Question:

The physical quantity \(P=\frac{a^{3}b^{2}}{c^{1/3}d^{1/4}}\), where the percentage error in a, b, c, and d are \(1\%\), \(\frac{3}{2}\%\), \(3\%\), and \(2\%\) respectively. Then the percentage error in P is

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For a product of powers, the maximum percentage error is the sum of each percentage error multiplied by its power.
Updated On: Nov 12, 2026
  • \(14\ \%\)
  • \(14.5\ \%\)
  • \(7\ \%\)
  • \(7.5\ \%\)
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept
For \(P=\dfrac{a^{3}b^{2}}{c^{1/3}d^{1/4}}\), the maximum percentage error adds each percentage error multiplied by the magnitude of its power.

Step 2: Calculation
\(\dfrac{\Delta P}{P}\times100=3(1)+2\left(\dfrac{3}{2}\right)+\dfrac{1}{3}(3)+\dfrac{1}{4}(2)\)
\(=3+3+1+0.5\)

Step 3: Result
The percentage error in P is \(7.5\%\).

Final Answer:
The percentage error in P is 7.5 percent. \[ \boxed{\text{(D) }7.5\ \%} \]
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