A physical quantity P-is related to four observations: a, b, c and d as follows: P = a³b²/√(cd). The percentage errors of measurement in a, b, c and dare 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?

The correct answer is 13%.

To find the percentage error in the physical quantity $P$, we analyze the standard relation:
$P=\frac{a^3{b}^2}{d\sqrt{c}}$
The relative error in $P$ is the sum of the relative errors of the individual measurements multiplied by their respective powers:
$\frac{\Delta P}{P}=3\left(\frac{\Delta a}{a}\right)+2\left(\frac{\Delta b}{b}\right)+\frac{1}{2}\left(\frac{\Delta c}{c}\right)+1\left(\frac{\Delta d}{d}\right)$

Expressing this in terms of percentage errors:
$\frac{\Delta P}{P}\times 100=3\left(\frac{\Delta a}{a}\times 100\right)+2\left(\frac{\Delta b}{b}\times 100\right)+\frac{1}{2}\left(\frac{\Delta c}{c}\times 100\right)+1\left(\frac{\Delta d}{d}\times 100\right)$
Substituting the given percentage errors of measurement for $a$ (1%), $b$ (3%), $c$ (4%), and $d$ (2%):
$\text{Percentage error in}P=3(1\%)+2(3\%)+\frac{1}{2}(4\%)+1(2\%)$
$\text{Percentage error in}P=3\%+6\%+2\%+2\%=13\%$

To determine the value to which the calculated result $P=3.763$ should be rounded off, we calculate the absolute error:
$\Delta P=3.763\times \frac{13}{100}=0.489\approx 0.5$
Since the error begins at the tenths place (the first decimal digit), the value of $P$ is only significant up to one decimal place.
Rounding off $3.763$ to one decimal place yields $3.8$.