If the formula is X = 3 Y Z^{2},where X and Z | vedclass

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(A) Given the formula $X = 3 Y Z^{2}$.We know the dimensions of capacitance $C$ in $MKSQ$ system are $[X] = [C] = [M^{-1} L^{-2} T^{2} Q^{2}]$.We know

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

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(A) Given the formula $X = 3 Y Z^{2}$.We know the dimensions of capacitance $C$ in $MKSQ$ system are $[X] = [C] = [M^{-1} L^{-2} T^{2} Q^{2}]$.We know the dimensions of magnetic induction $B$ in $MKSQ$ system are $[Z] = [B] = [M T^{-1} Q^{-1}]$.Rearranging the formula for $Y$,we get $Y = \frac{X}{3 Z^{2}}$.Ignoring the dimensionless constant $3$,the dimensional formula is $[Y] = \frac{[X]}{[Z^{2}]}$.Substituting the dimensions: $[Y] = \frac{[M^{-1} L^{-2} T^{2} Q^{2}]}{[M T^{-1} Q^{-1}]^{2}}$.$[Y] = \frac{[M^{-1} L^{-2} T^{2} Q^{2}]}{[M^{2} T^{-2} Q^{-2}]}$.$[Y] = [M^{-1-2} L^{-2} T^{2-(-2)} Q^{2-(-2)}] = [M^{-3} L^{-2} T^{4} Q^{4}]$.

If the formula is $X = 3 Y Z^{2}$,where $X$ and $Z$ have dimensions of capacitance and magnetic induction respectively,then the dimensions of $Y$ in the $MKSQ$ system are:

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