A rectangular open tank contains water upto a depth of 3m and above it an oil of specific gravity 0.80 for a depth of 1.5m. How much volume of oil is required to be added to the tank such that the pressure intensity at the interface of two liquids is 50% of the pressure intensity at the bottom of the tank. Calculate your answer in _____m3Correct answer is between '13.4,13.6'. Can you explain this answer?

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A rectangular open tank contains water upto a depth of 3m and above it an oil of specific gravity 0.80 for a depth of 1.5m. How much volume of oil is required to be added to the tank such that the pressure intensity at the interface of two liquids is 50% of the pressure intensity at the bottom of the tank. Calculate your answer in _____m³

Correct answer is between '13.4,13.6'. Can you explain this answer?

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A rectangular open tank contains water upto a depth of 3m and above it...

Initial volume of oil in the tank = L × B × H

V_initial = 3 × 2 × 1.5

V_initial = 9m³

Let height of oil required in the tank such that pressure at the interface of two liquids is 50% of the pressure at the bottom be “h”

Pressure at the interface of two liquids (P₁) = ρ_oil × g × h

Pressure at the bottom of the tank (P₂) = ρ_oil × g × h + ρ_water × g × 3

P₁ = 0.50 P₂

ρ_oil × g × h = 0.50 [ρ_oil × g × h + ρ_water × g × 3]

0.50 ρ_oil × g × h = 1.5 + ρ_water × g

ρ_oil = specific gravity ρ_w

ρ_oil = 0.80 × 1000

ρ_oil = 800 kg/m³

0.50 × 800 × g × h = 1.5 × 1000 × g

h = 3.75m

Final volume of oil in the tank = 3 × 2 × 3.75

V_final = 22.5m³

Volume of oil required to be added = V_final - V_initial

ΔV = 22.5 – 9

ΔV = 13.5m³

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