Question

If A, B and C are three different physical quantities with different dimensional formulae, then the combination which can never give a proper physical quantity is

• A. \( \frac{A}{BC} \)
• B. \( \frac{AB - C^2}{BC} \)
• C. \( \frac{A-C}{B} \)
• D. \( AC - B \)

Hint:

Addition and subtraction are the strictest operations in dimensional analysis. Always check terms separated by \( + \) or \( - \) first.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The principle of homogeneity states that we can only add or subtract physical quantities if they have the same dimensional formula.
Step 3: Detailed Explanation:
Let the dimensions be \([A], [B], [C]\). We are given that \( [A] \neq [B] \neq [C] \) and they are all different. Check options:
(A) \( \frac{A}{BC} \): Multiplication/Division is always allowed. This forms a new quantity. Valid.
(B) \( \frac{AB - C^2}{BC} \): For \( AB - C^2 \) to be valid, \( [AB] \) must equal \( [C^2] \). This is a possible specific condition, not strictly impossible for all sets of quantities (e.g., if A=Length, B=Length, C=Length is not allowed as dims are different, but maybe A=Force, B=Length/Force... wait. The question implies "different dimensional formulae". It is possible to find A, B, C such that \( [A][B] = [C]^2 \). So this operation is conditionally valid).
(C) \( \frac{A-C}{B} \): This requires \( A - C \). For subtraction, \([A]\) must equal \([C]\). But the problem states A and C have different dimensional formulae. Thus, \( A - C \) is never dimensionally valid. (D) \( AC - B \): Similar to (B), this requires \( [AC] = [B] \). This is possible to construct (e.g., A=Velocity, C=Time, B=Length).
Comparing (C) and (D): (C) demands \( [A] = [C] \), which directly contradicts the problem statement ("different dimensional formulae"). Therefore, (C) is strictly impossible.
Step 4: Final Answer:
The combination \( \frac{A-C}{B} \) is invalid.