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MHT CET 2024 (15 May, Shift 1) · Splitting & Merging of Drops

Splitting & Merging of Drops question from MHT CET 2024 (15 May, Shift 1)

Q27MHT CET 2024 · 15 MAY · SHIFT 1Medium

Splitting & Merging of Drops

When a mercury drop of radius ' $R$ ' splits up into 1000 droplets of radius ' $r$ ', the change in surface energy is ( $T =$ surface tension of mercury)

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Answer: (d)

$1000\, r^3 = R^3 \Rightarrow r = R/10$ .
Final area $= 1000 \cdot 4\pi \frac{R^2}{100} = 40\pi R^2$
$\Delta E = T(40\pi R^2 - 4\pi R^2) = 36\pi R^2 T$
So the answer is (d) $36\pi R^2 T$ .

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