Question

IfIf pressure at half the depth of a lake is equal to 3/4 pressure at the bottom of the lake then what is the depth of the lake ?

Answer 1

Hey.

Here is your answer.

Depth = d.
Atmospheric Pressure = P₀ = 10⁵ Pa.

Density of water at 4°C (ρ) = 10³ kg/m³

Pressure at half of depth
P = P₀ + (d/2)ρg
P = P₀ + 0.5dρg

Pressure at the bottom of lake
P’ = P₀ + dρg

Given that P = 3P’/4
P₀ + 0.5dρg = 3(P₀ + dρg)/4
P₀ + 0.5dρg = 3P₀/4 + 3dρg/4
P₀/4= dρg/4
d = P₀ / (ρg)
d = ( 10⁵ Pa) / (10³ kg/m³ × 10 m/s²)
d = 10 m

Depth of lake is 10 m

Thanks

Answer 2

Answer:

Let depth of the lake be h and pressure at bottom =P

Then P=Pa +ρgh →(1)    (Pa  = atmospheric pressure, ρ = density of water)

At half depth (h/2) pressure is 3P /4 then :

3P /4 = Pa + ρgh /2 →(2)

On subtracting equation 2 from 1 we get :

P /4 = ρg h/2

⇒P=2ρgh, substituting this value of P in equation 1:

2ρgh = Pa  + ρgh

⇒h=Pa/ ρg   → Depth of the lake

⇒h = 10^5 / 10^3 * 10 ( Pa = 10^5 pascal , ρ = 10^3 , g = 10m/s^2)

⇒h = 10 m