A cube of wood supporting $200\,\text{gm}$ mass just floats in water. When the mass is removed, the cube rises by $2\,\text{cm}$. What is the size of the cube?

Step 1: Define the variables and initial state.

Let $a$ be the edge length of the wooden cube. Initially, with the mass $m = 200\text{ gm}$ on top, the cube is fully submerged in water.

The initial submerged volume is:

$$V_1 = a^3$$

The equilibrium condition for the floating system is:

$$W_{\text{wood}} + mg = \rho g V_1$$

where $\rho = 1\text{ g/cm}^3$ is the density of water, and $g$ is the acceleration due to gravity.

Step 2: Analyze the state after the mass is removed.

When the mass is removed, the cube rises by $h = 2\text{ cm}$, leaving a submerged depth of $a - h$.

The new submerged volume is:

$$V_2 = a^2(a - h)$$

The new equilibrium condition is:

$$W_{\text{wood}} = \rho g V_2$$

Step 3: Relate the change in buoyant force to the removed mass.

Subtract the second equilibrium equation from the first.

$$mg = \rho g (V_1 - V_2)$$ $$mg = \rho g [a^3 - a^2(a - h)]$$ $$mg = \rho g a^2 h$$

Divide both sides by $g$.

$$m = \rho a^2 h$$

Step 4: Calculate the edge length of the cube.

Substitute the given values $m = 200\text{ gm}$, $\rho = 1\text{ g/cm}^3$, and $h = 2\text{ cm}$ into the equation.

$$200 = (1)(a^2)(2)$$ $$a^2 = 100$$ $$a = 10\text{ cm}$$

Answer: The size of the cube (edge length) is $10\text{ cm}$.