Question
A physical quantity P is related to four observables a, b, c and d as follows:$\text{P}=\text{a}^3\text{b}^3/\big(\sqrt{\text{c}}\text{ d}\big)$
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?

Answer

$\text{P}=\frac{\text{a}^3\text{b}^2}{(\sqrt{\text{c}}\text{ d})}$Maximum fractional error in P is given by
$\frac{\Delta\text{P}}{\text{P}}=\pm\Big[3\frac{\Delta\text{a}}{\text{a}}+2\frac{\Delta\text{b}}{\text{b}}+\frac{1}{2}\frac{\Delta\text{c}}{\text{c}}+\frac{\Delta\text{d}}{\text{d}}\Big]\\=\pm\Big[3\Big(\frac{1}{100}\Big)+2\Big(\frac{3}{100}\Big)+\frac{1}{2}\Big(\frac{4}{100}\Big)+\frac{2}{100}\Big]$
$=\pm\frac{13}{100}=\pm0.13$
Percentage error in $\text{P}=\frac{\Delta\text{P}}{\text{P}}\times100=\pm0.13\times100=\pm13\%$
Percentage error in P = 13%
Value of P is given as 3.763.
By rounding off the given value to the first decimal place, we get P = 3.8.

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