A wooden cube floating in water supports a mass 0.2 kg on its top. When the mass is removed the cube rises by 2cm. What is the side legnth of the cube ? Density of water `= 10^3 kg//m^3`

Options

A. 6 cm
B. 12 cm
C. 8 cm
D. 10 cm

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
When the wooden cube is floating in water with a mass of 0.2 kg on top, the total downward force (weight) is the sum of the weight of the cube and the weight of the mass. The buoyant force acting on the cube is equal to this total weight.

### Step 2: Set up the equations
Let:
- \( m \) = mass of the wooden cube
- \( a \) = side length of the cube
- Density of water \( \rho = 1000 \, \text{kg/m}^3 \)
- Gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \)

The total weight acting downwards when the mass is on the cube is:
\[ W_{\text{total}} = mg + 0.2g \]

The buoyant force \( F_b \) is given by:
\[ F_b = \rho g V_d \]
where \( V_d \) is the volume of water displaced. For a cube submerged to a depth \( h \):
\[ V_d = a^2 h \]

At equilibrium:
\[ F_b = W_{\text{total}} \implies \rho g (a^2 h) = mg + 0.2g \]

### Step 3: Analyze the situation after removing the mass
When the mass is removed, the cube rises by 2 cm (0.02 m). The new depth of the cube submerged in water is \( h - 0.02 \). The new buoyant force becomes:
\[ F_b' = \rho g (a^2 (h - 0.02)) \]

At this point, the weight of the cube alone is balanced by the new buoyant force:
\[ F_b' = mg \]

### Step 4: Set up the second equation
From the new equilibrium condition:
\[ \rho g (a^2 (h - 0.02)) = mg \]

### Step 5: Substitute the first equation into the second
From the first equation, we can express \( mg \):
\[ mg = \rho g (a^2 h) - 0.2g \]
Substituting this into the second equation:
\[ \rho g (a^2 (h - 0.02)) = \rho g (a^2 h) - 0.2g \]

### Step 6: Simplify the equation
Expanding and simplifying:
\[ \rho g (a^2 h - 0.02 a^2) = \rho g a^2 h - 0.2g \]
Cancelling \( \rho g a^2 h \) from both sides:
\[ -0.02 \rho g a^2 = -0.2g \]
Dividing both sides by \( -g \):
\[ 0.02 \rho a^2 = 0.2 \]

### Step 7: Solve for \( a^2 \)
Rearranging gives:
\[ a^2 = \frac{0.2}{0.02 \rho} \]
Substituting \( \rho = 1000 \, \text{kg/m}^3 \):
\[ a^2 = \frac{0.2}{0.02 \times 1000} = \frac{0.2}{20} = 0.01 \]

### Step 8: Find the side length \( a \)
Taking the square root:
\[ a = \sqrt{0.01} = 0.1 \, \text{m} = 10 \, \text{cm} \]

Thus, the side length of the cube is **10 cm**.