Question

A cubical box is to be constructed with iron sheets $ 1mm $ 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 $ = 1000kg\,{m^{ - 3}} $ .

Answer 1

Hint: In the given question, as we can see that the cubical box is not sinking in the water, so this is the situation of buoyant force in which an object floats on water instead of sinking in it. And in this situation, the force of buoyancy is equal to the weight or mass of an object. So here we will first find the volume of the iron with the help of the surface area of the iron in the box, and then we will calculate the mass of the iron. Again we will conclude the buoyant force with the help of the given value of density of water. Now, we can equate the mass of iron and the buoyant force.

Complete answer:
In the given question, we have to find the minimum value of the external edge of the cube. So, we first assume the minimum value of the external edge.
Let the required external edge be $ xcm $ .
And, as we know the formula of surface area of the cube or cubical box, i.e. $ 6{x^2}c{m^2} $ or,
 $ \because Surface\,area\,of\,the\,box = 6 \times edg{e^2} = 6{x^2}c{m^2} $ .
Now, as we can see in the question that the box is constructed with iron-
So, the volume of the iron in the box $ = Surface\,area \times thickness\,of\,iron $ i.e.
 $ \Rightarrow Volume\,of\,the\,iron\,in\,box = 6{x^2} \times 0.1c{m^3} = 0.60{x^2}c{m^3} $
Now, we will find the mass of the iron in the box, as we already concluded the volume of iron:
 $ \therefore Mass\,of\,the\,iron\,in\,the\,box = \dfrac{{volume\,of\,iron}}{{{{10}^6}}} \times Density\,of\,iron = \dfrac{{0.60{x^2}}}{{{{10}^6}{m^3}}} \times 8000kg.{m^{ - 3}} = \dfrac{{4.8{x^2}}}{{1000}}kg $
Now, the volume of the water displaced $ = {x^3}c{m^3} $
So, to find the external edge of the cubical box, first we will find the buoyancy force or force of buoyancy:
 $ \therefore The\,force\,of\,buoyancy = \dfrac{{{x^3}}}{{{{10}^6}}}{m^3} \times 1000kg.{m^{ - 3}} = \dfrac{{{x^3}}}{{1000}}kg $
According to the given condition in which the cube does not sink in water, the mass of the cubical box is equal to its force of buoyancy:
 $ \therefore Mass\,of\,the\,iron = Force\,of\,buoyancy \\
   \Rightarrow \dfrac{{4.8{x^2}}}{{1000}} = \dfrac{{{x^3}}}{{1000}} $
Denominators are equal on both sides, so they will cancel each other,
 $ \Rightarrow {x^3} = 4.8{x^2} \\
   \Rightarrow x = 4.8cm $
Hence, the required minimum value of the external edge of the given cube is $ 4.8cm $ .

Note:
An object will float if its average density is less than that of the surrounding fluid. The reason for this is that a fluid with a higher density has more mass and thus weight in the same volume. As a result, the buoyant force, which is equal to the weight of the fluid displaced, is more than the object's weight.