A metal wire has mass (0.4 pm 0.002),g, radiu | vedclass

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Question

$\rho=\frac{ M }{ V }$

$\rho =\frac{ M }{\pi r ^2 \ell}$

$\frac{\Delta \rho}{\rho} =\frac{\Delta M }{ M }+\frac{2 \Delta r }{ r }+\frac{\Delta l

Vedclass · Accepted Answer

$1.6$

Vedclass · Answer

$ho=\frac{ M }{ V }$

$ho =\frac{ M }{\pi r ^2 \ell}$

$\frac{\Delta ho}{ho} =\frac{\Delta M }{ M }+\frac{2 \Delta r }{ r }+\frac{\Delta l}{l}$

$\frac{\Delta ho}{ho} \%$ $=\left[\frac{0.002}{0.4}+\frac{2(0.001)}{(0.3)}+\frac{0.02}{5}ight] 	imes 100 \%$

$=\frac{1}{2} \%+\frac{2}{3} \%+\frac{2}{5} \%$

$=1.6 \%$

A metal wire has mass $(0.4 \pm 0.002)\,g$, radius $(0.3 \pm 0.001)\,mm$ and length $(5 \pm 0.02) \,cm$. The maximum possible percentage error in the measurement of density will nearly be $.......\%$

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