Physics And Measurement Question 110
Question: If velocity $ v $ , acceleration $ A $ and force $ F $ are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of $ v,A $ and $ F $ would be
Options:
A) $ F{{A}^{-1}}v $
B) $ Fv^{3}{{A}^{-2}} $
C) $ Fv^{2}{{A}^{-1}} $
D) $ F^{2}v^{2}{{A}^{-1}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ L\propto v^{x}A^{y}F^{z} $
$ \Rightarrow $ $ L=kv^{x}A^{y}F^{z} $
Putting the dimensions in the above relation $ [ML^{2}{{T}^{-1}}]=k{{[L{{T}^{-1}}]}^{x}}{{[L{{T}^{-2}}]}^{y}}{{[ML{{T}^{-2}}]}^{z}} $
therefore $ [ML^{2}{{T}^{-1}}]=k[M^{z}{{L}^{x+y+z}}{{T}^{-x-2y-2z}}] $
Comparing the powers of $ M,L $ and $ T $
$ z=1 $ -(i)
$ x+y+z=2 $ -(ii)
$ -x-2y-2z=-1 $ -(iii) On solving (i), (ii) and (iii) $ x=3,y=-2,z=1 $
So dimension of $ L $ in terms of $ v,A $ and $ f $
$ [L]=[Fv^{3}{{A}^{-2}}] $
