A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference P . The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled is

Options

A. `P`
B. `(3P)/(4)`
C. `(P)/(2)`
D. `(P)/(4)`

Answer

To solve the problem, we will use the principles of fluid mechanics, specifically the Hagen-Poiseuille equation, which describes the flow of a viscous liquid through a cylindrical pipe. The equation is given by:

\[ Q = \frac{\pi r^4 (P_1 - P_2)}{8 \eta L} \]

where:
- \( Q \) is the volumetric flow rate,
- \( r \) is the radius of the tube,
- \( P_1 - P_2 \) is the pressure difference,
- \( \eta \) is the dynamic viscosity of the liquid,
- \( L \) is the length of the tube.

### Step 1: Initial Conditions
Let the initial radius of the tube be \( r \) and the length be \( L \). The initial pressure difference is \( P \).

Using the Hagen-Poiseuille equation, the initial flow rate \( Q_1 \) can be expressed as:

\[ Q_1 = \frac{\pi r^4 P}{8 \eta L} \]

### Step 2: New Conditions
Now, we are doubling both the radius and the length of the tube. Therefore, the new radius \( r' \) is \( 2r \) and the new length \( L' \) is \( 2L \).

### Step 3: Calculate New Flow Rate
Substituting the new radius and length into the Hagen-Poiseuille equation gives us the new flow rate \( Q_2 \):

\[ Q_2 = \frac{\pi (2r)^4 P'}{8 \eta (2L)} \]

Simplifying this:

\[ Q_2 = \frac{\pi (16r^4) P'}{16 \eta L} = \frac{\pi r^4 P'}{ \eta L} \]

### Step 4: Relate the Flow Rates
We know from the problem statement that the new flow rate \( Q_2 \) is double the initial flow rate \( Q_1 \):

\[ Q_2 = 2Q_1 \]

Substituting the expressions for \( Q_1 \) and \( Q_2 \):

\[ \frac{\pi r^4 P'}{ \eta L} = 2 \left( \frac{\pi r^4 P}{8 \eta L} \right) \]

### Step 5: Simplify the Equation
Canceling common terms (\( \frac{\pi r^4}{\eta L} \)) from both sides gives:

\[ P' = \frac{2P}{8} = \frac{P}{4} \]

### Conclusion
Thus, the pressure difference required to double the flow rate when both the radius and length of the tube are doubled is:

\[ \boxed{\frac{P}{4}} \]