A 400 g solid cube having an edge of length 10 cm floats in water. How much volume of the cube is outside the water?

(Given: density of water = 1000 kg m^-3)

<p> First, let’s find the weight of the cube and set it equal to the upward buoyant force that keeps the cube floating. </p> <p> <b>Step 1: Set up the equation for floating</b><br> The weight of the cube (mass × gravity) is balanced by the buoyant force (density of water × volume of cube under water × gravity):<br> $$\mathrm{Mg} = \mathrm{F}_{\mathrm{B}} \Rightarrow (400 \times 10^{-3}) = 10^3 \times \mathrm{V}_{\mathrm{d}}$$ </p> <p> <b>Step 2: Solve for the volume under water</b><br> We find the volume of the cube under water ($\mathrm{V}_\mathrm{d}$):<br> $\mathrm{V}_{\mathrm{d}} = 400 \times 10^{-6}~\mathrm{m}^3$ </p> <p> <b>Step 3: Find the total volume of the cube</b><br> Each edge of the cube is 10 cm, so total volume is:<br> $(10 \times 10^{-2})^3$ </p> <p> <b>Step 4: Find the volume outside water</b><br> Subtract the volume under water from the total volume to get the volume outside water:<br> $(\text {Vol.})_{\text{outside}} = (10 \times 10^{-2})^3 - 400 \times 10^{-6}$ </p> <p> Calculate the answer:<br> $= 600 \times 10^{-6}~\mathrm{m}^3 = 600~\mathrm{cm}^3$ </p>