How many cubic centimetres of iron are 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 $cm$ of iron weighs $7.5 \, g$,find the weight of the box.

(N/A) Let the external length $(l) = 36 \, cm$,external breadth $(b) = 25 \, cm$,and external height $(h) = 16.5 \, cm$. The thickness of the iron is $x = 1.5 \, cm$.
External volume of the open box $= l \times b \times h = 36 \times 25 \times 16.5 = 14850 \, cm^3$.
For an open box,the internal dimensions are:
Internal length $(l_i) = l - 2x = 36 - 2(1.5) = 36 - 3 = 33 \, cm$.
Internal breadth $(b_i) = b - 2x = 25 - 2(1.5) = 25 - 3 = 22 \, cm$.
Internal height $(h_i) = h - x = 16.5 - 1.5 = 15 \, cm$ (since it is an open box,only the base thickness is subtracted).
Internal volume $= l_i \times b_i \times h_i = 33 \times 22 \times 15 = 10890 \, cm^3$.
Volume of iron required = External volume - Internal volume
$= 14850 - 10890 = 3960 \, cm^3$.
Weight of the box = Volume of iron $\times$ density
$= 3960 \, cm^3 \times 7.5 \, g/cm^3 = 29700 \, g = 29.7 \, kg$.