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 $$\pi = 3.14$$)
Cylinder
$$r=\dfrac{6}{2}=3cm$$
$$H=12cm$$
Cone
$$l=5cm$$
$$\therefore l^{2}=r^{2}+h^{2}$$ 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
$$=\pi r^{2}H+\dfrac{1}{3}\pi r^{2}h=\pi r^{2}\left [ H+\dfrac{1}{3}h \right ]$$
$$=3.15\times3\times3\left [ 12+\dfrac{1}{3}\times4 \right ]$$
$$=3.14\times9\left [ \dfrac{40}{3} \right ]=3.14\times3\times40=376.8cm^{3}$$.
$$\therefore$$ Volume of Rocket = $$376.8cm^{3}$$
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)]
$$=2\pi rH+\pi rl+\pi r^{2}=\pi r[2H+l+r]$$
$$=3.14\times3[2\times12+5+3]$$
$$=3.14\times3\times32$$
$$301.44cm^{2}$$
Hence, the surface area of the rocket is $$301.44cm^{2}$$.
A rocket is in the form of cylinder closed at lower end with a cone of same diameter attached to the top (see fig.). The cylinder is of diameter and height whereas conical top has slant height . Calculate the volume of the rocket.