A wooden cube supporting a mass of 100 g just floats in water. On removing the mass from the cube, the cube rises by 4 cm. The side of the cube is

Options

A. 5 cm
B. 8 cm
C. 7 cm
D. 9 cm

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the situation
A wooden cube is floating in water with a mass of 100 g on top of it. When the mass is removed, the cube rises by 4 cm. We need to find the side length of the cube.

### Step 2: Use the concept of buoyancy
When the cube is floating, it displaces a volume of water equal to the weight of the cube plus the weight of the mass on it. When the mass is removed, the cube displaces less water, and it rises by 4 cm.

### Step 3: Calculate the volume of water displaced
Let the side length of the cube be \( a \). The volume of the cube is \( V_{\text{cube}} = a^3 \).

When the mass is removed, the cube rises by 4 cm. The volume of water that corresponds to this rise is the area of the base of the cube multiplied by the height it rises:
\[ V_{\text{displaced}} = a^2 \times 4 \text{ cm} \]

### Step 4: Relate the displaced volume to the weight of the cube
The weight of the water displaced is equal to the weight of the cube when the mass is removed. The density of water is \( 1 \text{ g/cm}^3 \), so the weight of the displaced water is:
\[ \text{Weight of displaced water} = V_{\text{displaced}} \times \text{density of water} = (a^2 \times 4) \times 1 \text{ g/cm}^3 = 4a^2 \text{ g} \]

### Step 5: Set up the equation
When the mass is removed, the weight of the cube is equal to the weight of the displaced water:
\[ \text{Weight of the cube} = 100 \text{ g} = 4a^2 \]

### Step 6: Solve for \( a^2 \)
Rearranging the equation gives:
\[ a^2 = \frac{100 \text{ g}}{4} = 25 \text{ cm}^2 \]

### Step 7: Find the side length \( a \)
Taking the square root of both sides:
\[ a = \sqrt{25 \text{ cm}^2} = 5 \text{ cm} \]

### Conclusion
The side length of the cube is \( 5 \text{ cm} \).

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