JEE Main 2022 · 26 July · morning · Single correct · +4 / −1

A water drop of radius 1 cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/ cm, then the gain in surface energy upto first decimal place will be :
(Given π=3.14 )

  1. (a) 8.5 × 10⁻⁴ J
  2. (b) 8.2 × 10⁻⁴ J
  3. (c) 7.5 × 10⁻⁴ J
  4. (d) 5.3 × 10⁻⁴ J
Answer and solution

Answer (c) 7.5 × 10⁻⁴ J

Solution

Physics sir · 4 steps

Keep the volume fixed to get the small radius, then the gain is just T times the extra surface area.

  1. Volume is conserved. The 729 droplets hold the same water as the big drop, so 729 × 4/3π r³ = 4/3π R³. This gives r = R/∛729 = R/9.
  2. New surface area. Total area of the droplets is 729 × 4π (R/9)² = 9 × 4π R². That is 9 times the area of the original drop, so the extra area is 8 × 4π R² = 32π R².

    Δ U = T Δ A = 75 × 32 × 3.14 × (1)² = 7536 erg

  3. Convert to joules. Since 1 erg = 10⁻⁷ J, the gain is 7536 × 10⁻⁷ J = 7.536 × 10⁻⁴ J.
  4. So. To the first decimal place, the gain in surface energy is 7.5 × 10⁻⁴ J, option (c).

The idea to keep. Breaking one drop into n equal drops multiplies the surface area by n^(1/3). The energy gained is T × 4π R² (n^(1/3) - 1), so use only the extra area.

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 2022.