Question

An air bubble is lying just below the surface of water. The surface tension of water is $70\times {10}^{-3}N{m}^{-1}$ and atmospheric pressure is $1.013\times {10}^{5}N{m}^{-2}$. If the radius of bubble is $1mm$, then the pressure inside the bubble will be :

• A. $1.0270\times {10}^{5}Pa$
• B. $1.0160\times {10}^{5}Pa$
• C. $1.0144\times {10}^{5}Pa$
• D. $1.0131\times {10}^{5}Pa$

Answer 1

The correct option is C $1.0144\times {10}^{5}Pa$

Since, the bubble is just beneath the surface of water, we can safely assume that the pressure outside the bubble is same as atmospheric pressure.

However, the pressure inside the bubble is greater than that of the outside by,

$P_{in}-P_{out}=\frac{2T}{R}$

Therefore, $P_{in}=1.013\times {10}^5+\frac{140\times {10}^{-3}}{0.0001}$

Or, $P_{in}=1.0144\times {10}^5$