A solid iron pole having cylindrical portion 110 cm

We have to find the mass of a pole having a cylindrical base surmounted by a cone.

Radius of cone and cylinder $(r)=6 \mathrm{~cm}$

Height of cylinder $(h)=110 \mathrm{~cm}$

Height of cone $(l)=9 \mathrm{~cm}$

So volume of the pole is,

$=\pi r^{2} h+\frac{1}{3} \pi r^{2} l$

$=\pi r^{2}\left(h+\frac{1}{3} l\right)$

Put the values to get,

$=\left(\frac{22}{7}\right)(36)(110+3) \mathrm{cm}^{3}$

$=12785.14 \mathrm{~cm}^{3}$

Mass of $1 \mathrm{~cm}^{3}$ of iron is $8 \mathrm{gm}$.

Therefore mass of the iron,

$=(12785.14)(8) \mathrm{gm}$

$=102.2 \mathrm{~kg}$