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WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    A drop of some liquid of volume \[0.04\,c{{m}^{3}}\]is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area \[20\,c{{m}^{2}}\]between the surfaces of the two slides. To separate the slides a force of \[16\times {{10}^{5}}\]dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyne\[-\,c{{m}^{-1}}\])

    A)  60   

    B)  70    

    C)  80    

    D)  90

    Correct Answer: C

    Solution :

     Let, thickness of layer be \[x\] So, volume \[V=Area\,\,\times \,\,x\] \[V=A\times x\] \[(\because \,x=2r)\] \[x=V/A\] \[\therefore \] \[2r=\frac{V}{A}\Rightarrow r=\frac{V}{2A}\] ?(i) and \[\Delta P=\frac{T}{r}\] and \[\Delta P=\frac{T}{r}\] We know that \[F=\Delta P\times A\] \[=\frac{T}{r}\times A\] \[F=\frac{T}{\left( \frac{V}{2A} \right)}\times A\] [from Eq.(i)] \[T=\frac{F\times V}{2{{A}^{2}}}\] where \[F=16\times {{10}^{5}}\,dyne,\]\[V=0.04\,c{{m}^{3}},\] \[A=20\,c{{m}^{2}}=\frac{16\times {{10}^{5}}\times 0.04}{2\times {{20}^{2}}}\] \[=\frac{8\times {{10}^{5}}\times 4}{{{20}^{2}}\times 100}=\frac{8\times {{10}^{5}}\times 4}{400\times 100}\] \[=8\times {{10}^{5}}\times {{10}^{-4}}=80\,dyne/cm\] \[=80\,dyne\,c{{m}^{-1}}\]


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