Solve:
- (i) A ball of chapati dough of radius 6 cm is prepared. Estimate how many chapatis can be made from it?
- (ii) Cut a coconut/muskmelon in half and find out the approximate volume of edible coconut flesh/fruit by taking the necessary measurements.
Step-by-step solution
Idea: Volume of dough stays the same when it is rolled flat. A chapati is a very short cylinder (a disc). The flesh of a coconut is a thick spherical shell: outer sphere minus inner hollow.
(i) A ball of chapati dough of radius 6 cm is prepared. Estimate how many chapatis can be made from it?
- Volume of dough = 43π × 63 = 288π ≈ 905 cm3.1 mark
- Assume a chapati is a disc of diameter 16 cm (r = 8 cm) and thickness 2 mm (0.2 cm): volume = π × 64 × 0.2 ≈ 40.2 cm3.1 mark
- Number ≈ 905 ÷ 40.2 ≈ 22.5, so about 22 chapatis.½ mark
(ii) Cut a coconut/muskmelon in half and find out the approximate volume of edible coconut flesh/fruit by taking the necessary measurements.
- Measurements on the cut coconut: radius to the inside of the shell (outer edge of the flesh) R ≈ 5.5 cm; flesh thickness ≈ 1 cm, so the hollow has radius r ≈ 4.5 cm.1 mark
- Flesh = sphere of radius 5.5 minus sphere of radius 4.5: 43π(5.53 − 4.53) = 43 × 3.14 × (166.375 − 91.125) = 43 × 3.14 × 75.25 ≈ 315 cm3 (each half ≈ 158 cm3).1 mark
- So about 315 cm3 of edible coconut. (Muskmelon: inner radius 7.5 cm, seed hollow 3.5 cm ⇒ 43 × 3.14 × (421.875 − 42.875) ≈ 1590 cm3 of fruit.)½ mark
Check: (i) 22 chapatis × 40 cm3 = 880 cm3 ≈ 905 cm3 ✓. (ii) 315 cm3 of coconut weighs roughly 300 g, about what one coconut gives ✓.
Answer to write in the exam
(i)
Dough: V = 43π(6)3 = 288π ≈ 905 cm3
Chapati (disc): V = π(8)2(0.2) ≈ 40.2 cm3
∴ Number ≈ 905 ÷ 40.2 ≈ 22
(ii)
Coconut: R ≈ 5.5 cm, r ≈ 4.5 cm
Flesh = 43π(R3 − r3) = 43 × 3.14 × 75.25
∴ Flesh ≈ 315 cm3 (≈ 158 cm3 per half)
Common mistakes that cost marks
- In (i), using the dough’s surface area instead of its volume.
- In (ii), finding the volume of the whole sphere and forgetting to subtract the hollow (the water space or seed cavity).
- Using (5.5 − 4.5)3 = 1 instead of 5.53 − 4.53.
How this can come in the exam
A sweet shop makes laddoos from a cylindrical block of mixture of radius 7 cm and height 9 cm. Each laddoo is a sphere of radius 2.1 cm.
(i) Find the volume of the block. (ii) Find the volume of one laddoo. (iii) How many laddoos can be made? (iv) If each laddoo is made 10% bigger in radius, roughly how many can be made? (π = 227)
Show answer
(i) 227 × 49 × 9 = 1386 cm3 (1 mark). (ii) 43 × 227 × 9.261 = 38.808 cm3 (1 mark). (iii) 1386 ÷ 38.808 ≈ 35.7 ⇒ 35 laddoos (1 mark). (iv) Volume × 1.331 ≈ 51.65 cm3; 1386 ÷ 51.65 ≈ 26.8 ⇒ 26 laddoos (1 mark).Try one yourself
A metal ball of radius 3 cm is beaten into a flat circular plate 0.5 cm thick. Find the radius of the plate.
Show answer
43π × 27 = πR2 × 0.5 ⇒ R2 = 72 ⇒ R ≈ 8.49 cm.
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