સાબુના દ્રાવણનું $20 \,^oC$ તાપમાને પૃષ્ઠતાણ $2.50 \times 10^{-2}\, N\, m^{-1}$ આપેલ છે. $5.00\, mm$ ત્રિજ્યાના સાબુના દ્રાવણના પરપોટાની અંદરનું વધારાનું દબાણ કેટલું હશે ? જો આ જ પરિમાણનો હવાનો પરપોટો પાત્રમાંના સાબુના દ્રાવણની અંદર $40.0\, cm$ ઊંડાઈએ રચાય, તો તે પરપોટાની અંદરનું દબાણ કેટલું હશે ? ($1$ વાતાવરણ દબાણ $= 1.01 \times 10^5\, Pa$) 

Excess pressure inside the soap bubble is $20 Pa$

Pressure inside the air bubble is $1.06 \times 10^{5} Pa$

Soap bubble is of radius, $r=5.00 mm =5 \times 10^{-3} m$

Surface tension of the soap solution, $S=2$ $\times 10^{-2} Nm ^{-1}$

Relative density of the soap solution $=1.20$

$.$ Density of the soap solution, $\rho=1.2 \times 10^{3} kg / m ^{3}$

Air bubble formed at a depth, $h=40 cm =0.4 m$

Radius of the air bubble, $r=5 mm =5 \times 10^{-3} m$

$1$ atmospheric pressure $=1.01 \times 10^{5} Pa$

Acceleration due to gravity, $g=9.8 m / s ^{2}$

Hence, the excess pressure inside the soap bubble is given by the relation

$P=\frac{4 S}{r}$

$=\frac{4 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}$

$=20 Pa$

Therefore, the excess pressure inside the soap bubble is $20 Pa$.

The excess pressure inside the air bubble is given by the relation:

$P^{\prime}=\frac{2 S}{r}$

$=\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}$

$=10 Pa$

Therefore, the excess pressure inside the air bubble is $10 Pa$.

At a depth of $0.4 m ,$ the total pressure inside the air bubble

$=$ Atmospheric pressure $+h \rho g +P$

$=1.01 \times 10^{5}+0.4 \times 1.2 \times 10^{3} \times 9.8+10$

$=1.057 \times 10^{5} Pa$

$=1.06 \times 10^{5} Pa$

Therefore, the pressure inside the air bubble is $1.06 \times 10^{5} \;Pa$