Question

if pressure at half the depth of a lake is 2/3 pressure at the bottom of the lake, then waht is the depth of the lake?

Answer 1

20.67 m

h = depth of the lake
ρ = density of water in the lake = 1000 kg/m³
g = gravity = 9.8 m/s²

Pressure at half depth of the lake = P1 = ρ g h/2 + 1 atm
Pressure at the bottom of the lake = P2 = ρ g h + 1 atm
    (Atmospheric pressure = 1 atm = 1.013 * 10⁵ Pa)

   Given   P1 = P2 * 2/3
               ρgh/2 + 1.013 *10⁵  = [ ρ g h + 1.013 * 10⁵] 2 /3
               ρ g h * 1/6 = 1.013 * 10⁵ * 1/3

               h = 6 * 1.013 * 10⁵/ [ 3 * 1000 * 9.8 ] m
                  = 20.67 m
Answer... 20.67 m

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