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

Options

A. `FA^(1)v`
B. `Fv^(3)A^(-2)`
C. `Fv^(2)A^(-1)`
D. `F^(2)v^(2)A^(-1)`

Answer

B

Explanation

`Lpropv^xA^yF^zimpliesLkv^xA^yF^z`
putting the dimensions in the above relation
`[ML^2T^-1]=k[LT^-1]^x[LT^-2]^y[MLT^-2]^z`
`implies[ML^2T^-1]=k[M^zL^(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^3A^-2]`