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Question: If velocity \(v,\) acceleration A and force F are chosen as fundamental quantities, then the dimensi...

If velocity v,v, acceleration A and force F are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of v,Av,A and FF would be

A

FA−1FA^{- 1}v

B

Fv3A−2Fv^{3}A^{- 2}

C

Fv2A−1Fv^{2}A^{- 1}

D

F2v2A−1F^{2}v^{2}A^{- 1}

Answer

Fv3A−2Fv^{3}A^{- 2}

Explanation

Solution

Given, v = velocity = [LT−1]\lbrack LT^{- 1}\rbrack, A = Acceleration = [LT−2]\lbrack LT^{- 2}\rbrack, F = force = [MLT−2]\lbrack MLT^{- 2}\rbrack

By substituting, the dimension of each quantity we can check the accuracy of the formula

[Angular momentum] = Fv3A−2Fv^{3}A^{- 2}

[ML2T−1]=[MLT−2][LT−1]3[LT−2]−2\lbrack ML^{2}T^{- 1}\rbrack = \lbrack MLT^{- 2}\rbrack\lbrack LT^{- 1}\rbrack^{3}\lbrack LT^{- 2}\rbrack^{- 2}= [ML2T−1]\lbrack ML^{2}T^{- 1}\rbrack

L.H.S. = R.H.S. i.e., the above formula is Correct**.**