Answer
The absolute pressure at the bug's end of the feeding tube is $-107~kPa$
Work Step by Step
We can find the flow rate $Q$:
$Q = \frac{V}{t} = \frac{3.0\times 10^{-7}~m^3}{(25)(60~s)} = 2.0\times 10^{-10}~m^3/s$
We can use Poiseuille's law to find an expression for the flow rate:
$Q = \frac{\pi~\Delta P~r^4}{8~\eta~L}$
$Q$ is the flow rate ($m^3~s^{-1}$)
$\Delta P$ is the pressure difference ($Pa$)
$r$ is the radius ($m$)
$\eta$ is the fluid viscosity
$L$ is the length of the tube ($m$)
We can find the required pressure difference $\Delta P$:
$\Delta P = \frac{8~Q~\eta~L}{\pi~r^4}$
$\Delta P = \frac{(8)~(2.0\times 10^{-10}~m^3/s)~(0.0013~Pa~s)~(2.0\times 10^{-4}~m)}{(\pi)~(5.0\times 10^{-6}~m)^4}$
$\Delta P = 212~kPa$
Since the pressure difference is $212~kPa$ and the pressure in the arm is $105~kPa$, the absolute pressure at the bug's end of the feeding tube is $105~kPa-212~kPa$ which is $-107~kPa$.
