Physics - Error Question with Solution | TestHub

Given quantity is $Q=\frac{a^4{b}^3}{c^2}$.

The relative error in measuring $Q$ is given by

$\frac{\Delta Q}{Q}=4\frac{\Delta a}{a}+3\frac{\Delta b}{b}+2\frac{\Delta c}{c} ...\left(1\right)$

Hence, the required percentage error is given by

$\begin{array}{rcl}\frac{\Delta Q}{Q}\times 100& =& 4\left(\frac{\Delta a}{a}\times 100\right)+3\left(\frac{\Delta b}{b}\times 100\right)+2\left(\frac{\Delta c}{c}\times 100\right)\\ & =& 4\times 3\%+3\times 4\%+2\times 5\%\\ & =& 12\%+12\%+10\%\\ & =& 34\%\end{array}$