The maximum percentage error in the measurment of density of a wire is [Given, mass of wire $=(0.60 \pm 0.003) \mathrm{g}$ radius of wire $=(0.50 \pm 0.01) \mathrm{cm}$ length of wire $=(10.00 \pm 0.05) \mathrm{cm}]$

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{d}=\frac{\mathrm{m}}{	ext { vol. }}=\frac{\mathrm{m}}{\pi \mathrm{R}^2 \ell} \Rightarrow \frac{\mathrm{~d} ho}{ho}=\frac{\mathrm{dm}}{\mathrm{m}}+\frac{2 \mathrm{dR}}{\mathrm{R}}+\frac{\mathrm{d} \ell}{\ell} \ & \Rightarrow \frac{\mathrm{d} ho}{ho}=\left(\frac{0.003}{0.6}+\frac{2 	imes 0.01}{0.5}+\frac{0.05}{10}ight) 100=5 \%\end{aligned}$

The maximum percentage error in the measurment of density of a wire is [Given, mass of wire $=(0.60 \pm 0…

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