JEE Main 2023 · 31 January · morning · Single correct · +4 / −1
If 1000 droplets of water of surface tension 0.07 N / m, having same radius 1 mm each, combine to from a single drop. In the process the released surface energy is :-
( Take π = 22/7 )
- (a) 7 .92 × 10⁻⁴ J
- (b) 7 .92 × 10⁻⁶ J
- (c) 8 .8 × 10⁻⁵ J
- (d) 9 .68 × 10⁻⁴ J
Answer and solution
Answer (a) 7 .92 × 10⁻⁴ J
Solution
Physics sir · 3 steps
Volume stays the same but surface area shrinks. The energy released is T times the area lost.
- Radius of big drop. The volume is conserved, so 4/3π R³ = 1000 × 4/3π r³. That gives R = 10r = 10 mm = 10⁻² m.
- Area lost. The total area before is 1000 × 4π (10⁻³)² = 4π × 10⁻³ m². The area after is 4π (10⁻²)² = 4π × 10⁻⁴ m². So the area lost is Δ A = 4π × 9 × 10⁻⁴ = 36π × 10⁻⁴ m².
E = T Δ A = 0.07 × 36 × 22/7 × 10⁻⁴ J
- So. E = 0.07 × 36 × 22/7 × 10⁻⁴ = 7.92 × 10⁻⁴ J. That is option (a).
The idea to keep. When n drops merge, R = n^(1/3)r. The released energy is T × (initial area - final area). A drop has only one surface, so use a single factor of T.
Answer from the NTA final answer key. Solution by PrepWiser.