Under isothermal conditions, a single drop of radius R is formed from two mercury drops of radii R₁ and R₂. After combination, the relation between R, R₁ and R₂ is (a) R = (R₁ + R₂)/(2) (b) R³ = R₁³ + R₂³ (c) R² = R₁² + R₂² (d) R = R₁ + R₂

Answer: (b). When drops combine, the total volume stays the same. 4/3π R³ = 4/3π R₁³ + 4/3π R₂³ So the answer is (b) R³ = R₁³ + R₂³.

Under isothermal conditions, a single drop of radius R is formed from… | MHT CET 2026 (19 May, Shift 1)

Q5MHT CET 2026 · 19 MAY · SHIFT 1Easy

Excess Pressure: Drops & Bubbles

Under isothermal conditions, a single drop of radius $R$ is formed from two mercury drops of radii $R_1$ and $R_2$ . After combination, the relation between $R, R_1$ and $R_2$ is

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Answer: (b)

When drops combine, the total volume stays the same.
$\frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_1^3 + \frac{4}{3}\pi R_2^3$
So the answer is (b) $R^3 = R_1^3 + R_2^3$ .

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