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 $Z=\frac{A^4 B^{\frac{1}{3}}}{C^{\frac{3}{2}}}$ is

• A. $\frac{127}{2} \%$
• B. $\frac{127}{5} \%$
• C. $\frac{127}{6} \%$
• D. $\frac{127}{7} \%$

Answer 1

Answer: (C)

$\begin{array}{l}\text { Hint : } Z=\frac{A^4 B^{\frac{1}{3}}}{C D^{\frac{3}{2}}}=\frac{\Delta Z}{Z}=4 \frac{\Delta A}{A}+\frac{1}{3} \frac{\Delta B}{B}+\frac{\Delta C}{C}+\frac{3}{2} \frac{\Delta D}{D} \\ =4 \times 4 \%+\frac{1}{3} \times 2 \%+3 \%+\frac{3}{2} \times 1 \%=\left(16+\frac{2}{3}+3+\frac{3}{2}\right) \% \\ =\left(\frac{127}{6}\right) \%\end{array}$