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JEE Main & Advanced Physics Mathematical Tools, Units & Dimensions Question Bank Self Evaluation Test - Physical World, Units and Measurements

  • question_answer
    The momentum of an electron in an orbit is \[h/\lambda \] where A is a constant and \[\lambda \]f is wavelength associated with it. The nuclear magneton of electron of charge e and mass \[{{m}_{e}}\]is given as \[{{\mu }_{n}}\] \[=\frac{eh}{3672\pi {{m}_{e}}}\].The dimensions of \[{{\mu }_{n}}\] are \[(A\to current)\]

    A) \[[M{{L}^{2}}A]\]

    B)  \[[M{{L}^{3}}A]\]

    C)  \[[{{L}^{2}}A]\]

    D)  \[[M{{L}^{2}}]\]

    Correct Answer: C

    Solution :

    [c] \[{{\mu }_{n}}=\frac{ATM{{L}^{2}}{{T}^{-1}}}{M}={{L}^{2}}A\]


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