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

  • question_answer
    Force F is given in terms of time \[t\] and distance \[x\] by F =A sin C\[t\] + B cos\[Dx\]. Then the dimensions of \[A/B\] and \[C/D\] are

    A) \[[M_{{}}^{0}L_{{}}^{0}T_{{}}^{0}],[M_{{}}^{0}L_{{}}^{0}T_{{}}^{-1}]\]

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

    C) \[[M_{{}}^{0}L_{{}}^{0}T_{{}}^{0}],[M_{{}}^{0}L_{{}}^{{}}T_{{}}^{-1}]\]  

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

    Correct Answer: C

    Solution :

    [c] \[\frac{A}{B}=\frac{Force}{Force}=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\] Ct = angle\[\Rightarrow \]\[C=\frac{Angle}{Time}=\frac{1}{T}={{T}^{-1}}\] Dx = angle\[\Rightarrow \] \[D=\frac{Angle}{Dis\tan ce}=\frac{1}{L}={{L}^{-1}}\] \[\therefore \,\frac{C}{D}=\frac{{{T}^{-1}}}{{{L}^{-1}}}=[{{M}^{0}}L{{T}^{-1}}]\]


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