How many cubic centimetres of iron is required

Solution:

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 $(l)=36 \mathrm{~cm}$

external breadth of an open box $(b)=25 \mathrm{~cm}$

and external height of an open box $(h)=16.5 \mathrm{~cm}$

$\therefore$ External volume of an open box $=l b h$

$=36 \times 25 \times 16.5$

$=14850 \mathrm{~cm}^{3}$

Since, the thickness of the iron $(x)=1.5 \mathrm{~cm}$

So, internal length of an open box $(\zeta)=l-2 x$

$=36 \times 2 \times 1.5$

$=36-3=33 \mathrm{~cm}$

Therefore, internal breadth of an open box $\left(b_{2}\right)=b-2 x$

$=25-2 \times 1.5=25-3=22 \mathrm{~cm}$

and internal height of an open box $\left(h_{2}\right)=(h-x)$

$=16.5-1.5=15 \mathrm{~cm}$

So, internal volume of an open box $=(l-2 x) \cdot(b-2 x) \cdot(h-x)$

$=33 \times 22 \times 15=10890 \mathrm{~cm}^{3}$

Therefore required iron to construct an open box

$=$ External volume of an open box - Internal volume of an open box

$=14850-10890=3960 \mathrm{~cm}^{3}$

Hence, required iron to construct an open box is $3960 \mathrm{~cm}^{3}$.

Given that, $1 \mathrm{~cm}^{3}$ of iron weights $=7.5 \mathrm{~g}=\frac{7.5}{1000} \mathrm{~kg}=0.0075 \mathrm{~kg}$

$\therefore \quad 3960 \mathrm{~cm}^{3}$ of iron weights $=3960 \times 0.0075=29.7 \mathrm{~kg}$