Question
A physical quantity X is given by: $\text{X}=\frac{\text{P}^2\text{Q}^{\frac{3}{2}}}{\text{R}^4\text{S}^{\frac{1}{2}}}$ The percentage errors in P, Q, R and S are 1%, 2%, 4% and 2%. Calculate the percentage error in X.

Answer

The maximum fractional error in X is given by:$\frac{\Delta\text{X}}{\text{X}}=\pm\Big[\frac{2\Delta\text{P}}{\text{P}}+\frac{3}{2}\frac{\Delta\text{Q}}{\text{Q}}+\frac{4\Delta\text{R}}{\text{R}}+\frac{1}{2}\frac{\Delta\text{S}}{\text{S}}\Big]$
$\frac{\Delta\text{X}}{\text{X}}=\pm\Bigg[2\times\Big(\frac{1}{100}\Big)+\frac{3}{2}\Big(\frac{2}{100}\Big)+4\times\Big(\frac{4}{100}\Big)+\frac{1}{2}\times\Big(\frac{2}{100}\Big)\Bigg]$
$\frac{\Delta\text{X}}{\text{X}}=\pm\frac{2}{100}+\frac{3}{100}+\frac{16}{100}+\frac{1}{100}$
$=\pm\frac{22}{100}=\pm0.22$Percentage error in $\text{X}=\frac{\Delta\text{X}}{\text{X}}\times100$
$=\pm0.22\times100=\pm22%$

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