91Ó°ÊÓ

The depth is h = 250 m.

The diameter of glass window is d = 30 cm.

The net force on the circular glass window can be calculated by using the relationship between force, pressure and cross-sectional area.

The relation of gauge pressure can be written as:

P = pgh

Here, gis the gravitational acceleration and p is the density of seawater.

Substitute 250m for h,$1030\mathrm{kg}/{\mathrm{m}}^3$for p, and$9.80\mathrm{m}/{\mathrm{s}}^2$for g in the above relation.

$$\begin{gathered} P=(1030\mathrm{kg}/{\mathrm{m}}^3)(9.80\mathrm{m}/{\mathrm{s}}^2)\left(250\mathrm{m}\right) \\ P=2.5×{10}^6\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

Thus, the required pressure is $2.5×{10}^6\mathrm{N}/{\mathrm{m}}^2$.

The relation of net force can be written as:

$$\begin{gathered} F=PA \\ F=P\left(\frac{{\mathrm{Ï€»å}}^2}{4}\right) \end{gathered}$$

Here, Ais the cross-sectional area.

Substitute $2.5×{10}^6\mathrm{N}/{\mathrm{m}}^2$ for P, and 30 cm for d in the above relation.

$$\begin{gathered} F=(2.5×{10}^6\mathrm{N}/{\mathrm{m}}^2)\left(\frac{\mathrm{π}\left(30\mathrm{cm}{\displaystyle \frac{1\mathrm{m}}{100\mathrm{xcm}}}\right)}{4}\right) \\ \mathrm{F}=1.7×{10}^5\mathrm{N} \end{gathered}$$

Thus, the net force is $1.7×{10}^5\mathrm{N}$.