Question

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

The correct option is B $1.05\times {10}^5$ Pa

Here, $h=30cm=30\times {10}^{-2}m$,
$P_0=1.01\times {10}^{5}Pa$
${\rho }_{soap}=1.2\times {10}^{3}kg{m}^{-3}$
Excess pressure inside the air bubble,

$P^{\prime }=\frac{2S}{r}=\frac{2\times 2.5\times {10}^{-2}}{6\times {10}^{-3}}=8.3Pa$

Pressure inside the air bubble at depth h in soap solution

$P^{\prime }=P_0+h\rho g$

$=8.3+1.01\times {10}^5+30\times {10}^{-2}\times 1.2\times {10}^3\times 9.8$

$=1.05\times {10}^{5}Pa$