91Ó°ÊÓ

The height at which the blood is placed is \(h = 1.40\;{\rm{m}}\).

The length of the needle is \(L = 3.8\;{\rm{cm}}\).

The diameter is \(d = 0.40\;{\rm{mm}}\).

The flow rate is \(Q = 4.1\;{\rm{c}}{{\rm{m}}^{\rm{3}}}{\rm{/min}}\).

In this problem, the viscosity of the blood will be calculated by using the Poiseuille’s equation. Consider that the open end of the needle is at atmospheric pressure.

The relation fromPoiseuille’s equationis given by,

\(Q = \frac{{\pi {R^4}\left( {{P_2} - {P_1}} \right)}}{{8\eta L}}\)……... (i)

Here, \(R\) is the radius of the needle, \({P_1}\) and \({P_2}\) are the pressure at inside and outside and \(\eta \) is the required viscosity.

The relation of pressure difference is given by,

\(\left( {{P_{\rm{2}}} - {P_1}} \right) = \rho gh\)……... (ii)

Here, \(\rho \) is the density of blood and \(g\) is the gravitational acceleration.

On plugging the values of equation (ii) in the equation (i),

\(\begin{array}{l}Q = \frac{{\pi {{\left( {\frac{d}{2}} \right)}^4}\rho gh}}{{8\eta L}}\\\eta = \frac{{\pi {{\left( {\frac{d}{2}} \right)}^4}\rho gh}}{{8QL}}\\\eta = \frac{{\pi {{\left( {\frac{{0.40\;{\rm{mm}} \times \frac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}}}}{2}} \right)}^4}\left( {1.05 \times {{10}^3}\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {9.80\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {1.40\;{\rm{m}}} \right)}}{{8\left( {4.1\;{\rm{c}}{{\rm{m}}^3}{\rm{/min}} \times \frac{{{{10}^{ - 6}}\;{{\rm{m}}^3}{\rm{/s}}}}{{60\;{\rm{c}}{{\rm{m}}^3}{\rm{/min}}}}} \right)\left( {3.8\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)}}\\\eta = 3.5 \times {10^{ - 3}}\;{\rm{Pa}} \cdot {\rm{s}}\end{array}\)

Thus, the viscosity of this blood is \(3.5 \times {10^{ - 3}}\;{\rm{Pa}} \cdot {\rm{s}}\).