The rate of flow of liquid through a capillary tube of radius r is V, when the pressure difference across the two ends of the capillary is p. If pressure is increased by 3p and radius is reduced to r/2, then the rate of flow becomes

Options

A. V/9
B. 3V/8
C. V/4
D. V/3

Explanation

To solve the problem step by step, we will use the principles of fluid dynamics, particularly Poiseuille's law, which relates the flow rate of a liquid through a capillary tube to the pressure difference, the radius of the tube, and the viscosity of the liquid.

### Step 1: Understand the initial conditions
We are given:
- Initial flow rate \( V \)
- Initial pressure difference \( P \)
- Initial radius \( r \)

According to Poiseuille's law, the flow rate \( V \) through a capillary tube can be expressed as:
\[ V = k \cdot \frac{P}{r^4} \]
where \( k \) is a constant that depends on the viscosity of the liquid and the length of the tube.

### Step 2: Determine the new conditions
Now, the pressure is increased by \( 3P \), making the new pressure difference:
\[ P' = P + 3P = 4P \]
The radius is reduced to \( \frac{r}{2} \).

### Step 3: Write the new flow rate equation
Using the same formula for the new conditions, the new flow rate \( V' \) can be expressed as:
\[ V' = k \cdot \frac{P'}{(r')^4} \]
Substituting the new values:
\[ V' = k \cdot \frac{4P}{\left(\frac{r}{2}\right)^4} \]

### Step 4: Simplify the expression for the new radius
Calculating \( \left(\frac{r}{2}\right)^4 \):
\[ \left(\frac{r}{2}\right)^4 = \frac{r^4}{16} \]
Thus, we can rewrite the new flow rate as:
\[ V' = k \cdot \frac{4P}{\frac{r^4}{16}} = k \cdot \frac{4P \cdot 16}{r^4} = k \cdot \frac{64P}{r^4} \]

### Step 5: Relate the new flow rate to the initial flow rate
Since the initial flow rate \( V \) is given by:
\[ V = k \cdot \frac{P}{r^4} \]
We can express the new flow rate \( V' \) in terms of \( V \):
\[ V' = 64 \cdot \frac{P}{r^4} \cdot k = 64 \cdot V \]

### Step 6: Finalize the answer
Thus, the new flow rate \( V' \) is:
\[ V' = 64V \]

### Conclusion
The new rate of flow becomes \( 64V \).

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