Physics And Measurement Question 36
Question: The frequency of vibration of string is given by $ \nu =\frac{p}{2l}{{
[ \frac{F}{m} ]}^{1/2}} $ . Here p is number of segments in the string and l is the length. The dimensional formula for m will be [BHU 2004]
Options:
A) $ [M^{0}L{{T}^{-1}}] $
B) $ [ML^{0}{{T}^{-1}}] $
C) $ [M{{L}^{-1}}T^{0}] $
D) $ [M^{0}L^{0}T^{0}] $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \nu =\frac{P}{2l}{{[ \frac{F}{m} ]}^{1/2}} $
$ \Rightarrow {{\nu }^{2}}=\frac{P^{2}}{4l^{2}}[ \frac{F}{m} ]\therefore m\propto \frac{F}{l^{2}{{\nu }^{2}}} $
$ \Rightarrow [m]=[ \frac{ML{{T}^{-2}}}{L^{2}{{T}^{-2}}} ]=[M{{L}^{-1}}T^{0}] $
BETA
