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QJEEAdv09Pha07a: JEE Advanced 2009 Physics - Class 11 JEEAdv09Pha07a Solution: Two soap bubbles A and B are kept in a closed chamber where the …

Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m 2^2 . 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 nB/nAn_B/n_A , where nAn_A and nBn_B are the number of moles of air in bubbles A and B, respectively. (Neglect the effect of gravity.)

Step-by-Step Verified Solution

We have,

PA=P0+4SrA=8+[4(0.04)2×10−2]=16{P_A} = {P_0} + {{4S} \over {{r_A}}} = 8 + \left[ {{{4(0.04)} \over {2 \times {{10}^{ - 2}}}}} \right] = 16 N/m 2^2

PB=P0+4SrB=9+[4(0.04)4×10−2]=12{P_B} = {P_0} + {{4S} \over {{r_B}}} = 9 + \left[ {{{4(0.04)} \over {4 \times {{10}^{ - 2}}}}} \right] = 12 N/m 2^2

Now, PV=nRTPV = nRT

Therefore, nBnA=PBVBPAVA−PBrB3PArA3=1216×648=6{{{n_B}} \over {{n_A}}} = {{{P_B}{V_B}} \over {{P_A}{V_A}}} - {{{P_B}r_B^3} \over {{P_A}r_A^3}} = {{12} \over {16}} \times {{64} \over 8} = 6

IIT-JEE 2009 Paper 2 Offline Physics - Properties of Matter Question 12 English Explanation

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