Uttarakhand PMT Uttarakhand PMT Solved Paper-2008

  • question_answer
    Given that the displacement of an oscillating particle is given by\[y=Asin(B~+Ct+D)\]. The dimensional formula for (ABCD) is

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

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

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

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

    Correct Answer: B

    Solution :

     \[y=A\text{ }sin(Bx+Ct+D)\] 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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