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A physical quantity $X$ is related to four measurable quantities $a,\, b,\, c$ and $d$ as follows $X = a^2b^3c^{\frac {5}{2}}d^{-2}$. The percentange error in the measurement of $a,\, b,\, c$ and $d$ are $1\,\%$, $2\,\%$, $3\,\%$ and $4\,\%$ respectively. What is the percentage error in quantity $X$ ? If the value of $X$ calculated on the basis of the above relation is $2.763$, to what value should you round off the result.
Solution
Given, physical quantity, $\mathrm{X}=a^{2} b^{2} c^{\frac{5}{2}} d^{-2}$
Maximum percentage error in $\mathrm{X}$ is,
$\frac{\Delta \mathrm{X}}{\mathrm{X}} \times 100 =\left[2\left(\frac{\Delta a}{a} \times 100\right)+3\left(\frac{\Delta b}{b} \times 100\right)+\frac{5}{2}\left(\frac{\Delta c}{c} \times 100\right)+2\left(\frac{\Delta d}{d} \times 100\right)\right]$
$=\left[2(1)+3(2)+\frac{5}{2}(3)+2(4)\right] \%$
$=\left[2+6+\frac{15}{2}+8\right]=\pm 23.5 \%$
$\therefore$ Percentage error in $\mathrm{X}=23.5 \%$
Relative error in $\mathrm{X}=0.235=0.24$ (By rounding off upto two significant figures) The calculated value of $x$ should be round off upto two significant digits. $\therefore \mathrm{X}=2.8$
