An iron pillar has some part in the form of a right circular cylinder and the 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 cu. cm of iron weighs 7.8 grams.

Radius of base of cone (r) = 8 cm,

Radius of cylinder (r) = 8 cm

Height of cylindrical part (h₁) = 240 cm

Height of conical part (h₂) = 36 cm.

Volume of pillar (V) = Volume of cylinder + Volume of cone.

$V = πr^2h$1 + \dfrac{1}{3}πr^2h2 \\[1em] = πr^2(h1 + \dfrac{h2}{3}) \\[1em] = \dfrac{22}{7} \times (8)^2 \times \Big(240 + \dfrac{36}{3}\Big) \\[1em] = \dfrac{22}{7} \times 64 \times \Big(240 + 12\Big) \\[1em] = \dfrac{22}{7} \times 64 \times 252 \\[1em] = 22 \times 64 \times 36 \\[1em] = 50688 \text{ cm}^3

Weight of 1 cm³ of iron = 7.8 g

∴ Weight of 50688 cm³ of iron = 50688 × 7.8 = 395366.4 g

Converting it into Kg,

395366.4 g = $\dfrac{395366.4}{1000}$ Kg

= 395.3664 Kg

Hence, the weight of the pillar is 395.3664 kg.