Surface Areas and Volumes | Exercise 12.2

Question 6

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that $1\text{ cm}^3$ of iron has approximately $8\text{ g}$ mass. (Use $\pi = 3.14$ )

Solution

Understand the Question

The solid iron pole consists of two cylinders, one mounted on top of the other:
- Larger (bottom) cylinder: Height $h_1 = 220\text{ cm}$ , base diameter $d_1 = 24\text{ cm} \implies \text{radius } r_1 = 12\text{ cm}$ .
- Smaller (top) cylinder: Height $h_2 = 60\text{ cm}$ , radius $r_2 = 8\text{ cm}$ .
The total volume of the pole is the sum of the volumes of both cylinders: $V_{\text{total}} = V_1 + V_2 = \pi r_1^2 h_1 + \pi r_2^2 h_2$ .
The total mass is found by multiplying the volume by the density ( $8\text{ g/cm}^3$ ) and converting the result from grams to kilograms ( $1\text{ kg} = 1000\text{ g}$ ).

Step 1 · Calculate the Total Volume of the Pole

Diagram 1

For the larger cylinder:

r_1 = \dfrac{24}{2} = 12\text{ cm}, \quad h_1 = 220\text{ cm}

\begin{aligned} V_1 &= \pi r_1^2 h_1 \\[0.6em] &= \pi \times (12)^2 \times 220 \\[0.6em] &= \pi \times 144 \times 220 \\[0.6em] &= 31680\pi\text{ cm}^3 \end{aligned}

For the smaller cylinder:

r_2 = 8\text{ cm}, \quad h_2 = 60\text{ cm}

\begin{aligned} V_2 &= \pi r_2^2 h_2 \\[0.6em] &= \pi \times (8)^2 \times 60 \\[0.6em] &= \pi \times 64 \times 60 \\[0.6em] &= 3840\pi\text{ cm}^3 \end{aligned}

Total volume of the pole:

\begin{aligned} V_{\text{total}} &= V_1 + V_2 \\[0.6em] &= 31680\pi + 3840\pi \\[0.6em] &= 35520\pi\text{ cm}^3 \end{aligned}

Using $\pi = 3.14$ :

\begin{aligned} V_{\text{total}} &= 35520 \times 3.14 \\[0.6em] &= 111532.8\text{ cm}^3 \end{aligned}

Step 2 · Calculate the Mass of the Pole

Given that mass of $1\text{ cm}^3$ of iron $= 8\text{ g}$ :

\begin{aligned} \text{Total mass in grams} &= V_{\text{total}} \times 8 \\[0.6em] &= 111532.8 \times 8 \\[0.6em] &= 892262.4\text{ g} \end{aligned}

Converting to kilograms ( $1\text{ kg} = 1000\text{ g}$ ):

\begin{aligned} \text{Total mass in kg} &= \dfrac{892262.4}{1000} \\[0.6em] &= 892.262\text{ kg} \end{aligned}

Answer

$892.262\text{ kg}$

Common Mistakes

Diameter vs. Radius: Using the diameter $24\text{ cm}$ directly instead of halving it to get radius $r_1 = 12\text{ cm}$ .
Incorrect Value of $\pi$ : Using $\pi = \dfrac{22}{7}$ instead of the explicitly specified $\pi = 3.14$ .
Unit Conversion: Forgetting to divide the mass by $1000$ to convert grams into kilograms.

Question 6

More questions in Exercise 12.2