A cylindrical wire of mass (0.4 pm 0.01),g ha | vedclass

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Question

$\rho=\frac{ m }{\pi r ^2 l} \Rightarrow\left|\frac{ d \rho}{\rho}\right|_{\max }=\left|\frac{ dm }{ m }\right|+2\left|\frac{ dr }{ r }\right|+\left|\

Vedclass · Accepted Answer

$4$

Vedclass · Answer

$ho=\frac{ m }{\pi r ^2 l} \Rightarrow\left|\frac{ d ho}{ho}ight|_{\max }=\left|\frac{ dm }{ m }ight|+2\left|\frac{ dr }{ r }ight|+\left|\frac{ d l}{l}ight|$

$=\frac{0.01}{0.4}+\frac{2(0.03)}{6}+\frac{0.04}{8}$

$\Rightarrow \% 	ext { error in density }=\left(\frac{ d ho}{ho}ight) 	imes 100 \%$

$=(2.5+1+0.5) \%=4 \%$

A cylindrical wire of mass $(0.4 \pm 0.01)\,g$ has length $(8 \pm 0.04)\,cm$ and radius $(6 \pm 0.03)\,mm$.The maximum error in its density will be $......\,\%$

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