Question

Two small drops of mercury each of radius '\( R \)' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is

• A. \(\sqrt{2} : 1\)
• B. \(2^{2/3} : 1\)
• C. \(2^{1/3} : 1\)
• D. \(2 : 1\)

Hint:

When drops coalesce, total volume is conserved. For spheres, radius scales as \(R' = n^{1/3} R\) for n drops. Surface area scales as \(n^{2/3}\) times a single drop area. Here n=2, so final area = \(2^{2/3}\) times area of one drop, initial area = 2 times area of one drop. Ratio = \(2 / 2^{2/3} = 2^{1/3}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Two identical spherical mercury drops, each of radius R, coalesce to form a single larger drop. Surface energy is proportional to surface area. We need the ratio of total initial surface energy to final surface energy.

Step 2: Key Formula or Approach:
Volume conservation: \(2 \times \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R'^3\) ⇒ \(R'^3 = 2R^3\) ⇒ \(R' = 2^{1/3} R\).
Surface area of one drop: \(4\pi R^2\). Initial total surface area \(A_i = 2 \times 4\pi R^2 = 8\pi R^2\). Final surface area \(A_f = 4\pi R'^2 = 4\pi (2^{2/3} R^2) = 4\pi 2^{2/3} R^2\).
Ratio \(A_i : A_f = 8\pi R^2 : 4\pi 2^{2/3} R^2 = 2 : 2^{2/3} = 2^{1 - 2/3} = 2^{1/3}\). Thus ratio = \(2^{1/3} : 1\).

Step 3: Detailed Explanation:
Surface energy = surface tension × area, so the ratio of energies is same as ratio of areas.

Step 4: Final Answer:
Option (C).