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JEE Main 202125 JulyEvening Shift
Exam
JEE Main
Year
2021
Date
25 July
Shift
Evening Shift
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2021 · 25 July · Evening Shift · Physics · Kinetic Theory of Gases

Two spherical soap bubbles of radii r _ 1 and r _ 2 in vacuum...

Two spherical soap bubbles of radii r1{r}_{1} and r2{r}_{2} in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to:Two spherical soap bubbles of radii r _ 1 and r _ 2 in vacuum combine under isothermal conditions. The resulting bubble

Options

  1. Ar1r2r1+r2\dfrac{{r}_{1}{r}_{2}}{{r}_{1} + {r}_{2}}
  2. Br1r2\sqrt{{r}_{1}{r}_{2}}
  3. Cr12+r22\sqrt{{r}_{1}^{2} + {r}_{2}^{2}}
  4. Dr1+r22\dfrac{{r}_{1} + {r}_{2}}{2}

Correct answer

C. r12+r22\sqrt{{r}_{1}^{2} + {r}_{2}^{2}}

Step-by-step solution

Two spherical soap bubbles of radii r _ 1 and r _ 2 in vacuum combine under isothermal conditions. The resulting bubble

Number of moles is conserved.

n1+n2=n3{n}_{1} + {n}_{2} = {n}_{3}

P1V1+P2V2=P3V{P}_{1}{V}_{1} + {P}_{2}{V}_{2} = {P}_{3}V

4Sr1(43πr13)+4Sr2(43πr23)=4Sr3(43πr33)\dfrac{4S}{{r}_{1}}\left(\dfrac{4}{3}\pi{r}_{1}^{3}\right) + \dfrac{4S}{{r}_{2}}\left(\dfrac{4}{3}\pi{r}_{2}^{3}\right) = \dfrac{4S}{{r}_{3}}\left(\dfrac{4}{3}\pi{r}_{3}^{3}\right)

r12+r22=r32{r}_{1}^{2} + {r}_{2}^{2} = {r}_{3}^{2}

r3=r12+r22{r}_{3} = \sqrt{{r}_{1}^{2} + {r}_{2}^{2}}

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