If the velocity `(V),` acceleration `(A)` and force `(F)` are taken as fundamental quantities instead of mass `(M),` length `(L)` and time `(T),` the dimensions of young's modulus `(Y)` would be :

Options

A. `FA^(2)V^(-4)`
B. `FA^(2)V^(-5)`
C. `FA^(2)V^(-3)`
D. `FA^(2)V^(-2)`

Answer

A

Explanation

Let `Y=[V^(a)A^(b)F^(c)]`
`[ML^(-1)T^(-2)]=[LT^(-1)]^(a)[LT^(-2)]^(b)[MLT^(-2)]^(c)`
`ML^(-1)T^(-2)=M^(c )L^(a+b+c)T^(-a-2b-2c)`
`:.c=1,a+b+c=-1,-a-2b-2c=-2`
On solving, we get `a=-4,b=2` and `c=1`.
Hence the correct choice is `(a)`.