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. 2
B. 4
C. 6
D. 8

Answer

C

Explanation

The excess of pressure above atmospheric pressure, due to surface tension in a bubble`=(4T)/(r)`
The surronding pressure `P_(0)=8 N/m^(2)`
`therefore P_(A)` for `1^(st)` bubble `=P_(0)+(4T)/(r_(A))=8+(4xx0.04)/(0.02)`
`P_(A)=16N//m^(2)`
`P_(B)=P_(0)+(4T)/(r_(A))=12N//m^(2)`
`PV=nRT`
`(16)(4)/(3)pi(0.02)^(3)=n_(A)RT`
`(12)((4)/(3)pi(0.04)^(3))=n_(B)RT`
`(n_(B))/(n_(A))=6`