A wooden cube floats just inside the water, when a mass of x(in grams) is placed on it. If the mass is removed, the cube floats with a height `(x)/(100)` (cm) above the water surface. The length of the side of cube is (density of water is 1000 `kg//m^(3)`)

Options

A. 10 cm
B. 15 cm
C. 20 cm
D. 30 cm

Answer

To solve the problem, we need to understand the principles of buoyancy and how they apply to the floating wooden cube.

### Step 1: Understanding the situation
When the mass \( x \) (in grams) is placed on the wooden cube, it floats just inside the water. This means that the weight of the cube plus the weight of the mass \( x \) is equal to the weight of the water displaced by the submerged part of the cube.

### Step 2: Establishing the relationship
Let the side length of the cube be \( L \) (in meters). The volume of the cube is given by:
\[ V = L^3 \]
The weight of the water displaced when the cube is floating is equal to the weight of the cube plus the weight of the mass \( x \):
\[ \text{Weight of water displaced} = \text{Weight of cube} + \text{Weight of mass} \]
Using the density of water (\( \rho = 1000 \, \text{kg/m}^3 \)), the weight of the water displaced can be expressed as:
\[ \text{Weight of water displaced} = \text{Volume of submerged part} \times \text{Density of water} \times g \]
where \( g \) is the acceleration due to gravity.

### Step 3: Setting up the equations
When the mass \( x \) is placed on the cube:
\[ \text{Weight of cube} + x \cdot g = \text{Volume submerged} \cdot \rho \cdot g \]
When the mass is removed, the cube floats with a height of \( \frac{x}{100} \) cm above the water surface. This means the submerged height of the cube is:
\[ L - \frac{x}{100 \times 100} \text{ m} = L - \frac{x}{10000} \text{ m} \]
The volume submerged is:
\[ \text{Volume submerged} = L^2 \left(L - \frac{x}{10000}\right) \]

### Step 4: Equating the weights
Now we can set up the equations for both scenarios:
1. With mass \( x \):
\[ \text{Weight of cube} + x \cdot g = L^2 \cdot \left(L\right) \cdot 1000 \cdot g \]
2. Without mass \( x \):
\[ \text{Weight of cube} = L^2 \cdot \left(L - \frac{x}{10000}\right) \cdot 1000 \cdot g \]

### Step 5: Solving for \( L \)
From the first equation:
\[ \text{Weight of cube} = L^3 \cdot 1000 \cdot g - x \cdot g \]
From the second equation:
\[ \text{Weight of cube} = L^2 \cdot \left(L - \frac{x}{10000}\right) \cdot 1000 \cdot g \]
Equating both expressions for the weight of the cube gives:
\[ L^3 \cdot 1000 - x = L^2 \cdot \left(L - \frac{x}{10000}\right) \cdot 1000 \]
This can be simplified and solved for \( L \).

### Step 6: Final Calculation
After simplifying the equation, we can isolate \( L \) and find its value in terms of \( x \).