How many cubic cm of iron is required to construct open box whose external dimensions are 36 cm 25 cm and 16.5 cm provided the thickness of the iron is 1.5 cm lf one cubic cm of iron weights 7.5 gm then find the weight of the box.

Explanation

Let the length (l) , breadth (b) and height (h) be the external dimension of an open box and thickness be x .
Given that,
external length of an open box (l) = 36 cm
external breadth of an open box (h) = 16.5 cm
`therefore` External volume of an open box = lbh

`= 36 xx 25 xx 16.5`
`= 14850 cm^(3)`
Since, the thickness of the iron (x) = 1.5 cm
So, internal lenght of an open box `(l_(1)) = l - 2x`
`= 36 xx 2 xx 1.5 `
`= 36-3 = 33 cm`
Therefore, internal breadth lenght of an open box `(b_(2)) = b - 2x `
`= 25- 2xx 1.5 = 25- 3 = 22 cm `
and internal height of an open box `= (l- 2x) .(b - 2x) . (h- x)`
`= 33 xx 22xx 15 = 10890 cm^(3)`
Therefore required iron to construct an open box
= External volume of an open box - Internal volume of an open box
`= 14850-10890 = 3960 cm^(3)`
Hence, required iron to construct an open box is `3960 cm^(3)`.
Given that,` 1 cm^(3)` of iron weights `= 7.5 g = (7.5)/(1000)kg = 0.0075 Kg`
`therefore " "3960 cm^(3)` of iron weights`= 3960 xx 0.0075 = 29.7 kg`