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Standard XIII

Physics

Excess Pressure in Bubbles

Two soap bubb...

Question

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.
Given that for a gas, the relation between number of moles and pressure and volume is PV = kn, where k is constant at a particular temperature.
(IIT-JEE-2009)
[Neglect the effect of gravity.]

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Solution

The correct option is A 6
$P_{A} = P_{0} + \frac{4 T}{R_{A}} = 16 N / m^{2}$
$P_{B} = P_{0} + \frac{4 T}{R_{B}} = 12 N / m^{2}; \frac{n_{B}}{n_{A}} = \frac{P_{B}}{P_{A}} {(\frac{R_{B}}{R_{A}})}^{3} = 6$

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Similar questions

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The correct option is A 6PA=P0+4TRA=16N/m2 PB=P0+4TRB=12N/m2;nBnA=PBPA(RBRA)3=6

The correct option is A 6
$P_{A} = P_{0} + \frac{4 T}{R_{A}} = 16 N / m^{2}$
$P_{B} = P_{0} + \frac{4 T}{R_{B}} = 12 N / m^{2}; \frac{n_{B}}{n_{A}} = \frac{P_{B}}{P_{A}} {(\frac{R_{B}}{R_{A}})}^{3} = 6$