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Question 14
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 the slant height of the conical portion is 5 cm, find the total surface area and volume of the rocket [Use π=3.14]

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Solution

Since, a rocket is the combination of a right circular cylinder and a cone.
Given, diameter of the cylinder = 6 cm
Radius of the cylinder =62=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=l2r2

h=5232=259=16=4
Height of the cone, h =4 cm
Radius of the cone, r =3 cm
Now, volume of the cone
=13πr2h=13×3.14×(3)2×4=113.043=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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