State relation for pressure beneath a liquid surface to depth and density?
OR
Prove that (P = rgh)
Difficulty: Medium
Pressure in Liquids:
Consider a surface of area A in a liquid at a depth of h. The length of the cylinder of liquid over this surface will be h.
The force acting on this surface will be the weight w of the liquid above this surface. If r is the density of the liquid and m is the mass of liquid above the surface, then
Mass of the liquid cylinder = m = $volume\times density$
= $(A\times h) \times \rho$
Force acting on Area F = w = mg
= $A h \rho g$
As, Pressure P = $\frac{F}{A}$
= $\frac{(A h \rho g)}{A}$
Liquid pressure at depth h = P = $\rho g h$ ………(i)
Equation i) gives the pressure at a depth h in a liquid of density $\rho$. It shows that the pressure in a liquid increase with depth. $(P \alpha h)$
DO YOU KNOW?
The piston of the syringe is pulled out. This lowers the pressure in the cylinder. The liquid from the bottle enters the piston through the needle.
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