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: The set up.
A big water drop splits into 8 equal small droplets. We compare the excess pressure inside the big drop $\Delta P_B$ with that inside a small droplet $\Delta P_S$.
Step 2: Volume stays the same.
The water is only divided, not lost, so the big drop's volume equals 8 small drop volumes: \[ \frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3 \]
Step 3: Find the radius link.
Cancel the common parts: $R^3 = 8r^3$. Take the cube root: \[ R = 2r \] So the big drop has twice the radius of a small one.
Step 4: The excess pressure rule.
Inside a drop the excess pressure is \[ \Delta P = \frac{2T}{r} \] so it is larger for a smaller drop ($\Delta P \propto 1/r$).
Step 5: Take the ratio.
\[ \frac{\Delta P_B}{\Delta P_S} = \frac{2T/R}{2T/r} = \frac{r}{R} = \frac{r}{2r} = \frac{1}{2} \]
Step 6: Write the relation.
\[ \Delta P_B = \frac{1}{2}\Delta P_S \] The bigger drop has half the excess pressure. This is option (2). \[ \boxed{\Delta P_B = \tfrac{1}{2}\Delta P_S} \]