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:
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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:

G1 Physics · Accepted Answer

Volume conservation gives r = R/5. Initial area = 4πR². Final total area = 125×4π(R/5)² = 5×4πR². Increase in area = 4×4πR². Thus ΔE = SΔA = 0.45×16π×10⁻⁶ ≈ 2.26×10⁻⁵ J.