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.
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,
\({\rho _{Excess}} = h\rho g + \frac{{2s}}{r}\)
\( \Rightarrow 1100 = h \times {10^3} \times 9.8 + \frac{{2 \times 6 \times {{10}^{ - 2}}}}{{{{10}^{ - 3}}}}\)
\( \Rightarrow 1100 - 9800h + 120\)
\( \Rightarrow 9800h = 1100 - 120\)
\( \Rightarrow h = \frac{{980}}{{9800}} = 0.1\,m\)
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