Question

A physical quantity $P$ is described by the relation $p=a^{1/2}{b}^2{c}^3{d}^{-4}$
If the relative errors in the measurement of $a,b,c$ and $d$ respectively, are $2,1,3$ and $5$, then the relative error in $P$ will be :

• A. $8\%$
• B. $25\%$
• C. $12\%$
• D. $32\%$

Answer 1

The correct option is D $32\%$

$P=a^{1/2}{.b}^2.c^3.d^{-4}$

Taking log on both sides

$logP=\frac{1}{2}loga+2logb+3logc-4logd$

Now differentiating on both sides

$\frac{dP}{P}=\frac{da}{2a}+2\frac{db}{b}+3\frac{dc}{c}-4\frac{d\left(d\right)}{d}$

$or\frac{\Delta P}{P}=\frac{\Delta a}{2a}\pm 2\frac{\Delta b}{b}\pm 3\frac{\Delta c}{c}\pm 4\frac{\Delta d}{d}$

Now, given, $\frac{\Delta a}{a}\times 100=2,$

$\Rightarrow \frac{\Delta b}{b}\times 100=1,$

$\Rightarrow \frac{\Delta c}{c}\times 100=3,$

$\Rightarrow \frac{\Delta d}{d}\times 100=5,$

$\Rightarrow \frac{\Delta P}{P}\times 100=\frac{2}{2}\pm 2\times 1\pm 3\times 3\pm 4\times 5$

$\Rightarrow \frac{\Delta P}{P}\times 100$=32%