Question

Small air bubble of radius $r$ is found to form at a depth of $H$ from the open surface of the liquid contained in a beaker. If $S$ is the surface tension, $\rho$ is the density of the liquid and $P_0$is the atmospheric pressure, then the pressure inside the bubble is

• A. $\frac{4S}{r}+\rho gH+P_0$
• B. $\frac{2S}{r}-\rho gH+P_0$
• C. $\frac{4S}{r}-\rho gH+P_0$
• D. $\frac{2S}{r}+\rho gH+P_0$

Answer 1

The correct option is D $\frac{2S}{r}+\rho gH+P_0$

Let
$p^{{}^{\prime }}$ is the pressure just outside the bubble.
$P_{\text{in}}$ is the pressure inside the bubble.

Excess pressure inside the air bubble is given by

$P_{\text{in}}-p^{{}^{\prime }}=\frac{2S}{r}$,

$\Rightarrow p_{\text{in}}=\frac{2S}{r}+p^{{}^{\prime }}$

variation of pressure with depth of liquid is given by

$p^{{}^{\prime }}=P_0+\rho gH$,

Substituting

$\Rightarrow p_{\text{in}}=\frac{2S}{r}+P_0+\rho gH$