Question

If 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

pressure at depth h under the surface is p = ρgh + 1 atm 

Let the atmospheric pressure = 101300 N/m^2 
ρ = density = 1000 kg/m^3 
g = grav. accel. = 9.81 m/s^2 
h = depth 

Pressure at depth h = (ρgh + 101 300) N/m^2 
pressure at depth h/2 = ρg(h/2) + 101 300 = 1/2*ρgh + 101 300 

1/2*ρgh + 101 300 = 2/3(ρgh + 101 300) 
1/2*ρgh + 101 300 = 2/3*ρgh + 2/3*101 300 
1/3* 101 300 = 1/6*ρgh 
202 300 = 1000*9.81h 
h = 202 300/(1000*9.81) 
h = 20.65 m <--- ans. 

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