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Two capillaries of same length and radii...

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

A

9.4 m

B

4.9 m

C

0.49 m

D

0.94 m

Text Solution

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The correct Answer is:
To solve the problem, we need to find the pressure difference across the first capillary when two capillaries of the same length but different radii are connected in series. Let's break it down step by step. ### Step-by-Step Solution: 1. **Identify Given Data:** - Length of both capillaries, \( L_1 = L_2 = L \) (same for both). - Ratio of radii, \( R_1 : R_2 = 1 : 2 \) (let's denote \( R_1 = r \) and \( R_2 = 2r \)). - Total pressure difference across the combination, \( P = 1 \) m of water. 2. **Use Poiseuille's Law:** According to Poiseuille's law, the pressure drop across a capillary is given by: \[ \Delta P = \frac{8 \eta L Q}{\pi R^4} \] where \( \Delta P \) is the pressure difference, \( \eta \) is the viscosity of the liquid, \( L \) is the length of the capillary, \( Q \) is the flow rate, and \( R \) is the radius of the capillary. 3. **Express Pressure Drops:** For the first capillary (radius \( R_1 = r \)): \[ P_1 = \frac{8 \eta L Q}{\pi r^4} \] For the second capillary (radius \( R_2 = 2r \)): \[ P_2 = \frac{8 \eta L Q}{\pi (2r)^4} = \frac{8 \eta L Q}{\pi \cdot 16r^4} = \frac{1}{16} \cdot \frac{8 \eta L Q}{\pi r^4} = \frac{P_1}{16} \] 4. **Relate Pressure Drops:** Since the capillaries are in series, the total pressure drop is: \[ P = P_1 + P_2 \] Substituting \( P_2 \) in terms of \( P_1 \): \[ 1 = P_1 + \frac{P_1}{16} \] This simplifies to: \[ 1 = P_1 \left(1 + \frac{1}{16}\right) = P_1 \cdot \frac{17}{16} \] 5. **Solve for \( P_1 \):** Rearranging gives: \[ P_1 = \frac{16}{17} \text{ m of water} \] Converting this to decimal form: \[ P_1 \approx 0.94 \text{ m of water} \] 6. **Final Answer:** The pressure difference across the first capillary is approximately \( 0.94 \) m of water.

To solve the problem, we need to find the pressure difference across the first capillary when two capillaries of the same length but different radii are connected in series. Let's break it down step by step. ### Step-by-Step Solution: 1. **Identify Given Data:** - Length of both capillaries, \( L_1 = L_2 = L \) (same for both). - Ratio of radii, \( R_1 : R_2 = 1 : 2 \) (let's denote \( R_1 = r \) and \( R_2 = 2r \)). - Total pressure difference across the combination, \( P = 1 \) m of water. ...
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