Physics And Measurement Question 262
Question: If velocity (V), force (F) and energy (E) are taken as fundamental units, then dimensional formula for mass will be
Options:
A) $ {{V}^{-2}}F^{0}E^{3} $
B) $ V^{0}FE^{2} $
C) $ V{{F}^{-2}}E^{0} $
D) $ {{V}^{-2}}F^{0}E $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ Let(M)=V^{a}F^{b}E^{c} $
Putting the dimensions of V, F and E, we have $ (M)={{(L{{T}^{-1}})}^{a}}\times {{(ML{{T}^{-2}})}^{b}}\times {{(ML^{2}{{T}^{-2}})}^{c}} $
$ orM^{1}={{M}^{b+c}}{{L}^{a+b+2c}}{{T}^{-a-2b-2c}} $
Equating the powers of dimensions, we have $ b+c=1 $
$ a+b+2c=0;-a-2b-2c=0 $ which give a $ =-2,b=0andc=1. $
Therefore $ (M)=({{V}^{-2}}F^{0}E) $
BETA
