Eight droplets of water, each of radius 0.2 mm, coalesce into a single drop . Find the change in total surface energy. (Surface tension of water = 0.072 N/m)

Twenty seven droplets of water, each of radius 0.1 mm coalesce into a single drop. Find the change in surface energy. Surface tension of water is 0.072 N//m .

Twenty seven droplets of water , each of radius 1.1 mm coalesce into a single drop .Find the change in surface energy. Surface tension of water is 0.072 N/m.

Eight droplets of water each of radius 0.5 mm, when coalesce into single drop, then the change in surface energy will be,

Eight droplets of mercury each of radius 1 mm coalesces into a single drop. Find the change in the surface energy. Surface tension of mercury is 0.465 J//m^2

1000 droplets of water having 2 mm diameter each coalesce to form a single drop . Given the surface tension of water is 0.072 Nm^(-1) .the energy loss in the process is

An air bubble of radius 0.2 mm is situated just below the water surface. Calculate the gauge pressure. Surface tension of water = 7.2 xx 10^-2 N//m .

An air bubble of radius 0.2 mm is situated just below the water surface. Calculate the guage pressure. Surface tension of water = 7.2 xx 10^(-2) N/m .

If a million tiny droplets of water of the same radius coalesce into one larger drop, the ration of the surface energy of the large drop to the total surface energy of all the droplets will be