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A spherical liquid drop of radius R = 10⁻² m, with surface tension S = 0.1/(4*pi) N/m, breaks up into K identical smaller drops. The total increase in surface energy in this process is delta U = 10⁻³ J. If K = 10^alpha, find alpha.
- alpha = 6
- alpha = 9
- alpha = 12
- alpha = 3
Correct answer: alpha = 6
Solution
Volume conservation gives r = R*K^(-1/3). The new total area is K*4*pi*r² = 4*pi*R²*K^(1/3). The increase in surface energy is S*4*pi*R²*(K^(1/3) - 1). Setting this equal to delta U and solving for K (large, so K^(1/3) dominates) gives K = 10⁶, so alpha = 6.
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