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 $8cm$. The cylindrical part is $240cm$ high and the conical part is $36cm$ high. Find the weight of the pillar if one cubic cm of iron weighs $7.8$ grams.

Answer 1

Let $r_1$ cm and $r_2$ cm denote the radii of the base of the cylinder and cone respectively. Then,

$r_1=r_2=8$ cm

Let $h_1$ and $h_2$ cm be the heights of the cylinder and the cone respectively. Then

$h_1=240cm;h_2=36cm$

volume of Cylinder $=\pi r_1^2{h}_{1}36{cm}^3=(\pi \times 8\times 8\times 240){cm}^3=(\pi \times 64\times 240){cm}^3$

volume of Cone $=\frac{1}{3}\pi r_2^2{h}_2{cm}^3=(\frac{1}{3}\pi \times 8\times 8\times 36){cm}^3$

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

$=(\pi \times 64\times 240+\frac{1}{3}\pi \times 64\times 36){cm}^3=\frac{22}{7}\times 64\times 252{cm}^3=22\times 64\times 36{cm}^3$

total weight of the pillar = volume × Weight per $c{m}^3$

$(22\times 64\times 36)\times 7.8gms=395.3664kg$