A physical quantity P is related to four observables a, b, c and d are 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 ?

Given, $P=a^3{b}^2/\sqrt{cd}$

Maximum relative error in physical quantity P is given by

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

$\therefore$Maximum percentage error in P is given by

$\frac{\Delta P}{P}\times 100=\pm [3(\frac{\Delta a}{a}\times 100)+2(\frac{\Delta b}{b}\times 100)+\frac{1}{2}(\frac{\Delta c}{c}\times 100)+(\frac{\Delta d}{d}\times 100)]$

But $\frac{\Delta a}{a}\times 100=1\%,\frac{\Delta b}{b}\times 100=3\%$

$\frac{\Delta c}{c}\times 100=4\%,\frac{\Delta d}{d}\times 100=2\%$

$\therefore \frac{\Delta P}{P}\times 100=\pm [(3\times 1)+2\times \left(3\right)+\frac{1}{2}\times \left(4\right)+\left(2\right)]\pm [3+6+2]\%$

$=\pm 13\%$

As the result (13%) has two significant figures, therefore the value of P =3.763 should have only two significant figures. Rounding-off the value of P up to two significant figures, we get P =3.8