Question

A 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 $1c{m}^3$ of iron weighs 7.8 grams.

• A. 398.4 kg
• B. 397.4 kg
• C. 395.4 kg
• D. 393.4 kg

Answer 1

Answer: (C)

395.4 kg

Volume of the pillar $=$ Volume of the cylindrical part $+$ Volume of conical part
Volume of a Cylinder of Radius "R" and height "h" $=\pi R^{2}h$
Volume of a cone $=\frac{1}{3}\pi r^{2}h$ where r is the radius of the base of the cone and h is the height.
Hence, Volume of the pillar $=(\frac{22}{7}\times 8\times 8\times 240)+(\frac{1}{3}\times \frac{22}{7}\times 8^2\times 36)=50688{cm}^3$

If one cu cm wieighs $7.8$ grams, then $50688{cm}^3$ weighs $50688\times 7.8=395366.4$ grams or $395.37kg$