ExamBro
ExamBro
JEE Mains · Physics · STD 11 - Surface tension
  • JEE Mains 2014
  • Medium
  • MCQ

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 ....... \(m\) depth is the bubble below the surface of the liquid ? \((g\, = 9.8\, ms^{- 2})\)

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(A) \(0.1\)

Step-by-step Solution

Detailed explanation

Given: Radius of air bubble,

\(r = 0.1\,cm = {10^{ - 3}}m\)

Surface tension of liquid,

\(S = 0.06\,N/m = 6 \times {10^{ - 2}}\,N/m\)

Density of liquid, \(\rho  = {10^3}\,kg/{m^3}\)

Excess pressure inside the bubble,

\({\rho _{exe}} = 1100N{m^{ - 2}}\)

Depth of bubble below the liquid surface,

\(h=?\)

As we know,

\({\rho _{Excess}} = h\rho g + \frac{{2s}}{r}\)

\( \Rightarrow 1100 = h \times {10^3} \times 9.8 + \frac{{2 \times 6 \times {{10}^{ - 2}}}}{{{{10}^{ - 3}}}}\)

\( \Rightarrow 1100 - 9800h + 120\)

\( \Rightarrow 9800h = 1100 - 120\)

\( \Rightarrow h = \frac{{980}}{{9800}} = 0.1\,m\)

Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app