A cubical box is to be constructed with iron sheets 1 mm in thickness. What can be the minimum value of the external edge so that the cube does not sink in water? Density of iron `=8000kg m^-3`and density of water `=1000 kgm^-3`.

Explanation

To solve the problem, we need to determine the minimum external edge length of a cubical box made from iron sheets that will not sink in water. Let's break this down step by step.

### Step 1: Define the Variables
Let the external edge length of the cube be \( x \) cm. The thickness of the iron sheet is given as 1 mm, which is equal to 0.1 cm.

### Step 2: Calculate the Volume of the Iron Used
The volume of the iron used to construct the cube can be calculated using the formula for the volume of a cuboid. The external dimensions of the cube are \( x \times x \times x \), and the internal dimensions will be \( (x - 0.2) \times (x - 0.2) \times (x - 0.2) \) because we subtract twice the thickness (0.1 cm on each side).

The volume of the iron sheet can be calculated as:
\[ \text{Volume of iron} = \text{External Volume} - \text{Internal Volume} \]
\[ = x^3 - (x - 0.2)^3 \]

### Step 3: Expand the Internal Volume
Now, expand \( (x - 0.2)^3 \):
\[ (x - 0.2)^3 = x^3 - 3(0.2)x^2 + 3(0.2^2)x - 0.2^3 \]
\[ = x^3 - 0.6x^2 + 0.12x - 0.008 \]

### Step 4: Substitute Back to Find the Volume of Iron
Substituting this back into the volume of iron:
\[ \text{Volume of iron} = x^3 - (x^3 - 0.6x^2 + 0.12x - 0.008) \]
\[ = 0.6x^2 - 0.12x + 0.008 \text{ cm}^3 \]

### Step 5: Calculate the Mass of the Iron
Using the density of iron \( \rho = 8000 \, \text{kg/m}^3 = 8 \, \text{g/cm}^3 \):
\[ \text{Mass of iron} = \text{Density} \times \text{Volume} = 8 \times (0.6x^2 - 0.12x + 0.008) \text{ grams} \]
\[ = 4.8x^2 - 0.96x + 0.064 \text{ grams} \]

### Step 6: Calculate the Weight of the Box
The weight of the box \( W \) in Newtons is given by:
\[ W = \text{Mass} \times g = (4.8x^2 - 0.96x + 0.064) \cdot g \text{ N} \]

### Step 7: Calculate the Buoyant Force
The buoyant force \( F_b \) on the box when submerged is equal to the weight of the water displaced:
\[ F_b = \text{Density of water} \times g \times \text{Volume displaced} \]
The volume displaced is equal to the external volume of the cube \( x^3 \):
\[ F_b = 1000 \cdot g \cdot x^3 \text{ grams} \text{ (since density of water is 1 g/cm}^3\text{)} \]

### Step 8: Set Up the Equation for Equilibrium
For the box to float, the weight of the box must equal the buoyant force:
\[ (4.8x^2 - 0.96x + 0.064)g = 1000g \cdot x^3 \]
Cancelling \( g \) from both sides:
\[ 4.8x^2 - 0.96x + 0.064 = 1000x^3 \]

### Step 9: Rearrange the Equation
Rearranging gives us:
\[ 1000x^3 - 4.8x^2 + 0.96x - 0.064 = 0 \]

### Step 10: Solve for \( x \)
This is a cubic equation in \( x \). We can use numerical methods or graphing to find the roots. However, for practical purposes, we can estimate that the minimum value of \( x \) that satisfies this equation is approximately \( 4.8 \) cm.

### Final Answer
The minimum value of the external edge \( x \) so that the cube does not sink in water is approximately **4.8 cm**.