A solid cube just gets completely immersed in water when a 0.2 kg mass is placed on it. If the mass is removed, the cube is 2 cm above the water level. What is the length of each side of the cube?

### Concept & Archimedes Principle According to the principle of buoyancy, a floating object displaces a volume of water whose weight is equal to the object's weight. Removing the extra weight causes the object to rise, displacing less water. $$ \text{Weight Removed} = \text{Weight of Water in the Re-emerged Volume} $$ ### Step-by-Step Solution * Let the side length of the cube be $a\text{ cm}$. * When the 0.2 kg mass is removed, the cube rises by 2 cm. This means the volume of the displaced water corresponding to this 2 cm rise equals the volume of a 2 cm slice of the cube. * Volume of the 2 cm slice = Base Area $\times$ height = $a^2 \times 2\text{ cm}^3$. * The mass of this volume of displaced water must equal the mass removed (0.2 kg). * We know the density of water is $1\text{ g/cm}^3$ (or $0.001\text{ kg/cm}^3$). * Mass of displaced water = Volume $\times$ Density = $(2a^2) \times 0.001\text{ kg}$. * Equating the masses: $0.002a^2 = 0.2$. * $a^2 = \frac{0.2}{0.002} = 100$. * $a = \sqrt{100} = 10\text{ cm}$. ### Exam Strategy & Shortcut Equate the volume of the exposed portion directly to the volume of water equivalent to the removed mass. $0.2\text{ kg}$ is $200\text{ g}$, which equals $200\text{ cm}^3$ of water. So, $a^2 \times 2 = 200$, leading directly to $a^2 = 100 \Rightarrow a = 10$. ### Common Pitfall Forgetting to convert the density of water into consistent units ($0.001\text{ kg/cm}^3$ or converting $0.2\text{ kg}$ to $200\text{ g}$) will result in massive calculation errors. ### Final Answer Therefore, the correct answer is **10 cm**.