A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury 0.465 J m^-2.

23.4 μ J

Give, radius of the initial drop,

Area of the drop,

surface tension of mercury,

The volume of original drop,

Volume of each drop after division,

Let the new radius of each drop be

The total area of all droplets,

Therefore, increase in area

So, increase in energy,