Question

A solid iron pole consists of a cylinder of height 220 cm base diameter is 24 CM which is surmounted by another cylinder of height 60 cm radius 8 cm find the mass of the pole given that 1cm3 of iron approximately 8 gm mass use pie 3.14

Answer 1

Answer:

The Mass of the pole is 892.262 Kg.

Step-by-step explanation:

Given:

Radius of larger cylinder(R) = 24 / 2 = 12 cm

Height of larger cylinder(H) = 220 cm

Radius of smaller cylinder(r) = 8 cm

1 c$m^{3}$ Iron = 8 grams

Height of smaller cylinder(h) = 60 cm

Volume Of Pole = Volume Of Larger Cylinder + Volume of Smaller Cylinder

Formula to find the volume of Cylinder = $\pi r^{2} h$

Calculating:

Volume Of Pole = $\pi r^{2} h$  + $\pi R^{2} H$

= (8$)^{2}$(60) + (12$)^{2}$(220)

= (64) (60) + (144) (220)

= 3840 + 31680

= 35,520 * $\pi$ (was taken as common)

= 35,520 * 3.14

= 111532.8 c$m^{3}$

Mass of 111532.8 c$m^{3}$ of iron = 111532.8 * 8

= 892262.4 Grams

= 892262.4 / 1000

= 892.262 Kg

Therefore the Mass of the pole is 892.262 Kg.

Answer 2

Solution :-

Diameter of larger cylinder = 24cm

Radius of larger cylinder = 24/2 = 12cm

Height of the larger cylinder = 220 cm

Volume of the larger cylinder = πr²h

=> 3.14 × 12 × 12 × 220 cm³

=> 3.14 × 144 × 220 cm³

=> 99475.2 cm³

Radius of the smaller cylinder = 8 cm

Height of the smaller cylinder = 60 cm

volume of the smaller cylinder = πr²h

=> 3.14 × 8 × 8 × 60

=> 3.14 × 64 × 60

=> 12057.6 cm³

Volume of the iron pole :-

volume of smaller + volume of larger cylinder

=> 99475.2 cm³ + 12057.6 cm³

=> 111532.8 cm³

Mass in iron of 1 cm³ = 8g

Mass in iron of 111532.8 cm³ = 111532.8 × 8

=> 892,262.4 gm

=> $\frac{892,262.4}{1000}$

=> 892.26 kg

Hence, The mass of the iron pole is 892.26kg.

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