A solid iron pole consists of a solid cylinder of height 200 cm and base diameter 28 cm, which is surmounted by another cylinder of height 50 cm and radius 7 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass.
Generated by claude-sonnet-4-6 · 2026-06-15 10:33 · grounding rag
Model Answer
Given:
- Lower cylinder: height $h_1 = 200$ cm, base diameter $= 28$ cm $\Rightarrow r_1 = 14$ cm
- Upper cylinder: height $h_2 = 50$ cm, radius $r_2 = 7$ cm
- $\pi = \dfrac{22}{7}$
Volume of lower cylinder:
$$V_1 = \pi r_1^2 h_1 = \frac{22}{7} \times 14 \times 14 \times 200 = 123200 \text{ cm}^3$$
Volume of upper cylinder:
$$V_2 = \pi r_2^2 h_2 = \frac{22}{7} \times 7 \times 7 \times 50 = 7700 \text{ cm}^3$$
Total volume of pole:
$$V = V_1 + V_2 = 123200 + 7700 = 130900 \text{ cm}^3$$
Mass of pole:
$$\text{Mass} = 130900 \times 8 = 1047200 \text{ g} = 1047.2 \text{ kg}$$
∴ The mass of the iron pole is 1047.2 kg.
Source: Chapter 12, Section 12.3 – Volume of a Combination of Solids
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Explanation
- The pole is a combination of two cylinders, so total volume = sum of individual volumes (no part is shared/removed).
- Use $r_1 = 14$ cm (half of diameter 28 cm) carefully — a common error is using diameter directly in the formula.
- The textbook (Exercise 12.2, Q.6) has a similar problem with slightly different dimensions; the method is identical.
- Final answer must be converted from grams to kg (divide by 1000) for a sensible unit.
- Examiners award marks step-by-step: formula, substitution, each volume, total volume, and final mass — show all steps clearly.