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 the smaller droplet. (b) half of that for smaller droplet (c) (1/4) th of that for smaller droplet. (d) twice that for smaller droplet

Answer: (b). Volume conserved: R³ = 8r³ ⇒ R = 2r. Δ P = 2T/R ∝ 1/R Δ PbigΔ Psmall = r/2r = 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 2024 (16 May, Shift 2)

Q24MHT CET 2024 · 16 MAY · SHIFT 2Easy

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: $R^3 = 8r^3 \Rightarrow R = 2r$ .
$\Delta P = \frac{2T}{R} \propto \frac{1}{R}$
$\frac{\Delta P_{big}}{\Delta P_{small}} = \frac{r}{2r} = \frac{1}{2}$
So the answer is (b) half of that for a smaller droplet.

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