Question

An iron pillar has some part in the form of a right circular cylinder and remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if one cubic cm of iron weight is 7.8 grams.

Answer 1

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 = `πr_1^2h_1` cm³
= (π x 8 x 8 x 240 ) cm³
= (π x 64 x 240 ) cm³

Volume of the cone = `1/3 πr_2^2h_2` cm³
= `(1/3 π xx 8 xx 8 xx 36 )` cm³

= `(1/3 π xx 64 xx 36 )` cm³

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

= `(π xx 64 xx 240 + 1/3 π xx 64 xx 36)` cm³
= ` π xx 64 xx (240 + 12)` cm³

= `22/7 xx 64 xx 252` cm³

= 22 x 64 x 36 cm³

Hence, total weight of the pillar = Volume x weight per cm³
= ( 22 x 64 x 36 ) x 7.8 gms
= 395366.4 gms
= 395.3664 kg.