question_answer

To study the flow of a liquid through a narrow tube, the following formula is used : \[\eta =\frac{\pi }{8}\frac{{{\Pr }^{4}}}{Vl}\]where letters have their usual meaning. The values of P,r, V and l are 76 cm of Hg col., 0.28 cm, \[\text{1}.\text{2 c}{{\text{m}}^{\text{3}}}{{\text{s}}^{-\text{1}}}\]and 18.2 cm respectively. If these quantities are measured to the accuracy of 0.1 cm of Hg col., 0.01 cm, \[0.\text{1 c}{{\text{m}}^{\text{3}}}{{\text{s}}^{-\text{1}}}\]and 0.1 cm respectively, find the percentage error in the value of h.

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                                                                                                                                      \[\eta =\frac{\pi }{8}\frac{{{\Pr }^{4}}}{Vl}\] \[\frac{\Delta \eta }{\eta }=\frac{\Delta P}{P}+4\frac{\Delta r}{r}+\frac{\Delta V}{V}+\frac{\Delta l}{l}=\frac{0.1}{76}+\frac{4(0.01)}{0.28}+\frac{0.1}{1.2}+\frac{0.1}{18.2}\]        \[=0.00\text{13}+0.\text{1428}+0.0\text{833}+0.00\text{54}=0.\text{2328}\times \text{1}00=0.\text{2328}\]                                                 \[\frac{\Delta \eta }{\eta }\times \text{1}00=\text{23}.\text{28}%\]