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JEE Main & Advanced Physics Mathematical Tools, Units & Dimensions Question Bank Mock Test - Units Dimenstions and Measurment

  • question_answer
    Number of particles is given by \[n=-D\frac{{{n}_{2}}-{{n}_{1}}}{{{x}^{2}}-{{x}_{1}}}\] crossing a unit area perpendicular to X-axis in unit time, where\[n_{1}^{{}}\]and \[n_{2}^{{}}\] are number of particles per unit volume for the value of \[x_{{}}^{{}}\]meant to \[x_{2}^{{}}\] and \[x_{1}^{{}}\]. Find dimensions of D called as diffusion constant

    A) \[M_{0}^{{}}LT_{{}}^{2}\]  

    B) \[M_{{}}^{0}L_{{}}^{2}T_{{}}^{-4}\]

    C) \[M_{{}}^{0}L_{{}}^{{}}T_{{}}^{-3}\]      

    D) \[M_{{}}^{0}L_{{}}^{2}T_{{}}^{-1}\]

    Correct Answer: D

    Solution :

    [d] [n]= Number of particles crossing a unit area in unit time =\[[{{L}^{-2}}{{T}^{-1}}]\] \[[{{n}_{2}}]=[{{n}_{1}}]\]=number of particles per unit valume= \[[{{L}^{-3}}]\] \[[{{x}_{2}}]=[{{x}_{1}}]\]= positions \[\therefore \,\,D=\frac{[n][{{x}_{2}}-{{x}_{1}}]}{[{{n}_{2}}-{{n}_{1}}]}=\frac{[{{L}^{-2}}{{T}^{-1}}]\times [L]}{[{{L}^{-3}}]}=[{{L}^{2}}{{T}^{-1}}]\]


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