Question

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

Answer 1

Explanation:

Given If pressure at half death of a lake is equal to 3 by 4 times the pressure at its bottom then find the depth of the lake

• Let depth of the lake be h and pressure at bottom = P
• Then P = Po + ρgh ------------1
• where Po is atmospheric pressure, ρ = density of water)
• At half depth (h/2) pressure is 3P/4  
• So 3P/4 = Po + ρg h/2--------------2
• Subtracting equation 2 from 1 we get
• P / 4 = ρg h/2
• So we get P = 2ρ g h
• Substituting the value of P we get
• 2 ρ g h = Po + ρ g h
• Or 2 ρ g h – ρ g h = Po
• Or Po = ρ g h
• Or h = Po / ρ g is the depth of the lake
• Or h = 10^5 / 10^4
• Or h = 10 m

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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