Question

A rocket is in the shape of a cylinder with a cone attached to one end and a hemisphere attached to the other. All of them are of the same radius of $1.5m$. The total length of the rocket is $7m$ and height of the cone is $2m$. Calculate the volume of the rocket.

Answer 1

Hint: In this question we use some basic formulas like volume of cylinder$ = \pi {r^2}{h_2}$, volume of cone $ = \pi {r^2}\dfrac{{{h_1}}}{3}$, Volume of a hemisphere $ = \dfrac{2}{3}\pi {r^3}$ . Add all this to find the volume of a rocket.

Complete step-by-step answer:

According to the question it is given ;
Radius $\left( r \right)$$ = 1.5m$, Total length$ = 7m$
Height of the cone $\left( {{h_1}} \right) = 2m$
Height of hemisphere $ = $Radius of hemisphere $ = 1.5m$
Height of cylinder$\left( {{h_2}} \right) = $Total length $ - {h_1} - r = 7 - 2 - 1.5 = 3.5m$
Now, volume of the rocket$ = $Volume of the cylinder $ + $volume of a cone$ + $volume of a hemisphere ……(i)
We know that, Volume of the cylinder $ = \pi {r^2}{h_2}$
                                                                       $ = \pi {\left( {1.5} \right)^2} \times 3.5$
                                                                      $ = 7.875\pi {m^3}$
Volume of a cone $ = \pi {r^2}\dfrac{{{h_1}}}{3}$
                                  $
   = \pi {\left( {1.5} \right)^2} \times \dfrac{2}{3} \\
   = 1.5\pi {m^3} \\
 $
Volume of a hemisphere $ = \dfrac{2}{3}\pi {r^3}$
                                              $
   = \dfrac{2}{3}\pi \times {\left( {1.5} \right)^3} \\
   = 2.25\pi {m^3} \\
 $
From (i)
Volume of the rocket $ = 7.875\pi + 1.5\pi + 2.25\pi $
                                        $
   = 11.625\pi {m^3} \\
   = 11.625 \times \dfrac{{22}}{7}{m^3} \\
   = 36.521{m^3} \\
$

Note: In such types of questions where one shape is combined of two or more shapes then we will apply the concept of volume of combined shapes where the volumes of all the shapes will be added . So it is advisable to remember the basic formulas while involving in combined shapes questions.