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JEE Main & Advanced JEE Main Paper (Held On 11 April 2014)

  • question_answer
    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 \[1100N{{m}^{-2}}\] greater than the atmospheric pressure. At what depth is the bubble below the surface of the liquid? \[(g=9.8m{{s}^{-2}})\]   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

    A) 0.1m

    B) 0.15m

    C) 0.20m                   

    D) 0.25m

    Correct Answer: A

    Solution :

                    Given: Radius of air bubble, \[r=0.1cm={{10}^{-3}}m\] Surface tension of liquid, \[S=0.06N/m=6\times {{10}^{-2}}N/m\] 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, \[{{\rho }_{Excess}}=h\rho g+\frac{2s}{r}\] \[\Rightarrow \]\[1100=h\times {{10}^{3}}\times 9.8+\frac{2\times 6\times {{10}^{-2}}}{{{10}^{-3}}}\] \[\Rightarrow \]\[1100=9800h+120\] \[\Rightarrow \]\[9800h=1100-120\] \[\Rightarrow \]\[h=\frac{980}{9800}=0.1m\]                                


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