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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 42

  • question_answer
    If dimensions of critical velocity \[{{\upsilon }_{c}}\]of a liquid flowing through a tube are expressed as \[[{{\eta }^{x}}{{\rho }^{y}}{{r}^{x}}],\] where \[\eta ,\] \[\rho \] and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by:

    A) \[-1,-1,1\]            

    B)        \[-1,-1,-1\]

    C) \[1,1,1\]                

    D)        \[1,-1,-1\]

    Correct Answer: D

    Solution :

    Applying dimensional method: \[{{v}_{c}}={{\eta }^{x}}{{\rho }^{y}}{{r}^{z}}\] \[[{{M}^{0}}L{{T}^{-1}}]={{[M{{L}^{-1}}{{T}^{-1}}]}^{x}}{{[M{{L}^{-3}}{{T}^{0}}]}^{y}}{{[{{M}^{0}}L{{T}^{0}}]}^{z}}\] Equating powers both sides \[x+y=0;\] \[-x=-1\,\,\therefore x=1\] \[1+y=0\,\,\therefore \,\,y=-1\] \[-x-3y+z=1\] \[-1-3(-1)+z=1\] \[-1+3+z=1\] \[\therefore \,\,z=-1\]


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