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)

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