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

We know that the pressure exerted by a liquid is written as

P = hρg

now, let the pressure at the half depth of the lake be P1 = P0 + h1ρg

the pressure at the bottom of the lake is P2 = P0 + h2ρg

here, P0 is the atmospheric pressure

so, as given

P1 = (2/3)P2

or

P0 + h1ρg = (2/3)(P0 + h2ρg)

and h1 = h2/2

so,

(1/3)P0 = (1/6)h2ρg

solving further, we get

h2 = 2P0 / ρg

or h2 = (2 X 105) / 104

or the depth of the lake is

h2 = 20m



And when the pressure is 3/4

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

Answer 2

Let the pressure due to atmosphere be p'

The pressure at the depth h is p1=p'+hdg

The pressure at half the depth of the lake is p2= p'+(h/2)dg

According to the question p1=3/4 p2

p'+hdg=3/4(p'+h/2dg)

p'+h/2dg=3/4p'+3/4hdg

p'-3/4p'=(3/4-1/2)hdg

p'/4=(1/4)hdg

h=p'/(dg)=10mts