Tag: application of surface area and volume of solids

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Questions Related to application of surface area and volume of solids

Multiple choice 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

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

  1. $0\%$
  2. $25\%$
  3. $50\%$
  4. $75\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $25 \%$.
Multiple choice 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

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

  1. $12\pi\:cm^3$
  2. $18\pi\:cm^3$
  3. $24\pi\:cm^3$
  4. $36\pi\:cm^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

Multiple choice 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

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

  1. $\frac{11}{12}$
  2. $\frac{22}{11}$
  3. $\frac{11}{21}$
  4. $\frac{21}{11}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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$=\cfrac{8r^3}{\cfrac{4}{3}\pi r^3}$
$=\cfrac{8r^3}{\cfrac{4}{3}\times \cfrac{22}{7} r^3}\=\cfrac{21}{11}$

Multiple choice 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

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

  1. $15.24 $ kg
  2. $12.84 $ kg
  3. $14.64 $ kg

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

Multiple choice 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

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-

  1. $2:3$
  2. $3:4$
  3. $4:3$
  4. $3:2$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice 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

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

  1. $\cfrac{122\pi}{3}$
  2. $\cfrac{244\pi}{3}$
  3. $212$
  4. $257$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 }  $

Multiple choice 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

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.

  1. $43.25\;cm^3$
  2. $41.27\;cm^3$
  3. $42.25\;cm^3$
  4. $40.25\;cm^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
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) $

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

Multiple choice 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

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

  1. $114 cm$
  2. $76 cm$
  3. $38 cm$
  4. $19 cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $

Multiple choice 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

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?

  1. $3153.17 \space\ m^3$
  2. $4153.17 \space\ m^3$
  3. $2153.17 \space\ m^3$
  4. $153.17 \space\ m^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $
$=4153.17 \space\ m^3$

Multiple choice 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

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.

  1. $1875.62 \space\ cm^3$
  2. $875.62 \space\ cm^3$
  3. $2875.62 \space\ cm^3$
  4. $3875.62 \space\ cm^3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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


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

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

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

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