Answer
96.195 Pa
Work Step by Step
Let's apply the equation $A_{1}v_{1}=A_{2}v_{2}$ to find the velocity at point 1.
$A_{1}v_{1}=A_{2}v_{2}$ ; Let's plug known values into this equation
$\frac{A_{2}}{4}v_{1}=A_{2}(0.11\space m/s)=>v_{1}=0.44\space m/s$
Now, let's apply Bernoulli's equation to find the pressure difference.
$P+\frac{1}{2}\rho v^{2}+\rho gh=Constant$
$P_{1}+\frac{1}{2}\rho v_{1}^{2}+0=P_{1}+\frac{1}{2}\rho v_{2}^{2}+0$
$P_{2}-P_{1}=\frac{1}{2}(1060\space kg/m^{3})[(0.44\space m/s)^{2}-(0.11\space m/s)^{2}]=96.195\space Pa$
The density of blood is taken from table 11.1
