Question

A research submarine has a 10-cm-diameter window that is 8.4 cm thick. The manufacturer says the window can withstand forces up to 1.2×106 N . What is the submarine's maximum safe depth in salt water?
The pressure inside the submarine is maintained at 1.0 atm .
Express your answer using two significant figures.

Answer 1

Pnet = Po + dgh 

Density of saltwater = 1030 kg/m^3. 

Disregard the thickness. Assuming it's a circular window, then the area is pi(r^2). 

d = 20 cm = 0.2 m 
r = d/2 = 0.1 m 

A = pi(r^2) 
A = 3.14159265(.1^2) 
A = 0.0314159265 m^2 

p = F/A 
p = (1.1 x 10^6) / (0.0314159265) 
p = 35,014,087.5 Pa 

1 atm = 101,325 Pa 

P = Po + dgh 
h = (P - Po) / dg 
h = (35,014,087.5 - 101,325) / (1030 x 9.81) 
h = 3 455.23812 m 
h = 3.5 km

Answer 2

Answer: 15 km

Explanation:

The maximum pressure that the submarine can withheld would be equal to the maximum pressure that the window can withstand.

The maximum pressure that window can sustain is given by:

$Pressure\frac{force}{Area]$

Force = 1.2×10⁶ N

Area of the window = $\pi r^{2}= 7.85\times 10^{-5} m^2$

∵radius = (0.1 m)/2 = 0.05 m

⇒P = 1.2×10⁶ N ÷ 7.85 ×10⁻³ m² = 1.52 ×10⁸ N/m²

The pressure of the salt water at depth h is given by

P = ρ g h

where, ρ = 1027 kg/m³ is the density of the salt water, g is the acceleration due to gravity and h is the depth.

⇒1.52×10⁸ Pa =  (1027 kg/m³)(9.8 m/s²)(h)

⇒ h ≈ 15 km