JEE Main 2025 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 2 April 2025, Shift 2, Physics Q31

Question 31 of 75 in this shift, Physics question 6 of 25, Section A.

Two water drops each of radius

r

coalesce to form a bigger drop. If

T

is the surface tension, the surface energy released in this process is :

(1) $4\pi r^2 T\left[\sqrt{2}-1\right]$
(2) $4\pi r^2 T\left[2-2^{2/3}\right]$ Official answer
(3) $4\pi r^2 T\left[1+\sqrt{2}\right]$
(4) $4\pi r^2 T\left[2-2^{1/3}\right]$

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
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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.