Question

A physical quantity $P$ is related to four measurable quantities $a,b,c$ 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%. The percentage error in the quantity $P$ is:

• A. 10%
• B. 13%
• C. 5%
• D. 15%
• E. 17%

Answer 1

Answer: (B)

Given, $P=\frac{a^3{b}^2}{\sqrt{c}d}=a^3{b}^2{c}^{-1/2}{d}^{-1}$
The fractional error in P is given by
$\frac{\Delta P}{P}=3\frac{\Delta a}{a}+2\frac{\Delta b}{b}-\frac{1}{2}.\frac{\Delta c}{c}-\frac{\Delta d}{d}$
The maximum fractional error in P is
$\left(\frac{\Delta P}{P}\right)=3\frac{\Delta a}{a}+2\frac{\Delta b}{b}+\frac{1}{2}\frac{\Delta c}{c}+\frac{\Delta d}{d}$
The percentage error in P is
${\left(\frac{\Delta P}{P}\right)}_{\max }\times 100=3\left(\frac{\Delta a}{a}\times 100\right)$
$+2\left(\frac{\Delta b}{b}\times 100\right)+\frac{1}{2}\left(\frac{\Delta c}{c}\times 100\right)+\frac{\Delta d}{d}\times 100$
= 3 (% error in a) + 2 (% error in b) $+\frac{1}{2}$ (% error in c) + (% error in d) = 3 (1%) + 2 (3%) $+\frac{1}{2}$ (4%) + (2%) = 3% + 6% + 2% + 2% = 13%