Energy needed in breaking a drop of radius R into n drops of radius r, is
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a
4πr2n−4πR2T
b
4π3nπr2−43πR2T
c
4πR2−4πr2nT
d
4πR2−n4πr2T
answer is A.
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Detailed Solution
Energy needed = work done In splitting, the volume remains unchanged∴ 43πR3=43πr3×nor r=R/(n)1/3 Energy needed = W = surface tension x change in area =T4πr2n−4πR2