The rate of flow of a liquid through a capillary tube of radius r is under a pressure difference of P. Calculate the rate of flow when the diameter is reduced to half and the pressure difference is made 4 P?

Explanation

To solve the problem, we will use the principles of fluid dynamics, particularly Poiseuille's Law, which states that the rate of flow (V) of a liquid through a capillary tube is given by the equation:

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

Where:
- \( V \) = rate of flow
- \( P \) = pressure difference
- \( r \) = radius of the tube
- \( \eta \) = viscosity of the liquid
- \( L \) = length of the tube

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Let the initial radius of the capillary tube be \( r_1 \).
- The initial pressure difference is \( P_1 = P \).
- The initial rate of flow is \( V_1 \).

Using Poiseuille's Law:
\[ V_1 = \frac{\pi P r_1^4}{8 \eta L} \]

2. **Determine New Conditions:**
- The diameter is reduced to half, which means the new radius \( r_2 = \frac{r_1}{2} \).
- The new pressure difference is \( P_2 = 4P \).

3. **Calculate New Rate of Flow:**
Using the same formula for the new conditions:
\[ V_2 = \frac{\pi P_2 r_2^4}{8 \eta L} \]
Substituting \( P_2 \) and \( r_2 \):
\[ V_2 = \frac{\pi (4P) \left(\frac{r_1}{2}\right)^4}{8 \eta L} \]

4. **Simplify the Expression:**
\[ V_2 = \frac{\pi (4P) \left(\frac{r_1^4}{16}\right)}{8 \eta L} \]
\[ V_2 = \frac{\pi P r_1^4}{8 \eta L} \cdot \frac{4}{16} \]
\[ V_2 = \frac{\pi P r_1^4}{8 \eta L} \cdot \frac{1}{4} \]
\[ V_2 = \frac{1}{4} V_1 \]

5. **Conclusion:**
The new rate of flow \( V_2 \) is \( \frac{1}{4} \) of the initial rate of flow \( V_1 \).

### Final Answer:
The rate of flow when the diameter is reduced to half and the pressure difference is made \( 4P \) is \( \frac{1}{4} V_1 \).