Question

Two capillary tubes A and B are connected in series. The length and radius of the bore of tube A are twice those of tube B. The ratio of the pressure difference across the tubes A and B is:

• A. 8:1
• B. 1:8
• C. 4:1
• D. 1:4

Hint:

In viscous flow, pressure drop depends strongly on radius: \( \Delta P \propto \frac{1}{r^4} \), so small radius change drastically affects flow.

Answer 1

The Correct Option is B

Step 1: Use the electrical-circuit analogy for fluid flow instead of writing Poiseuille's pressure formula directly.
For steady viscous flow through a narrow tube, the tube behaves like a resistor: the flow resistance is \[ \mathcal{R} \propto \frac{L}{r^4} \] and the pressure drop across it plays the role of voltage, while the volume flow rate $Q$ plays the role of current. Two tubes joined end to end carry the same $Q$ through both, exactly like two resistors in series carrying the same current.
Step 2: Since the same $Q$ flows through both tubes, the pressure drops are proportional to the resistances.
\[ \frac{\Delta P_A}{\Delta P_B} = \frac{\mathcal{R}_A}{\mathcal{R}_B} = \frac{L_A/r_A^4}{L_B/r_B^4} \]
Step 3: Substitute the given dimensions.
Tube A has both length and radius twice that of tube B: $L_A = 2L_B$, $r_A = 2r_B$. \[ \frac{\mathcal{R}_A}{\mathcal{R}_B} = \frac{2L_B/(2r_B)^4}{L_B/r_B^4} = \frac{2L_B}{16r_B^4}\times\frac{r_B^4}{L_B} = \frac{2}{16} = \frac{1}{8} \]
Step 4: State the ratio.
\[ \Delta P_A : \Delta P_B = 1 : 8 \]
Final Answer:
\[ \boxed{1:8} \]