JEE Main 2023 · 1 February · morning · Single correct · +4 / −1

A mercury drop of radius 10⁻³ m is broken into 125 equal size droplets. Surface tension of mercury is 0.45 Nm⁻¹. The gain in surface energy is :

  1. (a) 28×10⁻⁵ J
  2. (b) 17.5×10⁻⁵ J
  3. (c) 5×10⁻⁵ J
  4. (d) 2.26×10⁻⁵ J
Answer and solution

Answer (d) 2.26×10⁻⁵ J

Solution

Physics sir · 3 steps

Volume stays the same when the drop breaks, so find the small radius first, then the extra surface area.

  1. Radius of droplets. Mercury is not lost, so the volume is conserved: 4/3π R³ = 125 × 4/3π r³. This gives r = R/5 = 2 × 10⁻⁴ m.
  2. Extra surface area. The big drop has area 4π R² = 4π × 10⁻⁶ m². The 125 droplets together have 125 × 4π × (2 × 10⁻⁴)² = 4π × 5 × 10⁻⁶ m². The increase is Δ A = 16π × 10⁻⁶ ≈ 5.03 × 10⁻⁵ m².

    Δ E = T Δ A = 0.45 × 16π × 10⁻⁶ J

  3. So. Δ E ≈ 2.26 × 10⁻⁵ J, which is option (d).

The idea to keep. When one drop splits into n equal droplets, each radius is R/n^(1/3) and the surface energy gained is 4π R² T(n^(1/3) - 1).

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 2023.