In an experiment four quantities $a$, $b$, $c$ and $d$ are measured with percentage error $1\%$, $2\%$, $3\%$

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 $P = \frac{a^3 b^2}{cd}$. The percentage error in $P$ is

Accepted Answer

Given the physical quantity $P = \frac{a^3 b^2}{cd}$.
The maximum percentage error in $P$ is determined by the sum of the percentage errors of the individual quantities multiplied by their respective powers.
The formula for percentage error is:
$\frac{\Delta P}{P} 	imes 100 = 3\left(\frac{\Delta a}{a} 	imes 100ight) + 2\left(\frac{\Delta b}{b} 	imes 100ight) + \left(\frac{\Delta c}{c} 	imes 100ight) + \left(\frac{\Delta d}{d} 	imes 100ight)$
Given the percentage errors are:
$\frac{\Delta a}{a} 	imes 100 = 1\%$
$\frac{\Delta b}{b} 	imes 100 = 2\%$
$\frac{\Delta c}{c} 	imes 100 = 3\%$
$\frac{\Delta d}{d} 	imes 100 = 4\%$
Substituting these values into the error equation:
$\% 	ext{ Error in } P = 3(1\%) + 2(2\%) + 1(3\%) + 1(4\%)$
$\% 	ext{ Error in } P = 3\% + 4\% + 3\% + 4\% = 14\%$.

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