An iron pillar consists of a cylindrical portion, 2.8 m high and 20 cm in diameter and a cone 42 cm high is surrounding it. Find the weight of the pillar, given that 1 cm³ of iron weighs 7.5 g.

Radius of cylindrical portion, r = $\dfrac{\text{diameter}}{2} = \dfrac{20}{2}$ = 10 cm

Height of the cylindrical portion, h = 2.8 m = 2.8 × 100 = 280 cm

Height of the conical portion, H = 42 cm

From figure,

Radius of the conical part = radius of the cylindrical portion = r = 10 cm

Volume of iron pole = Volume of cylindrical portion + Volume of conical portion

= πr²h + $\dfrac{1}{3}$ πr²H

$= \dfrac{22}{7} \times 10^2 \times 280 + \dfrac{1}{3} \times \dfrac{22}{7} \times 10^2 \times 42 \\[1em] = \dfrac{22}{7} \times 100 \times 280 + \dfrac{1}{3} \times \dfrac{22}{7} \times 100 \times 42 \\[1em] = \dfrac{22}{7} \times 100 \times 280 + \dfrac{22}{7} \times 100 \times 14 \\[1em] = \dfrac{616000}{7} + \dfrac{30800}{7} \\[1em] = \dfrac{616000 + 30800}{7} \\[1em] = \dfrac{646800}{7} \\[1em] = 92400 \text{ cm}^3.$

Given,

Weight of 1 cm³ of iron = 7.5 gm.

Total weight = 92400 × 7.5 = 693000 gm = $\dfrac{693000}{1000}$ kg = 693 kg.

Hence, the weight of the pillar is 693 kg.