A) 0.5 m/s
B) 0.125m/s
C) 1.25 m/s
D) 2.5 m/s
Correct Answer: B
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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