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QJEEM21Phf12e: JEE Main 2021 Physics - Class 11 JEEM21Phf12e Solution: Two spherical soap bubbles of radii r 1 and r 2 in vacuum…

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

Multiple Choice Options

  • A r1r2r1+r2{{{r_1}{r_2}} \over {{r_1} + {r_2}}}
  • B r1r2\sqrt {{r_1}{r_2}}
  • C r12+r22\sqrt {r_1^2 + r_2^2}
  • D r1+r22{{{r_1} + {r_2}} \over 2}

Step-by-Step Verified Solution

JEE Main 2021 (Online) 25th July Evening Shift Physics - Properties of Matter Question 165 English Explanation

no. of moles is conserved

n1 + n2 = n3

P1V1 + P2V2 = P3V

4Sr1(43πr13)+4Sr2(43πr23)=4Sr3(43πr33){{4S} \over {{r_1}}}\left( {{4 \over 3}\pi r_1^3} \right) + {{4S} \over {{r_2}}}\left( {{4 \over 3}\pi r_2^3} \right) = {{4S} \over {{r_3}}}\left( {{4 \over 3}\pi r_3^3} \right)

r12+r22=r32r_1^2 + r_2^2 = r_3^2

r3=r12+r22{r_3} = \sqrt {r_1^2 + r_2^2}

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