A liquid flows through two capillary tubes A and B connected in series. The length and radius of B are twice that of A. What is the ratio of the pressure difference across A to that across B ?

Options

A. 8
B. 4
C. 2
D. 1

Explanation

To solve the problem, we will use Poiseuille's law, which describes the flow of a liquid through a capillary tube. The flow rate (Q) through a tube is given by the formula:

\[ Q = \frac{\pi P r^4}{8 \eta L} \]

where:
- \( P \) is the pressure difference across the tube,
- \( r \) is the radius of the tube,
- \( \eta \) is the dynamic viscosity of the liquid,
- \( L \) is the length of the tube.

### Step-by-Step Solution:

1. **Identify the parameters for tubes A and B:**
- Let the radius of tube A be \( r_A = r \) and its length \( L_A = L \).
- For tube B, the radius is \( r_B = 2r \) (twice that of A) and the length is \( L_B = 2L \) (twice that of A).

2. **Write the flow rate equations for both tubes:**
- For tube A:
\[ Q_A = \frac{\pi P_A r_A^4}{8 \eta L_A} = \frac{\pi P_A r^4}{8 \eta L} \]
- For tube B:
\[ Q_B = \frac{\pi P_B r_B^4}{8 \eta L_B} = \frac{\pi P_B (2r)^4}{8 \eta (2L)} = \frac{\pi P_B \cdot 16r^4}{16 \eta L} = \frac{\pi P_B r^4}{\eta L} \]

3. **Set the flow rates equal to each other (since they are in series):**
\[ Q_A = Q_B \]
Therefore:
\[ \frac{\pi P_A r^4}{8 \eta L} = \frac{\pi P_B r^4}{\eta L} \]

4. **Cancel out common terms:**
- Cancel \( \pi \), \( r^4 \), and \( \eta L \) from both sides:
\[ \frac{P_A}{8} = P_B \]

5. **Rearranging gives the ratio of pressure differences:**
\[ P_A = 8 P_B \]
Therefore, the ratio of the pressure difference across A to that across B is:
\[ \frac{P_A}{P_B} = 8 \]

### Final Answer:
The ratio of the pressure difference across A to that across B is \( 8:1 \).