There were 100 droplets of mercury of 1 mm diameter on a glass plate. Subsequently they merged into one big drop. How will the energy of the surface layer change?

Explanation

The total surface area of all the droplets is `S_(0)=400 pi r^(2)=`
`100 pid^(2)` their total volume is `V_(0)= 100pi a^(3)//6`. After the droplets merge, the volume remains unchanged, but the surface area decreases:
`V=pi D^(3)//6=V_(0),S=piD^(2)`. From the condition of equality of the volumes, find the diameter of the large drop:
`(100 pi d^(3))/(6)=(piD^(3))/(6)`, where `D=d^(3)sqrt(100)`
The surface area of the large drop is `s=pid^(2)""^(3) sqrt(10^(4))`. The decrease in the surface layer energy corresponding to the decrease in the surface area of `DeltaS=S_(theta)-S= pid^(2)(10^(2)-10^(4//3))` is
`Delta_("sur")=alpha DeltaS~~ sigma DeltaS=pi sigma d^(2)(10^(2)-10^(4//3))`