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Question: \(X = 3YZ^{2}\) find dimension of \(Y\) in (MKSA) system, if \(X\) and \(Z\) are the dimension of ca...

X=3YZ2X = 3YZ^{2} find dimension of YY in (MKSA) system, if XX and ZZ are the dimension of capacity and magnetic field respectively

A

M−3L−2T−4A−1M^{- 3}L^{- 2}T^{- 4}A^{- 1}

B

ML−2ML^{- 2}

C

M−3L−2T4A4M^{- 3}L^{- 2}T^{4}A^{4}

D

M−3L−2T8A4M^{- 3}L^{- 2}T^{8}A^{4}

Answer

M−3L−2T8A4M^{- 3}L^{- 2}T^{8}A^{4}

Explanation

Solution

X=3YZ2X = 3YZ^{2}

∴\therefore [Y]=[X][Z2]=[M−1L−2T4A2][MT−2A−1]2=[M−3L−2T8A4]\lbrack Y\rbrack = \frac{\lbrack X\rbrack}{\lbrack Z^{2}\rbrack} = \frac{\lbrack M^{- 1}L^{- 2}T^{4}A^{2}\rbrack}{\lbrack MT^{- 2}A^{- 1}\rbrack^{2}} = \lbrack M^{- 3}L^{- 2}T^{8}A^{4}\rbrack.