JEE Main 2021 · 20 July · evening · Single correct · +4 / −1

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

  1. (a) 2^(1/3):1
  2. (b) 1:2^(1/3)
  3. (c) 2 : 1
  4. (d) 1 : 2
Answer and solution

Answer (a) 2^(1/3):1

Solution

Physics sir · 4 steps

Volume is conserved when drops merge, so find the new radius first, then compare the areas.

  1. Volume stays same. Mercury does not disappear when the drops join, so the big drop holds the volume of both small ones. If r is the new radius, 2 × 4/3π R³ = 4/3π r³.

    r³ = 2R³ ⇒ r = 2^(1/3) R

  2. Energy follows area. Surface energy is T × surface area, and T is the same before and after. Before: 2 × 4π R² T = 8π R² T. After: 4π r² T = 4π × 2^(2/3) R² T.
  3. Take the ratio. E_(before)/E_(after) = (8π R² T)/(4π × 2^(2/3) R² T) = 2/2^(2/3) = 2^(1/3).
  4. So. The ratio is 2^(1/3) : 1, option (a). The energy before is larger because merging reduces the total surface area.

The idea to keep. When drops merge or split, conserve volume to get the new radius, then compare r² for surface energy. n equal drops merging gives a radius n^(1/3) times larger.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Surface Tension

Asked about 50 times in JEE Main since 2020. This question is from 2021.