Surface Tension | JEE Main Solution

JEE MAIN 2025

24-01-2025 SHIFT-1

Question

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

Select the correct option:

0.05

0.1

0.02

0.25

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

T is surface tension
${\rm{P in air bubble }} = {{\rm{P}}_0} + \rho g{\rm{gh}} + \frac{{2\;{\rm{T}}}}{{\rm{R}}}$
${{\rm{P}}_{{\rm{in }}}} - {{\rm{P}}_0} = \rho g\;{\rm{g}} + \frac{{2\;{\rm{T}}}}{{\rm{R}}} = 2100$
$\frac{{2\;{\rm{T}}}}{{\rm{R}}} = 2100 - \rho {\rm{gh}}$
$\;{\rm{T}} = \frac{{\rm{R}}}{2}\left( {2100 - {{10}^3} \times 10 \times 0.2} \right)$
$ = \frac{1}{{20}}(2100 -2000) \times {10^{ - 2}}$
$ = 0.05$

Question Tags

JEE Main

Physics

Easy

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