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 . If the pressure inside the bubble is 2100 greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use g = 10

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
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 $kg / m^{3}$ . If the pressure inside the bubble is 2100 $N / m^{2}$ greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use g = 10 $m / s^{2}$ ) [2025]

1 0.02

2 0.1

3 0.25

4 0.05

Ans.
(4)

$P_{1} = P_{0} + p g h = P_{0} + 1000 \times 10 \times \frac{20}{100}$

$\Rightarrow P_{1} = P_{0} + 2000$

So, $P_{2} - P_{1} = \frac{2 T}{R} = (\frac{2 T}{1 \times 10^{- 3}})$

$\Rightarrow P_{2} = P_{0} + 2100$ (given)

So, $P_{0} + 2100 - P_{0} - 2000 = 2 T \times 10^{3}$

$\Rightarrow 100 = 2 T \times 10^{3}$

$\Rightarrow T = (\frac{1}{20}) = 0.05 N / m$

Want Us to Email you About Special Offers & Updates?