A solid iron pole having cylindrical portion $ 110 \mathrm{~cm} $ high and of base diameter $ 12 \mathrm{~cm} $ is surmounted by a cone $ 9 \mathrm{~cm} $ high Find the mass of the pole, given that the mass of $ 1 \mathrm{~cm}^{3} $ of iron is $ 8 \mathrm{gm} $.

Given:

A solid iron pole having cylindrical portion \( 110 \mathrm{~cm} \) high and of base diameter \( 12 \mathrm{~cm} \) is surmounted by a cone \( 9 \mathrm{~cm} \) high.

The mass of \( 1 \mathrm{~cm}^{3} \) of iron is \( 8 \mathrm{gm} \).

To do:

We have to find the mass of the pole.

Solution:

Diameter of the base $= 12\ cm$

This implies,

Radius of the base $r =\frac{12}{2}$

$ = 6\ cm$
Height of the cylindrical portion $h_1= 110\ cm$

Height of the conical portion $h_2 = 9\ cm$\

Therefore,

Total volume of the pole $=\pi r^{2} h_{1}+\frac{1}{3} \pi r^{2} h_{2}$

$=\pi r^{2}(h_{1}+\frac{1}{3} h_{2})$

$=\frac{22}{7} \times(6)^{2}(110+\frac{1}{3} \times 9)$

$=\frac{22}{7}+36(110+3)$

$=\frac{22}{7} \times 36 \times 113$

$=\frac{89496}{7} \mathrm{~cm}^{3}$
Mass of $1 \mathrm{~cm}^{3}$ of iron $=8 \mathrm{gm}$

Therefore,

Total mass of the pole $=\frac{89496}{7} \times 8 \mathrm{gm}$

$=\frac{89496 \times 8}{7 \times 1000} \mathrm{~kg}$

$=102.281 \mathrm{~kg}$

The mass of the pole is $102.281\ kg$.

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Updated on: 10-Oct-2022

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