A water drop is divided into 8 equal droplets. The pressure difference between the inner and outer side of the big drop will be (a) same as for smaller droplet (b) half of that for smaller droplet (c) 1th(1/4) of that for smaller droplet (d) twice that for smaller droplet

Answer: (b). Volume conserved: 4/3π R³ = 8 · 4/3π r³ ⇒ R = 2r. Excess pressure in a drop Δ P = 2T/R, so Δ P ∝ 1/R. Δ PbigΔ Psmall = r/R = 1/2 So the answer is (b) half of that for a smaller droplet.

A water drop is divided into 8 equal droplets. The pressure… | MHT CET 2026 (15 May, Shift 1)

Q1MHT CET 2026 · 15 MAY · SHIFT 1Easy

Excess Pressure: Drops & Bubbles

A water drop is divided into 8 equal droplets. The pressure difference between the inner and outer side of the big drop will be

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Answer: (b)

Volume conserved: $\frac{4}{3}\pi R^3 = 8 \cdot \frac{4}{3}\pi r^3 \Rightarrow R = 2r$ .
Excess pressure in a drop $\Delta P = \frac{2T}{R}$ , so $\Delta P \propto \frac{1}{R}$ .
$\frac{\Delta P_{big}}{\Delta P_{small}} = \frac{r}{R} = \frac{1}{2}$
So the answer is (b) half of that for a smaller droplet.

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Excess Pressure: Drops & Bubbles question from MHT CET 2026 (15 May, Shift 1)

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