There is a steady laminar blood flow in a tapered cylindrical vessel of circular cross-section, as shown in the figure below. Two cross-sections, labeled A and B in the figure, have radii \(0.5\) centimeter (cm) and \(0.45\) cm, respectively. If the average flow speed of the blood at cross-section A is \(1\) cm/second, the average flow speed at cross-section B is cm/second. (Round off to one decimal place).
Assume blood to be an incompressible fluid.
Show Hint
Use continuity for an incompressible fluid, \(A_A v_A = A_B v_B\), with \(A = \pi r^2\); the narrower cross-section must have the higher speed.
Step 1: Identify the governing principle.
Blood is treated as an incompressible fluid in steady laminar flow through a tapered vessel. For an incompressible fluid, the volume flow rate must stay the same at every cross-section along the vessel, since no fluid is created, destroyed, or stored anywhere inside. This is the continuity equation.
Step 2: Write the continuity equation.
For a circular cross-section of radius \(r\) carrying fluid at average speed \(v\), the volume flow rate is \(Q = A v = \pi r^2 v\). Since \(Q\) is the same at A and B:
\[ A_A v_A = A_B v_B \]
\[ \pi r_A^2 v_A = \pi r_B^2 v_B \]
Step 3: Substitute the given values.
Here \(r_A = 0.5\) cm, \(r_B = 0.45\) cm, and \(v_A = 1\) cm/s. The \(\pi\) cancels from both sides:
\[ r_A^2 v_A = r_B^2 v_B \]
\[ (0.5)^2 (1) = (0.45)^2 v_B \]
\[ 0.25 = 0.2025\, v_B \]
Step 4: Solve for \(v_B\).
\[ v_B = \frac{0.25}{0.2025} = 1.2346\ \text{cm/s} \]
Rounded to one decimal place, \(v_B \approx 1.2\) cm/s. This makes physical sense: since the vessel narrows from A to B, the same volume of blood must move faster through the smaller cross-section, so \(v_B > v_A\).
Final Answer:
The average flow speed at cross-section B is \(1.2\) cm/second.
\[ \boxed{v_B = 1.2\ \text{cm/s}} \]
