Physics · Mechanical Properties of Matter

JEE Main 2025 — 2 April, Evening Shift — Question 51

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

Option A:
$4 \pi r^{2} T[\sqrt{2}-1]$
Option B:
$4 \pi r^{2} T\left[2-2^{\frac{2}{3}}\right]$
Correct
Option C:
$4 \pi r^{2} T\left[2-2^{\frac{1}{3}}\right]$
Option D:
$4 \pi r^{2} T[1+\sqrt{2}]$

Answer: B

Step-by-step solution

$2 \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}$

\begin{aligned} & R=(2)^{1 / 3} r \\ & \Delta Q=E_{i}-E_{f} & \\ & =2 \times 4 \pi r^{2} T-4 \pi(2)^{2 / 3} r^{2} T \\ & =4 \pi R^{2} T\left(2-(2)^{2 / 3}\right) \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: Mechanical Properties of Matter
Topic: Surface Tension and Surface Energy