Exam

JEE Main

Year

2025

Date

24 January

Shift

Morning Shift

Subject

Physics

Chapter

Mechanical Properties of Fluids

JEE Main 2025 · 24 January · Morning Shift · Physics · Mechanical Properties of Fluids

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the...

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 $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Step-by-step solution

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