# A physical quantity $P$ is related to four observables $a, b, c$ and $d$ as follows:

Question.
A physical quantity $P$ is related to four observables $a, b, c$ and $d$ as follows:

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

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?

solution:

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

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

$\left(\frac{\Delta P}{P} \times 100\right) \%=\left(3 \times \frac{\Delta a}{a} \times 100+2 \times \frac{\Delta b}{b} \times 100+\frac{1}{2} \times \frac{\Delta c}{c} \times 100+\frac{\Delta d}{d} \times 100\right) \%$

$=3 \times 1+2 \times 3+\frac{1}{2} \times 4+2$

$=3+6+2+2=13 \%$

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 $P=3.8$.