Question

If the pressure at the half depth of lake is equal to 3 by 4 x to the pressure at its bottom then find the depth of the lake

Answer 1

here is your process
3P/4 =P° +density of liquid ×gravity ×depth/2..(1)
P =P° +density of liquid ×gravity×depth....(2)
where P°=atmospheric pressure
substitute (2) in (1)
then Height = P°/density of liquid×gravity
hope it is useful
ask if uou have problem

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