Mathematics · Physics

Sphere Geometry

157 Questions

Sphere geometry deals with calculating the volume and surface area of round objects. It includes problems on spherical shells, recasting spheres, and understanding radius variations. These mathematical formulas are essential for competitive exam preparation.

Volume of a sphereSurface area calculationsSpherical shellsRadius ratio variationsRecasting spheres

Sphere Geometry Questions

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 has inner diameter $4$ cm and outer diameter $8$ cm. Determine the volume of the shell.

  1. $204.45 \space\ cm^3$
  2. $134.45 \space\ cm^3$
  3. $234.45 \space\ cm^3$
  4. $334.45 \space\ cm^3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Outer radius, $R = 4$ cm
Inner radius, $r  = 2$ cm
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (4^3-2^3)$
= $\cfrac{4}{3}\pi (64-8)$
= $\cfrac{4}{3}\pi (56)$
= $\cfrac{224 \pi}{3}$
= $74.666\pi $
= $234.45 \space\ cm^3$

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 spherical shell has a outer radius $14$ m and inner radius $7$ m. What's the volume of the sphere?

  1. $\approx 9000 \space\ m^3$
  2. $\approx 8000 \space\ m^3$
  3. $\approx 10000 \space\ m^3$
  4. $\approx 7000 \space\ m^3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Outer radius, $R = 14$ cm
Inner radius, $r  = 7$ cm
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (14^3-7^3)$
= $\cfrac{4}{3}\pi (2744-343)$
= $\cfrac{4}{3}\pi (2401)$
= $\cfrac{9604 \pi}{3}$
= $3201.33\pi $
$\approx 10000 \space\ m^3$

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

What is the volume of material that is needed to form a spherical shell whose outer radius is $5$ ft and whose inner radius is $3$ ft?

  1. $610.293 \space\ ft^3$
  2. $510.293 \space\ ft^3$
  3. $450.293 \space\ ft^3$
  4. $410.293 \space\ ft^3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$R = 5$ ft
$r  = 3$ ft
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (5^3-3^3)$
= $\cfrac{4}{3}\pi (125-27)$
= $\cfrac{4}{3}\pi (98)$
= $\cfrac{392 \pi}{3}$
= $130.666\pi $
= $410.293 \space\ ft^3$

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

Calculate the volume of the material used in the shell to the nearest unit. The inside radius of a spherical metal shell is $2.5$ cm and the outer radius of the shell is $5$ cm.

  1. $257.91 \space\ cm^3$
  2. $457.91 \space\ cm^3$
  3. $417.91 \space\ cm^3$
  4. $357.91 \space\ cm^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$R = 5$ cm
$r  = 2.5$ cm
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (5^3-2.5^3)$
= $\cfrac{4}{3}\pi (125-15.625)$
= $\cfrac{4}{3}\pi (109.375)$
= $\cfrac{437.5 \pi}{3}$
= $1312.5\pi $
= $457.91 \space\ cm^3$

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

Determine the volume of a spherical shell which has an inner radius of $6$ cm and an outer radius of $24$ cm.

  1. $44972 \space\ cm^3$
  2. $56972 \space\ cm^3$
  3. $66972 \space\ cm^3$
  4. $56000 \space\ cm^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Outer radius, $R = 24$ cm
Inner radius, $r  = 6$ cm
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (24^3-6^3)$
= $\cfrac{4}{3}\pi (13824-216)$
= $\cfrac{4}{3}\pi (13608)$
= $\cfrac{54432 \pi}{3}$
= $18144\pi $
= $56972 \space\ cm^3$

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

Find the volume of material that is needed to form a spherical shell whose outer radius is $3.0$ inches and whose inner radius is $0.1$ inches.

  1. $103.035 \space\ in^3$
  2. $93.035 \space\ in^3$
  3. $123.035 \space\ in^3$
  4. $113.035 \space\ in^3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$R = 3.0$ in
$r  = 0.1$ in
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (3^3-0.1^3)$
= $\cfrac{4}{3}\pi (27-0.001)$
= $\cfrac{4}{3}\pi (26.999)$
= $\cfrac{107.996 \pi}{3}$
= $35.998\pi $
= $113.035 \space\ in^3$

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 spherical shell of lead, whose external diameter is $24$ cm, is melted and recast into a right circular cylinder, whose height is $12$ cm and diameter $16$ cm. Determine the internal diameter of the shell.

  1. $8(18)^{1/3}$ cm
  2. $10$ cm
  3. $12$ cm
  4. $18(18)^{1/3}$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

  

Outer radius of the spherical lead $ = \dfrac {24}{2} = 12 $ cm 
Radius of the cylinder $ = \dfrac {16}{2} = 8 $ cm  
Since the spherical lead is recasted into the cylinder, their volumes are equal. 
Volume of a hollow sphere of outer radius $R$ and inner radius $r$ $ = \dfrac { 4 }{ 3 } \pi ({R}^{3} -{ r }^{ 3 }) $
Volume of a Cylinder of Radius "$R$" and height "$h$" $ = \pi { R }^{ 2 }h $
Hence, $ \dfrac { 4 }{ 3 } \pi ({12}^{3} -{ r }^{ 3 }) = \pi { 8 }^{ 2 } \times 12 $ 
Thus $ 1728 - { r }^{ 3 } = 576 $
$\Rightarrow  { r }^{ 3 } = 1152 $
$\Rightarrow  r = \sqrt [3] {1152} = 4 \sqrt [3] {18}   $ cm 
Inner diameter of the spherical lead $ = 2 \times \ \text{radius }= 2 \times 4 \sqrt [3] {18} $ cm $= 8 \sqrt [3] {18} $ cm

Multiple choice maths how much does it weigh? define weight and units of weight using decimals in weight conversion of length measurement (length) basic operations with same units operations involving units of length

How many smaller solid balls of radius 2 cm can be made by melting a solid sphere of radius 8 cm?

  1. 128

  2. 512

  3. 64

  4. 32
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Number of balls made $ = \dfrac {Volume  of   sphere } {Volume  of 

each  spherical  ball} $
Volume of a sphere of radius 'r' $ = \dfrac { 4 }{ 3 } \pi { r }^{ 3 } $

Hence, number of balls made $ =\dfrac { \dfrac { 4 }{ 3 } \pi \times  { 8 }^{ 3 } }{ \dfrac { 4 }{ 3 } \pi \times { 2 }^{ 3 } } = 64  $

Multiple choice maths how much does it weigh? define weight and units of weight using decimals in weight conversion of length measurement (length) basic operations with same units operations involving units of length

How many spherical bullets can be made out of a solid cube of lead whose edge measures $44\space cm$, each bullet being $4\space cm$ in diameter.

  1. $2451$
  2. $2541$
  3. $2304$
  4. $2536$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Number of spherical bullets formed $ = \dfrac {Volume  of 

cube}{Volume   of   one   spherical  bullet} $





Volume of a cube of edge a $ = {a}^{3} $





 Volume of a sphere of radius 'r' $ = \dfrac { 4 }{ 3 } \pi { r }^{ 3 } $





As the diameter of the sphere is $4$ cm, its radius r $ = 2$ cm





Hence, number of spherical bullets formed $ = \dfrac {44 \times 44 \times 44}{\dfrac {4}{3}

\times \dfrac {22}{7} \times 2 \times 2 \times 2} = 2541 $

Multiple choice capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

If the circumferences of a sphere is $2\ m$, then capacitance of sphere in water would be:

  1. $2700\ pF$
  2. $2760\ pF$
  3. $2780\ pF$
  4. $2846\ pF$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Capacitance is given as

$C=\varepsilon _0\frac{A}{d}$
For a sphere placed in water, the capacitance will be,
$C=4\pi \varepsilon R$
Here, $\varepsilon$ os the permittivity of water 
In terms of permittivity of free space and dielectric constant of water, we get 
$C=4\pi \varepsilon _0kR$
It is given that circumference is 2m
Hence, $c=2\pi R$  
$\therefore R=\frac{1}{\pi}$
$C=4\pi \varepsilon _0k\frac{1}{\pi}=4\varepsilon _0k$
$C=4\times 8.85\times10^{-12}\times80.4$
$C=2846\times 10^{-12}F$
$C=2846 pF$

Multiple choice maths how big? how heavy? measuring volume volume of solids volume of cube and cuboid

A hemi-spherical depression is cutout from one face of the cubical wooden block such that the diameter of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid
Answer required

  1. $=\frac {l^2}{4}[25+\pi)sq.units$.
  2. $=\frac {l^2}{5}[24+\pi)sq.units$.
  3. $=\frac {l^2}{4}[24+\pi)sq.units$.
  4. $=\frac {l^2}{3}[24+\pi)sq.units$.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total surface area of the cube after hemispherical depression
$=$ T.S.A. of cube -Base area of hemisphere + C.S.A of hemisphere
$=6(edge)^2-\pi r^2+2\pi r^2$

Multiple choice maths how big? how heavy? measuring volume volume of solids volume of cube and cuboid

A hemispherical bowl of internal diameter $36$ cm is full of some liquid. This liquid is to be filled in cylindrical bottles of radius $3$ cm and height $6$ cm, then no. of bottles needed to empty the bowl

  1. $36$
  2. $72$
  3. $18$
  4. $144$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Radius of the bowl $ = \dfrac {36}{2} = 18  cm $

Volume of the bowl $ = \dfrac { 2 }{ 3 }

\pi { r }^{ 3 } = \dfrac {2}{3} \times \pi \times 18 \times 18 \times

18 {cm}^{3} $





Volume

of a Cylinder of Radius "R" and height "h" $ = \pi { R }^{

2 }h $





Hence, Volume of one cylindrical bottle, $ = \pi \times 3 \times 3 \times  6 $

Hence, number of bottled required $

= \dfrac {Volume  of  bowl} {Volume  of  each  bottle} = \dfrac{\dfrac {2}{3} \times \pi \times 18 \times 18 \times

18}{ \pi \times 3 \times 3 \times 

6} = 72 $

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

The diameter of a metallic sphere is $6 cm$. It was melted to make a wire of diameter $4 mm$. Find the length of the wire.

  1. $90mm$
  2. $90cm$
  3. $9cm$
  4. $9m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume of metallic sphere = $\dfrac{4}{3} \times \pi \times {r^3}$

                                            = $\dfrac{4}{3} \times \pi \times 6^3$
Volume of cylindrical wire = $ \pi \times r^{2} \times h$

Now,
Volume of metallic sphere = Volume of cylindrical wire
$\therefore \dfrac{4}{3} \times \pi \times 6^3$ = $\pi \times 0.02^2 \times h$
$\therefore h = 900 cm=9 m$
 

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

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.

  1. $1000$.
  2. $1200$.
  3. $1100$.
  4. $100$.
Reveal answer Fill a bubble to check yourself
A Correct answer
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 } $


                            $=$ $1000$

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

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

  1. $\dfrac{2\pi}{3}r^3$
  2. $\dfrac{2\pi}{3}$
  3. $\dfrac{8\pi}{3}$
  4. $\dfrac{\pi}{12}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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