A mercury drop of radius 10⁻³ m is broken into 125 equal size droplets. Surface tension of mercury is 0.45 N m⁻¹. The gain in surface energy is: (a) 28 × 10⁻⁵ J (b) 17.5 × 10⁻⁵ J (c) 5 × 10⁻⁵ J (d) 2.26 × 10⁻⁵ J

Answer: (d). Use Δ E = 4π R² T(n1/3 - 1). n = 125 ⇒ n1/3 = 5 Δ E = 4π (10⁻⁶)(0.45)(4) = 2.26 × 10⁻⁵ J So the answer is (d) 2.26 × 10⁻⁵ J.

A mercury drop of radius 10⁻³ m is broken into 125 equal size… | JEE Main 2023 (1 Feb, Shift 1)

Q37JEE Main 2023 · 1 FEB · SHIFT 1Medium

Splitting & Merging of Drops

A mercury drop of radius $10^{-3}\text{ m}$ is broken into 125 equal size droplets. Surface tension of mercury is $0.45\text{ N m}^{-1}$ . The gain in surface energy is:

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Answer: (d)

Use $\Delta E = 4\pi R^2 T\left(n^{1/3} - 1\right)$ .
$n = 125 \Rightarrow n^{1/3} = 5$
$\Delta E = 4\pi (10^{-6})(0.45)(4) = 2.26 \times 10^{-5}\text{ J}$
So the answer is (d) $2.26 \times 10^{-5}\text{ J}$ .

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