Question

Given: The percentage error in the measurements of A, B, C, and D are respectively 4%, 2%, 3%, and 1%. The relative error in $Z = \frac{A^4 B^{1/3}}{C D^{3/2}}$ is

• A. 127/2 %
• B. 127/5 %
• C. 127/6 %
• D. 127/7 %

Answer 1

The Correct Option is C

Given the expression for \( Z \) as: \[ Z = \frac{A^4 B^{1/3}}{C D^{3/2}}, \] the relative error in \( Z \) can be calculated by using the formula for propagation of errors in a product or quotient of quantities. The relative error in \( Z \) is the sum of the relative errors in the individual quantities, each multiplied by the corresponding power in the expression. The relative error in \( Z \) is given by: \[ \frac{\Delta Z}{Z} = 4 \frac{\Delta A}{A} + \frac{1}{3} \frac{\Delta B}{B} + \frac{\Delta C}{C} + \frac{3}{2} \frac{\Delta D}{D}. \] Substitute the given percentage errors: - The percentage error in \( A \) is 4%, - The percentage error in \( B \) is 2%, - The percentage error in \( C \) is 3%, - The percentage error in \( D \) is 1%. Thus, the total relative error is: \[ \frac{\Delta Z}{Z} = 4(4\%) + \frac{1}{3}(2\%) + 3\%(1) + \frac{3}{2}(1\%). \] Simplifying: \[ \frac{\Delta Z}{Z} = 16\% + \frac{2}{3}\% + 3\% + \frac{3}{2}\% = 16\% + 0.67\% + 3\% + 1.5\%. \] The total relative error is: \[ \frac{\Delta Z}{Z} = 16 + 0.67 + 3 + 1.5 = 20.17\% \approx 127/6\%. \] Hence, the relative error in \( Z \) is: \[ \boxed{127/6 \%}. \]