A water drop of radius 1 cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/ cm, then the gain in surface energy upto first decimal place will be: (Given π = 3.14) (a) 8.5 × 10⁻⁴ J (b) 8.2 × 10⁻⁴ J (c) 7.5 × 10⁻⁴ J (d) 5.3 × 10⁻⁴ J

Answer: (c). Use Δ E = 4π R² T(n1/3 - 1) with R = 0.01 m, T = 75 dyne/cm = 0.075 N/m. n = 729 ⇒ n1/3 = 9 Δ E = 4(3.14)(10⁻⁴)(0.075)(8) = 7.54 × 10⁻⁴ J So the answer is (c) 7.5 × 10⁻⁴ J.

A water drop of radius 1 cm is broken into 729 equal droplets. If… | JEE Main 2022 (26 Jul, Shift 1)

Q43JEE Main 2022 · 26 JUL · SHIFT 1Medium

Splitting & Merging of Drops

A water drop of radius $1\text{ cm}$ is broken into 729 equal droplets. If surface tension of water is 75 dyne/ $\text{cm}$ , then the gain in surface energy upto first decimal place will be:

(Given $\pi = 3.14$ )

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Answer: (c)

Use $\Delta E = 4\pi R^2 T\left(n^{1/3} - 1\right)$ with $R = 0.01\text{ m}$ , $T = 75\text{ dyne/cm} = 0.075\text{ N/m}$ .
$n = 729 \Rightarrow n^{1/3} = 9$
$\Delta E = 4(3.14)(10^{-4})(0.075)(8) = 7.54 \times 10^{-4}\text{ J}$
So the answer is (c) $7.5 \times 10^{-4}\text{ J}$ .

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