Answer:
Here, P = 76 cm of Hg col.
\[\Delta
\text{P}=0.\text{1 cm}\]
r = 0.28 cm
\[\Delta
\text{r}=0.0\text{1cm}\]
\[\text{V}=\text{l}.\text{2c}{{\text{m}}^{\text{3}}}{{\text{s}}^{-\text{1}}}\Delta
\text{V}=0.\text{1 c}{{\text{m}}^{\text{3}}}{{\text{s}}^{-\text{1}}}\]
l = 18.2 cm \[\Delta
\text{l}=0.\text{l cm}\]
From \[\eta
=\frac{\pi }{8}\frac{{{\Pr }^{4}}}{Vl}\]
\[\frac{\Delta \eta }{\eta }=\frac{\Delta
P}{P}+4\frac{\Delta r}{r}+\frac{\Delta V}{V}+\frac{\Delta
l}{l}=\frac{0.1}{76}+\frac{4(0.01)}{0.28}+\frac{0.1}{1.2}+\frac{0.1}{18.2}\]
\[=0.00\text{13}+0.\text{1428}+0.0\text{833}+0.00\text{54}=0.\text{2328}\times
\text{1}00=0.\text{2328}\]
\[\frac{\Delta
\eta }{\eta }\times \text{1}00=\text{23}.\text{28}%\]
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