BHU PMT BHU PMT (Screening) Solved Paper-2005

  • question_answer
    In a system of units if force (F) acceleration  and time (T) are taken as fundamental units, then the dimensional formula of energy is:

    A)  \[[F{{A}^{2}}T]\]                             

    B)  \[[FA{{T}^{2}}]\]

    C)  \[[{{F}^{2}}AT]\]             

    D) \[[FAT]\]

    Correct Answer: B

    Solution :

                      Key Idea: Every equation relating physical quantities should be in dimensional balance. To establish the relation, let kinetic energy depend on force (F) raised to the power a, on acceleration  raised to the power b and time (T) raised to the power c then, we have                 \[E\propto {{[F]}^{a}}{{[A]}^{b}}{{[T]}^{c}}\] Writing the dimensions on both sides, we have \[[M{{L}^{2}}{{T}^{-2}}]=k[ML{{T}^{-2}}]{{[L{{T}^{-2}}]}^{b}}{{[T]}^{c}}\] \[[M{{L}^{2}}{{T}^{-2}}]=k[{{M}^{a}}{{L}^{a+b}}{{T}^{-2a-2b+c}}]\] Taking dimensional balance of equation into consideration and equation the dimensions of both sides, we get \[a=1\]                                ...(1) \[a+b=2\]                             ...(2) \[-2a-2b-c=-2\]                ...(3) On solving Eqs. (1), (2) and (3), we get \[a=1,b=1,c=2\] Thus,    \[[E]=[{{F}^{-1}}{{A}^{1}}{{T}^{2}}]\]


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