Question

The maximum depth d_max that a diver can snorkel is set by the density of the water and the fact that human lungs can function against a maximum pressure difference (between inside and outside the chest cavity) of 0.050 atm. What is the difference in d_max for fresh water and the water of the Dead Sea (the saltiest natural water in the world, with a density of 1.5 × 10³kg/m³)?

Answer 1

Answer:

0.17214 m

Explanation:

P = Pressure difference between inside and outside the chest cavity = 0.05 atm

$\rho$ = Density of fresh water = 1000 kg/m³

$\rho_d$ = Density of water in the Dead Sea= 1500 kg/m³

h = Height

In fresh water

$P=\rho gh_1\\\Rightarrow h_1=\dfrac{P}{\rho g}\\\Rightarrow h_1=\dfrac{0.05\times 101325}{1000\times 9.81}\\\Rightarrow h_1=0.51643\ m$

In dead sea

$P=\rho_d gh_2\\\Rightarrow h_2=\dfrac{P}{\rho_d g}\\\Rightarrow h_2=\dfrac{0.05\times 101325}{1500\times 9.81}\\\Rightarrow h_2=0.34429\ m$

The difference in height is 0.51643-0.34429 = 0.17214 m

Answer 2

Answer:

17.18 m

Explanation:

density of fresh water = 1000 kg /m^3

density of dead sea = 1500 kg/m^3

Pressure, P = 5 atm = 5 x 1.01 x 10^5 = 5.05 x 10^5 Pa

Let h be the depth in fresh water and h' be the depth in sea water

P = h x density of fresh water x g

5.05 x 10^5 = h x 1000 x 9.8

h = 51.53 m

Now for dead sea

P = h' x density of dead sea x g

5.05 x 10^5 = h' x 1500 x 9.8

h' = 34.35 m

So, the difference in height

d = h - h'

d = 51.53 - 34.35

h' = 17.18 m