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NEET Physics Mathematical Tools, Units & Dimensions Question Bank Dimensions

  • question_answer
    The frequency of vibration \[f\] of a mass \[m\] suspended from a spring of spring constant \[K\]is given by a relation of this type \[f=C\,{{m}^{x}}{{K}^{y}}\]; where \[C\] is a dimensionless quantity. The value of \[x\] and \[y\] are                                                                                         [CBSE PMT 1990]

    A)    \[x=\frac{1}{2},\,y=\frac{1}{2}\] 

    B)     \[x=-\frac{1}{2},\,y=-\frac{1}{2}\]

    C)      \[x=\frac{1}{2},\,y=-\frac{1}{2}\]

    D)        \[x=-\frac{1}{2},\,y=\frac{1}{2}\]

    Correct Answer: D

    Solution :

                                By putting the dimensions of each quantity both the sides we get \[[{{T}^{-1}}]={{[M]}^{x}}{{[M{{T}^{-2}}]}^{y}}\] Now comparing the dimensions of quantities in both sides we get \[x+y=0\ \text{and }\,2y=1\] \[\therefore \] \[x=-\frac{1}{2},\,\,y=\frac{1}{2}\]


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