A physical quantity $P$ is related to four observables $a,b,c,$ and $d$ as $P=\frac{a^3{b}^2}{\sqrt{c}d}$. The percentage errors of measurement in $a,b,c,$ and $d$ are $1\%,3\%,4\%,$ and $2\%$ respectively. What is the percentage error in $P$?

Solution

1. Given the relationship $P=\frac{a^3{b}^2}{c^{1/2}d}$, the relative error is expressed as $\frac{\Delta P}{P}=3\frac{\Delta a}{a}+2\frac{\Delta b}{b}+\frac{1}{2}\frac{\Delta c}{c}+\frac{\Delta d}{d}$.
2. Substitute the given percentage errors: $\frac{\Delta P}{P}\times 100=3(1\%)+2(3\%)+\frac{1}{2}(4\%)+2\%$.
3. Calculate the sum: $3\%+6\%+2\%+2\%=13\%$.
4. Hence the answer is (B).