Tag: problems involving volume of combined solids

Questions Related to problems involving volume of combined solids

A metallic solid cone is melted and cast into a cylinder of the same base as that of the cone. If the height of the cylinder is $7\;cm$, what was the height of the cone?

maths surface areas and volumes volume of a sphere problems involving volume of combined solids application of surface area and volume of solids

  1. $20\;cm$

  2. $21\;cm$

  3. $22\;cm$

  4. $12\;cm$

Show answer
Correct Option: B
Explanation:

Volume of cone $=$ Volume of cylinder

$\dfrac { 1 }{ 3 } \pi { r } _{ 1 }^{ 2 }{ h } _{ 1 }=\pi { r } _{ 2 }^{ 2 }{ h } _{ 2 }$
$\Rightarrow \quad \dfrac { 1 }{ 3 } \times { r }^{ 2 }\times { h } _{ 1 }={ r }^{ 2 }\times 7\quad \left[ { r } _{ 1 }={ r } _{ 2 }=r \right] $
                  $\Rightarrow \quad { h } _{ 1 }=7\times 3=21cm$
Therefore, height of cone is $21$ cm.

A solid sphere of radius $6\;cm$ is melted and recast into small spherical balls each of diameter $1.2\;cm$. Find the number of balls, thus obtained.

maths surface areas and volumes volume of a sphere problems involving volume of combined solids application of surface area and volume of solids

  1. $1000$.

  2. $1200$.

  3. $1100$.

  4. $100$.

Show answer
Correct Option: A
Explanation:

Number of balls $=$ $\dfrac { Volume\quad of\quad solid\quad sphere }{ Volume\quad of\quad 1\quad small\quad ball } $


                            $=$ $\dfrac { \dfrac { 4 }{ 3 } \pi { R }^{ 3 } }{ \dfrac { 4 }{ 3 } \pi { r }^{ 3 } } =\dfrac { 6\times 6\times 6 }{ 0.6\times 0.6\times 0.6 } $


                            $=$ $1000$

If the circumference of base of a hemisphere is $2\pi$ then its volume is _________ $cm^3$.

maths surface areas and volumes volume of a sphere problems involving volume of combined solids application of surface area and volume of solids

  1. $\dfrac{2\pi}{3}r^3$

  2. $\dfrac{2\pi}{3}$

  3. $\dfrac{8\pi}{3}$

  4. $\dfrac{\pi}{12}$

Show answer
Correct Option: B
Explanation:

The circumference of base of a hemisphere $=2\pi$
$\therefore 2\pi r=2 \pi$
$\therefore r=1$
Its volume $= \dfrac{2}{3}\pi (r)^3$
$= \dfrac{2}{3}\pi (1)^3 =\dfrac{2}{3}\pi$