91Ó°ÊÓ

The given data can be listed below,

• The mass of Europa is,$\mathrm{m}=4.80×{10}^{22}\text{ k²µ}$
• The diameter of Europa is,$d=3120\text{ k³¾}$ .
• The area of the submarine is,$A=625{\text{ c³¾}}^2$ .
• The inward force on submarine is, $F_{âŠ\yen }=8750\text{ N/·É¾±²Ô»å´Ç·É}$.

Gravity acceleration is the acceleration created in a falling body due to gravity. The mass of the falling object has nothing to do with it.It is contingent on the planet's respective mass and radius.

The radius of Europa is given as,

$\begin{array}{c}r=\frac{3120}{2}\text{ k³¾}\\ =1560\text{ k³¾}\end{array}$

The acceleration due to gravity is given by,

$\mathrm{g}=\frac{\mathrm{Gm}}{{\mathrm{R}}^2}$

Here, G is the constant of gravitation whose value is $6.673×{10}^{−11}\text{ N}â‹\dots {\text{m}}^2/{\text{kg}}^2$, m is the mass of Europa, Ris the radius of Europa.

Substitute values in the above equation.

$\begin{array}{c}g=\frac{(6.673×{10}^{−11}\text{ N}â‹\dots {\text{m}}^2/{\text{kg}}^2)(4.80×{10}^{22}\text{ k²µ})}{{\left(1560\text{ k³¾}\right)}^2}{\left(\frac{1\text{ k³¾}}{1000\text{″¾}}\right)}^2\left(\frac{1\text{ k²µ}â‹\dots {\text{m/s}}^2}{1\text{ N}}\right)\\ =1.315{\text{″¾/s}}^2\end{array}$

The pressure due to inward force on windows is given by,

$\mathrm{p}=\frac{{\mathrm{F}}_{âŠ\yen }}{\mathrm{A}}$

Here, $F_{âŠ\yen }$ is the inward force on the windows of submarine, A is the area of the submarine.

Substitute values in the above equation.

$\begin{array}{c}p=\frac{8750\text{ N/·É¾±²Ô»å´Ç·É}}{625{\text{ c³¾}}^2}{\left(\frac{100\text{ c³¾}}{1\text{″¾}}\right)}^2\left(\frac{1\text{ P²¹}}{1{\text{ N/³¾}}^2}\right)\\ =1.40×{10}^5\text{ P²¹}\end{array}$

The pressure at depth h is given by,

$\begin{array}{c}p=\mathrm{ÒÏ}gh\\ h=\left(\frac{p}{\mathrm{ÒÏ}g}\right)\end{array}$

Here,$\mathrm{ÒÏ}$is the density of the water whose value is $1.00×{10}^3{\text{ k²µ/m}}^3$, gis the acceleration due to gravity on Europa, h is the depth of the submarine.

Substitute all the values in the above equation.

$\begin{array}{c}h=\left(\frac{1.40×{10}^5\text{ P²¹}}{(1.00×{10}^3{\text{ k²µ/m}}^3)\left(1.315{\text{″¾/s}}^2\right)}\right)\left(\frac{\text{1 k²µ}â‹\dots {\text{m/s}}^2}{1\text{ N}}\right)\left(\frac{1{\text{ N/³¾}}^2}{1\text{ P²¹}}\right)\\ =106\text{″¾}\end{array}$

Thus, the depth at which the submarine is safely dive is $106\text{″¾}$.