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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.
Solution
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.
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