In the equation [X+frac{a}{Y^2}][Y-b]= RT,X i | vedclass

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(A) According to the principle of homogeneity of dimensions,only physical quantities of the same dimension can be added or subtracted.In the term $[X

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Energy

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(A) According to the principle of homogeneity of dimensions,only physical quantities of the same dimension can be added or subtracted.In the term $[X + \frac{a}{Y^2}]$,$X$ is pressure,so $\frac{a}{Y^2}$ must also have the dimensions of pressure.$[X] = [ML^{-1}T^{-2}]$$[Y] = [L^3]$$[\frac{a}{Y^2}] = [ML^{-1}T^{-2}] \implies [a] = [ML^{-1}T^{-2}] 	imes [L^3]^2 = [ML^5T^{-2}]$In the term $[Y - b]$,$Y$ is volume,so $b$ must have the dimensions of volume.$[b] = [L^3]$Now,the ratio $\frac{a}{b}$ has dimensions:$\frac{[a]}{[b]} = \frac{[ML^5T^{-2}]}{[L^3]} = [ML^2T^{-2}]$These are the dimensions of energy (or work/torque).

List-$I$	List-$II$
$A$. Meter $(L)$	$I$. $\sqrt{\frac{hc}{G}}$
$B$. Second $(S)$	$II$. $\sqrt{\frac{Gh}{c^5}}$
$C$. Kilogram $(M)$	$III$. $\sqrt{\frac{L^2c^3}{Gh}}$
$D$. Kelvin $(K)$	$IV$. $\sqrt{\frac{Gh}{c^3}}$

In the equation $[X+\frac{a}{Y^2}][Y-b]= RT$,$X$ is pressure,$Y$ is volume,$R$ is universal gas constant and $T$ is temperature. The physical quantity equivalent to the ratio $\frac{a}{b}$ is

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