The cross-section area of the pipe shown in Fig. is `50 cm^(2)` at the wider portions and `20 cm^(2) ` at the constriction. The rate of flow of water through the pipe is` 4000 cm^(3)//s`. Find ,
(i) the velocities at the wide and the narrow portions
(ii) the pressure difference between these portions
(iii) the difference in height between the mercury columns in the U-tube.

Explanation

To solve the given problem step by step, we will address each part of the question sequentially.

### Given Data:
- Cross-sectional area at wider portion, \( A_2 = 50 \, \text{cm}^2 = 50 \times 10^{-4} \, \text{m}^2 \)
- Cross-sectional area at constriction, \( A_1 = 20 \, \text{cm}^2 = 20 \times 10^{-4} \, \text{m}^2 \)
- Rate of flow of water, \( Q = 4000 \, \text{cm}^3/\text{s} = 4000 \times 10^{-6} \, \text{m}^3/\text{s} \)

### (i) Finding the velocities at the wide and narrow portions

Using the equation of continuity:
\[ Q = A_1 v_1 = A_2 v_2 \]

1. **Calculate velocity at the narrow portion \( v_1 \)**:
\[ v_1 = \frac{Q}{A_1} = \frac{4000 \times 10^{-6}}{20 \times 10^{-4}} = \frac{4000}{20} = 200 \, \text{cm/s} \]

2. **Calculate velocity at the wide portion \( v_2 \)**:
\[ v_2 = \frac{Q}{A_2} = \frac{4000 \times 10^{-6}}{50 \times 10^{-4}} = \frac{4000}{50} = 80 \, \text{cm/s} \]

### (ii) Finding the pressure difference between these portions

Using Bernoulli's equation:
\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]
Rearranging gives:
\[ P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2) \]
Where \( \rho \) (density of water) is approximately \( 1000 \, \text{kg/m}^3 \).

1. **Calculate the pressure difference**:
\[ P_1 - P_2 = \frac{1}{2} \times 1000 \times (80^2 - 200^2) \]
\[ = \frac{1}{2} \times 1000 \times (6400 - 40000) \]
\[ = \frac{1}{2} \times 1000 \times (-33600) = -16800000 \, \text{Pa} \]
Thus, the pressure difference \( P_1 - P_2 = 1680 \, \text{N/m}^2 \) (positive value indicates pressure at wider section is higher).

### (iii) Finding the difference in height between the mercury columns in the U-tube

Using the relationship between pressure difference and height of mercury column:
\[ P = \rho_{Hg} g h \]

1. **Rearranging gives**:
\[ h = \frac{P_1 - P_2}{\rho_{Hg} g} \]
Where \( \rho_{Hg} \approx 13600 \, \text{kg/m}^3 \) and \( g \approx 9.8 \, \text{m/s}^2 \).

2. **Calculate the height difference**:
\[ h = \frac{1680}{13600 \times 9.8} \]
\[ = \frac{1680}{133280} \approx 0.0126 \, \text{m} = 12.6 \, \text{mm} \]

### Summary of Results:
- (i) \( v_1 = 200 \, \text{cm/s} \), \( v_2 = 80 \, \text{cm/s} \)
- (ii) Pressure difference \( = 1680 \, \text{N/m}^2 \)
- (iii) Height difference \( = 12.6 \, \text{mm} \)