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}}] $



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