Question

A beaker is filled with a liquid of density ρ1 up to a certain height. A is a point, h m downwards from the free surface such that the pressure at A is P. If the liquid is replaced by equal volume of another liquid of density ρ2, at what distance from the free surface will the pressure be P now?

I have been asked this question in a national level competition.

My question is taken from Pressure Distribution in a Fluid topic in chapter Pressure and Its Measurement of Fluid Mechanics

Answer 1

Problem Explanation:

The problem involves two liquids with different densities (ρ1\rho_1ρ1​ and ρ2\rho_2ρ2​) and requires determining the depth from the free surface where the pressure remains the same (PPP) after replacing the liquid.

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Key Concepts:

1. Hydrostatic Pressure Formula:

   P=Patm+ρghP = P_{\text{atm}} + \rho g hP=Patm​+ρgh

   where:

   • PPP: Pressure at a depth hhh.
   • PatmP_{\text{atm}}Patm​: Atmospheric pressure acting on the free surface.
   • ρ\rhoρ: Density of the liquid.
   • ggg: Acceleration due to gravity.
   • hhh: Depth from the free surface.

2. In both cases, PPP remains the same, and PatmP_{\text{atm}}Patm​ is constant.

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Given:

1. In the first liquid (ρ1\rho_1ρ1​):

   P=Patm+ρ1gh1P = P_{\text{atm}} + \rho_1 g h_1P=Patm​+ρ1​gh1​

   where h1=hh_1 = hh1​=h (depth of point AAA).

2. In the second liquid (ρ2\rho_2ρ2​): Let the depth from the free surface at which the pressure is PPP be h2h_2h2​. The pressure equation becomes:

   P=Patm+ρ2gh2P = P_{\text{atm}} + \rho_2 g h_2P=Patm​+ρ2​gh2​

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Solution:

Since the pressure PPP and PatmP_{\text{atm}}Patm​ are the same in both cases:

ρ1gh1=ρ2gh2\rho_1 g h_1 = \rho_2 g h_2ρ1​gh1​=ρ2​gh2​

Canceling ggg (as gravity is constant) and substituting h1=hh_1 = hh1​=h:

ρ1h=ρ2h2\rho_1 h = \rho_2 h_2ρ1​h=ρ2​h2​

Solving for h2h_2h2​:

h2=ρ1ρ2hh_2 = \frac{\rho_1}{\rho_2} hh2​=ρ2​ρ1​​h

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Final Answer:

The depth from the free surface where the pressure is PPP after replacing the liquid is:

h2=ρ1ρ2hh_2 = \frac{\rho_1}{\rho_2} hh2​=ρ2​ρ1​​h

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Notes:

1. If ρ2>ρ1\rho_2 > \rho_1ρ2​>ρ1​, h2<hh_2 < hh2​<h (the point moves closer to the surface).
2. If ρ2<ρ1\rho_2 < \rho_1ρ2​<ρ1​, h2>hh_2 > hh2​>h (the point moves deeper into the liquid).