• # question_answer The flow of blood in a large artery of an anesthetised dog is diverted through a venturimeter. The wider part of the meter has a Cross-sectional area equal to that of the artery. $A =8 m{{m}^{2}}$. The narrower part has an area $a=4m{{m}^{2}}$ and density of blood. i.e., $\rho =~1.06\times {{10}^{3}} kg\,{{m}^{-3}}$. The pressure drop in the artery is 24 Pa. What is the speed of the blood in the artery? A) 0.5 m/s B) 0.125m/s         C) 1.25 m/s D) 2.5 m/s

Solution :

 Given, wides part of the meter cross-sectional area, $A=8m{{m}^{2}}$ and narrow part, $a=4m{{m}^{2}}$ and density of blood i.e., $\rho$is $1.06\times {{10}^{3}}kg-{{m}^{-3}}$. The ratio of the area is $\left( \frac{A}{a} \right)=2.$ so, the speed of the blood in the artery i.e., $v=\sqrt{\frac{2\rho \,mgh}{\rho }}{{\left[ {{\left( \frac{A}{a} \right)}^{2}}-1 \right]}^{1/2}}$$\Rightarrow$$v=\sqrt{\frac{2\times 24Pa}{1060kg-{{m}^{-3}}\times ({{2}^{2}}-1)}}$$=0.125m{{s}^{-1}}$

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