JEE Physics 2023 Question with Solution

Given: $\left( P + \frac{a}{V^2} \right) (V - b) = RT$

Find: The physical quantity having the same dimensional formula as $\frac{b^2}{a}$.

From the equation, $b$ is subtracted from $V$, so both must have the same dimensions:

$$[b] = [V]$$

Also, $P$ is added to $\frac{a}{V^2}$, so

$$\left[\frac{a}{V^2}\right] = [P]$$

Hence,

$$[a] = [P][V^2]$$

Therefore,

$$\left[\frac{b^2}{a}\right] = \frac{[V^2]}{[P][V^2]} = \frac{1}{[P]}$$

Now, compressibility has dimensions equal to the reciprocal of pressure:

$$[K] = \frac{1}{[P]}$$

Therefore, the physical quantity is Compressibility. The correct option is C.

The solution lists the label as B, but the worked statement identifies Compressibility, which matches option C in the given options.

Given: $\left( P + \frac{a}{V^2} \right) (V - b) = RT$

Find: Which option matches the dimensions of $\frac{b^2}{a}$.

Use dimensional consistency of addition and subtraction:

1. In $V - b$, both terms must have identical dimensions, so $[b] = [V]$.
2. In $P + \frac{a}{V^2}$, both terms must have identical dimensions, so

$$\left[\frac{a}{V^2}\right] = [P]$$

This gives

$$[a] = [P][V^2]$$

Now evaluate:

$$\left[\frac{b^2}{a}\right] = \frac{[b]^2}{[a]} = \frac{[V]^2}{[P][V^2]} = [P]^{-1}$$

Among the given physical quantities, compressibility is the reciprocal of bulk modulus, and hence has dimensions of inverse pressure.

So the correct option is C (Compressibility).