Tag: solids

Questions Related to solids

From a solid sphere of radius $R$, a concentric solid sphere of radius $\dfrac{R}{2}$ is removed. The total surface area increases by

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

  1. $0\%$

  2. $25\%$

  3. $50\%$

  4. $75\%$

Show answer
Correct Option: B
Explanation:
Solution:- (B) $25 \%$
Initial area of sphere $ \left( {A} _{1} \right) = 4 \pi {R}^{2}$
New area of sphere $\left( {A} _{2} \right) = 4 \pi {R}^{2} + 4 \pi {\left( \cfrac{R}{2} \right)}^{2} = 5 \pi {R}^{2}$
$\therefore$ Increase in area $= \cfrac{{A} _{2} - {A} _{1}}{{A} _{1}} \times 100$
$\Rightarrow$ Increase in area $= \cfrac{5 \pi {R}^{2} - 4 \pi {R}^{2}}{4 \pi {R}^{2}} \times 100 = 25 \%$
Hence the area will be increased by $25 \%$.

If the circumference of the inner edge of a hemispherical bowl is $\displaystyle\frac{132}{7}:cm$, then what is the capacity?

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

  1. $12\pi:cm^3$

  2. $18\pi:cm^3$

  3. $24\pi:cm^3$

  4. $36\pi:cm^3$

Show answer
Correct Option: B
Explanation:

$\displaystyle2\pi r=\frac{132}{7}\implies r=3$

$\therefore$ Capacity $\displaystyle=\frac{2}{3}\pi r^3=18\pi$

The side of a cube is equal to diameter of the sphere. The ratio of volumes 
of cube and sphere is

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

  1. $\frac{11}{12}$

  2. $\frac{22}{11}$

  3. $\frac{11}{21}$

  4. $\frac{21}{11}$

Show answer
Correct Option: D
Explanation:

Let the sides of cube be $s$ and radius of sphere be $r$

then $s=2r=(2r)^3=8r^3$
Volume of a cube$=s^3$
Volume of a sphere$=\cfrac{4}{3}\pi r^3$
Ratio of volume of cube and sphere$=\cfrac{8r^3}{\cfrac{4}{3}\pi r^3}$
$=\cfrac{8r^3}{\cfrac{4}{3}\times \cfrac{22}{7} r^3}\=\cfrac{21}{11}$

A hollow spherical shell is made of metal of density $4.8$ g/cm$^3$. If its internal and external radii are $10$ cm and $12$ cm respectively, find the weight of the shell

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

  1. $15.24 $ kg

  2. $12.84 $ kg

  3. $14.64 $ kg

  4. None of these

Show answer
Correct Option: C
Explanation:

Volume of spherical shell
$= \displaystyle \frac{4 \pi}{3} (R^3 - r^3) = \frac{4 \pi}{3} (12^3 - 10^3)$
$=\displaystyle \frac{4}{3} \times \pi \times (12 -10) (12^2 +12 \times 10 +10^2)$
$=\displaystyle \frac{4}{3} \times \pi \times 2 \times 264 cm^3$
$Weight = volume \times density = \displaystyle \frac{4}{3}\times \pi \times 364 \times 4.8 = 14.64 kg$

The radius of a sphere is r and radius of base of a cylinder is r and height is 2r. The ratio of their volumes will be-

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

  1. $2:3$

  2. $3:4$

  3. $4:3$

  4. $3:2$

Show answer
Correct Option: A
Explanation:

Given,

The radius of the sphere $=r$

The radius of the cylinder $=r$

The height of the cylinder $=2r$

 

We know that the volume of the sphere

${{V} _{1}}=\dfrac{4}{3}\pi {{r}^{3}}$

 

We know that the volume of the cylinder

$ {{V} _{2}}=\pi {{r}^{2}}h $

$ {{V} _{2}}=\pi {{r}^{2}}\left( 2r \right) $

$ {{V} _{2}}=2\pi {{r}^{3}} $

 

Therefore, the required ratio

$ \dfrac{{{V} _{1}}}{{{V} _{2}}}=\dfrac{\dfrac{4}{3}\pi {{r}^{3}}}{2\pi {{r}^{3}}} $

$ \dfrac{{{V} _{1}}}{{{V} _{2}}}=\dfrac{2\pi {{r}^{3}}}{3\pi {{r}^{3}}} $

$ \dfrac{{{V} _{1}}}{{{V} _{2}}}=\dfrac{2}{3} $

$ {{V} _{1}}:{{V} _{2}}=2:3 $

 

Hence, this is the answer.

The volume of a spherical shell whose internal and external diameters are $8cm$ and $10cm$ respectively (in cubic cm) is:

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

  1. $\cfrac{122\pi}{3}$

  2. $\cfrac{244\pi}{3}$

  3. $212$

  4. $257$

Show answer
Correct Option: B
Explanation:

Given, internal diameter $=8cm$ and external diameter $=10cm$
Volume of a hollow sphere of outer Radius R and inner radius r $ = \frac { 4 }{ 3 } \pi ({R}^{2} -{ r}^{ 3 }) $
Inner radius of the spherical shell $ = \frac {8}{2} = 4 cm $
Outer radius of the spherical shell $ = \frac {10}{2} = 5  cm $
Hence, volume of spherical shell $ = \frac { 4 }{ 3 } \times \pi \times ({5}^{3} - {4}^{3}) = \frac { 4 }{ 3 } \times \pi  \times 61 = \frac {244\pi}{3}  { cm }^{ 3 }  $

A metallic hemispherical bowl is $0.25\;cm$ thick. The inside radius of the bowl is $5\;cm$. Find the volume of steel used in making the bowl.

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

  1. $43.25\;cm^3$

  2. $41.27\;cm^3$

  3. $42.25\;cm^3$

  4. $40.25\;cm^3$

Show answer
Correct Option: B
Explanation:

Volume of speed used $=$ $\dfrac { 2 }{ 3 } \pi \left( { r } _{ 1 }^{ 3 }-{ r } _{ 2 }^{ 3 } \right) $

${ r } _{ 1 }=5+0.25=5.25cm$
${ r } _{ 2 }=5cm$
$\therefore \quad $ Volume $=$ $\dfrac { 2 }{ 3 } \times \dfrac { 22 }{ 7 } \times \left( { \left( 5.25 \right)  }^{ 3 }-{ \left( 5 \right)  }^{ 3 } \right) $

                       $= \dfrac { 44 }{ 21 } \times \left( 144.70-125 \right) $

                       $= 41.27$ ${ cm }^{ 3 }$

A metallic spherical shell of internal and external diameters $8 cm$ and $12 cm$, respectively is melted and recast into the form of a cone of base diameter $8 cm$. The height of the cone is

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

  1. $114 cm$

  2. $76 cm$

  3. $38 cm$

  4. $19 cm$

Show answer
Correct Option: C
Explanation:

Volume of a hollow sphere of outer Radius R and inner radius r $ =

\cfrac { 4 }{ 3 } \pi ({R}^{3} -{ r }^{ 3 }) $

Inner radius of the spherical shell $ = \cfrac {8}{2} = 4 cm $

Outer radius of the spherical shell $ = \cfrac {12}{2} = 6  cm $

Volume of a cone $ = \cfrac { 1 }{ 3 } \pi { r }^{ 2 }h $  where r is the

radius of the base of the cone and h is the height.
Radius of the cone $ = \cfrac{8}{2} = 4  cm $

Now, Volume of the hollow sphere $ = $ Volume of cone
$ => \cfrac { 4 }{ 3 } \pi ({6}^{3} -{ 4 }^{ 3 }) = \cfrac { 1 }{ 3 } \pi { 4 }^{ 2 }h $
$ => 4 \times(216-64) = 16h $
$ h = 38  cm $

The radius of the smaller circle is $2$ m and the radius of the larger circle is $10$ m. What is the volume of the of the spherical shell inscribed in the larger circle?

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

  1. $3153.17 \space\ m^3$

  2. $4153.17 \space\ m^3$

  3. $2153.17 \space\ m^3$

  4. $153.17 \space\ m^3$

Show answer
Correct Option: B
Explanation:

$R = 10 m$$
$r  = 2 m$
Volume $=\cfrac{4}{3}\pi (R^3-r^3)$
$=\cfrac{4}{3}\pi (10^3-2^3)$

$=\cfrac{4}{3}\pi (1000-8)$
$=\cfrac{4}{3}\pi (992)$
$=\cfrac{3968 \pi}{3}$
$=1322.66\pi $
$=4153.17 \space\ m^3$

The radius of the smaller circle is 4 cm and the radius of the larger circle is 8 cm. Find the volume of the of the spherical shell inscribed in the larger circle.

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

  1. $1875.62 \space\ cm^3$

  2. $875.62 \space\ cm^3$

  3. $2875.62 \space\ cm^3$

  4. $3875.62 \space\ cm^3$

Show answer
Correct Option: A
Explanation:

R $= 8\ cm$
r  $= 4\ cm$
Volume = $\dfrac{4}{3}\pi (R^3-r^3)$


= $\dfrac{4}{3}\pi (8^3-4^3)$

= $\dfrac{4}{3}\pi (512-64)$

= $\dfrac{4}{3}\pi (448)$

= $\dfrac{1792 \pi}{3}$
= $1875.62 \space\  cm^3$