Question

A physical quantity $z$ depends on four observables $a,b,c$ and $d$ , as $z=\frac{a^2{b}^{\left(2/3\right)}}{\sqrt{c}{d}^3}$. The percentages of error in the measurement of $a,b,c$ and $d$ are $2\%,1.5\%,4\%$ and $2.5\%$ respectively. The percentage of error in $z$ is

• A. $12.25\%$
• B. $16.5\%$
• C. $13.5\%$
• D. $14.5\%$

Answer 1

The correct option is D $14.5\%$

Given: $z=\frac{a^2{b}^{2/3}}{\sqrt{c}{d}^3}$
Percentage error in $Z$,
$\frac{\Delta Z}{Z}=\frac{2\Delta a}{a}+\frac{2}{3}\frac{\Delta b}{b}+\frac{1}{2}\frac{\Delta c}{c}+\frac{3\Delta d}{d}$
$=2\times 2+\frac{2}{3}\times 1.5+\frac{1}{2}\times 4+3\times 2.5=14.5\%$