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 cylinder. The diameter and height of cylinder are $6cm$ and $12cm$, respectively. If the slant height of the conical portion is $5cm$, then find the total surface area and volume of rocket. (Use $\pi =3.14$)

Answer 1

Cylinder

$r=\frac{6}{2}=3cm$

$H=12cm$

Cone
$l=5cm$

$r=3cm$

$\therefore l^2=r^2+h^2$ or $h^2=l^2-r^2$

$=5^2-3^2=25-9=16$

$\Rightarrow h=\sqrt{16}=4cm$

Now, volume of rocket

= Volume of cylinder + Volume of cone

$=\pi r^{2}H+\frac{1}{3}\pi r^{2}h=\pi r^2[H+\frac{1}{3}h]$

$=3.15\times 3\times 3[12+\frac{1}{3}\times 4]$

$=3.14\times 9\left[\frac{40}{3}\right]=3.14\times 3\times 40=376.8c{m}^3$.

$\therefore$ Volume of Rocket = $376.8c{m}^3$

Total surface area of rocket = Curved surface area of cylinder + Curved surface area of cone + Area of base of cylinder [As it is closed (Given)]

$=2\pi rH+\pi rl+\pi r^2=\pi r[2H+l+r]$

$=3.14\times 3[2\times 12+5+3]$

$=3.14\times 3\times 32$

$301.44c{m}^2$

Hence, the surface area of the rocket is $301.44c{m}^2$.