In question number 88, if an air bubble of the same dimensions were formed at a depth of 30 cm inside a container containing the soap solution of relative density 1.20, then the pressure inside the bubble is ( Take 1 atm = `1.01 xx 10^(5)` Pa)

Options

A. `1.01 xx 10^(4)` Pa
B. `1.05 xx 10^(5)` Pa
C. `2.01 xx 10^(4)` Pa
D. `3.01 xx 10^(4)` Pa

Answer

B

Explanation

Here, `h=30 cm=30xx10^(-2)m,`
`P_(0)=1.01xx10^(5)`Pa,
`rho_("soap")=1.2xx10^(3) kg m^(-3)`
Excess pressure inside the air bubble,
`P'=(2S)/(r)=(2xx2.5xx10^(-2))/(6xx10^(-3))=8.3` Pa
Pressure inside the air bubble at depth `h` in soap solution
`=P'+P_(0)+hrhog`
`=8.3++1.01xx10^(5)+30xx10^(-2)xx1.2xx10^(3)xx9.8`
`=1.05xx10^(5)`Pa