Question

A physical quantity P is related to four observables a, b, c and d as follows :
P $a^3{b}^2$ / $\sqrt{c}$ d
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 1

$P=\frac{a^3{b}^2}{\sqrt{c}d}$

Maximum fractional error in P is given by

$\frac{\Delta P}{P}=\pm [3\frac{\Delta a}{a}+2\frac{\Delta b}{b}+\frac{1}{2}\frac{\Delta c}{c}+\frac{\Delta d}{d}]$

On putting above value:

$=\pm \frac{13}{100}=0.13$

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 $P=3.8$.