A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then

Answer

`5.86 xx 10^(-6) J`

Explanation

`27xx ( 4//3) pi r^(3) = ( 4//3) pi R^(3), R= 3 xx 0.6 = 1.8 xx 10^(-3)m`. Loss of energy = Decrease in surface area `xx` surface tension `= ( 27 xx 4 pi r^(2)- 4 pi R^(2)) sigma = (122.1 - 40.69) xx 10^(-6) xx 72 xx 10^(-3) = 5.86 xx 10^(-6)J`