Gate Pillar: Paint Budget and a Change of Top
The art club of a school in Haryana is making a decorative pillar to stand at the gate for its annual day. The design is a solid cylinder of radius 14 cm and height 1.5 m with a solid hemisphere of the same radius fixed on top; the flat base of the cylinder rests on the ground and is not painted. Only the exposed curved surfaces of the cylinder and the hemisphere will be painted, at ₹5 per 100 cm², and the club has collected exactly ₹700 for paint. A second, cheaper-looking design replaces the hemisphere with a solid cone of the same radius and height 21 cm, which the carpenter says will 'hold more or less the same amount of material'. The club needs the exact painted area and cost of the first design, the volume of material in it in litres, and a ruling on whether the cone version would cost less and hold more. There is no money for a third design.
1. Find the total painted area and the paint cost of the first design, and compare with the ₹700 collected.
Read the worked answer
r = 14 cm, h = 150 cm.
Painted surfaces = curved surface of the cylinder + curved surface of the hemisphere. The cylinder's base sits on the ground and the hemisphere's flat face is glued to the cylinder top, so neither is painted.
Cylinder CSA = 2πrh = 2 × (22/7) × 14 × 150 = 13,200 cm². Hemisphere CSA = 2πr² = 2 × (22/7) × 196 = 1,232 cm². Total = 14,432 cm².
Cost = (14,432/100) × 5 = ₹721.60.
The collection is ₹700, so the design is ₹21.60 over budget. Each 1 cm of cylinder height adds 2πr = 88 cm² of paint, i.e. ₹4.40; removing 21.60/4.40 ≈ 4.9 cm of height (to about 145 cm) brings the cost to ₹700 exactly.
2. Find the volume of material in the first design, in litres, and say what the figure is actually useful for.
Read the worked answer
Cylinder volume = πr²h = (22/7) × 196 × 150 = 92,400 cm³.
Hemisphere volume = (2/3)πr³ = (2/3) × (22/7) × 2744 = (2/3) × 8,624 = 5,749.33 cm³.
Total = 98,149.33 cm³ = 98.15 litres (1 litre = 1000 cm³).
Reading the result: the pillar is decorative, not a container, so the volume is really a material and weight estimate - about 0.098 m³ of concrete or plaster mix, light enough to be cast on site and lifted by two people. It also shows the hemisphere adds only about 5.9% of the total volume, so the top is a small part of the material cost.
3. Evaluate the carpenter's claim by comparing the cone version (r = 14 cm, h = 21 cm) with the hemisphere version in both volume and paint cost.
Read the worked answer
Slant height of the cone: l = √(14² + 21²) = √637 ≈ 25.24 cm.
Volume of cone = (1/3)πr²h = (1/3) × (22/7) × 196 × 21 = 4,312 cm³, against 5,749.33 cm³ for the hemisphere. The cone holds 1,437.33 cm³ less - about 25% less, not 'more or less the same'. The claim is wrong.
Paint: cylinder CSA is unchanged at 13,200 cm²; cone CSA = πrl = (22/7) × 14 × 25.24 ≈ 1,110.5 cm², slightly less than the hemisphere's 1,232 cm². Total ≈ 14,310.5 cm², cost ≈ ₹715.53 - about ₹6 cheaper, but still ₹15.53 over the ₹700 budget.
Verdict: the cone saves a little paint and loses a quarter of the volume; neither design fits the budget, so the club must cut the cylinder height by about 3.5 cm or negotiate a lower paint rate.