Question

A physical quantity, $y=\frac{a^4{b}^2}{(c{d}^4{)}^{1/3}}$ has four quantities $a,b,c$ and $d$. The percentage error in each quantity are $2\%,3\%,4\%$ and $5\%$ respectively. The error in $y$ will be:

• A. $6\%$
• B. $11\%$
• C. $12\%$
• D. $22\%$

Answer 1

The correct option is D $22\%$

$\begin{array}{c}\text{Given}\to \\ \text{physical quantity}y=\frac{a^4\cdot b^2}{{(c\cdot d^4)}^{1/3}}\\ \text{Percentage error in}a=2\%,b=3\%,c=40\%\\ d=5\%\end{array}$

$\text{Percuntage error in}y=?$

$\begin{array}{c}\text{Calculating error with differentiation method}\\ \text{Physical qty}y=\frac{a^4\cdot b^2}{{(c\cdot d^4)}^{1/3}}\\ \text{Rewriting}y=\frac{a^4\cdot b^2}{c^{1/3}\cdot d^{4/3}}\end{array}$

$\begin{array}{c}\text{On Taking logarithm both sides we get-}\\ \log y=4\cdot \log a+2\log b-\frac{1}{3}\log c-\frac{4}{3}\log d\end{array}$

$\begin{array}{c}\text{for maximum permissible error in}y\text{rewrite as}\\ \frac{dy}{y}=4\cdot \frac{da}{a}+\frac{2\cdot db}{b}+\frac{1}{3}\cdot \frac{dc}{c}+\frac{4}{3}\cdot \frac{d\left(d\right)}{d}\end{array}$

$\begin{array}{cc}& \text{Now}\frac{da}{a}=2\%\frac{db}{b}=3\%\frac{dc}{c}=4\%\frac{d\left(d\right)}{d}=5\%\text{(given)}\\ & \begin{array}{cc}\text{So we get}& \frac{dy}{y}=4\cdot \left(2\%\right)+2\left(3\%\right)+\frac{1}{3}\cdot \left(4\%\right)+\frac{4}{3}\cdot \left(5\%\right)\\ \frac{dy}{y}& =22\%\end{array}\\ & \text{which is the percentage error in}y\text{.}\end{array}$