Question

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 = 8000 kg m−3 and density of water = 1000 kg m−3.

Answer 1

The minimum outer edge value = 44.4 mm .

Explanation:

Step 1:

Let the outer edge minimum value be x mm, so the inner edge is (x -2) mm

Therefore, total cubic box volume = eternal volume-internal volume

$= [ x^3 - (x - 2)^3] mm^3$

Cubic Box Mass = V \times density \times g

$= [x^3 - (x - 2)^3 ] \times 8000 \times g$

Water volume displaced through box  $= x^3 mm^3$

Step 2:

Mass of water displaced through box = $V \times density' \times g$

$= x^3 \times 1000 \times g$

Don't fall for the cube,

Step 3:

Cubic box mass = box mass of water displaced,

$[x^3 - (x - 2)^3 ] \times 8000 \times g == x^3 \times 1000 \times g$

$8[x^3 - (x -2)^3 ] = x^3$

$8x^3 - 8(x -2)^3 = x^3$

$7x^3 = 8(x -2)^3$

$\sqrt[3]{7} \ x = 2(x -2)$

$1.91x = 2x - 4$

$2x - 1.91x = 4 =>,$

$0.09x = 4$

$x = 44.4 mm$

Therefore, minimum outer edge value = 44.4 mm