If D1 and D2 are depths of water upstream and downstream of a hydraulic jump, the loss of head at the jump, is A. $$\frac{{{{\left( {{{\text{D}}_2} – {{\text{D}}_1}} \right)}^3}}}{{{{\text{D}}_1}{{\text{D}}_2}}}$$ B. $$\frac{{{{\left( {{{\text{D}}_2} – {{\text{D}}_1}} \right)}^3}}}{{2{{\text{D}}_1}{{\text{D}}_2}}}$$ C. $$\frac{{{{\left( {{{\text{D}}_2} – {{\text{D}}_1}} \right)}^3}}}{{3{{\text{D}}_1}{{\text{D}}_2}}}$$ D. $$\frac{{{{\left( {{{\text{D}}_2} – {{\text{D}}_1}} \right)}^3}}}{{4{{\text{D}}_1}{{\text{D}}_2}}}$$

The correct answer is $\frac{{{{\left( {{{\text{D}}_2} – {{\text{D}}_1}} \right)}^3}}}{{2{{\text{D}}_1}{{\text{D}}_2}}}$.

A hydraulic jump is a sudden change in the depth of a fluid, typically a liquid, caused by a change in the flow conditions. The jump is characterized by a sudden increase in the water depth and a decrease in the water velocity. The loss of head at the jump is the difference in the water energy between the upstream and downstream of the jump.

The loss of head at a hydraulic jump can be calculated using the following equation:

$$\Delta h = \frac{{{{\left( {{{\text{D}}_2} – {{\text{D}}_1}} \right)}^3}}}{{2{{\text{D}}_1}{{\text{D}}_2}}}$$

where:

• $\Delta h$ is the loss of head at the jump,
• $D_1$ is the depth of water upstream of the jump,
• $D_2$ is the depth of water downstream of the jump.

The equation shows that the loss of head is proportional to the cube of the difference in the water depths. This means that a small change in the water depth can cause a large change in the loss of head.

The loss of head at a hydraulic jump can be reduced by increasing the upstream depth of water. This can be done by using a weir or a dam to raise the water level. The loss of head can also be reduced by decreasing the downstream depth of water. This can be done by using a sluice gate or a diffuser to lower the water level.