Question

A physical quantity \(\mathrm{P}=\frac{\mathrm{A}^3 \sqrt{\mathrm{~B}}}{\mathrm{C}^{-4} \mathrm{D}^{3 / 2}}\). Out of the quantities \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and D , the quantity which brings maximum percentage error in P is

• A. A
• B. B
• C. C
• D. D

Hint:

The percentage error in a product of powers is the sum of each power times its own percentage error. The quantity with the largest power magnitude contributes the most.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For \(P = A^a B^b C^c D^d\), the maximum fractional error is \(\dfrac{\Delta P}{P} = a\dfrac{\Delta A}{A} + b\dfrac{\Delta B}{B} + c\dfrac{\Delta C}{C} + d\dfrac{\Delta D}{D}\), using the magnitudes of the powers.

Step 2: Rewrite the Expression
\(P = \dfrac{A^3 \sqrt{B}}{C^{-4} D^{3/2}} = A^{3} B^{1/2} C^{4} D^{-3/2}\)

Step 3: Compare the Powers
Magnitudes of the powers: \(A: 3\), \(B: \tfrac{1}{2}\), \(C: 4\), \(D: \tfrac{3}{2}\).
The percentage errors are multiplied by these powers.

Step 4: Conclusion
The largest power is 4, which belongs to \(C\). So \(C\) contributes the maximum percentage error in \(P\).

Final Answer:
The quantity C brings the maximum percentage error in P. \[ \boxed{\text{(C) }\text{C}} \]