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Question: An air bubble of radius \(0.1cm\) is in a liquid having surface tension \(0.06\dfrac{N}{m}\) and den...

An air bubble of radius 0.1cm0.1cm is in a liquid having surface tension 0.06Nm0.06\dfrac{N}{m} and density 103kgm3{10^3}\,\dfrac{{kg}}{{{m^3}}} . The pressure inside the bubble is 1100Nm21100N{m^{ - 2}} greater than the atmospheric pressure. At what depth is the bubble below the surface of the liquid? (g=9.8m/s)\left( {g\, = \,9.8m/s} \right)
(A) 0.20m0.20m
(B) 0.15m0.15m
(C) 0.10m0.10m
(D) 0.25m0.25m

Explanation

Solution

Hint
You can easily see that the question is from the liquids topic and we can use the relation equation of Pressure inside the bubble with atmospheric pressure, gravity, surface tension and depth. The equation is as follows:
Pinside=Patmospheric+ρgh+2TR{P_{inside}}\, = \,{P_{atmospheric}}\, + \,\rho gh\, + \,\dfrac{{2T}}{R}

Complete step by step answer
We will solve the question just as told in the hint, this is the simplest and conceptual method. Solving questions with this method will also help us in strengthening our concepts of the topic.
First, have a look at what is given to us in the question.
Radius of the bubble (R)=0.1cm\left( R \right)\, = \,0.1cm
Surface Tension of the liquid (T)=0.06Nm\left( T \right)\, = \,0.06\dfrac{N}{m}
Density of the liquid (ρ)=103kgm3\left( \rho \right)\, = \,{10^3}\,\dfrac{{kg}}{{{m^3}}}
Difference in the pressure inside the bubble and atmospheric pressure (PinsidePatmospheric)=1100Nm2\left( {{P_{inside}}\, - \,{P_{atmospheric}}} \right)\, = \,1100N{m^{ - 2}}
The equation that we will be using is:
Pinside=Patmospheric+ρgh+2TR{P_{inside}}\, = \,{P_{atmospheric}}\, + \,\rho gh\, + \,\dfrac{{2T}}{R}
We can clearly see that the question has already given us the difference between the pressure inside the bubble and atmospheric pressure, so we can transpose and simplify the equation as:
PinsidePatmospheric=ρgh+2TR{P_{inside}}\, - \,{P_{atmospheric}}\, = \,\rho gh\, + \,\dfrac{{2T}}{R}
We have to find the value of hh , so we first need to do some transposing:
ρgh=PinsidePatmospheric2TR\rho gh\, = \,{P_{inside}}\, - \,{P_{atmospheric}}\, - \,\dfrac{{2T}}{R}
The question has already given us the values of ρ,g,T\rho ,\,g,\,T and RR .
All that is left for us to do is to put in the values and find the answer. So let’s do this:
(103×9.8)h=11002×0.060.1×102\left( {{{10}^3}\, \times \,9.8} \right)h\, = \,1100\, - \,\dfrac{{2\, \times \,0.06}}{{0.1\, \times \,{{10}^{ - 2}}}}
After solving and further transposing the equation, we will deduce the value of hh :
h=0.1mh\, = \,0.1m
Hence, the option (C) is the correct answer.

Note
This is the simplest and fastest method to solve such questions, so do remember the way around it like we did. Also, many students confuse the equation with minor errors, always remember that Pinside{P_{inside}} is on one side of the equation and the sum of everything else is on the other side of the equation.