A solid uniform cylinder of height 0.20\text{ m} and cross-sectional area 5.0 \times 10^{-… — JAMB UTME

A solid uniform cylinder of height $0.20\text{ m}$ and cross-sectional area $5.0 \times 10^{-3}\text{ m}^2$ floats vertically at the boundary between oil of density $800\text{ kg/m}^3$ and water of density $1000\text{ kg/m}^3$ . If a height of $0.08\text{ m}$ of the cylinder extends into the water layer while the remaining upper portion is completely covered by the oil layer, what is the mass of the cylinder in kilograms?

Answer: 0.88 kg

Answer

The mass of the cylinder is

0.88\text{ kg}

According to the Law of Flotation, a floating body displaces its own weight of fluid. When floating at the interface of two immiscible liquids, the total mass of the body equals the sum of the masses of the displaced liquids. Displaced water mass is

\rho_w A h_w = 0.40\text{ kg}

and displaced oil mass is

\rho_o A h_o = 0.48\text{ kg}

, giving a total cylinder mass of

0.88\text{ kg}

Step-by-Step Solution

Find the height of the cylinder submerged in the oil layer.

h_o = 0.20\text{ m} - 0.08\text{ m} = 0.12\text{ m}

The total cylinder height is

0.20\text{ m}

, and

0.08\text{ m}

is submerged in water.

Calculate the volumes of water and oil displaced by the cylinder.

V_w = 5.0 \times 10^{-3} \times 0.08 = 4.0 \times 10^{-4}\text{ m}^3

;

V_o = 5.0 \times 10^{-3} \times 0.12 = 6.0 \times 10^{-4}\text{ m}^3

Volume displaced in each fluid equals cross-sectional area multiplied by the submerged height in that fluid.

Calculate the mass of the floating cylinder using the Law of Flotation.

m = \rho_w V_w + \rho_o V_o = (1000 \times 4.0 \times 10^{-4}) + (800 \times 6.0 \times 10^{-4}) = 0.40\text{ kg} + 0.48\text{ kg} = 0.88\text{ kg}

For a floating object in static equilibrium, its mass equals the total mass of the fluids displaced by its submerged parts.

Key Concept

Law of Flotation in Layered Liquids

Estimated Time:1m 30s

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