Chapter 21 : Surface Areas and Volume of a Sphere - Rd Sharma Solutions for Class 9 Maths CBSE

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Chapter 21 - Surface Areas and Volume of a Sphere Excercise Ex. 21.1

Question 1
Solution 1
Question 2
Solution 2
Question 3

Solution 3
Question 4
Solution 4
Question 5
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 4 per 100 cm2.
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8


Question 9

The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Solution 9

Question 10

A hemi-spherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 cm, find the cost of painting, if given the cost of painting is Rs 5 per 100 cm2.

Solution 10


Question 11

Solution 11



Question 12

Solution 12



Question 13
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the given figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.
 
                                      
Solution 13

Chapter 21 - Surface Areas and Volume of A Sphere Excercise Ex. 21.2

Question 1
Solution 1
Question 2
Solution 2

Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6

Solution 6

Question 7
Solution 7
Question 8
Solution 8

Question 9
If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere?
Solution 9

Question 10

A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively. Find the height to which the water will rise in the cylinder.

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

The diameter of a coper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter.

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

A cube of side 4 cm contained a sphere touching its sides. Find the volume of the gap in between.

Solution 21

Question 22

A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Solution 22
Inner radius (r1) of hemispherical tank  = 1 m
     Thickness of hemispherical tank       = 1 cm = 0.01 m
Outer radius (r2) of hemispherical tank = (1 + 0.01) m = 1.01 m
Volume of iron used to make the tank  =
                                                           
Question 23

A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?

Solution 23
Radius (r) of capsule  
Volume of spherical capsule
                                       
Thus, approximately 22.46 mm3 of medicine is required to fill the capsule.

Question 24

The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Solution 24
    Let diameter of earth be d. So, radius earth will be  .
    Then, diameter of moon will be  . So, radius of moon will be  .
    Volume of moon =    
    Volume of earth =   
      
    Thus, the volume of moon is  of volume of earth.

Question 25

Solution 25

Question 26

A cylinderical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball?

Solution 26

Question 27

A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Use = 22/7)

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Chapter 21 - Surface Areas and Volume of a Sphere Excercise 21.26

Question 1

In a sphere the number of faces is

(a) 1

(b) 2

(c) 3

(d) 4

Solution 1

Sphere has only one surface i.e. curved surface, so number of faces = 1

Hence, correct option is (a).

Question 2

The total surface area of a hemisphere of radius r is

(a) ∏r2

(b) 2∏r2

(c) 3∏r2

(d) 4∏r2

Solution 2

A hemisphere has two surfaces: one top surface and other curved surface.

T.S.A. = 2∏r2 + (∏r2)   {Area of Top-face = ∏r2}

         = 3∏r2

Question 3

The ratio of the total surface area of a sphere and a hemisphere of same radius is

(a) 2 : 1

(b) 3 : 2

(c) 4 : 1

(d) 4 : 3

Solution 3

begin mathsize 12px style Total space Surface space Area space of space Sphere space equals 4 πr squared space
Total space surface space Area space of space Hemisphere equals space 3 πr squared
therefore space Required space Ratio space equals space fraction numerator 4 πr squared space over denominator 3 πr squared space end fraction equals 4 over 3 equals space 4 space colon space 3
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 4

A sphere and a cube are of the same height. The ratio of their volumes is

(a) 3 : 4

(b) 21 : 11

(c) 4 : 3

(d) 11 : 21

Solution 4

begin mathsize 12px style Height space of space sphere space equals space Diameter space equals space 2 straight r
Height space of space cube space equals space Side space of space cube space equals space Height space of space sphere space equals space 2 straight r
Volume space of space sphere space equals space 4 over 3 πr cubed
Volume space of space cube space equals space left parenthesis 2 straight r right parenthesis cubed space equals space 8 straight r cubed
Ratio space of space their space volumes space equals space fraction numerator begin display style 4 over 3 πr cubed end style over denominator 8 straight r cubed end fraction equals straight pi over 6 equals fraction numerator up diagonal strike 22 to the power of 11 over denominator 7 cross times up diagonal strike 6 subscript 3 end strike end fraction equals space 11 over 21 equals 11 space colon space 21
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 5

The largest sphere is cut off from a cube of side 6 cm. The volume of the sphere will be

(a) 27∏ cm3

(b) 36∏ cm3

(c) 108∏ cm3

(d) 12∏ cm3

Solution 5

begin mathsize 12px style The space largest space sphere space that space can space be space cut space from space straight a space cube space of space side space 6 space cm space will space have space
its space diameter space equals space side space of space cube
straight i. straight e. space 2 straight r space equals space 6 space cm space rightwards double arrow space straight r space equals space 3 space cm
Volume space of space that space sphere space equals space 4 over 3 πr cubed space equals fraction numerator 4 over denominator up diagonal strike 3 end fraction straight pi cross times up diagonal strike 3 cross times 3 cross times 3 equals space 36 straight pi space cm cubed
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. space end style

Question 6

A cylindrical rod whose height is 8 times of its radius is melted and recast into spherical balls of same radius. The number of balls will be

(a) 4

(b) 3

(c) 6

(d) 8

Solution 6

begin mathsize 12px style Volume space of space cylindrical space rod space equals space πr squared straight h
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space πr squared space left parenthesis 8 straight r right parenthesis space space space left square bracket straight h space equals space 8 straight r space left parenthesis given right parenthesis right square bracket
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 8 πr cubed
Now comma space if space spherical space balls space have space same space radius comma space then space the space volume space of space one space ball space equals space 4 over 3 πr cubed
therefore space No. space of space balls space equals space fraction numerator Volume space of space Cylindrical space Rod over denominator Volume space of space one space Rod end fraction equals fraction numerator 8 up diagonal strike πr cubed end strike over denominator begin display style 4 over 3 up diagonal strike πr cubed end strike end style end fraction equals space 6
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 7

If the ratio of volumes of two sphere is 1 : 8, then the ratio of their surface areas is

(a) 1 : 2

(b) 1 : 4

(c) 1 : 8

(d) 1 : 16

Solution 7

begin mathsize 12px style Volume space of space sphere space equals space 4 over 3 πr cubed space equals space straight V
straight V subscript 1 over straight V subscript 2 equals fraction numerator up diagonal strike begin display style 4 over 3 straight pi end style end strike straight r subscript 1 superscript 3 over denominator up diagonal strike 4 over 3 straight pi end strike straight r subscript 2 superscript 3 end fraction equals fraction numerator straight r subscript 1 superscript 3 over denominator straight r subscript 2 superscript 3 end fraction equals 1 over 8
rightwards double arrow straight r subscript 1 over straight r subscript 2 equals 1 half
Now comma space Surface space Area space of space Sphere space equals space 4 πr squared space equals space straight S
straight S subscript 1 over straight S subscript 2 equals fraction numerator 4 πr subscript 1 superscript 2 over denominator 4 πr subscript 2 superscript 2 end fraction equals open parentheses straight r subscript 1 over straight r subscript 2 close parentheses squared space equals space open parentheses 1 half close parentheses squared space equals space 1 fourth space equals space 1 space colon space 4
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 8

If the surface area of a sphere is 144∏ m2, then its volume (in m3) is

(a) 288 ∏

(b) 316 ∏

(c) 300 ∏

(d) 188 ∏

Solution 8

begin mathsize 12px style Surface space Area space of space Sphere space rightwards double arrow space 4 πr squared space equals space 144 straight pi
rightwards double arrow straight r squared space equals space 36 space rightwards double arrow space straight r space equals space 6
Volume space of space Sphere equals 4 over 3 πr cubed space equals space 4 over 3 straight pi open parentheses 6 close parentheses cubed space equals space 288 straight pi space straight m cubed
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. space end style

Question 9

If a solid sphere of radius 10 cm is moulded into 8 spherical solid balls of equal radius, then the surface area of each ball (in sq. cm) is

(a) 100∏

(b) 75∏

(c) 60∏

(d) 50∏

Solution 9

begin mathsize 12px style Volume space of space solid space sphere space equals 4 over 3 straight pi space open parentheses 10 close parentheses cubed space equals space fraction numerator 4000 straight pi over denominator 3 end fraction space cm cubed
Volume space 8 space solid space spheres space of space radius space left parenthesis say right parenthesis space straight r space equals space 8 space cross times 4 over 3 πr cubed equals fraction numerator 32 πr cubed over denominator 3 end fraction space cm cubed
Now comma space fraction numerator 32 up diagonal strike straight pi straight r cubed over denominator up diagonal strike 3 end fraction equals fraction numerator 4000 up diagonal strike straight pi over denominator up diagonal strike 3 end fraction
rightwards double arrow straight r space equals space open parentheses 1000 over 8 close parentheses to the power of bevelled 1 third end exponent space equals space 10 over 2 space equals space 5 space cm
Surface space Area space of space each space small space ball space equals space 4 πr squared space equals space 4 straight pi space open parentheses 5 close parentheses squared space equals 100 straight pi space cm squared
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Chapter 21 - Surface Areas and Volume of a Sphere Excercise 21.27

Question 1

If a sphere is inscribed in a cube, then the ratio of the ratio of the volume of the sphere to the volume of the cube is

(a) ∏ : 2

(b) ∏ : 3

(c) ∏ : 4

(d) ∏ : 6

Solution 1

begin mathsize 12px style Edge space of space cube space equals space straight a
rightwards double arrow Volume space of space cube space equals space straight a cubed
If space Sphere space is space inscribed space inside space cube space then space straight a space equals space 2 straight r space rightwards double arrow straight r space equals space straight a over 2
Volume space of space sphere space equals space 4 over 3 πr cubed equals space 4 over 3 straight pi open parentheses straight a over 2 close parentheses cubed space equals space straight pi over 6 straight a cubed
Ratio space of space volume space of space sphere space to space volume space of space cube space equals space fraction numerator begin display style straight pi over 6 straight a cubed end style over denominator straight a cubed end fraction equals straight pi over 6
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 2

If a solid sphere of radius r is melted and cast into the shape of a solid cone of height r, then the radius of the base of the cone is

(a) 2r

(b) 3r

(c) r

(d) 4r

Solution 2

begin mathsize 12px style Volume space of space sphere space equals space 4 over 3 πr cubed
Sphere space casted space into space straight a space cone space of space height space straight r.
Let space the space radius space of space cone space equals space straight R
therefore space Volume space of space cone space equals space 1 third πR squared open parentheses straight r close parentheses
Volume space of space cone space equals space Volume space of space sphere
rightwards double arrow 1 third πR squared straight r space space equals space 4 over 3 πr cubed
rightwards double arrow straight R squared space equals space 4 straight r squared
rightwards double arrow straight R space equals space 2 straight r
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 3

begin mathsize 12px style straight A space sphere space is space placed space inside space straight a space right space circular space cylinder space so space as space to space touch space the space top comma space base space and space lateral space surface space of space the space cylinder.
If space the space radius space of space the space sphere space is space straight r comma space then space the space volume space of space the space cylinder space is
left parenthesis straight a right parenthesis space 4 straight pi space straight r cubed
left parenthesis straight b right parenthesis space 8 over 3 straight pi space straight r cubed
left parenthesis straight c right parenthesis space 2 straight pi space straight r cubed
left parenthesis straight d right parenthesis space 8 straight pi space straight r cubed end style

Solution 3

Radius of sphere = r

Sphere touches cylinder at Top, Base and Lateral Surface.

Then,

2r = height of cylinder = h

r = Radius of cylinder

Volume of cylinder = ∏r2h

                         =∏r2(2r)

                         = 2∏r3

Hence, correct option is (c).


Question 4

The ratio between the volume of a sphere and volume of a circumscribing right circular cylinder is

(a) 2 : 1

(b) 1 : 1

(c) 2 : 3

(d) 1 : 2

Solution 4

begin mathsize 12px style Volume space of space sphere space of space radius space straight r space equals space 4 over 3 πr cubed space equals space straight V subscript 1 space space space space space space space.... open parentheses 1 close parentheses
If space straight a space cylinder space is space circumscribing space the space sphere comma space then
diameter space of space cylinder space equals space diameter space of space sphere
height space of space cylinder space equals space diameter space of space sphere
rightwards double arrow Radius space of space cylinder space equals space Radius space of space sphere
Height space of space cylinder space equals space 2 straight r
Volume space of space cylinder space equals space straight V subscript 2 space equals space πr squared straight h
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space πr squared space left parenthesis 2 straight r right parenthesis
rightwards double arrow straight V subscript 2 equals space 2 πr cubed space space space space.... open parentheses 2 close parentheses
Dividing space equation space open parentheses 1 close parentheses space and space open parentheses 2 close parentheses
straight V subscript 1 over straight V subscript 2 equals fraction numerator begin display style bevelled 4 over 3 space πr cubed end style over denominator 2 πr cubed end fraction
rightwards double arrow straight V subscript 1 over straight V subscript 2 equals 2 over 3
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 5

begin mathsize 12px style straight A space cone space and space straight a space hemisphere space have space equal space bases space and space equal space volumes. space The space ratio space of space their
heights space is
left parenthesis straight a right parenthesis space 1 space colon space 2
left parenthesis straight b right parenthesis space 2 space colon space 1
left parenthesis straight c right parenthesis space 4 space colon space 1
left parenthesis straight d right parenthesis space square root of 2 space colon space 1 end style

Solution 5

begin mathsize 12px style Let space the space radius space of space cone space equals space straight r
Then comma space radius space of space hemisphere space equals space straight r space space space space left parenthesis Both space have space equal space bases right parenthesis
straight V subscript cone space equals space 1 third πr squared straight h
straight V subscript Hemisphere space equals 2 over 3 πr cubed
straight V subscript cone space equals space straight V subscript Hemisphere
rightwards double arrow fraction numerator 1 over denominator up diagonal strike 3 end fraction up diagonal strike straight pi straight r squared straight h space equals space fraction numerator 2 over denominator up diagonal strike 3 end fraction up diagonal strike straight pi straight r cubed space space space space space
rightwards double arrow straight h space equals space 2 straight r
rightwards double arrow straight h over straight r equals 2 over 1 space space space space open curly brackets straight r space equals space height space of space hemisphere close curly brackets
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 6

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is

(a) 1 : 2 : 3

(b) 2 : 1 : 3

(c) 2 : 3 : 1

(d) 3 : 2 : 1

Solution 6

begin mathsize 12px style If space all space of space these space have space equal space bases comma space then space their space radii space are space equal.
Their space heights space are space same. space space space left parenthesis given right parenthesis
straight r space equals space straight h subscript 1 space equals space straight h subscript 2
straight V subscript cone space equals 1 third πr squared straight h subscript 1 equals 1 third πr squared open parentheses straight r close parentheses equals 1 third πr cubed
straight V subscript hemisphere equals space 2 over 3 πr cubed
straight V subscript cylinder space equals πr squared space straight h subscript 2 space equals space πr squared space open parentheses straight r close parentheses space equals πr cubed
straight V subscript cone space space colon space straight V subscript hemisphere space space colon space straight V subscript cylinder space equals 1 third space colon space 2 over 3 space colon space 1 space equals space 1 space colon space 2 space colon space 3
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

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