Two capillaries of same length and radii in the ratio `1 : 2` are connected in series. A liquid flows through them in streamlined condition. If the pressure across the two extreme ends of the combination is 1 m of water, the pressure difference across first capillary is

Options

A. `9.4 m`
B. `4.9 m`
C. `0.49 m`
D. `0.94 m`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have two capillaries of the same length but different radii, connected in series. The radii are in the ratio of 1:2. We need to find the pressure difference across the first capillary when the total pressure difference across both capillaries is 1 meter of water.

### Step 2: Define the Variables
Let:
- \( r_1 \) = radius of the first capillary
- \( r_2 = 2r_1 \) = radius of the second capillary (since the ratio is 1:2)
- \( L \) = length of both capillaries (same for both)
- \( \Delta P_1 \) = pressure difference across the first capillary
- \( \Delta P_2 \) = pressure difference across the second capillary

### Step 3: Use the Relationship of Pressure Drop in Capillaries
According to Poiseuille's law for laminar flow in a capillary, the pressure drop (\( \Delta P \)) across a capillary is given by:

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

Where:
- \( \eta \) = viscosity of the liquid
- \( Q \) = volumetric flow rate

### Step 4: Establish the Flow Rate Equality
Since the capillaries are in series, the flow rate through both capillaries is the same:

\[ Q_1 = Q_2 = Q \]

### Step 5: Write the Pressure Drop Equations
For the first capillary:

\[ \Delta P_1 = \frac{8 \eta L Q}{\pi r_1^4} \]

For the second capillary:

\[ \Delta P_2 = \frac{8 \eta L Q}{\pi r_2^4} = \frac{8 \eta L Q}{\pi (2r_1)^4} = \frac{8 \eta L Q}{16 \pi r_1^4} = \frac{1}{2} \cdot \frac{8 \eta L Q}{\pi r_1^4} = \frac{1}{2} \Delta P_1 \]

### Step 6: Relate the Total Pressure Drop
The total pressure drop across both capillaries is given as:

\[ \Delta P_1 + \Delta P_2 = 1 \text{ meter of water} \]

Substituting \( \Delta P_2 \):

\[ \Delta P_1 + \frac{1}{2} \Delta P_1 = 1 \]

This simplifies to:

\[ \frac{3}{2} \Delta P_1 = 1 \]

### Step 7: Solve for \( \Delta P_1 \)
Now, we can solve for \( \Delta P_1 \):

\[ \Delta P_1 = \frac{2}{3} \text{ meters of water} \]

### Step 8: Convert to Meters
Since we need the answer in meters, we can convert:

\[ \Delta P_1 = \frac{2}{3} \approx 0.67 \text{ meters of water} \]

However, we need to ensure the total pressure drop matches the given total of 1 meter of water.

### Step 9: Correct Calculation
Since we derived \( \Delta P_2 = \frac{1}{2} \Delta P_1 \), we can express \( \Delta P_1 \) in terms of the total pressure drop:

Let \( \Delta P_1 = x \) and \( \Delta P_2 = \frac{1}{2}x \):

\[ x + \frac{1}{2}x = 1 \]

This gives:

\[ \frac{3}{2}x = 1 \implies x = \frac{2}{3} \text{ meters of water} \]

### Final Answer
Thus, the pressure difference across the first capillary is:

\[ \Delta P_1 = \frac{2}{3} \text{ meters of water} \approx 0.67 \text{ meters} \]