VMMC VMMC Medical Solved Paper-2008

  • question_answer
    Given that the displacement of an oscillating particle is given by y = A sin (Bx + Ct + D). 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. \[\therefore \]  \[[ABCD]=[L][{{L}^{-1}}][{{T}^{-1}}][1]\]                 \[=[{{M}^{0}}{{L}^{0}}{{T}^{-1}}]\]


You need to login to perform this action.
You will be redirected in 3 sec spinner