Question

A bubble of air has radius $0.2\text{mm}$ in water. If the bubble had been formed $20\text{cm}$ below the water surface on a day when the atmospheric pressure was $1.013\times {10}^5\text{Pa}$, then what would have been the total pressure inside the bubble? (surface tension of water = $73\times {10}^{-3}\text{N/m}$)

• A. $1.052\times {10}^5\text{Pa}$
• B. $1.039\times {10}^5\text{Pa}$
• C. $1.022\times {10}^5\text{Pa}$
• D. $1.152\times {10}^5\text{Pa}$

Answer 1

The correct option is B $1.039\times {10}^5\text{Pa}$

The pressure inside the bubble will be $\frac{2T}{r}$ more than that of the outside.
Hence
$P_{in}=P_{atm}+\rho gh+\frac{2T}{r}$
$P_{in}=101300+1960+730$
$P_{in}=103990\text{Pa}$
$P_{in}=1.0399\times {10}^5\text{Pa}$