Mathematics · Physics

Sphere Geometry

183 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 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

The capacitance of a spherical condenser is $1mF$. If the spacing between the two spheres is $1mm$, then the radius of the outer sphere is

  1. $30cm$
  2. $6m$
  3. $5cm$
  4. $3m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} From\, \, the\, question \\ C=\dfrac { { 4\pi { \varepsilon _{ 0 } } } }{ { \left[ { \dfrac { 1 }{ { { r _{ in } } } } -\dfrac { 1 }{ { { r _{ out } } } }  } \right]  } } =\dfrac { { 4\pi \varepsilon  } }{ { \left[ { \dfrac { 1 }{ { { r _{ 1 } } } } -\dfrac { 1 }{ { { r _{ 1 } }+0.001 } }  } \right]  } }  \\ 1\times { 10^{ -6 } }=\dfrac { { 4\times 3.14\times 8.854\times { { 10 }^{ -12 } } } }{ { \left[ { \dfrac { 1 }{ { { r _{ 1 } } } } -\dfrac { 1 }{ { { r _{ 1 } }+0.001 } }  } \right]  } }  \\ \left[ { \dfrac { 1 }{ { { r _{ 1 } } } } -\dfrac { 1 }{ { { r _{ 1 } }+0.001 } }  } \right] =\dfrac { { 4\times 3.14\times 8.854\times { { 10 }^{ -12 } } } }{ { 1\times { { 10 }^{ -6 } } } }  \\ \dfrac { { \left[ { \left( { { r _{ 1 } }+0.001 } \right) -{ r _{ 1 } } } \right]  } }{ { { r _{ 1 } }\times \left( { { r _{ 1 } }+0.001 } \right)  } } =4\times 3.14\times 8.854\times { 10^{ -6 } } \\ { r _{ 1 } }\times \left( { { r _{ 1 } }+0.001 } \right) =\dfrac { { 4\times 3.14\times 8.854\times { { 10 }^{ -6 } } } }{ { 0.001 } }  \\ r _{ 1 }^{ 2 }+0.001{ r _{ 1 } }-4\times 3.14\times 8.854\times { 10^{ -3 } }=0 \\ r _{ 1 }^{ 2 }+0.001{ r _{ 1 } }-0.1112=0 \\ { r _{ 1 } }=0.333m\, \, \, or\, \, { r _{ 1 } }=-334m \\ Since,\, it\, cannot\, be\, negative \\ Thereforem\, radius\, \, of\, outer\, \, sphere\, ={ r _{ 1 } }+0.001 \\ { r _{ outer } }=0.334m \\ or,\, { r _{ 1 } }=33.4cm \\  \end{array}$

Hence, the option $A$ is the correct answer.

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

The capacitance of a spherical condenser is $1mF$. If the spacing between the two spheres is $1mm$, then the radius of the outer space is

  1. $30cm$
  2. $6m$
  3. $5cm$
  4. $3m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\begin{array}{l} C=\frac { { 4\pi { E _{ 0 } } } }{ { \left( { \frac { 1 }{ { { r _{ i } } } }  } \right) -\left( { \frac { 1 }{ { { r _{ 0 } } } }  } \right)  } } .....................\left( 1 \right)  \ According\, \, to\, \, the\, \, question:- \ { r _{ 0 } }-{ r _{ 1 } }=0.001\, m \ C=0.00000\, 1F..............\left( 2 \right)  \ Putting\, \, \left( 2 \right) \, \, in\, \, \, \left( 1 \right)  \ \therefore r _{ 0 }^{ 2 }-{ r _{ 0 } }\left( { 0.001 } \right) -\left( { 9\times 000000000\times 0.000000001 } \right) =0 \ therefore\, \, { r _{ 0 } }=3\, \, m \end{array}$

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

The capacitance (C) for an isolated conducting sphere of radius(a) is given by $4\pi \varepsilon _0a$. If the sphere is enclosed with an earthed concentric sphere, the ratio of the radii of the spheres being $\dfrac{n}{(n-1)}$ then the capacitance of such a sphere will be increased by a factor?

  1. $n$
  2. $\dfrac{n}{(n-1)}$
  3. $\dfrac{(n-1)}{n}$
  4. $an$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Capacitance of isolated sphere C1 = 4 * pi * epsilon_0 * a. Capacitance of spherical capacitor C2 = 4 * pi * epsilon_0 * a * b / (b - a). Given b/a = n/(n-1), then b = a * n / (n-1). Substituting gives C2 = C1 * n.

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 botany cell theory, cell specialization, and cell replacement number and shape of cells unicellular and multicellular organisms diversity and classification in animals

When a cell of $2\mu$$m$ diameter grows to double its diameter, its surface area: volume relationship will _____________.

  1. Remain the same.

  2. Become double.

  3. Reduce to half.

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

Surface area (SA) is proportional to r^2 and volume (V) to r^3. The ratio SA/V is proportional to 1/r. If the diameter (and thus radius) doubles, the ratio becomes 1/2 of the original.

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 $= \dfrac{2}{3}\pi (r)^3$
$= \dfrac{2}{3}\pi (1)^3 =\dfrac{2}{3}\pi$