A cubical box of wood of side 30 cm and mass 21.6 kg floats on water with two faces horizontal. The length of immersed part in water is :
`(rho_("wood")=0.8g//c c)`

Options

A. 30 cm
B. 12 cm
C. 6 cm
D. 24 cm

Answer

To find the length of the immersed part of a cubical wooden box floating in water, we can follow these steps:

### Step 1: Understand the Problem
We have a cubical wooden box with a side length of 30 cm and a mass of 21.6 kg. The box floats on water, and we need to find the length of the part that is immersed in water.

### Step 2: Use the Principle of Buoyancy
According to the principle of buoyancy (Archimedes' principle), the weight of the fluid displaced by the submerged part of the object is equal to the weight of the object itself when it is floating.

### Step 3: Calculate the Weight of the Wooden Box
The weight (W) of the wooden box can be calculated using the formula:
\[ W = mg \]
where:
- \( m = 21.6 \, \text{kg} \)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 21.6 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 211.6 \, \text{N} \]

### Step 4: Calculate the Volume of Water Displaced
Let \( h \) be the height of the immersed part of the box in meters. The volume of the displaced water (V) can be expressed as:
\[ V = \text{Area} \times h \]
The area of one face of the cube is:
\[ \text{Area} = \text{side}^2 = (0.3 \, \text{m})^2 = 0.09 \, \text{m}^2 \]

Thus, the volume of displaced water becomes:
\[ V = 0.09 \, \text{m}^2 \times h \]

### Step 5: Relate Weight to Buoyancy Force
The buoyancy force (B) is equal to the weight of the displaced water:
\[ B = \rho_{\text{water}} \times V \times g \]
where:
- \( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \)

Substituting for V:
\[ B = 1000 \, \text{kg/m}^3 \times (0.09 \, \text{m}^2 \times h) \times 9.81 \, \text{m/s}^2 \]

### Step 6: Set Weight Equal to Buoyancy Force
Setting the weight equal to the buoyancy force:
\[ 211.6 \, \text{N} = 1000 \times 0.09 \times h \times 9.81 \]

### Step 7: Solve for h
Rearranging the equation to solve for \( h \):
\[ h = \frac{211.6}{1000 \times 0.09 \times 9.81} \]
Calculating the denominator:
\[ 1000 \times 0.09 \times 9.81 = 882.9 \]
Thus,
\[ h = \frac{211.6}{882.9} \approx 0.239 \, \text{m} \]

### Step 8: Convert to Centimeters
To convert \( h \) from meters to centimeters:
\[ h \approx 0.239 \, \text{m} \times 100 = 23.9 \, \text{cm} \]

### Step 9: Final Answer
Rounding to the nearest whole number, the length of the immersed part in water is approximately:
\[ \text{Length of immersed part} \approx 24 \, \text{cm} \]

### Conclusion
Thus, the length of the immersed part of the wooden box is **24 cm**.

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