The correct answer is A. 1.084.
The discharge over a trapezoidal crest is greater than the discharge over a rectangular crest of identical parameters. This is because the trapezoidal crest has a larger cross-sectional area, which allows for more water to flow through it.
The discharge over a trapezoidal crest can be calculated using the following equation:
$Q = \frac{1}{2}bh\sqrt{2gh}$
where:
$Q$ is the discharge,
$b$ is the width of the base of the trapezoid,
$h$ is the height of the trapezoid,
$g$ is the acceleration due to gravity, and
$h$ is the head of water.
The discharge over a rectangular crest can be calculated using the following equation:
$Q = bwh$
where:
$Q$ is the discharge,
$b$ is the width of the rectangle,
$h$ is the height of the rectangle,
and
$h$ is the head of water.
Comparing the two equations, we can see that the discharge over a trapezoidal crest is greater than the discharge over a rectangular crest for the same values of $b$, $h$, and $g$. This is because the trapezoidal crest has a larger cross-sectional area.
The ratio of the discharge over a trapezoidal crest to a rectangular crest of identical parameters is given by the following equation:
$\frac{Q_{trapezoid}}{Q_{rectangle}} = \frac{1}{2} \left( \frac{b_{trapezoid} + b_{rectangle}}{b_{rectangle}} \right) \sqrt{2gh}$
Substituting in the values of $b_{trapezoid} = 1$ and $b_{rectangle} = 1$, we get:
$\frac{Q_{trapezoid}}{Q_{rectangle}} = \frac{1}{2} \left( \frac{1 + 1}{1} \right) \sqrt{2gh} = 1.084$