How many cubic centimetres of iron is required to construct an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm provided the thickness of the iron is 1.5 cm. If one cubic centimeter of iron weights 7.5 g, then find the weight of the box.

Let the length breath and height be the external dimension of an open box and thickness be

The volume of metal used in box = Volume of external box - Volume of internal box

For external box,

Length, l = 36 cm

Breadth, b = 25 cm

Height, h = 16.5 cm

We know that,

Volume of cuboid = lbh

Where, l, b and h are the length, breadth and height of tank respectively

Volume of external box = 36(25)(16.5) = 14850 cm³

Now,

As the box is open from top,

For internal box,

Length, l' = Length of external box - 2(thickness of box) = 36 - 2(1.5) = 33 cm

[i.e. the thickness of two sides is deduced]

Breadth, b' = Breadth of external box - 2(thickness of box) = 25 - 2(1.5) = 22 cm

[i.e. the thickness of two sides is deduced]

Height, h' = Height of external box - thickness of box = 16.5 - 1.5 = 15 cm

[i.e. the thickness of bottom is reduced]

Volume of internal box = 33(22)(15) = 10890

So,

Volume of metal in box = 14850 - 10890 = 3960 cm³

Also,

1 cm³ weighs 7.5 g

3960 cm³ weighs 3960(7.5) = 29,700 g

So,

The weight of box is 29,700 g i.e. 29.7 kg