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Q10:- A container is filled up to height of 5 m with a liquid density 4000 kg/m^3 and above that there is a liquid
of density 2000 kg/m^3.Find the pressure at the bottom (g = 10 m/s^2,atmospheric pressure 10^5 Pa.​

Answer 1

Answer:

hp(g+a)

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Answer 2

Therefore the pressure exerted at the bottom is 4 × 10⁵ Pa or 4 bar.

Given:

Let the fluid at the top of the container be fluid-1:

The density of the first fluid = ρ₁ = 2000 kg/m³

Height of the first fluid = h₁ = 5 m

Let the fluid at the bottom of the container be fluid-2:

The density of the second fluid = ρ₂ = 4000 kg/m³

Height of the second fluid = h₂ = 5 m

Acceleration due to gravity = g = 10 m/s²

Atmospheric pressure = 10⁵ Pa

To Find:

The pressure is exerted at the bottom.

Solution:

The given question can be solved as shown below.

The pressure exerted by the first fluid:

⇒ P₁ = ρ₁gh₁ = 2000 × 10 × 5 = 10⁵ Pa

The pressure exerted by the second fluid:

⇒ P₂ = ρ₂gh₂ = 4000 × 10 × 5 = 2 × 10⁵ Pa

Total pressure at the bottom = Atmospheric pressure + Pressure exerted by fluid-1 + Pressure exerted by fluid-2

⇒ Total pressure at the bottom = P = 10⁵ + 10⁵ + 2 × 10⁵

⇒ Total pressure at the bottom = P = 10⁵ (1 + 1 + 2)

⇒ Total pressure at the bottom = P = 10⁵ × 4

Therefore the pressure exerted at the bottom is 4 × 10⁵ Pa or 4 bar.

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