A closed tank with length 10m, breadth 8m and depth 6m is filled with water to the top. If $g = 10m{s^{ - 2}}$ and density of water is $1000kg{m^{ - 2}}$, then the thrust on the bottom is:

Hint:We can calculate the hydrostatic pressure of the liquid in a tank as the force per area for the area of the bottom of the tank as given by pressure = force/area units. In this case, the force would be the weight the liquid exerts on the bottom of the tank due to gravity.

Formula Used:The pressure of a liquid at a certain height is given by the mathematical expression given below:
\[P = {P_0} + \rho gh\]
In this mathematical expression, ${P_0}$ is the pressure at the surface of the liquid.
$\rho $ is the density of the liquid.
$h$ is the depth of the point from the surface of water.

Complete step by step solution:
Now, as in the numerical problem it is given that the tank is closed, thus the pressure at the surface of the liquid is equal to zero (That is ${P_0} = 0$).
Thus, putting this in the mathematical expression for pressure give above, we get:
$P = 1000 \times 6 \times 10$
Now, we know that area of the rectangular tank is the product of the length and breadth of the tank.
Thus, Area$ = 8 \times 10 = 80{m^2}$
Now, the thrust at the bottom of the rectangular tank can be mathematically defined as the product of the pressure of the tank and the area of the tank. Thus we can write:
$F = P \times A = (1000 \times 6 \times 10 \times 80)N$
This is the expression for the thrust at the bottom of the tank.

Note:This gives you a rough way of determining the forces between particles for the liquid in the tank, but it assumes that the force due to gravity is an accurate measure of the force between particles that causes pressure.