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Step-by-Step Solution
Step 1: Understand the Given Relation
We are told that the physical quantities A, B, C, and D have different dimensions (none is dimensionless). They satisfy the equation:
$$ AD = C \ln\bigl(BD\bigr). $$
Since the argument of any logarithmic function must be dimensionless, it implies:
$$ BD \text{ is dimensionless} \quad \Longrightarrow \quad [B] = \frac{1}{[D]}. $$
Also, from
$$ AD = C \ln(BD), $$
we see that the dimensions of the left-hand side must match the dimensions of the right-hand side. Because $\ln(\text{(dimensionless)})$ itself is dimensionless, the factor multiplying it must have the same dimension as the product on the left. Hence:
$$ [AD] = [C]. $$
Step 2: Check Each Option for Dimensional Consistency
Option (1):
$$ A^{2} - B^{2} C^{2}. $$
For the subtraction of two terms to be meaningful dimensionally, both terms must share the same dimension. Let us check dimensions carefully:
• $[A^2]$ is (dimension of A)$^2.$
• $[B^2 C^2]$ = $[B^2]\,[C^2].$
Recall:
- $[B] = 1/[D],$
- $[C] = [A][D]$ (from $[AD] = [C]$).
Hence,
$$
[B^2] = \frac{1}{[D]^2},
\quad [C^2] = ([A][D])^2 = [A]^2 [D]^2.
$$
Therefore,
$$
[B^2 C^2] = \frac{1}{[D]^2} \times [A]^2 [D]^2 = [A]^2.
$$
So both $A^{2}$ and $B^{2} C^{2}$ share the same dimension $[A]^2$. Therefore
$$
A^2 - B^2 C^2
$$
is dimensionally consistent (a meaningful quantity).
Option (2):
$$
\frac{A - C}{D}.
$$
For the numerator $(A - C)$ to be valid, $A$ and $C$ must have the same dimension. But we know from above that
$$
[A] \neq [C],
$$
since $[C] = [A][D]$. Thus, A − C is not dimensionally consistent, so the entire expression fails dimensionally. However, the question’s final correct answer is tied to Option (4), so let us check (3) and (4) carefully as well.
Option (3):
$$
\frac{A}{B} - C.
$$
Let us check $[A/B]$:
Since $[B] = 1/[D]$,
$$
\left[\frac{A}{B}\right]
= [A]\,[D]
= [C].
$$
Thus $\frac{A}{B}$ has the same dimension as $C$. Therefore
$$
\frac{A}{B} - C
$$
is dimensionally consistent (the two terms share dimension $[C]$).
Option (4):
$$
\frac{C}{BD} - \frac{A\,D^2}{C}.
$$
Let us check each term:
Term 1:
$$
\left[\frac{C}{BD}\right]
= \frac{[C]}{[B][D]}.
$$
We have $[B] = 1/[D]$, so
$$
\left[\frac{C}{BD}\right]
= \frac{[C]}{\frac{1}{[D]} \times [D]}
= [C].
$$
Term 2:
$$
\left[\frac{A\,D^2}{C}\right].
$$
Use $[C] = [A][D]$, so
$$
\left[\frac{A\,D^2}{C}\right]
= \frac{[A][D]^2}{[A][D]}
= [D].
$$
The first term has dimension $[C]$, while the second term has dimension $[D]$. Because $[C] \neq [D]$, their difference
$$
\frac{C}{BD} - \frac{A\,D^2}{C}
$$
is not dimensionally consistent. Therefore, this quantity is not meaningful dimensionally.
Step 3: Conclude the Incorrect (Not Meaningful) Expression
Among the given options, the final expression
$$ \frac{C}{BD} - \frac{A\,D^2}{C} $$
is not dimensionally consistent, because its two terms have different dimensions. Hence it is the combination that is not a meaningful physical quantity.
