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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2011

  • question_answer
    The equation of a wave is given by \[y=\]A\[\sin \,\omega \left( \frac{x}{v}=k \right),\]where \[\omega \] is angular velocity, v be linear velocity. Then dimension off c is

    A) \[\left[ LT \right]\]                                          

    B) \[\left[ T \right]\]

    C) \[\left[ {{T}^{1}} \right]\]

    D) \[\left[ {{T}^{2}} \right]\]

    Correct Answer: B

    Solution :

    \[y=\sin \omega \left( \frac{x}{v}-k \right)\] Comparing this equation with \[y=A\,\sin \,\omega \left( \frac{x}{v}-t \right)\] Dimension of k = dimension of t \[=[T]\]


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