Question

an iron pillar has some part I the form of a right circular cylinder and the remaining in the form of a right circular cone . the radius of the base of cylinder and cone is 8cm . the cylinder part is 240cm high and conical part is 36cm high . find the weight of the pillar , if 1cu. cm of iron weights 7.5 grams.

Answer 1

The volume of the cylindrical part = πr²h
                                                      = 22/7 x 8² x 240 = 48274 2/7 cm³

The volume of the conical part = 1/3 x base area x height
                                                 = 1/3 x 22/7 x 8² x 36 
                                                 = 2413  5/7 cm³

Total volume = 48274 2/7  + 2413  5/7 = 50688 cm³

Mass = Density x Volume
          = 7.5 x 50688
          = 380160g
          = 380.16 kg

If you want the weight in Newtons, we assume that the gravitational force g = 10N/kg

The weight = 380.16 x 10 = 3801.6 N

Answer 2

Answer:

Let r1 cm and r2 cm denote the radii of the base of the cylinder and cone respectively. Then,

r1 = r2 = 8 cm

Let h1 and h2 cm be the height of the cylinder and the cone respectively. Then

h1 = 240 cm and h2 = 36 cm

Now, volume of the cylinder = πr12h1 cm3

= (π × 8 × 8 × 240)cm3

= (π × 64 × 240) cm3

Volume of the cone = 1/3πr22h2 cm3

= (1/3π × 8 × 8 × 36) cm3

= (1/3π × 64 × 36) cm3

∴ Total volume of iron = Volume of the cylinder + Volume of the cone

= (π × 64 × 240 + 1/3π × 64 × 36) cm3

= π × 64 × (240 + 12) cm3

= 22/7 × 64 × 252 cm3

= 22 × 64 × 36 cm3

Hence, total weight of the pillar = Volume Weight per cm3

= (22 × 64 × 36) × 7.8 gms

= 395366.4 gms

= 395.3664 kg

..........HOPE IT HAS HELPED U...........

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