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A shuttle cock, used for playing badminton, has the shape of a frustum of a cone mounted on a hemisphere. If the external diameters of the frustum are $5 cm$ and 2 $cm$ and height of the entire shuttle cock is $7 cm$ , then what is its external surface area?

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74.26 cm 2

74 cm 2

76 cm 2

75.16 cm 2

answer is A.

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Detailed Solution

Given that the external diameters of the frustum are

5 cm

and 2

cm

and height is

7 cm

.
We have the following formulae,
Curved surface area of a Cone =

2 π r 2

Curved surface area of a hemisphere =

π (R + r) l

The radius of smaller circular end will be,

= 22 = 1 cm

The radius of the larger end will be,

= 52 = 2.5 cm

Radius of hemisphere will be equal to the radius of smaller circular end of frustum,

= 1 cm

Height of frustum will be,
= Total height - Height of hemisphere

(h) = 7 − 1 = 6 cm

The slant height of frustum will be,

l 2 = (R − r) 2 + h 2

l = (R − r) 2 + h 2

= (2.5 − 1) 2 + (6) 2

= 38.25

= 6.18 cm

The External surface area of shuttle

= C . S . A

of frustum

+ C . S . A

of hemisphere

= π [2 r 2 + (R + r) l]

= 227 2 (1) 2 + (2.5 + 1) 6.18

= 74.26 cm 2

Thus, the surface area of the shuttlecock is

74.26 cm 2

.
Therefore the correct option is 1.

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