Question

A water drop is divided into eight equal droplets the pressure difference between the inner and outer side of the big drop will be –

• A. Same as for smaller droplet
• B. 1/2 of that for smaller droplet
• C. 1/4 of that for smaller droplet
• D. Twice that for smaller droplet

Answer 1

Answer: (B)

1/2 of that for smaller droplet

Answer 2

When break into eight drop radius of small drop is r = $\frac{R}{n^{1/3}}$ Ž r = $\frac{R}{8^{1/3}}$ = $\frac{R}{2}$

For large drop DP₁ = $\frac{2T}{R}$

for small drop DP₂ = $\frac{2T}{r}$ = $\frac{2T}{(R/2)}$= 2$\left( \frac{2T}{R} \right)$