JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Fluid Mechanics

Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure

8 N / m^{2}

. 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.]

Answer: 6

Solution

The pressure inside a soap bubble is

P = P_{0} + \frac{4 T}{R}

. For bubble A:

P_{A} = 8 + \frac{4 \times 0.04}{0.02} = 8 + 8 = 16

N/m². For bubble B:

P_{B} = 8 + \frac{4 \times 0.04}{0.04} = 8 + 4 = 12

N/m². Using the ideal gas law at constant temperature:

\frac{n _{B}}{n _{A}} = \frac{P _{B} V _{B}}{P _{A} V _{A}} = \frac{P _{B}}{P _{A}} (\frac{R _{B}}{R _{A}})^{3} = \frac{12}{16} \times (\frac{4}{2})^{3} = \frac{3}{4} \times 8 = 6

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Integer answer