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MCQEasyJEE 2024Surface Tension & Capillarity

JEE Physics 2024 Question with Solution

The excess pressure inside a soap bubble is three times the excess pressure inside a second soap bubble. The ratio between the volume of the first and the second bubble is:

  • A

    1:91:9

  • B

    1:31:3

  • C

    1:811:81

  • D

    1:271:27

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The excess pressure inside the first soap bubble is three times that inside the second bubble.

Find: The ratio of the volumes of the first and second bubbles.

For a soap bubble, the excess pressure is

ΔP=4TR\Delta P = \frac{4T}{R}

where TT is the surface tension and RR is the radius of the bubble.

According to the question,

ΔP1=3ΔP2\Delta P_1 = 3\Delta P_2

Substituting the excess-pressure formula,

4TR1=3×4TR2\frac{4T}{R_1} = 3 \times \frac{4T}{R_2}

Cancelling common terms,

1R1=3R2\frac{1}{R_1} = \frac{3}{R_2}

So,

R2=3R1R_2 = 3R_1

Volume Ratio Step

The volume of a spherical bubble is

V=43πR3V = \frac{4}{3}\pi R^3

Hence,

V1=43πR13V_1 = \frac{4}{3}\pi R_1^3

and

V2=43πR23=43π(3R1)3V_2 = \frac{4}{3}\pi R_2^3 = \frac{4}{3}\pi (3R_1)^3

Therefore,

V2=27(43πR13)=27V1V_2 = 27 \left(\frac{4}{3}\pi R_1^3\right) = 27V_1

Thus,

V1V2=127\frac{V_1}{V_2} = \frac{1}{27}

Therefore, the ratio of the volumes of the first and second bubbles is 1:271:27. The correct option is D.

Common mistakes

  • Using the excess-pressure formula for a liquid drop, ΔP=2TR\Delta P = \frac{2T}{R}, instead of a soap bubble formula, ΔP=4TR\Delta P = \frac{4T}{R}. A soap bubble has two surfaces, so use 4TR\frac{4T}{R}.

  • Assuming pressure is directly proportional to radius. Since ΔP∝1R\Delta P \propto \frac{1}{R}, a larger excess pressure means a smaller radius. First convert the pressure ratio into a radius ratio.

  • Comparing volumes linearly with radii. Volume of a sphere varies as R3R^3, so after finding R1:R2=1:3R_1:R_2 = 1:3, cube the ratio to get the volume ratio.

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