What should be the radius of water drop so that excess pressure inside it is 72 Nm ⁻²? (The surface tension of water 7.2 × 10 ⁻² Nm ⁻¹) (a) 1 mm (b) 2 mm (c) 8 mm (d) 4 mm

Answer: (b). Water drop (one surface): Δ P = 2T/r ⇒ r = (2T)/(Δ P). r = 2 × 7.2 × 10⁻²72 = 2 × 10⁻³ m So the answer is (b) 2 mm.

What should be the radius of water drop so that excess pressure… | MHT CET 2021 (20 Sep, Shift 2)

Q44MHT CET 2021 · 20 SEP · SHIFT 2Easy

Excess Pressure: Drops & Bubbles

What should be the radius of water drop so that excess pressure inside it is 72 Nm ${}^{-2}$ ? (The surface tension of water 7.2 $\times$ 10 ${}^{-2}$ Nm ${}^{-1}$ )

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Answer: (b)

Water drop (one surface): $\Delta P = \frac{2T}{r} \Rightarrow r = \frac{2T}{\Delta P}$ .
$r = \frac{2 \times 7.2 \times 10^{-2}}{72} = 2 \times 10^{-3}$ m
So the answer is (b) 2 mm.

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