A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is α×10⁻⁶ J. The value of α is (take the surface tension of the liquid =0.08 N m⁻¹)

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(2) 7. Given: parent drop of diameter 2 mm, so R=10⁻³ m; n=512 droplets; T=0.08 N m⁻¹. Volume is conserved, so each droplet has radius r=R/(n^1/3)=R/8. Total droplet area =n(4π r²)=n^1/3(4π R²), so the new surface created costs Δ E=4π R²T(n^1/3-1). Δ E=4π(10⁻³)²(0.08)(8-1)=4π(5.6×10⁻⁷)=7.04×10⁻⁶ J. Hence α≈7.

A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is α×10⁻⁶ J. The value of α is (take the surface tension of the liquid =0.08 N m⁻¹)

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