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 area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

A hollow cylindrical pipe is $21 \ cm$ long. If its outer and inner diameters are $10 \ cm$ and $6 \ cm$ respectively, them the volume of the metal used in making the pipe is $\displaystyle \left(Take\, \pi\, =\, \frac{22}{7}\right)$

  1. $1048\, cm^{3}$
  2. $1056\, cm^{3}$
  3. $1060\, cm^{3}$
  4. $1064\, cm^{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The pipe is in the shape of a hollow cylinder.
Volume of a hollow Cylinder of outer Radius "R", inner Radius ""r" and height "h" $ = \pi ({ R }^{ 2 }-{ r }^{ 2 })h $
Outer Radius $ = \frac {10}{2} = 5  cm $
Inner Radius $ = \frac {6}{2} = 3  cm $
Hence, volume of the pipe $ = \frac { 22 }{ 7 } \times ({ 5 }^{ 2 }-{ 3 }^{ 2 })\times 21 = 1056  {cm}^{3} $

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

A rectangular sheet of width $14$ m is rolled along its width and is converted to form a cylinder. Find the radius of cylinder.

  1. $\displaystyle \frac { 22 }{ 49 } $
  2. $\displaystyle \frac { 44 }{ 29 } $
  3. $\displaystyle \frac { 49 }{ 22 } $

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

Curved surface area of cylinder$=100m^2$

$\therefore 2 \pi r h = 100$
Here, width of the rectangle = height of the cylinder.
$\therefore h=14m$

$\therefore 2 \times \dfrac {22}{7} \times r \times 14 = 100$

$ \therefore r = \dfrac {100 \times 7}{2 \times 22 \times 14}$

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

What will be the Inner surface area of a spherical shell of inner radius $15\ cm$ and outer radius $16\ cm$? (Correct upto 2 decimal places)

  1. $706.86\ {cm^2}$
  2. $804.25\ {cm^2}$
  3. $2827.43\ {cm^2}$
  4. $3216.99\ {cm^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Inner surface area of a spherical shell = 4 * pi * r^2. With r = 15, Area = 4 * 3.14159 * 225 = 2827.43 cm^2.

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

The surface area of a solid sphere is always greater than the surface area of a hemisphere for the same value of radius.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let radius of solid sphere=radius of hemisphere$=r$
Then, S.A of solid sphere$=4\pi { r }^{ 2 }$
S.A of hemisphere$=2\pi { r }^{ 2 }$
$\therefore $S.A. of solid sphere$>$ S.A of hemisphere
Hence statement is true
Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

A hemispherical bowl has inner radius $5cm$ and outer radius $6cm$. What will be the volume of solid enclosed between the two hemispheres? (Correct upto 2 decimal places)

  1. $904.78 \ {cm}^{3}$
  2. $523.60 \ {cm}^{3}$
  3. $381.18 \ {cm}^{3}$
  4. $190.59 \ {cm}^{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The volume of hemispherical shell$= \dfrac{2}{3}\pi*R^{3}-\dfrac{2}{3}\pi*r^{3}$
where R and r are the outer and inner radius of the hemisphere
On solving the equation we get Volume$= 190.59 \ {cm}^{3}$

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

The surface area of a solid spherical ball of diameter $10\ cm$ is equal to :

  1. $25\pi\ {cm}^2$
  2. $50\pi\ {cm}^2$
  3. $100\pi\ {cm}^2$
  4. $200\pi\ {cm}^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given: Diameter of the sphere $= 10\ cm$
Hence, Radius ($r$) of the sphere will be $5\ cm$

We know that, 
Surface area of the sphere is $4\pi r^2$
Therefore, Area will be $4\pi (5)^2 = 100\pi\ {cm}^2$

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

What will be the Inner and Outer radius of a spherical shell of inner surface area $452.39\ {cm}^2$ and outer surface area $804.25\ {cm}^2$ ? (Surface areas are accurate upto 2 decimal places)

  1. $6 \ cm, 7 \ cm$
  2. $7 \ cm, 8 \ cm$
  3. $6 \ cm, 8 \ cm$
  4. $7 \ cm, 9\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Inner surface area$=4\pi { \left( inner\quad radius \right)  }^{ 2 }$
$\Rightarrow 4\pi { \left( { r } _{ 1 } \right)  }^{ 2 }=452.39cm^{2}\Rightarrow { r } _{ 1 }^{ 2 }=\cfrac { 452.39\times 7 }{ 4\times 22 } =35.99$
$\Rightarrow { r } _{ 2 }=\sqrt { 35.99 } =5.99\approx 6cm$
Outer surface area$=4\pi { \left( outer\quad radius \right)  }^{ 2 }$
$\Rightarrow 4\pi { \left( { r } _{ 2 } \right)  }^{ 2 }=804.25㎠\Rightarrow { r } _{ 2 }^{ 2 }=\cfrac { 804.25\times 7 }{ 4\times 22 } =63.97$
$\Rightarrow { r } _{ 2 }=\sqrt { 63.97 } =7.99\approx 8cm$
Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

The value of radius for which the numerical value of total surface area of a sphere and the volume of sphere are equal, will be:(Consider the units of volume and surface area as ${cm}^3\  \text{and}\ {cm}^2$)

  1. $1cm$
  2. $2cm$
  3. $3cm$
  4. $4cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let radius of sphere be $'r'㎝$, then
TSA of sphere=volume of sphere
$\Rightarrow 4\pi { r }^{ 2 }=\cfrac { 4 }{ 3 } \pi { r }^{ 3 }\Rightarrow r=3cm$
Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

The increase in the total surface area of a sphere of Radius ${R}$ when it is cut to make two hemispheres of same Radius will be equal to:

  1. $5\ \pi{R}^2$
  2. $4\ \pi{R}^2$
  3. $3\ \pi{R}^2$
  4. $2\ \pi{R}^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Total surface area of sphere$=4\pi { R }^{ 2 }$
TSA of hemisphere=CSA of hemisphere+CSA of circle
$=2\pi { R }^{ 2 }+\pi { R }^{ 2 }=3\pi { R }^{ 2 }$
$\therefore $TSA of two hemisphere$=2\times 3\pi { R }^{ 2 }=6\pi { R }^{ 2 }$
Therefore, increase in TSA$=6\pi { R }^{ 2 }-4\pi { R }^{ 2 }=2\pi { R }^{ 2 }$
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The radius of the sphere is measured as $ \left( {10 \pm 0.02} \right)cm$. The error in the measurement of its volume is 

  1. $25.1 cc$
  2. $25.21 cc$
  3. $2.51 cc$
  4. $251.2 cc$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $r$ be the radius of the sphere.


$\Rightarrow$  $r=10$


Error in the measurement of radius $=\Delta r$

$\therefore$  $\Delta r=0.02\,m$

$\Rightarrow$  Volume of the sphere $(V)=\dfrac{4}{3}\pi r^3$

We need to find error in calculating the volume that is $\Delta V$

$\Delta V=\dfrac{dv}{dr}\times \Delta r$

         $=\dfrac{d\left(\dfrac{4}{3}\pi r^3\right)}{dr}\times \Delta r$

         $=\dfrac{4}{3}\pi\dfrac{d(r^3)}{dr}\times \Delta r$

         $=\dfrac{4}{3}\pi(3r^2)\times (0.0.2)$

         $=4\pi r^2\times 0.02$

         $=4\times 3.14\times (10)^3\times 0.02$

         $=251.2\,cm^3$ i.e. $251.2\,cc$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If there is an error of $0.01 cm$ in the diameter of a sphere then percentage error in surface area when the radius $= 5 cm$, is

  1. $0.005\%$
  2. $0.05\%$
  3. $0.1\%$
  4. $0.2\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Surface area of sphere $S=4\pi r^{2}$
$\displaystyle S=\pi D^{2}$
$\Rightarrow S=100\pi$
Also, $ \displaystyle \frac{dS}{dD}=2\pi D=20\pi$
Approximate error in S is $\displaystyle dS=(\frac{dS}{dD})\Delta D$
                                         $ =20\pi (0.01)$
                                          $=\dfrac{1}{500} S$
                                           $=0.2$% of S
Percentage error in $S=0.2%$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If the radius of a sphere is measured as $9 \ cm$ with an error of $ 0.03 \ cm$ then, find the approximate error in calculating its volume.

  1. $\displaystyle 9.72\pi\:\: cm^{3}$
  2. $\displaystyle 7.92\pi\:\: cm^{3}$
  3. $\displaystyle 8.72\pi\:\: cm^{3}$

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

Given, $r=9 cm, \Delta r=0.03cm$

We know Volume of sphere with radius 'r' is $V=\cfrac{4}{3}\pi r^3$
$\therefore \Delta V=4\pi r^2\Delta r$
$\Rightarrow \Delta V=4\pi\times 81\times .03=9.72\pi  cm^3 $(using given values)

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

The rotational kinetic energy of a hollow spherical shell 2.5 J. If its frequency of rotation is made 10 times, then new kinetic energy will be -

  1. $0.25 J$
  2. $2.5\times { 10 }^{ 2 }J$
  3. $2500 J$
  4. $2.5 J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The rotational kinetic energy is given by
$K=\dfrac{1}{2}mr^2\omega ^2=2.5 J$. . . . .(1)
If frequency of rotation made 10 times , then the new rotational kinetic energy is
$K'=\dfrac{1}{2}mr^2 \times 10^2\omega ^2$
$K'=100K$ 
$K'=2.5\times 10^2J$
The correct option is B.
Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

Two spheres are having the ratio of their densities 1:3 and masses in 3:4 respectively. Find the ratio of their volumes. i.e $\dfrac{V _1}{V _2}$

  1. 1:3

  2. 3:1

  3. 4:9

  4. 9:4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle \frac{d _1}{d _2} = \frac{1}{3} ; \frac{m _1}{m _2} = \frac{3}{4}$
$d = m/v$
$\therefore \displaystyle \frac{V _1}{V _2}  = \frac{m _1}{m _2} \times \frac{d _2}{d _1} = \frac{3}{4} \times \frac{3}{1} = 9:4$