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A physical quantity $P$ is related to four observables $a,b,c$ and $d$ as follows $P=\large\frac{\large a^3b^2}{(\sqrt {cd})}$. The percentage error in measuring $a,b,c,d$ are $1 \%,3 \%,4 \%$ and $2 \%$ respectively. What is percentage error in calculation of $P$

\[(a)\;13 \% \quad (b)\;12 \%\quad (c)\;10 \% \quad (d)\;1.0\%\]

1 Answer

$P=\large\frac{a^3b^2}{(\sqrt {cd})}$
$\large\frac{\Delta p}{p}$$=3 \large\frac{\Delta b}{b}$$+2 \large\frac{\Delta b}{b}+\frac{1}{2} \frac{\Delta c}{c}+\frac{1}{2} \frac{\Delta d}{d}$
$\%$ error in $P=3 \%$ error in a+$2 \%$ error in b+$\large\frac{1}{2} \%$ error in c+$\large\frac{1}{2} \%$ error in d
$\qquad\qquad=3 \%+6 \%+2 \%+1 \%$
$\qquad\qquad=12 \%$
Hence the percentage of error in calculation of $P$ is $12\%$
Hence b is the correct answer. 
answered Mar 23, 2014 by balaji.thirumalai

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