Physics

Physics

Problem

A vertical tube open to the atmosphere holds two immiscible fluids: 0.18 m of kerosene (ρ=820 kg m−3\rho=820\,\text{kg\,m}^{-3}) above 0.30 m of water. Use g=9.8 m s−2g=9.8\,\text{m\,s}^{-2} and atmospheric pressure P0=1.01×105 PaP_0=1.01\times10^{5}\,\text{Pa}. (a) Calculate the gauge pressure at the bottom of the water layer. (b) Determine the absolute pressure at that same point.
Intuitive explanation

The key is recognizing that pressure increases with depth in a fluid column, and at any horizontal level inside the tube, the pressure must be continuous. Start at the top (open to atmosphere) and work downward, adding the pressure contribution from each fluid layer using ΔP=ρgh\Delta P = \rho g h. The pressure at the water surface equals atmospheric pressure plus the weight of kerosene above it; the pressure at the bottom equals that plus the weight of water. This layered approach lets you find the pressure at any depth.

Intuitive explanation

The key is recognizing that pressure increases with depth in a fluid column, and at any horizontal level inside the tube, the pressure must be continuous. Start at the top (open to atmosphere) and work downward, adding the pressure contribution from each fluid layer using ΔP=ρgh\Delta P = \rho g h. The pressure at the water surface equals atmospheric pressure plus the weight of kerosene above it; the pressure at the bottom equals that plus the weight of water. This layered approach lets you find the pressure at any depth.

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