Water stands upto a height $h$ behind the vertical wall of a dam. What is the net horizontal force pushing the dam down by the stream, if the width of the dam is $\sigma $ ? ($\rho $ is the density of water).

Hint As we have to find the horizontal force and the force is equal to the pressure multiplied by area. So simply we just have to find the pressure and the area using the given values and then we can find the force by multiplying them.

Complete step by step solution
A force is defined as a push or pull upon an object resulting from the object's interaction of one object with another. When there is an interaction between two objects, there is a force upon each of the objects. On the other hand when the interaction ceases, the two objects no longer experience the force.
Here given :
Height of the water level is $h$
And
Width of the dam be $\sigma $
Given density of the water as $\rho $
Now as we know, exerting a force on an object changes the object's speed, direction of movement or shape (in some cases).And Pressure is a measure of how much force is acting upon an area. Pressure can be found using the equation:
$pressure = \dfrac{{force}}{{area}}$
Which gives us
$force = pressure \times area$
Hence more is the pressure or area, the more will be the force acting.
Now as $area = height \times width = h\sigma $
And we know $pressure = \rho g \times height$
Here height would be taken from centre of mass
So desired pressure becomes
$pressure = \rho g\dfrac{h}{2}$
Which gives us
$Force = pressure \times area = \rho g\dfrac{h}{2} \times h\sigma = \dfrac{{\rho g\sigma {h^2}}}{2}$
Hence desired answer is

$Force = \dfrac{{\rho g\sigma {h^2}}}{2}$

Note While finding the pressure do not forget to use height carefully. And remember if force is acting over a smaller area then it will create more pressure as both are inversely proportional to each other.