JEE Mains · Physics · STD 11 - Surface tension
- JEE Mains 2025
- Hard
- MCQ
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 \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(0.05\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& P_1=P_0+\rho g h=P_0+1000 \times 10 \times \frac{20}{100} \\
& \Rightarrow P_1=P_0+2000
\end{aligned}\)
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)
\(\begin{aligned}
& \text { So, } P_0+2100-P_0-2000=2 S \times 10^3 \\
& \Rightarrow 100=2 S \times 10^3 \\
& \Rightarrow s=\left(\frac{1}{20}\right)=0.05
\end{aligned}\)
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