Question

At a depth of 10.9 km, the Challenger Deep in the Marianas Trench of the Pacific Ocean is the deepest site in any ocean. Yet, in 1960, Donald Walsh and Jacques Piccard reached the Challenger Deep in the bathyscaphe Trieste. Assuming that seawater has a uniform density of 1024 \[kg/{m^3}\], approximate the hydrostatic pressure (in atmospheres) that the Trieste had to withstand. (Even a slight defect in the Trieste structure would have been disastrous.)

Answer 1

Hint: We are required to find the total pressure exerted on him because that is the amount of pressure he had to withstand. The total pressure will be the sum of that exerted by water as well as atmosphere. Pressure exerted by the atmosphere has a constant value i.e. \[1.01 \times {10^5}pa\] but that exerted by water which is a fluid depends upon other physical quantities.

Complete Step by step answer:
It is given that the density of water $\left( \rho \right)$ is 1024 kg/m³, and the height (h) he covered (depth) is 10.9 km.
The pressure that he had to withstand is the pressure that is exerted by the water as well as the atmosphere. So the total pressure is given as:
\[{P_{total}} = {P_{atm}} + {P_{hydro}}\] ____ (1)
As water is a fluid, the pressure exerted by it will be equal to the fluid pressure, so it can be given as:
\[{P_{hydro}} = \rho gh\] and here,
Density $\left( \rho \right)$ = 1024 kg/m³ (given)
Height (h) = 10.9 km or \[\left( {10.9 \times {{10}^3}} \right)m\] [As 1 km = 1000 m]
Acceleration due to gravity (g) = \[9.8m/{s^2}\]
$ \Rightarrow {P_{hydro}} = 1024 \times 9.8 \times \left( {10.9 \times {{10}^3}} \right)$
And the known value of atmospheric pressure is :
\[{P_{atm}} = 1.01 \times {10^5}pa\]

Substituting the values in (1), we get:
\[
  {P_{total}} = 1.01 \times {10^5}pa + 1024 \times 9.8 \times \left( {10.9 \times {{10}^3}} \right) \\
 \Rightarrow {P_{total}} = {10^3}\left( {1.01 \times {{10}^2} + 1024 \times 9.8 \times 10.9} \right) \\
 \Rightarrow {P_{total}} = {10^3} \times 109596.296 \\
\therefore {P_{total}} = 1.09 \times {10^8} \\
 \]
The SI unit of pressure is Pascal (pa) so:
\[{P_{total}} = 1.09 \times {10^8}pa\]

Therefore, the approximate the hydrostatic pressure (in atmospheres) that the Trieste had to withstand was \[1.09 \times {10^8}pa\]

Note: As the density was given in \[kg/{m^3}\], we could not take the height in km and thus converted the same to m so as to keep units like.
For calculations, remember that for the same base number, the powers will be added when multiplied and subtracted when divided.Any substance possessing the property to flow and has viscosity (thickness) is called as fluid, it can be liquid, gas or even plasma Always answer the question with appropriate SI Units, SI Unit of pressure is pascal (pa)