JEE Main 2025 · 2 April · evening · Single correct · +4 / −1

Two water drops each of radius ' r ' coalesce to form a bigger drop. If ' T ' is the surface tension, the surface energy released in this process is :

  1. (a) 4 π r² T[2-2^(1 / 3)]
  2. (b) 4 π r² T[1+√2]
  3. (c) 4 π r² T[2-2^(2 / 3)]
  4. (d) 4 π r² T[√2-1]
Answer and solution

Answer (c) 4 π r² T[2-2^(2 / 3)]

Solution

Physics sir · 3 steps

Volume stays the same when drops merge, but surface area shrinks, and that lost area releases energy.

  1. Volume is conserved. The water in the two small drops makes up the big drop, so 2 × 4/3π r³ = 4/3π R³. This gives R³ = 2r³, so R = 2^(1/3) r.
  2. Compare surface areas. Before merging, the total area is 2 × 4π r². After merging, it is 4π R² = 4π (2^(1/3) r)² = 4π r² × 2^(2/3).

    Δ E = T(Aᵢₙᵢₜᵢₐₗ - A_(final)) = 4π r² T[2 - 2^(2/3)]

  3. So. Surface energy is T times area. The area falls by 4π r²[2 - 2^(2/3)], which is about 0.41 × 4π r², so the energy released is 4π r² T[2 - 2^(2/3)]. That is option (c).

The idea to keep. When drops merge, fix the new radius from volume conservation first. Then energy released is T times the drop in total surface area. Radius goes as the cube root, area as its square.

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 2025.