JEE Main 2024 · 29 January · evening · Single correct · +4 / −1
A small liquid drop of radius R is divided into 27 identical liquid drops. If the surface tension is T, then the work done in the process will be:
- (a) 4 π R² T
- (b) 8 π R² T
- (c) 1/8 π R² T
- (d) 3 π R² T
Answer and solution
Answer (b) 8 π R² T
Solution
Physics sir · 3 steps
The volume stays the same, the surface area grows, and you pay T for every new square metre of surface.
- Volume is conserved. The liquid only splits, so its total volume does not change. If each small drop has radius r, then 27 × 4/3π r³ = 4/3π R³. That gives r³ = R³/27, so r = R/3.
- Compare the areas. The big drop starts with area 4π R². The 27 small drops together have 27 × 4π (R/3)² = 12π R². The new surface created is 12π R² - 4π R² = 8π R².
W = T Δ A = T(12π R² - 4π R²) = 8π R² T
- So. The work done is 8π R² T, which is option (b).
The idea to keep. Splitting one drop into n equal drops multiplies the total area by n^(1/3). The work done is always T times the extra area, final minus initial.
Answer from the NTA final answer key. Solution by PrepWiser.