large spherical fluid drop of radius $R$ breaks apart into 8 smaller identical droplets, each of radius $r$. We need to find the mathematical relationship connecting the internal excess pressure of the large drop ($\Delta P_B$) to that of the small droplets ($\Delta P_S$).

Step 2: Key Formula or Approach:
1.

Volume Conservation: The total volume before splitting must equal the combined volume after splitting: $$V_{\text{Big}} = 8 \cdot V_{\text{small}} \implies \frac{4}{3}\pi R^3 = 8 \cdot \left(\frac{4}{3}\pi r^3\right)$$ 2.

Excess Pressure ($\Delta P$): The excess pressure inside a spherical liquid drop with surface tension $T$ is inversely proportional to its radius: $$\Delta P = \frac{2T}{r} \implies \Delta P \propto \frac{1}{r}$$

Step 3: Detailed Explanation:
First, find the relationship between the radii by simplifying the volume equation: $$R^3 = 8r^3$$ Take the cube root of both sides: $$R = \sqrt[3]{8} \cdot r \implies R = 2r$$ Now, write out the excess pressure formulas for both drop sizes: $$\Delta P_B = \frac{2T}{R}$$ $$\Delta P_S = \frac{2T}{r}$$ Set up the ratio between the two pressures: $$\frac{\Delta P_B}{\Delta P_S} = \frac{\frac{2T}{R}}{\frac{2T}{r}} = \frac{r}{R}$$ Substitute $R = 2r$ into the fraction: $$\frac{\Delta P_B}{\Delta P_S} = \frac{r}{2r} = \frac{1}{2}$$ Cross-multiplying to isolate the large drop's pressure gives: $$\Delta P_B = \frac{1}{2} \Delta P_S$$

Step 4: Final Answer:
The correct relationship is $\Delta P_B = \frac{1}{2} \Delta P_S$, which corresponds to option (B).