Question

1 - A submarine has a circular porthole with a radius of 0.10 m. If a force of more than 2x106 N is applied, the porthole will fail. What is the maximum pressure that this window can support? To what maximum depth can the submarine dive, assuming that the interior is maintained at atmospheric pressure?

Answer 1

In order to solve this problem, it is necessary to apply the concepts related to the Pressure according to the Force and the Area as well as to the pressure depending on the density, gravity and height.

In the first instance we know that the pressure can be defined as

$P = \frac{F}{A}$

Where

F= Force

A =  Mass

In the second instance the pressure can also be defined as

$P = \rho gh$

Where,

$\rho=$Density of Fluid at this case Water

g = Gravitational Acceleration

h = Height

If we develop the problem to find the pressure then,

$P = \frac{F}{A}$

$P = \frac{F}{\pi r^2}$

$P = \frac{2*10^6}{\pi (0.1)^2}$

$P = 6.37*10^{7} Pa$

Through the second equation we can find the depth to which it can be submerged,

$P = \rho gh$

Re-arrange to find h

$h = \frac{P}{\rho g}$

$h = \frac{6.37*10^{7}}{1000*9.8}$

$h = 6499.4m$