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

`L prop v^(x)A^(y)F^(z) rArr L = kv^(x)A^(y)F^(z)`
Putting the dimensions in the above relation
`[ML^(2)T^(-1)] = k[LT^(-1)]^(x)[LTG^(-2)]^(y)[MlT^(-2)]^(z)`
`rArr[ML^(2)T^(-1)] = k[M^(2)L^(x+y+z) T^(-x -2y-2z)]`
X 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)A6(-2)]`