The energy needed for breaking a liquid drop of radius ' R ' into 216 droplets, each of radius ' r ' is ' x ' times TR². The value of ' x ' is [ T = surface tension of the liquid]. (a) 4π (b) 12π (c) 180π (d) 20π

Answer: (d). 216 r³ = R³ ⇒ r = R/6. Final area = 216 · 4π R²/36 = 24π R² W = T(24π R² - 4π R²) = 20π T R² So the answer is (d) 20π.

The energy needed for breaking a liquid drop of radius ' R ' into 216… | MHT CET 2025 (23 Apr, Shift 2)

Q18MHT CET 2025 · 23 APR · SHIFT 2Medium

Splitting & Merging of Drops

The energy needed for breaking a liquid drop of radius ' $R$ ' into 216 droplets, each of radius ' $r$ ' is ' $x$ ' times $TR^2$ . The value of ' $x$ ' is [ $T =$ surface tension of the liquid].

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Answer: (d)

$216\, r^3 = R^3 \Rightarrow r = R/6$ .
Final area $= 216 \cdot 4\pi \frac{R^2}{36} = 24\pi R^2$
$W = T(24\pi R^2 - 4\pi R^2) = 20\pi T R^2$
So the answer is (d) $20\pi$ .

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