(B) Given the formula $X = 5YZ^2$,we can express $Y$ as $Y = \frac{X}{5Z^2}$.The dimension of capacitance $X$ is $[M^{-1} L^{-2} T^4 A^2]$.The dimensi

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(B) Given the formula $X = 5YZ^2$,we can express $Y$ as $Y = \frac{X}{5Z^2}$.The dimension of capacitance $X$ is $[M^{-1} L^{-2} T^4 A^2]$.The dimension of magnetic field $Z$ is $[M^1 L^0 T^{-2} A^{-1}]$.Substituting these into the expression for $Y$:$Y = \frac{[M^{-1} L^{-2} T^4 A^2]}{[M^1 L^0 T^{-2} A^{-1}]^2}$$Y = \frac{[M^{-1} L^{-2} T^4 A^2]}{[M^2 L^0 T^{-4} A^{-2}]}$$Y = [M^{-1-2} L^{-2-0} T^{4-(-4)} A^{2-(-2)}]$$Y = [M^{-3} L^{-2} T^8 A^4]$.

Class 11 Physics Units and Measurements Dimensional Analysis, Uses and Limitations

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The formula is $X = 5YZ^2$,where $X$ and $Z$ have dimensions of capacitance and magnetic field,respectively. What are the dimensions of $Y$ in $SI$ units?

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A
$[M^{-2} L^0 T^{-4} A^{-2}]$
B
$[M^{-3} L^{-2} T^8 A^4]$
C
$[M^{-2} L^{-2} T^6 A^3]$
D
$[M^{-1} L^{-2} T^4 A^2]$

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Dimensional Analysis, Uses and Limitations

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The formula is $X = 5YZ^2$,where $X$ and $Z$ have dimensions of capacitance and magnetic field,respectively. What are the dimensions of $Y$ in $SI$ units?

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