Question

IDASIC STANDARD
A cylindrical iron pillar of diameter 14 dm and height 14 dm is surmounted by a cone,
9 dm high. Find the weight of the iron pillar, if I cubic dm of iron weigh 9.6 g
Basics​

Answer 1

Given :

• Diameter of cylindrical iron pillar = 14 dm
• Height of cylindrical iron pillar = 14 dm
• Height of cone surmounted over cylindrical iron pillar = 9 dm
• Density of iron = 9.6 g/dm³

To find :

Weight of iron pillar

Formula used :

• Radius = Diameter ÷ 2
• Volume of cylinder = π (radius)² × height of cylinder
• Volume of cone = (1/3) × π (radius)² × height of cone
• Density = mass / volume

Solution :

First of all we need to find radius of base of iron pillar

➝ Radius = Diameter/2

➝ Radius = 14 dm/2

➝ Radius = 7dm

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Now we will find Volume of iron pillar

➝ Volume of iron pillar = Volume of cylindrical part + Volume of cone

➝ Volume of iron pillar = [ π × (7 dm)² × 14 dm ] + [ (1/3) × (7dm)² × (9dm) ]

$\sf \implies \: Volume \: of \: iron \: pillar \: = \: \bigg( \dfrac{22}{7} \times 49 \times 14 \: {dm}^{3} \bigg) \: + \: \bigg( \dfrac{1}{3} \times \dfrac{22}{7} \times 49 \times 9\: {dm}^{3} \bigg)$

$\sf \implies \: Volume \: of \: iron \: pillar \: = \: \bigg( 22 \times 49 \times 2\: {dm}^{3} \bigg) \: + \: \bigg( 22 \times 7 \times 3\: {dm}^{3} \bigg)$

$\sf \implies \: Volume \: of \: iron \: pillar \: = \: 2156 \: {dm}^{3} + \: 462\: {dm}^{3}$

$\sf \implies \: Volume \: of \: iron \: pillar \: = \: 2618\: {dm}^{3}$

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Now ,

Weight of iron pillar = Volume of iron pillar × density of iron

➝ Weight of iron pillar = 2618 dm³ × 9.6 g/dm³

➝ Weight of iron pillar = 25,132.8 g

➝ Weight of iron pillar = 25.1328 kg

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ANSWER :

Weight = 25,132.8 g = 25.1328 kg