QJEEM23Ph50ed: JEE Main 2023 Physics - Class 11 JEEM23Ph50ed Solution: In the equation [ X + a Y 2 ] [ Y − b ] = R T , X is pressure,…

Given that, $$[X+\frac{a}{Y^2}][Y-b]=\mathrm{R}T$$

$$\therefore$$ $X$ and $\frac{a}{Y^2}$ have the same dimensions and $Y$ and $b$ have the same dimensions, let's analyze the dimensions of $\frac{a}{b}$ .

Since $X$ represents pressure, it has dimensions of $[M L^{-1} T^{-2}]$ .

Since $X$ and $\frac{a}{Y^2}$ have the same dimensions, we have:

$$\left[\frac{a}{Y^2}\right]=[M{L}^{-1}{T}^{-2}]$$

Then, the dimensions of $a$ are:

$$[a]=[M{L}^{-1}{T}^{-2}][Y^2]=[M{L}^5{T}^{-2}]$$

Now, since $Y$ and $b$ have the same dimensions and $Y$ represents volume, we have:

$$[b]=[L^3]$$

Now, let's find the dimensions of the ratio $\frac{a}{b}$ :

$$\frac{[a]}{[b]}=\frac{[M{L}^5{T}^{-2}]}{[L^3]}=[M{L}^2{T}^{-2}]$$

Indeed, the dimensions of $\frac{a}{b}$ are $[M L^2 T^{-2}]$ , which corresponds to the dimensions of energy.