Solution to 2008 Problem 32

Let v_1 and P_1 denote the pressure in the constriction. The fluid is incompressible, so
\begin{align*}\mbox{(cross-sectional area)} \cdot \mbox{(velocity)} = \mbox{constant} \Longrightarrow v_1 = v_0 \frac{\pi r^2...

Also, from Bernoulli's equation, we have that

\begin{align*}\frac{v_0^2}{2} +  \frac{P_0}{\rho} = \frac{v_1^2}{2} +  \frac{P_1}{\rho}\end{align*}
This implies that
\begin{align*}P_1 = \rho \left(\frac{v_0^2}{2} - \frac{v_1^2}{2} \right) + P_0 = \boxed{\frac{-15 \rho v_0^2}{2} + P_0}\end{a...
Therefore, answer (A) is correct.

return to the 2008 problem list

return to homepage

Please send questions or comments to where X = physgre.