In the formula `X=3YZ^(2),X` and `Z` have the dimensions of capacitance and magnetic induction respectively. The dimensions of `Y` in `MKS` system are :

Options

A. `[M^(-3)L^(-2)T^(-2)A^(-4)]`
B. `ML^(-2)overset(@)(A)`
C. `[M^(-3)L^(-2)T^(8)A^(4)]`
D. `[M^(-3)L^(-2)T^(4)A^(-4)]`

Answer

C

Explanation

Here `X=(Q)/(V).` But `V=(W)/(Q)`
`:.X=(Q^(2))/(W).` Now `Z=B=(F)/(IL)`
`:.Y=(X)/(3Z^(2))=(1)/(3)(Q^(2))/(W)xx(I^(2)L^(2))/(F^(2))=(A^(2)T^(2)A^(2)L^(2))/([ML^(2)T^(-2)][M^(2)L^(2)T^(-4)])`
`=[M^(-3)L^(-2)T^(8)A^(4)]`
Hence correct choice is `(c ).`