Find the volume of aluminium required to make an open box whose outer dimensions are 36cm×25×16.5 cm the box being 1.5cm thick throughout. If 1cm cube of aluminium weight 4.5 grams,find the weight of an empty box in kilogram. ..
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46
given dimension of the box - 36cm×25cm×16.5cm.
thickness = 1.5cm
internal dimension o the box= 36 - 2× 1.5 = 33cm
breadth = 25 -2 × 1.5 = 22cm
height = 16.5 -1.5 = 15cm
external volume of the box = (36×25×16.5) cm³
= 14,850 cm³
internal volume = 33 × 22×15 cm³
=10,890 cm³
vole used t make the box = 14850 - 10890 cm³
= 3960 cm³
weight of 1 cm³ of iron = 4.5 g
∴ weight of 3960 of iron = 4.5 × 3960 g
= 17820 g
= 17820/1000 kg = 17.82 kg
thickness = 1.5cm
internal dimension o the box= 36 - 2× 1.5 = 33cm
breadth = 25 -2 × 1.5 = 22cm
height = 16.5 -1.5 = 15cm
external volume of the box = (36×25×16.5) cm³
= 14,850 cm³
internal volume = 33 × 22×15 cm³
=10,890 cm³
vole used t make the box = 14850 - 10890 cm³
= 3960 cm³
weight of 1 cm³ of iron = 4.5 g
∴ weight of 3960 of iron = 4.5 × 3960 g
= 17820 g
= 17820/1000 kg = 17.82 kg
Answered by
2
Internal dimension of the box = 36-2×1.5=33cm
Breadth = 25-2×1.5=22cm
Height = 16.5-1.5=15cm
External Vol. of the box = (36×25×16.5)cm³
= 14,850cm²
Internal Vol. of the box = (33×22×15)cm³
= 10,890cm³
Weight of 1 cm³ of aluminium = 4.5g
Weight of 3,960 of aluminium = (4.5×3960)g
= 17820g
17820g =(17820/1000) kg = 17.82 kg.
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