The pressure at the depth h in the ocean is given by (Stevin's law)
[tex]p= p_0 + \rho g h[/tex]
where
[tex]p_0 = 1.0 \cdot 10^5 Pa[/tex] is the atmospheric pressure
and [tex]\rho g h[/tex] is the pressure exerted by the column of water of height h=4267 m, with [tex]\rho = 1000 kg/m^3[/tex] being the water density and [tex]g=9.81 m/s^2[/tex].
Substituting, we find
[tex]p=1.0 \cdot 10^5 Pa + (1000 kg/m^3)(9.81 m/s^2)(4267 m)=4.20 \cdot 10^7 Pa[/tex]
We want to convert this into atmospheres: we know that 1 atm corresponds to the atmospheric pressure at sea level, so [tex]1 atm=1.0\cdot 10^5 Pa[/tex], therefore we just need to divide by this number:
[tex]p= \frac{4.20 \cdot 10^7 Pa}{1.0 \cdot 10^5 Pa/atm} =420 atm[/tex]
