91Ó°ÊÓ

1. Surface area is$A=77{\mathrm{m}}^2$
2. Applied force is$F=480\mathrm{N}$
3. Pressure outside the container = atmospheric pressure is${\mathrm{ÒÏ}}_0=1.0×{10}^{5}Pa$

A fluid is a substance that can flow; it conforms to the boundaries of its container because it cannot withstand shearing stress. It can, however, exert a force perpendicular to its surface. If theforce is uniform over a flat area, then the force is described in terms of pressure p as,

$\mathit{p}\mathbf{=}\frac{\mathbf{F}}{\mathbf{A}}$

In the given problem, the magnitude of the force$\left(F\right)$required to open the lid of the container is equal to the difference in the outer pressure$\left(P_0\right)$and inner pressure$\left(P_i\right)$multiplied by an area$\left(A\right)$of the lid.

Using the relation between internal pressure, external pressure, and force, we can write an expression internal pressure.Then, we can find the external pressure from the applied force and the given surface area.

Formulae:

The pressure at a point in a fluid at static equilibrium,

$F=\left(p_0-p_i\right)A$ …(¾±)

Using equations (i), we get

$$\begin{gathered} p_o-p_i=\frac{F}{A} \\ {p}_i=p_0-\frac{F}{A} \\ =1.0×{10}^{5}Pa-\frac{480N}{77m^2}.........................\left(\mathrm{Substituting}\mathrm{the}\mathrm{given}\mathrm{values}\right) \\ =9.99×{10}^4\mathrm{Pa} \end{gathered}$$

Therefore, the internal air pressure of container is$9.99×{10}^4\mathrm{Pa}$ .