Painting the Society Water Tank
A housing society is installing a new overhead water tank. The tank is an upright cylinder of radius 1.4 m and height 2.5 m, closed at the bottom, with a hemispherical dome of the same radius welded on top. The flat base rests on a concrete platform, so the bottom face cannot be painted and does not need paint. The society has decided to paint the entire exposed surface - the curved wall of the cylinder and the curved outer surface of the dome - with two coats of an anti-corrosion paint. The paint tin states that one litre covers 8 m² per coat. The painter estimates that about 10% of the paint is lost to spillage, roller wastage and the inside of the container. Paint is sold only in 1-litre cans costing ₹420 each. The secretary wants an exact figure before releasing money, and the treasurer asks what would happen to the bill if a larger tank of the same shape but double the radius were ordered.
1. Calculate the total area to be painted (two coats, including wastage) and the number of litres of paint required before rounding to whole cans. Take π = 22/7.
Read the worked answer
Step 1 - one-coat area.
- Curved surface of the cylinder: 2πrh = 2 × (22/7) × 1.4 × 2.5 = 22.0 m².
- Curved surface of the hemispherical dome (not a full sphere): 2πr² = 2 × (22/7) × 1.4² = 2 × (22/7) × 1.96 = 12.32 m².
- The flat bottom is on the platform, so it is excluded.
- One coat: 22.0 + 12.32 = 34.32 m².
Step 2 - two coats. 2 × 34.32 = 68.64 m².
Step 3 - wastage. 68.64 × 1.10 = 75.504 m² of paint must actually be bought.
Step 4 - litres. 75.504 ÷ 8 = 9.438 litres.
Note the two traps: using 4πr² (the whole sphere) for the dome, and painting the base that stands on the platform.
2. How many 1-litre cans must be bought and what will the paint cost? Explain why the answer cannot be rounded down, and how much paint capacity is left over.
Read the worked answer
The requirement is 9.438 litres, but paint is sold only in whole 1-litre cans.
- 9 cans give 9 × 8 = 72 m², which is less than the 75.504 m² needed → the second coat would run out.
- 10 cans give 80 m² of coverage, which is enough.
Cost = 10 × ₹420 = ₹4,200.
Rounding down is not merely careless, it is physically impossible: a can cannot be bought in parts, and stopping short leaves part of the tank with a single coat, defeating the purpose of the anti-corrosion treatment.
Left over: 10 − 9.438 = 0.562 litres, about 4.5 m² of coverage - enough for touching up welds and the dome seam later, so it is not pure waste. If the painter could reduce wastage to 5%, the requirement would fall to about 9.0 litres, but 10 cans would still be needed.
3. The treasurer asks: if a tank of the same shape but double the radius (2.8 m) and the same height is ordered, does the paint bill simply double? Answer with calculations.
Read the worked answer
No - it grows faster than doubling, because one part of the surface scales with r and the other with r².
For r = 2.8 m, h = 2.5 m:
- Cylinder: 2 × (22/7) × 2.8 × 2.5 = 44.0 m² (this doubled, from 22.0, since it is proportional to r).
- Dome: 2 × (22/7) × 2.8² = 2 × (22/7) × 7.84 = 49.28 m² (this quadrupled, from 12.32, since it is proportional to r²).
- One coat: 93.28 m², versus 34.32 m² - a factor of about 2.72.
Two coats: 186.56 m²; with 10% wastage: 205.216 m²; litres: 205.216 ÷ 8 = 25.65 → 26 cans → ₹10,920 instead of ₹4,200.
So the bill rises by 2.6 times, not 2. The dome dominates at large sizes - a useful reminder that area scales differently from length.