Water flows along horizontal pipe whose cross-section is not constant. The pressure is 1 cm of Hg, where the velocity is 35 cm/s . At a point where the velocity is 65cm/s then pressure will be

Options

A. a. 0.89 cm of Hg
B. b. 8.9 cm of Hg
C. c. 0.5 cm of Hg
D. d. 1 cm of Hg

Explanation

To solve the problem, we will apply Bernoulli's theorem, which relates the pressure, velocity, and height of a fluid flowing through a pipe. The steps are as follows:

### Step 1: Understand the given data
- Pressure at point 1 (P1) = 1 cm of Hg
- Velocity at point 1 (V1) = 35 cm/s = 0.35 m/s
- Velocity at point 2 (V2) = 65 cm/s = 0.65 m/s

### Step 2: Convert pressure from cm of Hg to Pascals
To convert the pressure from cm of Hg to Pascals, we use the formula:
\[ P = h \cdot \rho \cdot g \]
Where:
- \( h \) = height of the mercury column = 1 cm = 0.01 m
- \( \rho \) = density of mercury = 13600 kg/m³
- \( g \) = acceleration due to gravity = 9.81 m/s² (approx. 10 m/s² for simplicity)

Calculating P1:
\[ P1 = 0.01 \, \text{m} \cdot 13600 \, \text{kg/m}^3 \cdot 10 \, \text{m/s}^2 = 1360 \, \text{Pa} \]

### Step 3: Apply Bernoulli's equation
According to Bernoulli's equation for horizontal flow:
\[ P1 + \frac{1}{2} \rho V1^2 = P2 + \frac{1}{2} \rho V2^2 \]

Rearranging the equation to solve for P2:
\[ P2 = P1 + \frac{1}{2} \rho V1^2 - \frac{1}{2} \rho V2^2 \]

### Step 4: Substitute values into the equation
We know:
- \( \rho \) (density of water) = 1000 kg/m³
- \( V1 = 0.35 \, \text{m/s} \)
- \( V2 = 0.65 \, \text{m/s} \)

Now substituting the values:
\[ P2 = 1360 \, \text{Pa} + \frac{1}{2} \cdot 1000 \cdot (0.35^2) - \frac{1}{2} \cdot 1000 \cdot (0.65^2) \]

Calculating the kinetic energy terms:
\[ \frac{1}{2} \cdot 1000 \cdot (0.35^2) = 0.5 \cdot 1000 \cdot 0.1225 = 61.25 \, \text{Pa} \]
\[ \frac{1}{2} \cdot 1000 \cdot (0.65^2) = 0.5 \cdot 1000 \cdot 0.4225 = 211.25 \, \text{Pa} \]

### Step 5: Calculate P2
Now substituting these values back into the equation:
\[ P2 = 1360 + 61.25 - 211.25 \]
\[ P2 = 1360 - 150 = 1210 \, \text{Pa} \]

### Step 6: Convert P2 back to cm of Hg
To convert from Pascals back to cm of Hg:
\[ P2 = \frac{1210}{13600} \, \text{cm of Hg} \]
\[ P2 \approx 0.089 \, \text{cm of Hg} \]

### Final Answer
Thus, the pressure at the point where the velocity is 65 cm/s is approximately **0.089 cm of Hg**.

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