Question

The surface tension of soap solution at a temperature $20^{\circ}C $ is $2.5 \times 10^{-2} \, N \, m^{-1}$. The excess pressure inside a bubble of soap solution of radius $6\, mm$ is, 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 $1\, atm = 1.01 \times 10^{5 }\,Pa)$

• A. $1.01 \times 10^{4}\, Pa$
• B. $1.05 \times 10^{5} \, Pa$
• C. $2.01 \times 10^{4} \, Pa$
• D. $3.01 \times 10^{4} \, Pa$

Answer 1

Answer: (B)

Here, $h = 30 \,cm = 30 \times 10^{-2} \, m$
$P_{0} = 1.01 \times 10^{5} \, Pa$,
$\rho_{soap}=1.2 \times10^{3}\, kg \, m^{-3}$
Excess pressure inside the air bubble,
$P'=\frac{2S}{r}=\frac{2\times2.5\times10^{-2}}{6 \times10^{-3}}=8.3 \, Pa$
Pressure inside the air bubble at depth $h$ in soap solution
$=P' + P_{0}+h\rho g$
$=8.3+1.01\times10^{5}+30\times10^{-2}\times1.2\times10^{3}\times9.8$
$=1.05\times10^{5}\, Pa$