Let ' n ' is the number of liquid drops, each with surface energy ' E '. These drops join to form single drop. In this process (a) some energy will be absorbed (b) energy absorbed is [E(n - n2/3)] (c) energy released will be [E(n - n2/3)] (d) energy released will be [E(22/3 - 1)]

Answer: (c). Each small drop: E = 4π r² T. Big drop radius R = n1/3 r, so its energy = n2/3 E. Total area decreases on merging, so energy is released (not absorbed). Energy released = nE - n2/3E = E(n - n2/3) So the answer is (c).

Let ' n ' is the number of liquid drops, each with surface energy ' E… | MHT CET 2024 (9 May, Shift 2)

Q31MHT CET 2024 · 9 MAY · SHIFT 2Medium

Splitting & Merging of Drops

Let ' $n$ ' is the number of liquid drops, each with surface energy ' $E$ '. These drops join to form single drop. In this process

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Answer: (c)

Each small drop: $E = 4\pi r^2 T$ . Big drop radius $R = n^{1/3} r$ , so its energy $= n^{2/3} E$ .
Total area decreases on merging, so energy is released (not absorbed).
Energy released $= nE - n^{2/3}E = E\left(n - n^{2/3}\right)$
So the answer is (c).

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Splitting & Merging of Drops question from MHT CET 2024 (9 May, Shift 2)

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