If the radius of a spherical liquid (of surface tension S) drop increases from r to r + Delta r, the corresponding increase in the surface energy is

If the radius of a spherical liquid (of surface tension S) drop increa

1.	If the radius of a spherical liquid (of surface tension S) drop increases from r to r + Delta r, the corresponding increase in the surface energy is
Answer» `8 pi r Delta r S` `4 pi r Delta r S` `16 pi r Delta r S` `2 pi r Delta r S` Solution :The required CHANGE in surface ENERGY `=[4pi (r + Delta r)^(2) - 4pi r^(2)] S = [4pi (Delta r)^(2) + 8pi r Delta r]S` If we take `Delta r` to be small, `(Delta r)^(2)` is still SMALLER and can be neglected. Thus INCREASE in surface energy `=8 pi r Delta RS`