If the velocity is zero over half of the cross-sectional area and is uniform over the remaining half, then the momentum correction factor is A. 1 B. $$\frac{4}{3}$$ C. 2 D. 4

The correct answer is $\frac{4}{3}$.

The momentum correction factor is a dimensionless quantity that is used to account for the fact that the velocity is not uniform across the cross-sectional area of a pipe. The momentum correction factor is defined as follows:

$$f = \frac{2}{3} \left( 1 + \frac{2}{3} \frac{A_1}{A} \right)$$

where $A_1$ is the area of the half of the cross-sectional area where the velocity is zero and $A$ is the total cross-sectional area of the pipe.

In this case, the velocity is zero over half of the cross-sectional area and is uniform over the remaining half. Therefore, $A_1 = \frac{1}{2} A$ and $A = 2 A_1$. Substituting these values into the equation for the momentum correction factor, we get:

$$f = \frac{2}{3}

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