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MHT CET 2025 (22 Apr, Shift 2) · Splitting & Merging of Drops

Splitting & Merging of Drops question from MHT CET 2025 (22 Apr, Shift 2)

Q19MHT CET 2025 · 22 APR · SHIFT 2Medium

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: (b)

Volume conserved: $n r^3 = R^3 \Rightarrow n = R^3/r^3$ .
$W = T(n \cdot 4\pi r^2 - 4\pi R^2) = 4\pi T\left(\frac{R^3}{r} - R^2\right)$
$W = 4\pi T R^2\left[\frac{R}{r} - 1\right]$
So the answer is (b).

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