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: 893 Kg (approx.)

Step-by-step explanation:

Given, the height of the big cylinder (H) = 220 cm

Radius of the base (R) = 24/12 = 12 cm  

So, the volume of the big cylinder = πR2H

= π(12)2 × 220 cm3

= 99565.8 cm3

Now, the height of smaller cylinder (h) = 60 cm

Radius of the base (r) = 8 cm

So, the volume of the smaller cylinder = πr2h  

= π(8)2×60 cm3

= 12068.5 cm3

∴ Volume of iron = Volume of the big cylinder+ Volume of the small cylinder

= 99565.8 + 12068.5  

=111634.5 cm3

We know,

Mass = Density x volume

So, mass of the pole = 8×111634.5  

= 893 Kg (approx.)

Thanks

I hope this helps you friend :)

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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