3.Derive the relation for liquid pressure : P = h d g , where h=depth ,d=density g= acc. Due to gravity.
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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.
_____________________
Hope it helps.
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.
_____________________
Hope it helps.
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