Question

In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and Dare \(1\%, 2\%, 3\%\) and\( 4\%\) respectively. Then the maximum percentage of error in the measurement \(X\), where \(X =\frac{A^2B^{1/2}}{C^{1/3}D^3}\) will be:

• A. \((\frac{3}{13})\%\)
• B. \(16\%\)
• C. \(-10\%\)
• D. \(10\%\)

Answer 1

The Correct Option is B

To find the maximum percentage error in the measurement of \( X \), where \( X = \frac{A^2B^{1/2}}{C^{1/3}D^3} \), we will use the formula for error propagation in products and divisions:

The percentage error in a derived quantity \( X \) of the form \( X = A^nB^m \cdots \) is given by:

\[ \frac{\Delta X}{X} \times 100\% = n \left( \frac{\Delta A}{A} \times 100\% \right) + m \left( \frac{\Delta B}{B} \times 100\% \right) + \cdots \]

In this problem, we have:

• \( n = 2 \) for \( A \)
• \( m = \frac{1}{2} \) for \( B \)
• \( \frac{-1}{3} \) for \( C \)
• \( -3 \) for \( D \)

The percentage errors in \( A, B, C, \text{ and } D \) are 1%, 2%, 3%, and 4% respectively.

Substituting these values into the error formula, we get:

\[ \frac{\Delta X}{X} \times 100\% = 2 \times 1\% + \frac{1}{2} \times 2\% + \left| \frac{-1}{3} \right| \times 3\% + 3 \times 4\% \]

Calculating each term:

• \( 2 \times 1\% = 2\% \)
• \( \frac{1}{2} \times 2\% = 1\% \)
• \( \left| \frac{-1}{3} \right| \times 3\% = 1\% \)
• \( 3 \times 4\% = 12\% \)

Add all these contributions:

\[ 2\% + 1\% + 1\% + 12\% = 16\% \]

Thus, the maximum percentage of error in the measurement of \( X \) is \( 16\% \).

The correct answer is: \( 16\% \).