Question

experimenter measures a,b,c and then finds x= ab²/c³ if error in a,b,c are +-1%,+-7%,+-4% respectively find error in x​

Answer 1

Explanation:

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Answer 2

Explanation :

Given :

$\bf x=\dfrac{ab^2}{c^3}$ = $\sf a^1b^2c^{-3}$

Taking logarithm on both sides,

$\sf\log x=\log a+2\log b-3\log c$

Differentiating both sides,

$\sf\dfrac{\Delta x}{x}=\dfrac{\Delta a}{a}+2\dfrac{\Delta b}{b}-3\dfrac{\Delta c}{c}$

The percentage error in x is,

$\\ \sf\dfrac{\Delta x}{x}\times100=\left(\dfrac{\Delta a}{a}\times100\right)+2\left(\dfrac{\Delta b}{b}\times100\right)+3\left(\dfrac{\Delta c}{c}\times100\right)$

$\\ \sf =1(\%\ error\ in\ a)+2(\%\ error\ in\ b)+3(\%\ error\ in\ c)$

$\\ \sf =\pm\left[1(1\%)+2(7\%)+3(4\%)\right]$

$\\ \sf =\pm\left[1\%+14\%+12\%\right]$

$\\ \sf =\pm\ 27\%$

$\mapsto\boxed{\bf Error\%\ in\ x=\pm\ 27\%.}$

The percentage error in x is ± 27%.