Question

Two small drops of mercury, each of radius 'r', coalesce to form a single large drop. The ratio of the total surface energies before and after the change is:

(a) 1:2
(b) 2:1
(c) 3:1
(d) 1:3

Answer 1

When two small drops of mercury coalesce to form a single large drop, the ratio of the total surface energies before and after the change is 1:2.

Step-by-step explanation:

When two small drops of mercury coalesce to form a single large drop, the total surface area decreases because the smaller drops have more surface area than the larger drop. The total surface energy is directly proportional to the surface area of the drops.

Therefore, if we assume that the radius of each small drop is 'r', then the original total surface area is 4πr^2 + 4πr^2 = 8πr^2.

After the coalescence, the radius of the larger drop is 2r, so the total surface area becomes 4π(2r)^2 = 16πr^2. The ratio of the total surface energies before and after the change is therefore 8πr^2 : 16πr^2, which simplifies to 1:2.