Question:
A solid iron pole having 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 the mass of 1 cm3 of iron is 8 gm.
Solution:
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,
$=(12785.14)(8) \mathrm{gm}$
$=102.2 \mathrm{~kg}$