Question

At a depth of $1000m$ in an ocean, find out the force acting on the window of an area $20cm\times 20cm$ of a submarine at this depth, the interior of which is maintained at sea-level atmospheric pressure. Given that the density of seawater is $1.03\times {{10}^{3}}kg{{m}^{-3}}$ and acceleration due to gravity is $g=10m{{s}^{-2}}$.
$\begin{align}
  & A.3.2\times {{10}^{8}}N \\
 & B.4.12\times {{10}^{5}}N \\
 & C.8.3\times {{10}^{2}}N \\
 & D.3.1\times {{10}^{-5}}N \\
\end{align}$

Answer 1

Hint: The pressure outside the submarine will be the sum of pressure inside the submarine and the product of density, acceleration due to gravity and depth of the submarine in the ocean. The pressure difference on the window should be calculated first and then calculate the force on it. The force on the window will be the product of pressure difference and the area of the window. These all may help you to solve this question.
Formula used:
The pressure at the outer surface is,
${{P}_{o}}={{P}_{i}}+\rho gh$

Complete answer:
The pressure at the inner surface of the submarine is given as ${{P}_{i}}$,
The pressure at the outer surface can be written as,
${{P}_{o}}={{P}_{i}}+\rho gh$
From this we can find the pressure difference between the inner and outer surface on the window of the submarine,
$\Delta P={{P}_{o}}-{{P}_{i}}=\rho gh$
In the question it is already mentioned that,
The density of seawater is,
$\rho =1.03\times {{10}^{3}}$
The acceleration due to gravity is
$g=10m{{s}^{-2}}$
And the depth of the submarine inside the ocean is,
$h=1000m$
Substituting this values in the equation will give,
$\begin{align}
  & \Delta P={{P}_{o}}-{{P}_{i}}=1.013\times {{10}^{5}}\times 10\times 1000 \\
 & \Delta P=103\times {{10}^{5}}Pa \\
\end{align}$
Now let us find out the force acting on the window,
It can be written in the form of mathematical equation,
$F=\Delta P\times A$
Where $A$be the area of the window which is mentioned in question as,
$A=0.04{{m}^{2}}$
Substituting this in the equation will give,
\[\begin{align}
  & F=103\times {{10}^{5}}Pa\times 0.04 \\
 & F=4.12\times {{10}^{5}}N \\
\end{align}\]

Therefore the correct answer is option B.

Note: There are two types of pressure, absolute pressure and gauge pressure. The absolute pressure is defined by its usage of absolute zero as its zero point. The gauge pressure is explained by the usage of atmospheric pressure as its zero point. Gauge pressure measurement is not always precise because of the varying atmospheric pressure, while absolute pressure is always definite.