Question

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 $\pi =3.14]$

Answer 1

Since, a rocket is the combination of a right circular cylinder and a cone.
Given, diameter of the cylinder = 6 cm
$\therefore$ Radius of the cylinder =$\frac{6}{2}=3cm$
And height of the cylinder = 12 cm.
$\therefore$ Volume of the cylinder =$\pi r^{2}h=3.14\times (3{)}^2\times 12$
$=339.12c{m}^3$
And curved surface area =$2\pi rh$
$=2\times 3.14\times 3\times 12=226.08$
Now, in right angled $\Delta AOC$,
$h=\sqrt{l^2-r^2}$
$h=\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4$
$\therefore$ Height of the cone, h =4 cm
Radius of the cone, r =3 cm
Now, volume of the cone
$=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times 3.14\times (3{)}^2\times 4=\frac{113.04}{3}=37.68c{m}^3$
and curved surface area = $\pi rl=3.14\times 3\times 5=47.1$
Hence, total volume of the rocket
$=339.12+37.68=376.8c{m}^3$
and total surface area of the rocket = CSA of cone + CSA of cylinder + Area of base of cylinder
=47.1+226.08 + 28.26
$=301.44c{m}^2$