The surface energy of a liquid drop of radius r is proportional to(a) r³(b) r²(c) r(d) 1/r
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option b is right
r^2
since, we know that surface energy = surface tension * area
so,
area =4*pi*r^2
now,
surface energy is proportional to 4 pi *r^2
hence,
r^2 is the right answer
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