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 $........... \%$
JEE MAIN 2017, Medium
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Give, $p = {a^{1/2}}{b^2}{c^3}{d^{ - 4}},$
Maximum relative error,
$\frac{{\Delta p}}{p} = \frac{1}{2}\frac{{\Delta a}}{a} + 2\frac{{\Delta b}}{b} + 3\frac{{\Delta c}}{c} + 4\frac{{\Delta d}}{d}$
$= \frac{1}{2} \times 2 + 2 \times 1 + 3 \times 3 + 4 \times 5$
$= 32\%$
Maximum relative error,
$\frac{{\Delta p}}{p} = \frac{1}{2}\frac{{\Delta a}}{a} + 2\frac{{\Delta b}}{b} + 3\frac{{\Delta c}}{c} + 4\frac{{\Delta d}}{d}$
$= \frac{1}{2} \times 2 + 2 \times 1 + 3 \times 3 + 4 \times 5$
$= 32\%$
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