Question

Find the volume of iron required to make an open box whose external
dimensions are 36 cm x 25 cm x 16.5 cm, the box being 1.5 cm thick
throughout. If 1 cm of iron weighs 8.5 grams, find the weight of the empty
box in kilograms.​

Answer 1

Answer:

The weight of the empty box is 33.66 kg.

Step-by-step explanation:

$\textbf{\underline{\underline{Given : }}}$

$\sf{External\:Dimensions = 36\:cm \times 25\:cm \times 16.5\:cm}$

$\textsf{Thickness of the box = 1.5 cm }$

$\sf{1 cm^{3}= 8.5\:grams}$

$\textbf{\underline{\underline{To find :}}}$

$\textsf{Weight of the empty box}$

$\textbf{\underline{\underline{Solution :}}}$

$\textsf{Let the volume of external cuboid be}$ $\sf{V_1}$ $\textsf{and of internal cuboid be}$$\sf{V_2}$

$\rule{300}{1.5}$

✯ $\textsf{\underline{\underline{Volume of the cuboid - }}}$

• $\textsf{Height = 16.5 cm}$
• $\textsf{Length = 36 cm}$
• $\textsf{Breadth = 25 cm}$

$\boxed{\mathsf{Volume \: of \: Cuboid = Length \times Breadth \times Height}}$

$\sf{\longrightarrow} \: 36 \times 25 \times 16.5 \\ \\ \sf{\longrightarrow} \: 14850 \: {cm}^{3} \: ......... \: \gray{(V_1)}$

$\rule{300}{1.5}$

✯ $\textsf{\underline{\underline{Volume of the internal cuboid -}}}$

$\textsf{The thickness of the iron is 1.5 cm.}$

$\sf{\longrightarrow} \: Length \: = 36 - (1.5 + 1.5) = 36 - 3 = 33 \: cm \\ \\ \sf{\longrightarrow} \: Breadth = 25 - (1.5 + 1.5) = 25 - 3 = 22 \: cm \\ \\ \sf{\longrightarrow} \: Height = 16.5 - 1.5 = 15 \: cm$

$\textsf{Length = 33 cm}$

$\textsf{Breadth = 22 cm }$

$\textsf{Height = 15 cm}$

The 1.5 cm would be subtracted only once from the height as the box is open.

$\boxed{\mathsf{Volume \: of \: Cuboid = Length \times Breadth \times Height}}$

$\sf{\longrightarrow} \: 33 \times 22 \times 15 \\ \\ \sf{\longrightarrow} \: 10890 \: {cm}^{3} ........ \: \gray{(V_2)}$

$\rule{300}{1.5}$

✯ $\textsf{\underline{\underline{Volume of the iron used -}}}$

$\boxed{\sf{Volume \: of \: iron \: used = V_1 - V_2}}$

$\sf{\longrightarrow} \: 14850 \: {cm}^{3} - 10890 \: {cm}^{3} \\ \\ \sf{\longrightarrow} \: 3690 \: {cm}^{3}$

$\textsf{Volume of the iron used = 3690 cm}$³.

$\rule{300}{1.5}$

✯ $\textsf{\underline{\underline{Weight of the box -}}}$

Multiply the volume of the iron with weight of each 1cm³.

$\boxed{\sf{1cm^{3} = 8.5\:grams}}$

$\sf{\longrightarrow} \: 3690 \times 8.5 \\ \\ \sf{\longrightarrow} \: 33660 \: grams$

Convert grams into kilograms -

$\boxed{\sf{1kg= 1000\:grams}}$

$\sf{\longrightarrow} \: \dfrac{33660}{1000} \\ \\ \sf{\longrightarrow} \: 33.66 \: kg$

Weight of the box = 33.66 kg.

$\therefore$ The weight of the empty box is 33.66 kg.

Answer 2

Answer:

Volume = 3960 cm³

Weight = 33.66 kg

Step-by-step explanation:

Volume of iron required = External volume - Internal volume

= > ( 36 × 25 × 16.5 ) - ( 33 - 22 × 15 )    [ Subtracted thick from external volume ]

= > 14850 - 10890

=  > 3960 cm³

Now :

1 cm³ of iron weight = 8.5 g

3960 cm³ f iron weight = 8.5 g × 3960

= >  33660 g

= > 33.66 kg

Therefore , Volume of iron required is 3960 cm³ and weight of iron is 33.66 kg.