Question

Derive the relation for the liquid pressure where h=hdg,h= depth the d= density is g=ACC due to gravity

Answer 1

Derivation: P=hdg

              Let us consider a vessel with cross sectional area "A" is filled with a liquid with density "d" , volume V and mass "m" upto a height of "h"

As we know the definition of Pressure :P=F/A

P=mg/A [ since F=mg]

P=vxdxg/A [ since density=mass/ volume]

P=Ahdg/A [ Volume= Ah]

P=hdg where g is the acceleration due to gravity of that place.

This equation is used to calculate the pressure exerted by any fluid at certain depth h.

Answer 2

Hello Dear.


To Prove ⇒ Pressure = hρg

Proof ⇒

Refers to the attachment for the Question.

Consider a Vessel containing the Liquid of density ρ. Let us considers that the liquid is stationary. 
For Calculate the Pressure at the height h, considers the Horizontal circular surface PQ of Area A at depth h below the fee surface XY of the liquid.

Now, The Thrust Exerted on the Surface PQ .
 = Weight of the liquid Column PQRS.
= Volume of the Column PQRS × Density × Acceleration due to gravity.
= (Area of the Base PQ × Depth h) × ρ × g.
= Ahρg.

∵ The thrust is exerted on the surface PQ of the of the Area A.

From the Formula,
Pressure = Thrust/Area.
⇒ Pressure = Ahρg/A
⇒ Pressure = hρg


Hence Proved.

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Hope it helps.