What is the excess pressure inside a bubble of soap solution of radius $5.00 \; mm$,given that the surface tension of soap solution at the temperature $(20 \; ^{\circ}C)$ is $2.50 \times 10^{-2} \; N m^{-1}$? 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? ($1$ atmospheric pressure is $1.01 \times 10^{5} \; Pa$).

(A) Excess pressure inside the soap bubble is given by $P = \frac{4S}{r}$.
Given $S = 2.50 \times 10^{-2} \; N m^{-1}$ and $r = 5.00 \times 10^{-3} \; m$.
$P = \frac{4 \times 2.50 \times 10^{-2}}{5.00 \times 10^{-3}} = 20 \; Pa$.
For an air bubble at depth $h = 0.40 \; m$,the excess pressure is $P' = \frac{2S}{r} = \frac{2 \times 2.50 \times 10^{-2}}{5.00 \times 10^{-3}} = 10 \; Pa$.
The total pressure inside the air bubble is $P_{total} = P_{atm} + h\rho g + P'$.
Here $\rho = 1.20 \times 1000 = 1200 \; kg/m^3$ and $g = 9.8 \; m/s^2$.
$P_{total} = 1.01 \times 10^5 + (0.40 \times 1200 \times 9.8) + 10$.
$P_{total} = 101000 + 4704 + 10 = 105714 \; Pa \approx 1.06 \times 10^5 \; Pa$.