Calculate Mass of Metal Coin Added to Floating Wooden Cube

To find the mass of the metal coin, we apply Archimedes' principle, which states that a floating body displaces a weight of fluid equal to its total weight.

Step 1: Base Area of the Cube The dimensions of the wooden cube are $10\text{ cm} \times 10\text{ cm} \times 10\text{ cm}$. The base area $A$ of the cube is: $A = 10\text{ cm} \times 10\text{ cm} = 100\text{ cm}^2$

Step 2: Initial Equilibrium (Cube Alone) When only the wooden cube is floating on the water, the upward buoyant force balances the weight of the cube: $F_{b,1} = W_{\text{cube}}$ $\rho_w A h_1 g = m_{\text{cube}} g$ where:

• $\rho_w = 1\text{ g/cm}^3$ is the density of water,
• $h_1$ is the initial depth of the cube submerged in water,
• $g$ is the acceleration due to gravity.

Step 3: Final Equilibrium (Cube + Metal Coin) When the metal coin of mass $m_c$ is placed on top of the wooden cube, the submerged depth increases to $h_2 = h_1 + \Delta h$, where $\Delta h = 3.87\text{ cm}$. The total weight is now balanced by the new buoyant force: $F_{b,2} = W_{\text{cube}} + W_{\text{coin}}$ $\rho_w A h_2 g = (m_{\text{cube}} + m_c) g$

Step 4: Deriving the Mass of the Coin Subtracting the initial equilibrium equation from the final equilibrium equation: $\rho_w A (h_2 - h_1) g = m_c g$ $\rho_w A \Delta h = m_c$

Step 5: Calculation Substitute the given values into the equation: $m_c = 1\text{ g/cm}^3 \times 100\text{ cm}^2 \times 3.87\text{ cm}$ $m_c = 387\text{ g}$

Thus, the mass of the metal coin is $387$ grams.