The difference of pressure between two points along a horizontal pipe through which water is flowing is `14 cm` of `Hg`. If due to non-uniform cross section the speed of flow at a point of greater cross section is `60 cm//s`, speed of flow at the other point is

Options

A. `2m//s`
B. less than `60 cm//s`
C. not affected by non-uniform cross section
D. greater than `60 cm//s `

Explanation

To solve the problem, we will use Bernoulli's equation and the concept of fluid flow in a pipe. Here are the steps to find the speed of flow at the other point in the horizontal pipe:

### Step 1: Understand the Given Information
We are given:
- Pressure difference (ΔP) between two points = 14 cm of Hg
- Speed of flow at the point of greater cross-section (V1) = 60 cm/s

### Step 2: Convert Pressure Difference to SI Units
We need to convert the pressure difference from cm of Hg to Pascals (Pa).
1. The density of mercury (ρ_Hg) = 13,600 kg/m³.
2. The acceleration due to gravity (g) = 9.81 m/s².
3. Convert 14 cm of Hg to meters: 14 cm = 0.14 m.

Using the formula for pressure:
\[ \Delta P = \rho_{Hg} \cdot g \cdot h \]
\[ \Delta P = 13600 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2 \cdot 0.14 \, \text{m} \]

Calculating ΔP:
\[ \Delta P = 13600 \cdot 9.81 \cdot 0.14 \]
\[ \Delta P \approx 18850.4 \, \text{Pa} \]

### Step 3: Apply Bernoulli's Equation
For a horizontal pipe, Bernoulli's equation states:
\[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \]

Rearranging gives us:
\[ P_1 - P_2 = \frac{1}{2} \rho V_2^2 - \frac{1}{2} \rho V_1^2 \]
\[ \Delta P = \frac{1}{2} \rho (V_2^2 - V_1^2) \]

### Step 4: Substitute Known Values
We know ΔP and V1. We can express V2 in terms of V1:
\[ 18850.4 = \frac{1}{2} \cdot 1000 \cdot (V_2^2 - (0.6)^2) \]
(Note: The density of water (ρ) = 1000 kg/m³, and V1 = 60 cm/s = 0.6 m/s)

### Step 5: Solve for V2
Rearranging the equation:
\[ 18850.4 = 500 (V_2^2 - 0.36) \]
\[ 37.7 = V_2^2 - 0.36 \]
\[ V_2^2 = 37.7 + 0.36 \]
\[ V_2^2 = 38.06 \]

Taking the square root:
\[ V_2 = \sqrt{38.06} \]
\[ V_2 \approx 6.17 \, \text{m/s} \]

### Step 6: Convert to cm/s
Convert V2 back to cm/s:
\[ V_2 \approx 617 \, \text{cm/s} \]

### Final Answer
The speed of flow at the other point is approximately **617 cm/s**.

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