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, the thickness being 1.5 cm? If 1 cm³ of iron weighs 7.5 g, find the weight of the box.
Iron volume = 3960 cm³; Weight = 29.7 kg
Step 1: Understand the problem.
The box is made of iron. It is hollow inside (open box). To find the volume of iron used, we subtract the inside (internal) volume from the outside (external) volume.
Step 2: External dimensions of the box.
Length = 36 cm, Breadth = 25 cm, Height = 16.5 cm
So, external volume = \(36 \times 25 \times 16.5 = 14850\,\text{cm}^3\).
Step 3: Internal dimensions.
The thickness of the iron sheet = 1.5 cm. This thickness reduces the size on both sides of each dimension.
(For height, we subtract only once, because the top is open.)
Step 4: Internal volume.
Internal volume = \(33 \times 22 \times 15 = 10890\,\text{cm}^3\).
Step 5: Volume of iron used.
Iron volume = External volume − Internal volume
= \(14850 − 10890 = 3960\,\text{cm}^3\).
Step 6: Weight of the box.
Given, 1 cm³ of iron weighs 7.5 g.
So, weight = \(3960 \times 7.5 = 29700\,\text{g}\).
Convert grams to kilograms: \(29700 \div 1000 = 29.7\,\text{kg}\).
Final Answer: Iron volume = 3960 cm³, Weight of the box = 29.7 kg.