A candle of diameter d floats in a liquid in a cylindrical container of diameter D (D ≫ d), as shown in the figure. Initially, the length above the liquid and the length submerged are each L. If the candle burns at 2 cm/hour, how does the top of the candle move? [Image]
Options

Question

A candle of diameter d floats in a liquid in a cylindrical container of diameter D (D ≫ d), as shown in the figure. Initially, the length above the liquid and the length submerged are each L. If the candle burns at 2 cm/hour, how does the top of the candle move? [Image]

Options

G1 Physics · Accepted Answer

Step 1: By Archimedes' principle, buoyant force equals weight: ρₗAℓₛg = ρ𝚌Aℓg. Step 2: Initially ℓₛ = ℓ/2, so ρ𝚌/ρₗ = 1/2. Since densities remain constant, the submerged fraction stays 1/2 as the candle burns. Step 3: Height of the candle above the liquid is ℓₐ = ℓ − ℓₛ = ℓ/2. Step 4: Candle length decreases by 2 cm/hour, so height above the liquid decreases by (1/2) × 2 = 1 cm/hour. Step 5: D ≫ d, so changes in liquid level are negligible. Final answer: The top falls at 1 cm/hour (Option B).