When a submarine dives to a depth of 5.0 × 102 m, how much pressure, in pa, must its hull be able to withstand? how many times larger is this pressure than the pressure at the surface?

Depth = 5.0 Ă— 10^2 m
 Density of sea water = 1.025 x 10^3
 Pd = Po + pgh
Atmospheric pressure is standard Patm = 1.01325 x 10^5 Pa
  Since the normal pressure is retained in the hull, no need to bother about Po Pd = pgh = 1.025 x 10^3 x 9.8 x 5.0 x 10^2 = 50.225 x 10^5
 So now Pd / Patm = 50.225 x 10^5 / 1.01325 x 10^5 = 49.56
 So it is 49.56 times larger.

Pressure is force over surface.
 p=F/S.
 In this case the force is given by the weight of the water G, so
 p = G/S; 
 G=m*g;
 m=ρ*V,
 V=S*h
 since we are interested only in the volume of water which presses on the surface of the submarine S, at depth h.
 So:
 p= ρ*S*h*g/S = ρ*g*h.
 This is the pressure at a certain depth h, exerted only by the water so we must add the atmospheric pressure p0.
 Therefore the total pressure is pt = p0 + ρ*g*h.
 We can see that when h=0, pt=p0 which is correct since at the surface we have only atmospheric pressure.
 The ratio pt/p0 = 1+ ρ*g*h/p0
 p0=101325 Pa 
 h = 500 m 
 g= 9.8 m/s^2 
 ρ = 1.025*10^3 kg/m^3. 
 We substitute in the equation and get pt/p0 = 50.56 .
 So the pressure at 500m depth is 50.56 times greater than at the surface.