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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.

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Let the length (l), breath (b), and height (h) be the external dimension of an open box and thickness be x.

The volume of metal used in box = Volume of external box – Volume of internal box

Consider external box,

Length, l = 36 cm

Breadth, b = 25 cm

Height, h = 16.5 cm

We know that the equation of the volume of cuboid is given by,

Volume of cuboid = lbh, where, l, b and h are the length, breadth and height of tank respectively

Volume of external box = 36(25)(16.5) = 14850 cm3

Since the box is open from top,

Consider internal box,

The thickness of two sides is reduced as follows,

Length, l’ = Length of external box – 2(thickness of box) = 36 – 2(1.5) = 33 cm

Breadth, b’ = Breadth of external box – 2(thickness of box) = 25 – 2(1.5) = 22 cm

Height, h’ = Height of external box – thickness of box = 16.5 – 1.5 = 15 cm

Volume of internal box = 33(22)(15) = 10890

And,

Volume of metal in box = 14850 – 10890 = 3960 cm3

We know that,

1 cm3 weighs 7.5 g

So, 3960 cm3 weighs 3960(7.5) = 29,700 g

Therefore, the weight of box is 29,700 g i.e. 29.7 kg

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