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Question
A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6 cm and 12 cm, respectively. If the slant height of the conical portion is 5 cm, find the total surface area and volume of the rocket [Use π = 3.14].
Solution
Since, rocket is the combination of a right circular cylinder and a cone.
Given, diameter of the cylinder = 6 cm
∴ Radius of the cylinder = `6/2` = 3 cm
And height of the cylinder = 12 cm
∴ Volume of the cylinder = πr2h
= 3.14 × (3)2 × 12
= 339.12 cm3
And curved surface area = 2πrh
= 2 × 3.14 × 3 × 12
= 226.08
Now, In right angled ΔAOC,
h = `sqrt(5^2 - 3^2)`
= `sqrt(25 - 9)`
= `sqrt(16)`
= 4
∴ Height of the cone, h = 4 cm
Radius of the cone, r = 3 cm
Now, volume of the cone
= `1/3 pir^2h`
= `1/3 xx 3.14 xx (3)^2 xx 4`
= `113.04/3`
= 37.68 cm3
And curved surface area = πrl
= 3.14 × 3 × 5
= 47.1
Hence, total volume of the rocket
= 339.12 + 37.68
= 376.8 cm3
And total surface area of the rocket
= CSA of cone + CSA of cylinder + Area of base of cylinder
= 47.1 + 226.08 + 28.26
= 301.44 cm2
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