Question 100 of 192

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density $1000 \mathrm{~kg} / \mathrm{m}^3$. If the pressure inside the bubble is $2100 \mathrm{~N} / \mathrm{m}^2$ greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

A$0.1$B$0.05$C$0.02$D$0.25$

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Answer: (B) $0.05$

$\begin{aligned}
& P_1=P_0+\rho g h=P_0+1000 \times 10 \times \frac{20}{100} \\
& \Rightarrow P_1=P_0+2000
\end{aligned}$
So, $P_2-P_1=\frac{2 S}{R}=\left(\frac{2 S}{1 \times 10^{-3}}\right)$
$\Rightarrow P_2=P_0+2100$
(given)
$\begin{aligned}
& \text { So, } P_0+2100-P_0-2000=2 S \times 10^3 \\
& \Rightarrow 100=2 S \times 10^3 \\
& \Rightarrow s=\left(\frac{1}{20}\right)=0.05
\end{aligned}$