Question

A physical quantity $Q$ is calculated according to the expression
$Q=\frac{A^3{B}^3}{C\sqrt{D}}$
If percentage errors in $A,B,C,D$ are $2\%,1\%,3\%$ and $4\%$ respectively. What is the maximum percentage error in $Q$?

• A. $\pm 8\%$
• B. $\pm 10\%$
• C. $\pm 14\%$
• D. $\pm 12\%$

Answer 1

The correct option is C $\pm 14\%$

$\begin{array}{c}\text{Given - * Physical Quantity}Q=\frac{A^3\cdot B^3}{C\cdot \sqrt{D}}\\ \ast \%\text{error in}A=20/0,B=10/0,C=3\%,D=4\%\\ \text{we hove}Q=\frac{A^3\cdot B^3}{C\cdot \sqrt{D}}\end{array}$

$\begin{array}{c}\text{By error analysis method -}\\ \text{- Taking logarithm both sides -}\\ \log Q=3\cdot \log A+3\cdot \log B-\log C-\frac{1}{2}\cdot \log D\end{array}$

$\begin{array}{c}\text{For maximum permissible value-}\\ \log \varphi =3\cdot \log A+3\cdot \log B+\log C+\frac{1}{2}\cdot \log D\end{array}$

$\begin{array}{c}\text{Differentiating both Sides -}\\ \frac{dQ}{Q}=3\cdot \frac{dA}{A}+3\cdot \frac{dB}{B}+\frac{dC}{C}+\frac{1}{2}\cdot \frac{d\left(D\right)}{D}\end{array}$

$\begin{array}{c}=3\cdot \left(20/0\right)+3\cdot \left(10/0\right)+3010+\frac{1}{2}\times \left(4\%\right)\\ =14\%\end{array}$

$\Rightarrow \text{Maximum}\%\text{error in}\phi \text{is}=\pm 14\%$