Two spherical soap bubbles of… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Thermodynamics

2021mcqhard

Two spherical soap bubbles of radii ${r}_{1}$ and ${r}_{2}$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to:

Official previous-year question

Held on 25 Jul 2021 · Verified 6 Jul 2026.

Options

A
$\frac{{r}_{1}{r}_{2}}{{r}_{1}+{r}_{2}}$
B
$\sqrt{{r}_{1}{r}_{2}}$
C
$\sqrt{{r}_{1}^{2}+{r}_{2}^{2}}$
D
$\frac{{r}_{1}+{r}_{2}}{2}$

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Solution

JEE Main 2021 Physics, Kinetic Theory of Gases — solution figure

Number of moles is conserved.

${n}_{1}+{n}_{2}={n}_{3}$

${P}_{1}{V}_{1}+{P}_{2}{V}_{2}={P}_{3}V$

$\frac{4S}{{r}_{1}}(\frac{4}{3}\pi {r}_{1}^{3})+\frac{4S}{{r}_{2}}(\frac{4}{3}\pi {r}_{2}^{3})=\frac{4S}{{r}_{3}}(\frac{4}{3}\pi {r}_{3}^{3})$

${r}_{1}^{2}+{r}_{2}^{2}={r}_{3}^{2}$

${r}_{3}=\sqrt{{r}_{1}^{2}+{r}_{2}^{2}}$

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Solution

JEE Main 2021 Physics, Kinetic Theory of Gases — solution figure

Number of moles is conserved.

${n}_{1}+{n}_{2}={n}_{3}$

${P}_{1}{V}_{1}+{P}_{2}{V}_{2}={P}_{3}V$

$\frac{4S}{{r}_{1}}(\frac{4}{3}\pi {r}_{1}^{3})+\frac{4S}{{r}_{2}}(\frac{4}{3}\pi {r}_{2}^{3})=\frac{4S}{{r}_{3}}(\frac{4}{3}\pi {r}_{3}^{3})$

${r}_{1}^{2}+{r}_{2}^{2}={r}_{3}^{2}$

${r}_{3}=\sqrt{{r}_{1}^{2}+{r}_{2}^{2}}$

Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

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Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics