A School Water Tank: Cylinder with a Hemispherical Dome
A village panchayat in Ludhiana district has sanctioned a mild-steel water tank for the government senior secondary school. The fabricator's drawing shows a cylindrical body standing on flat ground with a hemispherical dome welded on top; there is no separate lid and no opening at the base. The internal radius is 1.4 m for both the cylinder and the dome, and the cylindrical portion is 3.5 m tall. Only the exposed outer surfaces - the curved wall of the cylinder and the dome - will be painted; the circular base that rests on the ground is left bare. Paint and labour together cost ₹85 per square metre, and the painter has quoted a lump sum of ₹4,000. The tank must be installed on an existing square concrete platform whose side is 2.5 m. At the meeting, the headmaster says the tank will hold 'roughly 30,000 litres', while the fabricator's bill states 27,000 litres.
1. Calculate the capacity of the tank in litres and decide whose figure - the headmaster's 30,000 L or the fabricator's 27,000 L - is closer to the truth.
Read the worked answer
Capacity = volume of cylinder + volume of hemisphere.
Cylinder: V₁ = πr²h = (22/7) × (1.4)² × 3.5 = (22/7) × 1.96 × 3.5 1.96 × 3.5 = 6.86, and (22/7) × 6.86 = 22 × 0.98 = 21.56 m³
Hemisphere: V₂ = (2/3)πr³ = (2/3) × (22/7) × (1.4)³ (1.4)³ = 2.744; (22/7) × 2.744 = 22 × 0.392 = 8.624; × 2/3 = 5.7493 m³
Total = 21.56 + 5.7493 = 27.3093 m³. Since 1 m³ = 1000 litres, capacity ≈ 27,309 litres.
Comparison: |27,309 − 27,000| = 309 L, while |30,000 − 27,309| = 2,691 L. The fabricator's 27,000 L is far closer; the headmaster has overstated the capacity by nearly 10% (about 270 buckets of 10 L).
Two cautions: the 1.4 m is an internal radius, so 27,309 L is the true holding capacity - an external radius would overstate it. Also, in practice the tank is never filled into the top of the dome, so usable capacity is slightly less again.
2. Find the area that must actually be painted and decide whether the painter's lump sum of ₹4,000 is more or less than the fair measured cost.
Read the worked answer
Painted area = curved surface of cylinder + curved surface of dome. The flat base circle on the ground is not painted - this is the common trap.
Cylinder wall: 2πrh = 2 × (22/7) × 1.4 × 3.5 2 × 1.4 × 3.5 = 9.8; (22/7) × 9.8 = 22 × 1.4 = 30.80 m²
Dome: 2πr² = 2 × (22/7) × 1.96 = 2 × 22 × 0.28 = 12.32 m²
Total painted area = 30.80 + 12.32 = 43.12 m²
Fair cost = 43.12 × 85 = ₹3,665.20
The quote of ₹4,000 is ₹334.80 higher, about 9% above the measured rate. That is a small, defensible margin for scaffolding and wastage, but it is not the measured price; the panchayat could ask for ₹3,665 plus agreed extras.
If the base circle were wrongly included, πr² = (22/7) × 1.96 = 6.16 m² would add ₹523.60 to the bill and would make the ₹4,000 quote look cheaper than it is.
3. Will the tank stand safely on the 2.5 m × 2.5 m platform? Support your answer with a calculation and explain why comparing areas alone gives the wrong answer.
Read the worked answer
Compare widths, not just areas.
Tank base is a circle of radius 1.4 m, so its diameter is 2 × 1.4 = 2.8 m.
The largest circle that fits inside a square of side 2.5 m has diameter 2.5 m (radius 1.25 m). Equivalently, a circle of diameter 2.8 m needs a square of side at least 2.8 m, i.e. area 7.84 m².
Area comparison: tank base = πr² = 6.16 m², platform = 2.5 × 2.5 = 6.25 m². Areas suggest the tank fits with 0.09 m² spare - but this is misleading. The circle's diameter (2.8 m) exceeds the platform's side (2.5 m), so the rim would overhang by (2.8 − 2.5)/2 = 0.15 m on every side, and the dome would overhang too. The load would sit partly on unsupported edge and the tank would be unstable.
Verdict: No, the tank must not be placed on this platform. Either cast a new platform of side at least 2.8 m (3 m is safer, giving working room and keeping the load inside the edges) or reduce the tank radius to 1.25 m or less.