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} $
Show Answer
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} $
BETA
