If 1000 droplets of water of surface tension 0.07 N m⁻¹, having same radius 1 mm each, combine to form a single drop, the released surface energy is: [Take π = 22/7]
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If 1000 droplets of water of surface tension 0.07 N m⁻¹, having same radius 1 mm each, combine to form a single drop, the released surface energy is: [Take π = 22/7]

G1 Physics · Accepted Answer

Volume conservation gives R = 10r = 10 mm. Released energy = S[1000(4πr²) − 4πR²] = 0.07×4π×(1000−100)×10⁻⁶ = 7.92×10⁻⁴ J.