Management Aptitude Test Feb 2012 Paper

Question : 114 of 200

Marks: +1, -0

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

67.98 $cm^2$
74.26 $cm^2$
70 $cm^2$
72 $cm^2$

Validate

Solution:

(2) External surface area of the shuttlecock = External surface area of frustrum + External surface area of hemisphere

= π(R + r )l = 2π

r^

= π (R + r)

(√{h^2+(R+r)^2})

+ 2π

r^

where R = 2.5 cm, r = 1 cm, h = 6 cm
∴ Required surface area
=

22/7

x 3.5

√{36 + 1.5^2}

+ 2 x

22/7

x 1

= 22 x 0.5

√38.25

+ 6.28
= 11 x 6.18 + 6.28 = 74.26

cm^2

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