AMU Medical AMU Solved Paper-2008

  • question_answer
    Given that the displacement of an oscillating particle is given by \[\text{y}=\text{A sin }\left( \text{Bx}+\text{Ct}+\text{D} \right).\]The dimensional formula for (ABCD) is

    A) \[\left[ {{M}^{0}}{{L}^{-1}}{{T}^{0}} \right]\]       

    B)  \[\left[ {{M}^{0}}{{L}^{0}}{{T}^{-1}} \right]\]

    C) \[\left[ {{M}^{0}}{{L}^{-1}}{{T}^{-1}} \right]\]                     

    D)  \[\left[ {{M}^{0}}{{L}^{0}}{{T}^{0}} \right]\]

    Correct Answer: B

    Solution :

                     \[y=A\sin \,(Bx+Ct+D)\] As each term inside the bracket is dimensionless, so                 \[A=y=[{{L}^{1}}]\]                 \[B=\frac{1}{x}=[{{L}^{-1}}]\]                 \[C=\frac{1}{t}=[{{T}^{-1}}]\] and D is dimensionless.                 \[\therefore \]  \[[ABCD]=[L][{{L}^{-1}}][{{T}^{-1}}][1]\]                                 \[=[{{M}^{0}}{{L}^{0}}{{T}^{-1}}]\]          


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