An air bubble of radius 0.1 ~cm is in a liquid having surface tension 0.06 ~N / m and density 10^3 ~kg / m^3 . The pressure inside the bubble is 1100 Nm^ -2 greater than the atmospheric pressure. At what depth is the

Correct answer: A. 0.1 ~m Given: Radius of air bubble, r=0.1 ~cm= 10^ -3 ~m Surface tension of liquid, S=0.06 ~N / m= 6 × 10^ -2 ~N / m Density of liquid, =10^3 ~kg / m^3Excess pressure inside the bubble, _ exe =1100 Nm^ -2 Depth of bubble below the liquid surface, h= ? As we know, aligned & _ Excess =h g+(2 s)/(r) & 1100=h ×10^3 ×9.8+ 2 × 6 × 10^ -2 10^ -3 & 1100=9800 h+120 & 9800 h=1100-120 & h=(980)/(9800)=0.1 ~m aligned

JEE Main 201411 AprilOnline

Exam: JEE Main
Year: 2014
Date: 11 April
Mode: Online
Subject: Physics
Chapter: Mechanical Properties of Fluids

JEE Main 2014 · 11 April · Online · Physics · Mechanical Properties of Fluids

An air bubble of radius 0.1 ~cm is in a liquid having surface...

An air bubble of radius

0.1 \mathrm{~cm}

is in a liquid having surface tension

0.06 \mathrm{~N} / \mathrm{m}

and density

10^3 \mathrm{~kg} / \mathrm{m}^3

. The pressure inside the bubble is

1100 \mathrm{Nm}^{-2}

greater than the atmospheric pressure. At what depth is the bubble below the surface of the liquid?

\left(\mathrm{g}=9.8 \mathrm{~ms}^{-2}\right)

Options

A $0.1 \mathrm{~m}$
B $0.15 \mathrm{~m}$
C $0.20 \mathrm{~m}$
D $0.25 \mathrm{~m}$

Correct answer

A. $0.1 \mathrm{~m}$

Step-by-step solution

Given: Radius of air bubble, r=0.1

\mathrm{~cm}=

10^{-3}

\mathrm{~m}

Surface tension of liquid, S=0.06

\mathrm{~N}

\mathrm{m}=

\times

10^{-2}

\mathrm{~N}

\mathrm{m}

Density of liquid,

\rho=10^3 \mathrm{~kg} / \mathrm{m}^3Excess pressure inside the

bubble, \rho_

{\text{exe }}

=1100

\mathrm{Nm}^{-2}

Depth of bubble below the liquid surface,

h=

? As we know,

\begin{aligned}&\rho_{\text{Excess }}=h \rho g+\frac{2 s}{r}\\ &\Rightarrow \quad 1100=h \times10^3 \times9.8+\frac{2 \times 6 \times 10^{-2}}{10^{-3}}\\ &\Rightarrow \quad 1100=9800 h+120 \\ &\Rightarrow \quad 9800 h=1100-120 \\ &\Rightarrow \quad h=\frac{980}{9800}=0.1 \mathrm{~m} \end{aligned}

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