A cubical block of side a carries a coin of mass m and floats in a liquid of density ρ with immersed depth x₁. After the coin is removed, the immersed depth becomes x₂. Find x₁ - x₂. Image
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A cubical block of side a carries a coin of mass m and floats in a liquid of density ρ with immersed depth x₁. After the coin is removed, the immersed depth becomes x₂. Find x₁ − x₂. Image

G1 Physics · Accepted Answer

Step 1: Removing the coin reduces the required buoyant force by mg. Step 2: The decrease in displaced volume is a²(x₁−x₂), so ρg a²(x₁−x₂)=mg. Hence x₁−x₂ = m/(ρa²). Final answer: A.