JEE Main 2026 · 22 January · evening · Single correct · +4 / −1
Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is :
- (a) 1:3^(2/3)
- (b) 3^(2/3):1
- (c) 1:3^(1/3)
- (d) 1:2^(2/3)
Answer and solution
Answer (a) 1:3^(2/3)
Solution
Physics sir · 3 steps
Volume and charge both add up when drops merge. Find the new radius first, then the potential follows.
- Volume is conserved. Let each small bubble have radius r and charge q. Three of them make one bubble of radius R, so 4/3π R³ = 3 × 4/3π r³, which gives R = 3^(1/3) r.
- Charge is conserved. The total charge on the big bubble is Q = 3q. The potential of a charged sphere is V = kQ/R, so the small bubble has v = kq/r.
V = k(3q)/(3^(1/3) r) = 3^(2/3) kq/r = 3^(2/3) v
- So. The ratio of the potential of one small bubble to that of the big bubble is v : V = 1 : 3^(2/3), which is option (a).
The idea to keep. When n identical charged drops merge, the radius grows as n^(1/3) and the charge as n. So the potential becomes n^(2/3) times the potential of one drop.
Answer from the NTA final answer key. Solution by PrepWiser.