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Question
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 cm of iron weighs 7.5 g, find the weight of the box.
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 cm
External breadth of an open box (b) = 25 cm
And external height of an open box (h) = 16.5 cm
∴ External volume of an open box
= lbh
= 36 × 25 × 16.5
= 14850 cm3
Since, the thickness of the iron (x) = 1.5 cm
So, internal length of an open box (l1)
= l – 2x
= 36 × 2 × 1.5
= 36 – 3
= 33 cm
Therefore, internal breadth of an open box (b2)
= b – 2x
= 25 – 2 × 1.5
= 25 – 3
= 22 cm
And internal height of an open box (h2)
= (h – x)
= 16.5 – 1.5
= 15 cm
So, internal volume of an open box
= (l – 2x) × (b – 2x) × (h – x)
= 33 × 22 × 15
= 10890 cm3
Therefore, required iron to construct an open box
= External volume of an open box – Intemal volume of an open box
= 14850 – 10890
= 3960 cm3
Hence, required iron to construct an open box is 3960 cm3.
Given that, 1 cm3 of iron weights
= 7.5 g
= `7.5/1000` kg
= 0.0075 kg
∴ 3960 cm3 of iron weights
= 3960 × 0.0075
= 29.7 kg
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