When a liquid moves steadily under some pressure through a horizontal tube, it moves in the form of cylindrical layers coaxial to the ends of the tube. The velocity of different layers is different. The velocity of the layer is maximum along the axis of the tube and it decreases as one moves towards the walls of the tube. According to Poiseuille, the rate of flow of liquid through a horizontal capillary tube varies as the relation `V alpha (pr^(4))/(eta l)` where `l` and `r` are the length and radius of the tube and `p` is the pressure different between the ends of the tube and is the constant of proportionality.
When two capillary tubes through which the liquid is flowing are joined in series then the equivalent liquid resistance is

Options

A. `R_(s)=R_(1)+R_(2)`
B. `Rs=R_(1)-R_(2)`
C. `1/(Rs)=1/R_(1)+1/R_(2)`
D. `(K eta)/(pr)=1/R_(1)-1/R_(2)`

Answer

To determine the equivalent liquid resistance when two capillary tubes are joined in series, we can follow these steps:

### Step 1: Understand the Poiseuille's Law
According to Poiseuille's Law, the volume flow rate \( V \) through a capillary tube is given by:
\[ V \propto \frac{p r^4}{\eta l} \]
where:
- \( p \) is the pressure difference between the ends of the tube,
- \( r \) is the radius of the tube,
- \( \eta \) is the viscosity of the liquid,
- \( l \) is the length of the tube.

### Step 2: Define Resistance in Fluid Flow
From the Poiseuille equation, we can derive the fluid resistance \( R \) as:
\[ R = \frac{8 \eta l}{\pi r^4} \]
This shows that the resistance is dependent on the length \( l \) and radius \( r \) of the tube.

### Step 3: Consider Two Tubes in Series
When two capillary tubes (let's call them Tube 1 and Tube 2) are connected in series, the total pressure drop across both tubes is the sum of the pressure drops across each tube:
\[ P_1 - P_3 = (P_1 - P_2) + (P_2 - P_3) \]
Here, \( P_1 \) is the pressure at the start of Tube 1, \( P_2 \) is the pressure at the end of Tube 1 and the start of Tube 2, and \( P_3 \) is the pressure at the end of Tube 2.

### Step 4: Apply Ohm's Law Analogy
Using the analogy to Ohm's Law, we can express the pressure drops in terms of the flow rate \( V \) and resistances \( R_1 \) and \( R_2 \) of the two tubes:
\[ P_1 - P_2 = V \cdot R_1 \]
\[ P_2 - P_3 = V \cdot R_2 \]

### Step 5: Combine the Equations
By substituting the equations from Step 4 into the equation from Step 3:
\[ P_1 - P_3 = V \cdot R_1 + V \cdot R_2 \]
This can be simplified to:
\[ P_1 - P_3 = V \cdot (R_1 + R_2) \]

### Step 6: Determine the Equivalent Resistance
From the above equation, we can see that the equivalent resistance \( R \) of the two tubes in series is:
\[ R = R_1 + R_2 \]

### Final Result
Thus, the equivalent liquid resistance when two capillary tubes are joined in series is:
\[ R_{\text{eq}} = R_1 + R_2 \]

---