IIT-JEE 2009 · Paper 2 · Physics Q52

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PhysicsInteger+4 / -1Question 52 of 57

Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m². The radii of bubbles A and B are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04 N/m. Find the ratio $n_B/n_A$ , where $n_A$ and $n_B$ are the number of moles of air in bubbles A and B, respectively. [Neglect the effect of gravity.]

Your answer

Answer and worked solution

Official answer: 6

Pressure inside a soap bubble. $P=P_0+\dfrac{4T}r$ : for A, $8+\dfrac{0.16}{0.02}=16$ N/m²; for B, $8+\dfrac{0.16}{0.04}=12$ N/m².

Moles $n\propto PV\propto Pr^3$ (same temperature):
$\frac{n_B}{n_A}=\frac{12\times4^3}{16\times2^3}=\frac{768}{128}=6.$

Answer: 6

Topics:Surface tension Ideal gas equation

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IIT-JEE 2009 · Paper 2, Physics question 52