Question

The flow of blood in a large artery of an anesthetised dog is diverted through a venturi meter. The wider part of the meter has a cross-sectional area equal to that of the artery $A=8m{m}^2$ .The narrower part has an area $a=4m{m}^2$.The pressure drop in the artery is 24 Pa. What is the speed of blood in the artery?Density of blood is $1.06\times {10}^{3}kg/m^3$.

• A. $13.5\times {10}^{-2}m{s}^{-1}$
• B. $12.5\times {10}^{-2}m{s}^{-1}$
• C. $26.5\times {10}^{-3}m{s}^{-1}$
• D. $32.5\times {10}^{-}m{s}^{-1}$

Answer 1

The correct option is B $12.5\times {10}^{-2}m{s}^{-1}$

Given : $A_a=8m{m}^2$

$A_n=4m{m}^2$

$P_a-P_n=24$ Pa

Let the speed of blood in artery and narrow part be $v_a$ and $v_n$ respectively.

From continuity equation : $A_a{v}_a=A_n{v}_n$

$\therefore$ $8v_a=4v_n$

$⟹v_n=2v_a$

Using Bernoulli's equation :

$P_a+\rho g{h}_a+\frac{1}{2}\rho v_a^2=$ $P_n+\rho g{h}_n+\frac{1}{2}\rho v_n^2$

where $h_n=h_n$

$\therefore$ $P_a-P_n=\frac{1}{2}\rho (v_n^2-v_a^2)$

$\therefore$ $24=\frac{1}{2}\times 1.06\times {10}^3(4v_a^2-v_a^2)$

$⟹$ $v_a=12.5\times {10}^{-2}$ $m/s$