A liquid of density `rho` flows along a horizontal pipe of uniform cross-section `A` with a velocity v through a right angled bend as shown in fig. What force has to be exerted at the bend to hold the pipe in equilibrium ? .

Options

A. `2arhov^2`
B. `arhov^2//sqrt2`
C. `sqrt2 arhov^2`
D. `arhov^2`

Answer

To find the force that must be exerted at the bend of a horizontal pipe to hold it in equilibrium, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Situation**:
- We have a liquid flowing through a horizontal pipe with a uniform cross-section and a right-angled bend. The liquid has a density \( \rho \) and flows with a velocity \( v \).

2. **Identify the Change in Momentum**:
- As the liquid flows through the bend, it changes direction. The change in momentum is crucial to determine the force required to keep the pipe in equilibrium.

3. **Calculate the Mass Flow Rate**:
- The mass flow rate \( \frac{dm}{dt} \) can be calculated using the formula:
\[ \frac{dm}{dt} = \rho A v \]
- Here, \( A \) is the cross-sectional area of the pipe, and \( v \) is the velocity of the liquid.

4. **Determine the Force**:
- The force exerted by the liquid due to the change in momentum can be expressed as:
\[ F = \frac{dm}{dt} \cdot v \]
- Substituting the mass flow rate into this equation gives:
\[ F = (\rho A v) \cdot v = \rho A v^2 \]

5. **Consider the Resultant Force at the Bend**:
- At the bend, the forces acting on the pipe can be resolved into two components (one vertical and one horizontal). The resultant force \( R \) at the bend can be calculated as:
\[ R = \sqrt{F^2 + F^2} = \sqrt{2F^2} = \sqrt{2}F \]
- Since both components are equal, we can express this as:
\[ R = \sqrt{2} \cdot (\rho A v^2) \]

6. **Final Expression for the Force**:
- Therefore, the total force that must be exerted at the bend to hold the pipe in equilibrium is:
\[ F = \rho A v^2 \]

### Conclusion:
The force that has to be exerted at the bend to hold the pipe in equilibrium is \( F = \rho A v^2 \).