Twenty seven droplets of water each of radius 0.1 mm merge to form a single drop then the energy released is (a) 1.6 × 10⁻³ J (b) 1.6 J (c) 1600 J (d) 1.6 × 10⁻⁷ J

Answer: (d). 27 r³ = R³ ⇒ R = 3r = 3 × 10⁻⁴ m. Energy released = 4π T(27 r² - R²) = 4π T(27 - 9)(10⁻⁸) = 2.26 × 10⁻⁶ T With T ≈ 0.07 N/m for water: E ≈ 1.6 × 10⁻⁷ J So the answer is (d) 1.6 × 10⁻⁷ J.

Twenty seven droplets of water each of radius 0.1 mm merge to form a… | MHT CET 2023 (13 May, Shift 1)

Q39MHT CET 2023 · 13 MAY · SHIFT 1Medium

Splitting & Merging of Drops

Twenty seven droplets of water each of radius $0.1\text{ mm}$ merge to form a single drop then the energy released is

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Answer: (d)

$27\, r^3 = R^3 \Rightarrow R = 3r = 3 \times 10^{-4}\text{ m}$ .
Energy released $= 4\pi T(27 r^2 - R^2) = 4\pi T(27 - 9)(10^{-8}) = 2.26 \times 10^{-6}\, T$
With $T \approx 0.07\text{ N/m}$ for water: $E \approx 1.6 \times 10^{-7}\text{ J}$
So the answer is (d) $1.6 \times 10^{-7}\text{ J}$ .

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