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MHT CET 2026 (19 Apr, Shift 2) · Splitting & Merging of Drops

Splitting & Merging of Drops question from MHT CET 2026 (19 Apr, Shift 2)

Q1MHT CET 2026 · 19 APR · SHIFT 2Easy

Splitting & Merging of Drops

The energy needed for breaking a liquid drop of radius 'R' into 'n' droplets each of radius 'r' is [ $T$ = surface tension of the liquid]

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Answer: (a)

Volume conserved: $n r^3 = R^3 \Rightarrow n = R^3/r^3$ .
Energy $= T\Delta A = 4\pi T (n r^2 - R^2) = 4\pi T\left(\dfrac{R^3}{r} - R^2\right)$
$E = 4\pi T R^2\left[\dfrac{R}{r} - 1\right]$
So the answer is (a).

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