JEE Main 2026 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 5 April 2026, Shift 2, Physics Q33

Question 33 of 75 in this shift, Physics question 8 of 25, Section A.

Eight mercury drops, each of radius

r

, coalesce to form a bigger drop. The surface energy released in this process is __________. (S is the surface tension of mercury).

(1) $8\pi r^2 S$
(2) $16\pi r^2 S$ Official answer
(3) $64\pi r^2 S$
(4) $4\pi r^2 S$

Official answer

Option 2

NTA final key.

Chapter: Mechanical Properties of Fluids
Topic: Surface Tension
Idea tested: Splitting and Coalescence

Same idea in other shifts

Asked 4× in all

2 Apr 2026, Shift 1 · Q45A liquid drop of diameter $2$ mm breaks into $512$ droplets. The change in surface energy is $\alpha\times10^{-6}$ J. The value of…EasySingle correct
2 Apr 2025, Shift 2 · Q31Two water drops each of radius $r$ coalesce to form a bigger drop. If $T$ is the surface tension, the surface energy released in this…EasySingle correct
30 Jan 2024, Shift 2 · Q53A big drop is formed by coalescing 1000 small identical drops of water. If $E_1$ be the total surface energy of 1000 small drops of water…EasyNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.