A shuttle cock used for playing badminton has the shape of a frustum of a cone mounted on a hemisphere. The two diameters of the frustum are 5 cm and 2 cm, the height of the entire shuttle cock is 7 cm. Find the external surface area.

Question

Question

This question was previously asked in

RRB ALP Previous Paper 10 (Held On: 10 Aug 2018 Shift 1)

Answer 1

F2 Savita Railways 10-4-24 D1

r = 1 cm, R = 2.5 cm, h = 7 cm

height of frustum = 7 - 1 = 6 cm

⇒ l = √[h² + (R – r)]²

⇒ l = √[(6)² + (2.5 – 1)²]

⇒ l = √(36 + 2.25)

⇒ Slant height, l = 6.2 cm

External Surface area = (curved surface area of frustum) + (curved surface area of hemisphere)

⇒ External Surface area = 22/7 × (r + R) × l + 2 × (22/7) × r²

⇒ External Surface area = 22/7 × [(1 + 2.5)6.2 + 2 × (1)²]

⇒ External Surface area = 22/7 × (3.5 × 6.2 + 2)

⇒ External Surface area = 22/7 × (21.7 + 2) = 74.29 cm²