What is the excess pressure inside a bubble of soap solution of radius 5.00mm, given that the surface tension of soap solution at the temperature `(20^(@)C)` is `2.50xx10^(-2)Nm^(-1)`? If an air bubble of the same dimension were formed at a depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1atm. is `1.01 xx 10^(5)Pa`).

Options

A. `2 xx 10^(4) Pa`
B. `1.06 xx 10^(5) Pa`
C. `3 xx 10^(4) Pa`
D. `5 xx 10^(3) Pa`

Answer

B

Explanation

Excess pressure inside a soap bubble is given by
`Delta p = (4S)/(r )`
where, S = surface tension of the soap solution and r = radius of the bubble
Given, surface tension of soap solution `(S ) = 2.5 xx 10^(-2) Nm^(-1)`
Density of soap solution `(rho) = 1.2 xx 10^(3) kg m^(-3)`
Radius of soap bubble `(r) = 5.00 mm = 5.0 xx 10^(-3) m`
Atmospheric pressure `(p_(0)) = 1.01 xx 10^(5) Pa`
Excess pressure inside the soap bubble
`= (4S)/(r ) = (4 xx 2.5 xx 10^(-2))/(5.0 xx 10^(-3)) = 20 Pa`
Excess pressure inside the air bubble
`= (2S)/(R ) = (2 xx 2.5 xx 10^(-2))/(5.0 xx 10^(-3)) = 10 Pa`
`:.` Pressure inside the air bubble = Atmospheric pressure + Pressure due to 40 cm of soap solution column + Excess pressure inside the bubble
`= (1.01 xx 10^(5)) + (0.40 xx 1.2 xx 10^(3) xx 9.8) + 10`
`= (1.01 xx 10^(5)) + 4.704 xx 10^(3) + 10`
`= 1.01 xx 10^(5) + 0.04704 xx 10^(5) + 0.00010 xx 10^(5)`
`= 1.05714 xx 10^(5) Pa = 1.06 xx 10^(5) Pa`