Physics And Measurement Question 66

Question: 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=Cm^{x}K^{y} $ ; where $ C $ is a dimensionless quantity. The value of $ x $ and $ y $ are

[CBSE PMT 1990]

Options:

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} $

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Answer:

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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