JEE Main 2022 · 28 June · morning · Single correct · +4 / −1
A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be :
- (a) 2.8 × 10⁻⁴ J
- (b) 1.5 × 10⁻³ J
- (c) 1.9 × 10⁻⁴ J
- (d) 9.4 × 10⁻⁵ J
Answer and solution
Answer (a) 2.8 × 10⁻⁴ J
Solution
Physics sir · 4 steps
Volume stays the same, surface area grows: find the new radius first, then multiply the extra area by T.
- Volume is conserved. The big drop has radius R = 1 cm = 0.01 m. The total volume does not change, so 64 × 4/3π r³ = 4/3π R³. That gives r³ = R³/64, so r = R/4.
- Compare the areas. The original area is 4π R². The droplets together have area 64 × 4π (R/4)² = 4 × 4π R². The area has become four times larger, so the increase is 3 × 4π R².
Δ U = T Δ A = T × 4π R²(n^(1/3)-1)
- Put in numbers. Δ A = 12π × (0.01)² = 3.77 × 10⁻³ m². Then Δ U = 0.075 × 3.77 × 10⁻³ = 2.83 × 10⁻⁴ J.
- So. The gain in surface energy is about 2.8 × 10⁻⁴ J, which is option (a).
The idea to keep. When one drop splits into n equal drops, each radius becomes R/n^(1/3) and the total area grows by a factor n^(1/3). Energy gained is T times the extra area only.
Answer from the NTA final answer key. Solution by PrepWiser.