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Physics Question on Units and measurement

In the formula X=3YZ2X = 3YZ^2, XX and ZZ have dimensions of capacitance and magnetic induction respectively. The dimensions of YY in MKSQMKSQ system are

A

[M−3L−2T4Q4]\left[M^{-3}L^{-2}T^{4}Q^{4}\right]

B

[M−2L−1T5Q3]\left[M^{-2}L^{-1}T^{5}Q^{3}\right]

C

[M−1L−2T4Q4]\left[M^{-1}L^{-2}T^{4}Q^{4}\right]

D

[M−3L−1T4Q4]\left[M^{-3}L^{-1}T^{4}Q^{4}\right]

Answer

[M−3L−2T4Q4]\left[M^{-3}L^{-2}T^{4}Q^{4}\right]

Explanation

Solution

[X]=[C]=[M−1L−2T2Q2]\left[X\right]=\left[C\right]=\left[M^{-1}L^{-2}T^{2}Q^{2}\right], [Z]=[B]=[MT−1Q−1]\left[Z\right]=\left[B\right]=\left[MT^{-1}Q^{-1}\right] ∴[Y]=[X][3Z2]=[M−1L−2T2Q2][MT−1Q−1]2\therefore \left[Y\right]=\frac{\left[X\right]}{\left[3Z^{2}\right]}=\frac{\left[M^{-1}L^{-2}T^{2}Q^{2}\right]}{\left[MT^{-1}Q^{-1}\right]^{2}} =[M−3L−2T4Q4]=\left[M^{-3}L^{-2}T^{4}Q^{4}\right]