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 and $\rho$ , the density of the liquid and ${\rho }_0$, the atmospheric pressure 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

$\frac{2s}{r}+\rho gH+p_0$

As we know that

$P_{\text{in}}-p^{{}^{\prime }}=\frac{2s}{r}$, where $p^{{}^{\prime }}$ is the pressure at some depth inside the fluid.

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

As $p^{{}^{\prime }}=p_0+\rho gH$,

$\Rightarrow p_{\text{in}}=\frac{2s}{r}+p_0+\rho gH$