A sliver wire has mass (0.6 pm 0.006) ; g, ra | vedclass

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Question

$M=(0.6 \pm 0.006) \; g$

$r=(0.5 \pm 0.005) \; mm$

$l=(4 \pm 0.04) \; cm$

$\rho=\frac{m}{V}$

$\Rightarrow \frac{\Delta \rho}{\rho}=\frac{\

Vedclass · Accepted Answer

$4$

Vedclass · Answer

$M=(0.6 \pm 0.006) \; g$

$r=(0.5 \pm 0.005) \; mm$

$l=(4 \pm 0.04) \; cm$

$ho=\frac{m}{V}$

$\Rightarrow \frac{\Delta ho}{ho}=\frac{\Delta m}{m}+\frac{2 \Delta r}{r}+\frac{\Delta l}{l}$

(Volume of cylinder $=\pi r^{2} l$ )

$=\frac{0.006}{0.6}+\frac{2 	imes 0.005}{0.5}+\frac{0.04}{4}$

$100 	imes \frac{\Delta ho}{ho}=4 	imes 10^{-2} 	imes 100$

$\frac{\Delta ho}{ho} 	imes 100=4 \%$

A sliver wire has mass $(0.6 \pm 0.006) \; g$, radius $(0.5 \pm 0.005) \; mm$ and length $(4 \pm 0.04) \; cm$. The maximum percentage error in the measurement of its density will be $......\,\%$

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