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For $X = 3 Y Z 2$ , find dimension of $Y$ in (MKSA) system, if $X$ and $Z$ are the dimension of capacity and magnetic field respectively.

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$M − 3 L − 2 T − 4 A − 1$

M L − 2

M − 3 L − 2 T 4 A 4

M − 3 L − 2 T 8 A 4

answer is D.

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Detailed Solution

$Y = \frac{X}{3 Z^{2}} = \frac{[M^{- 1} L^{- 2} T^{4} A^{2}]}{{[M T^{- 2} A^{- 1}]}^{2}} = [M^{- 3} L^{- 2} T^{8} A^{4}]$

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