An iron pole consisting of a cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that 1 cm³ of iron has 8 gm of mass (approx). (Take π = $\dfrac{355}{113}$)

Given,

Diameter of cylindrical portion = 12 cm

Radius of cylindrical portion (r) = $\dfrac{12}{2}$ = 6 cm

Height of the cylindrical part (H) = 110 cm

Height of the conical part (h) = 9 cm

From figure,

Radius of conical part = Radius of cylindrical portion = r = 6 cm.

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

$= πr^2H + \dfrac{1}{3}πr^2h \\[1em] = πr^2\Big(H + \dfrac{1}{3}h\Big) \\[1em] = \dfrac{355}{113} \times 6^2 \times \Big(110 + \dfrac{1}{3} \times 9\Big) \\[1em] = \dfrac{355}{113} \times 36 \times 113 \\[1em] = 12780 \text{ cm}^3.$

Given,

Weight of 1 cm³ of iron = 8 gm.

Total weight = 12780 x 8 gm = 102240 gm

= $\dfrac{102240}{1000}$ kg = 102.24 kg.

Hence, mass of pole = 102.24 kg