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NEET NEET SOLVED PAPER 2015 (Re)

  • question_answer
    If dimensions of critical velocity \[{{\text{v}}_{\text{c}}}\] of a liquid flowing through a tube are expressed as \[[{{\eta }^{x}}{{\rho }^{y}}{{r}^{z}}]\], 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: A

    Solution :

    Key Concept According to principle of homogeneity of dimension states that, a physical quantity equation will be dimensionally correct, if the dimensions of all the terms occurring on both sides of the equations are same. Given critical velocity of liquid flowing through a tube are expressed as \[{{v}_{c}}\propto {{\eta }^{n}}{{\rho }^{y}}{{r}^{z}}\] Coefficient of viscosity of liquid, \[\eta =[M{{L}^{-1}}{{T}^{-1}}]\] Density of liquid, \[\rho =[M{{L}^{-3}}]\] Radius of a tube, r = [L] Critical velocity of liquid \[{{v}_{c}}=[{{M}^{0}}L{{T}^{-1}}]\]                 \[\Rightarrow \,\,\,[{{M}^{0}}{{L}^{-1}}{{T}^{-1}}]={{[M{{L}^{-1}}{{T}^{-1}}]}^{x}}.{{[M{{L}^{-3}}]}^{y}}.{{[L]}^{z}}\]                                 \[[{{M}^{0}}{{L}^{1}}{{T}^{-1}}]=[{{M}^{x+y}}.{{L}^{x-3y+z}}{{T}^{x}}]\] Comparing exponents of M, L and L, we get \[x+y=0,-x-3y+z=1,-x=-1\]                 Þ         \[z=-1,x=1,y=-1\]


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