`X=3YZ^(2)` find dimension of `Y` in (MKSA) system, if `X` and `Z` are the dimension of capacity and magnetic field respectively

Options

A. `M^(-3) L^(-2) T^(-4) A^(-1)`
B. `ML^(-2)`
C. `M^(-3) L^(-2) T^(4) A^(4)`
D. `M^(-3) L^(-2) T^(8) A^(4)`

Answer

D

Explanation

`X = 3YZ^2`
`Y=(X)/(3Z^2) = ([M^(-1)L^(-2)T^(4)A^2])/([MT^(-2)A^(-1)]^2)=[M^(-2)L^(-2)T^(8) A^4]`
Capacity or capacitance
`[X] =[C] =[M^(-1)L^(-2)T^(4) A^2]`
Magnetic field [Z] `=[MT^(-2)A^(-1)]`