Question

A small air bubble of $0.1mm$ diameter is formed just below the surface of water. If surface tension of water is $0.072N{m}^{-1}$, the pressure inside the air bubble in $kPa$ is (Atmospheric pressure $=1.01\times {10}^{5}Pa$):

• A. $28.9$
• B. $0.289$
• C. $0.0289$
• D. $103.88$

Answer 1

Answer: (D)

$103.88$

$P_{in}=P_o$ +$\frac{2T}{R}$

$\frac{2T}{R}=\frac{2\times 0.072}{\frac{0.1\times {10}^{-3}}{2}}=4\times 720$$=$ $2880Pa$

$\Rightarrow P_{in}=101+2.88kPa$$=$ $103.88kPa$