A beaker is filled with a liquid of density ρ1 up to a certain height. The pressure at the base of the beaker id Pb. If the liquid is replaced by an equal volume of another liquid of density ρ2, what will be the pressure at the base of the beaker now?

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1 Answer

ridin · Answer 1 · 2024-11-19T08:26:19+0000

To determine the pressure at the base of the beaker after replacing the liquid, let’s solve the problem step by step.

Hydrostatic Pressure Formula:
Pb=Patm+ρghP_b = P_{\text{atm}} + \rho g hPb=Patm+ρgh
where:
- PbP_bPb: Pressure at the base.
- PatmP_{\text{atm}}Patm: Atmospheric pressure acting on the free surface.
- ρ\rhoρ: Density of the liquid.
- ggg: Acceleration due to gravity.
- hhh: Height of the liquid column.
The pressure at the base depends only on the density of the liquid (ρ\rhoρ) and the height of the column (hhh).

The liquid of density ρ1\rho_1ρ1 creates a pressure at the base:
Pb=Patm+ρ1ghP_b = P_{\text{atm}} + \rho_1 g hPb=Patm+ρ1gh
The liquid is replaced with another liquid of density ρ2\rho_2ρ2 but with the same volume. Since the volume remains constant and the cross-sectional area of the beaker is the same, the height hhh remains unchanged.

The pressure at the base with the second liquid will be:

Pb′=Patm+ρ2ghP'_b = P_{\text{atm}} + \rho_2 g hPb′=Patm+ρ2gh

If ρ2>ρ1\rho_2 > \rho_1ρ2>ρ1: The pressure at the base increases (Pb′>PbP'_b > P_bPb′>Pb).
If ρ2<ρ1\rho_2 < \rho_1ρ2<ρ1: The pressure at the base decreases (Pb′<PbP'_b < P_bPb′<Pb).

The pressure at the base after replacing the liquid is:

Pb′=Patm+ρ2ghP'_b = P_{\text{atm}} + \rho_2 g hPb′=Patm+ρ2gh

The pressure at the base depends only on the density of the new liquid (ρ2\rho_2ρ2) and the unchanged height hhh.