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 $......\,\%$
A$1$
B$3.5$
C$4$
D$5$
JEE MAIN 2023, Medium
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C$4$
c $\rho=\frac{ m }{\pi r ^2 l} \Rightarrow\left|\frac{ d \rho}{\rho}\right|_{\max }=\left|\frac{ dm }{ m }\right|+2\left|\frac{ dr }{ r }\right|+\left|\frac{ d l}{l}\right|$
$\Rightarrow \% \text { error in density }=\left(\frac{ d \rho}{\rho}\right) \times 100 \%$
$=(2.5+1+0.5) \%=4 \%$
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