91Ó°ÊÓ

1. Height of aquarium, H=5.00m
2. Depth of aquarium, h=2.0m
3. Thickness of the plastic, w=8.00m

Using the formula of pressure in terms of force per unit area and the formula for pressure in terms of density, gravitational acceleration, and depth, we can find the increase in the total force on the wall.

Formulae:

Force applied on a body in terms of pressure, F = pA (i)

Pressure applied on a fluid surface, p =$\mathrm{ÒÏ}gh$ (ii)

Differentiating equation (i) with respect to area A, we get

dF = pdA

$$\begin{gathered} ∫dF=∫pdA\left(âˆ\mu \mathrm{To}\mathrm{get}\mathrm{force}\mathrm{in}\mathrm{that}\mathrm{given}\mathrm{area},\mathrm{integrating}\mathrm{both}\mathrm{sides}\right) \\ F=∫pdA...................\left(1\right) \end{gathered}$$

The force at depth, h=2.0cm,as

$$\begin{gathered} dA=w×dh, \\ P=\mathrm{ÒÏ}gh \end{gathered}$$,

Substituting the above values in the equation (1), we get

$$\begin{gathered} F=∫\mathrm{ÒÏ}dh\left(w×dh\right) \\ =\mathrm{ÒÏ}gw∫\left(hdh\right) \\ =\mathrm{ÒÏ}gw\left[\frac{h^2}{2}\right] \\ =\left(1000\frac{kg}{m^3}\right)\left(9.8\frac{m}{s^2}\right)\left(8\right)\left[\frac{2^2}{2}\right] \\ =1.5×{10}^{5}N..............\left(2\right) \end{gathered}$$

Similarly, the force at depth h=4.0cm,

localid="1657253614722" $$\begin{gathered} F\text{'}=\mathrm{ÒÏ}gw\left[\frac{h^2}{2}\right] \\ F\text{'}\left(1000\frac{kg}{m^3}\right)\left(9.8\frac{m}{s^2}\right)\left(8\right)\left[\frac{4^2}{2}\right] \\ =6.272×{10}^{5}N....................\left(3\right) \end{gathered}$$

Therefore, the total force using equations (2) and (3) can be given as:

$$\begin{gathered} ∆F=F\text{'}-F \\ =\left(6.272×{10}^{5}N\right)-\left(1.5×{10}^{5}N\right) \\ =4.69×{10}^{5}N \end{gathered}$$

Therefore, the increase in the total force on the wall, if the aquarium is filled to a depth of$2.00m$, is$4.69×{10}^{5}N$