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.]

Options

A. 9
B. 6
C. 4
D. 3

Answer

B

Explanation

Pressure inside bubble A:
`P_(A)=P_(0)+(4T)/(r)=8+(4 xx 0.04)/(0.02) =16N//m^(2)`
Pressure inside bubbe B:
`P_(B)=P_(0)+(4T)/(r)=8+(4 xx 0.04)/(0.04) =12 N//m^(2)`
`PV=nRT rArr P(4/3 pir^(3)) =nRT`
`rArr n alpha Pr^(3)`
`rArr n_(B)/n_(A)=(P_(B))/(P_(A)) r_(B)^(2)/r_(A)^(3)=12/16 xx (64)/8=6`