A water bed is 2.00m on a side and 30.0 cm deep. Find the pressure that the water bed exerts on the floor. Assume that the entire lower surface of the bed makes contact with the floor. Density of water, $\rho = 1000\dfrac{{kg}}{{{m^3}}}$;$g = 10\dfrac{m}{{{s^2}}}$.
A) 2000 Pa.
B) 3000 Pa.
C) 3500 Pa.
D) 4000 Pa.

Hint: The pressure is defined as the ratio of perpendicular force on the area to the area of cross section. The pressure on the lowest point is equal to the product of density, acceleration due to gravity and height from the top surface. The pressure increases with increase in height.

Formula used:
The formula of the pressure in the liquid at the bottom surface is given by,
$ \Rightarrow P = \rho gh$
Where pressure is P, the height is h and the acceleration due to gravity.
Step by step solution:
In the problem the pressure at the bottom of the surface of the bed is asked, if the side is 2 m and 3 cm is the depth of the water bed.
The pressure on the bottom of the liquid is due to the liquid which is filled above the bottom surface.
The formula of the pressure in the liquid at the bottom surface is given by,
$ \Rightarrow P = \rho gh$
Where pressure is P, the height is h and the acceleration due to gravity.
The value of$g = 10\dfrac{m}{{{s^2}}}$,$\rho = 1000\dfrac{{kg}}{{{m^3}}}$,$h = 0 \cdot 3m$.
The pressure on the bed is equal to,
$ \Rightarrow P = \rho gh$
Converting the height from cm to m.
$ \Rightarrow P = 1000 \times 10 \times 0 \cdot 3$
$ \Rightarrow P = 3000\dfrac{N}{{{m^2}}}$
The unit $\dfrac{N}{{{m^2}}}$ is also known as Pascal. The pressure on the water bed is equal to,
$ \Rightarrow P = 3000\dfrac{N}{{{m^2}}}$
$ \Rightarrow P = 3000Pa$.
The pressure on the bed surface is equal to $P = 3000Pa$.

The correct answer for this problem is option B.

Note: The students are advised to understand and remember the formula of the pressure of any liquid at some depth. The units of the physical quantities should be converted and all the physical quantities should be similar. The unit Pascal is equal to $\dfrac{N}{{{m^2}}}$.