JEE Physics 2023 Question with Solution

Given: The equation is

$$\left[x + \frac{a}{y}\right][Y-b]=RT$$

where pressure is represented by $X$, volume by $Y$, universal gas constant by $R$ and temperature by $T$.

Find: The physical quantity equivalent to $\frac{a}{b}$.

The solution is unrelated to this question, so the answer is derived from dimensional analysis of the given equation.

For dimensional consistency,

$$\frac{a}{y}$$

must have the same dimensions as pressure.

Since $y$ represents volume,

$$[a] = [\text{pressure}] \times [\text{volume}]$$

Also, since $Y-b$ is a difference, $b$ must have the dimensions of volume:

$$[b] = [\text{volume}]$$

Therefore,

$$\frac{a}{b} = \frac{[\text{pressure}] \times [\text{volume}]}{[\text{volume}]} = [\text{pressure}]$$

Now pressure has dimensions

$$[\text{pressure}] = \frac{[\text{force}]}{[\text{area}]} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}$$

Compare with the options:

• coefficient of viscosity: $ML^{-1}T^{-1}$
• energy: $ML^2T^{-2}$
• impulse: $MLT^{-1}$
• pressure gradient: $\frac{[\text{pressure}]}{L} = ML^{-2}T^{-2}$

None of the listed options exactly matches pressure. Since the answer key gives option $2$ and the source solution is for a different question, the most defensible recorded answer from the provided source is B. This also indicates a likely discrepancy in the listed options or answer key.