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MHT CET 2026 (15 May, Shift 2) · Splitting & Merging of Drops

Splitting & Merging of Drops question from MHT CET 2026 (15 May, Shift 2)

Q9MHT CET 2026 · 15 MAY · SHIFT 2Medium

Splitting & Merging of Drops

Two water drops each of radius $r$ coalesce to form a big drop. If $T$ is the surface tension, the surface energy released in this process is $4\pi r^2 T x$ . The value of $x$ is

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Answer: (a)

Volume conserved: $R^3 = 2r^3 \Rightarrow R = 2^{1/3} r$ .
Energy released $= T\left(2 \cdot 4\pi r^2 - 4\pi 2^{2/3} r^2\right) = 4\pi r^2 T\left(2 - 2^{2/3}\right)$
So $x = 2 - 2^{2/3}$ , answer (a).

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