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NEET
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Physics
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Mechanical Properties of Fluids

NEET · Physics · Mechanical Properties of Fluids

An open cylindrical vessel contains two immiscible liquids. A layer...

An open cylindrical vessel contains two immiscible liquids. A layer of water of height 22 m is at the bottom, and a layer of oil of height 33 m is on top. What is the absolute pressure at the bottom of the vessel?
(Given: density of water =1000= 1000 kg m−3^{-3}, density of oil =800= 800 kg m−3^{-3}, atmospheric pressure =1.01×105= 1.01 \times 10^5 Pa, and g=10g = 10 m/s2^2)

Options

  1. A0.44×1050.44 \times 10^5 Pa
  2. B1.21×1051.21 \times 10^5 Pa
  3. C1.45×1051.45 \times 10^5 Pa
  4. D1.25×1051.25 \times 10^5 Pa

Correct answer

C. 1.45×1051.45 \times 10^5 Pa

Step-by-step solution

The absolute pressure at the bottom of the vessel is the sum of the atmospheric pressure and the pressure exerted by the two fluid columns.

Pressure due to the oil layer:
Poil=ρoilghoilP_{\text{oil}} = \rho_{\text{oil}} g h_{\text{oil}}
Poil=800×10×3=24,000P_{\text{oil}} = 800 \times 10 \times 3 = 24,000 Pa

Pressure due to the water layer:
Pwater=ρwaterghwaterP_{\text{water}} = \rho_{\text{water}} g h_{\text{water}}
Pwater=1000×10×2=20,000P_{\text{water}} = 1000 \times 10 \times 2 = 20,000 Pa

Total gauge pressure at the bottom:
Pgauge=Poil+Pwater=24,000+20,000=44,000P_{\text{gauge}} = P_{\text{oil}} + P_{\text{water}} = 24,000 + 20,000 = 44,000 Pa =0.44×105= 0.44 \times 10^5 Pa

Absolute pressure at the bottom:
Pabs=Patm+PgaugeP_{\text{abs}} = P_{\text{atm}} + P_{\text{gauge}}
Pabs=1.01×105+0.44×105=1.45×105P_{\text{abs}} = 1.01 \times 10^5 + 0.44 \times 10^5 = 1.45 \times 10^5 Pa.

Answer: 1.45×1051.45 \times 10^5 Pa
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