Physics · Mechanical Properties of Matter

JEE Main 2024 — 29 January, Shift 2 — Question 40

A small liquid drop of radius R is divided into 27 identical liquid drops. If the surface tension is $T$ , then the work done in the process will be :

Option A:
$8 \pi R^{2} T$
Correct
Option B:
$3 \pi R^{2} T$
Option C:
$\frac{1}{8} \pi R^{2} T$
Option D:
$4 \pi R^{2} T$

Answer: A

Step-by-step solution

Volume will be constant

$\frac{4}{3} \pi \mathrm{R}^{3}=27 \times \frac{4}{3} \times \pi \mathrm{r}^{3}$

$\mathrm{R}^{3}=27 \mathrm{r}^{3}$

$\mathrm{R}=3 \mathrm{r}$

$r=\frac{R}{3}$

$\mathrm{r}^{2}=\frac{\mathrm{R}^{2}}{9}$

Work done $=\mathrm{T} . \Delta \mathrm{A}$

$=27 \mathrm{~T}\left(4 \pi \mathrm{r}^{2}\right)-\mathrm{T} 4 \pi \mathrm{R}^{2}$

$=27 \mathrm{~T} 4 \pi \frac{\mathrm{R}^{2}}{9}-4 \pi \mathrm{R}^{2} \mathrm{~T}$

$=8 \pi \mathrm{R}^{2} \mathrm{~T}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Physics
Chapter: Mechanical Properties of Matter
Topic: Surface Tension and Surface Energy

A wire of length L and radius r is clamped at one end. If its other end is pulled by a force F, its length increases by l . If the radius…

Answer: A

Step-by-step solution

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