Given radius of cylindrical part(r) = \(\frac{12}2\) = 6 cm
Height of cylinder (h) = 110 cm
Length of cone (l) = 9 cm
Volume of cylinder = πr2h
V1= π (6)2 110 cm3
Volume of cone = \(\frac{1}3\)πr2l
V2 = \(\frac{1}3\)π(6)29
=108 π cm3
Volume of pole = v1 + v2
= π (6)2110 +108 π
= 12785.14 108 cm3
Given mass of 1 cm3 of iron = 8 gm
Mass of 1275.14 cm3 of iron = 12785.14 × 8
= 102281.12
= 102.2 kg
∴ Mass of pole for 12785.14 cm3 of iron is 102.2 kg