Question

In an experiment four quantities $a,b,c$ and $d$ are measured with percentage error 1%, 2%, 3% and 4% respectively. Quantity $P$ is calculated as follows : $P=\frac{a^3{b}^2}{cd}$. % error in $P$ is:

• A. 14%
• B. 10%
• C. 7%
• D. 4%

Answer 1

The correct option is A 14%

$y=P=\frac{a^3{b}^2}{cd}a=1\%;b=2\%;c=3\%;d=4\%$

$\begin{array}{cc}\frac{\Delta P}{P}\times 100\%& =\frac{3\Delta a}{a}+2\frac{\Delta b}{b}+\frac{\Delta C}{C}+\frac{\Delta d}{d}\\ & =3\left[1\%\right]+2\left[2\%\right]+3\%+4\%\\ & =3\%+4\%+3\%+4\%\\ & =14\%\end{array}$

$\text{Error in}P=\pm 14\%$