Surface Energy | JEE Main 2025 Solution

JEE MAIN 2025

02-04-2025 SHIFT-2

Question

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 :

Select the correct option:

$4 \pi r^{2} T[\sqrt{2}-1]$

$4 \pi r^{2} T\left[2-2^{2 / 3}\right]$

$4 \pi r^{2} T\left[2-2^{1 / 3}\right]$

$4 \pi r^{2} T[1+\sqrt{2}]$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\quad 2 \times \frac{4}{3} \pi R^{3}=\frac{4}{3} \pi r^{3} \Rightarrow r=2^{1 / 3} R$
$U_{i}=2 \times 4 \pi R^{2} T$
$U_{f}=4 \pi r^{2} T=4 \pi R^{2} T 2^{2 / 3}$
∴ Heat lost $=u_{i}-u_{f}=4 \pi R^{2} T\left[2-2^{2 / 3}\right]$

Question Tags

JEE Main

Physics

Medium

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