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In the formula X =3Y $Z^{2}$ , X and Z have dimensions of capacitance and magnetic induction. The dimensions of Y in MKSQ system are

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$M^{- 3} L^{- 2} T^{4} Q^{4}$

${ML}^{2} T^{4} Q^{4}$

$M^{- 2} L^{- 3} T^{2} Q^{4}$

$M^{- 2} L^{- 2} {TQ}^{2}$

answer is A.

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Detailed Solution

$[X] = [C] = [M^{- 1} L^{- 2} T^{2} Q^{2}]$
and $[Z] = [B] = [{MT}^{- 1} Q^{- 1}]$
$[Y] = \frac{[X]}{[Z^{2}]} = \frac{[M^{- 1} L^{- 2} T^{2} Q^{2}]}{{[{MT}^{- 1} Q^{- 1}]}^{2}}$
$= [M^{- 3} L^{- 2} T^{4} Q^{4}]$

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