Answer
(a) $1.71 \times 10^4 \ Pa$
(b) $1.61 N$
Work Step by Step
For (a):
We assume the density $\rho$ of blood is $1.06 \times 10^3 kg/m^3$. The pressure difference between two points is $\Delta p = \rho g \Delta h$, which in this case is $(1.06 \times 10^3)(9.8)(1.65) = 1.71\times10^4 \ Pa$.
For (b):
The surface area $A$ of the segment is $2\pi r h$. We're given $2r = 1.50 \times 10^{-3} m$ and $h = 2 \times 10^{-2} m$.
Therefore, by direct calculation, $\Delta F = A\Delta P \implies \Delta F = (\pi \times 1.50 \times 10^{-3} \times 2 \times 10^{-2})(1.71 \times 10^4) = 1.61 N$
