Chapter 20 : Surface Areas and Volume of A Right Circular Cone - Rd Sharma Solutions for Class 9 Maths CBSE

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Chapter 20 - Surface Areas and Volume of A Right Circular Cone Excercise Ex. 20.1

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Find its curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.
Solution 6
Question 7
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution 7

Question 8
Solution 8
Question 9
The curved surface area of a cone is 4070 cm2 and its diameter is 70 cm. What is its slant height? (Use = 22/7).
Solution 9
Question 10
Solution 10
Question 11
A joker's cap is in the form of right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Solution 11
Question 12
Find the ratio of the curved surface area of two cones if their diameters of the bases are equal and slant heights are in the ratio 4:3.
Solution 12
Question 13
Solution 13
Question 14
Solution 14
Question 15
Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.
Solution 15
(i)    Slant height of cone = 14 cm
    Let radius of circular end of cone be r.
    CSA of cone =

    
    Thus, the radius of circular end of the cone is 7 cm.

(ii)    Total surface area of cone = CSA of cone + Area of base
                     =

Thus, the total surface area of the cone is 462 .
    

                      


Question 16
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs 210 per 100 m2.
Solution 16
Question 17
A conical tent is 10 m high and the radius of its base is 24 m. Find slant height of the tent. If the cost of 1 m2 canvas is Rs 70, find the cost of the canvas required to make the tent,
Solution 17
Question 18

Solution 18

Question 19

What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use  = 3.14).

Solution 19
Height (h) of conical tent = 8 m
    Radius (r) of base of tent = 6 m
    Slant height (l) of tent =     
    CSA of conical tent =  = (3.14  10)  = 188.4

Let length of tarpaulin sheet required be L.
As 20 cm will be wasted so, effective length will be (L - 0.2 m)
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
    [(L - 0.2 m)  3] m = 188.4
    L - 0.2 m = 62.8 m
    L = 63 m
   
    Thus, the length of the tarpaulin sheet will be 63 m.
Question 20

Solution 20

Question 21

A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, show that the radius of each to the height of each is 3:4.

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Chapter 20 - Surface Areas and Volume of A Right Circular Cone Excercise Ex. 20.2

Question 1
Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm

(ii) radius 3.5 cm, height 12 cm

(iii) height 21 cm and slant height 28 cm.
Solution 1
(i)    Radius (r) of cone = 6 cm
       Height (h) of cone = 7 cm
       Volume of cone   

(ii)   Radius (r) of cone = 3.5 cm
       Height (h) of cone = 12 cm
       Volume of cone   
(iii)
Question 2
Find the capacity in litres of a conical vessel with
(i)    radius 7 cm, slant height 25 cm
(ii)   height 12 cm, slant height 13 cm

Solution 2
(i)    Radius (r) of cone = 7 cm
       Slant height (l) of cone = 25 cm
       Height (h) of cone   
       Volume of cone 
       Capacity of the conical vessel =  litres= 1.232 litres
(ii)    Height (h) of cone = 12 cm
        Slant height (l) of cone = 13 cm
        
        Radius (r) of cone
        Volume of cone   = 314.28 cm3
        Capacity of the conical vessel = litres =  litres.

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
A heap of wheat is in the form of a cone of diameter 9 m and height is 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? (use = 3.14).
Solution 9
Question 10
Solution 10
Question 11
A right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn round on the longer side. Find the volume of the solid, thus generated. Also, find its curved surface area.
Solution 11
Question 12
Solution 12
Question 13
The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find

(i) height of the cone

(ii) slant height of the cone

(iii) curved surface area of the cone
Solution 13
(i)    Radius of cone =   =14 cm
       Let height of cone be h.
       Volume of cone = 9856 cm3
       
       h = 48 cm
       Thus, the height of the cone is 48 cm.
 
(ii)   Slant height (l) of cone  
           
       Thus, the slant height of the cone is 50 cm.
 
(iii)    CSA of cone = rl =
Question 14

A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Solution 14
Radius (r) of pit = 
Depth (h) of pit = 12 m
Volume of pit =
Capacity of the pit = (38.5  1) kilolitres = 38.5 kilolitres

Question 15
Monica has a piece of canvas whose area is 551 m2. She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 1 m2, find the volume of the tent that can be made with it.
Solution 15

Chapter 20 - Surface Areas and Volume of A Right Circular Cone Excercise 20.24

Question 1

The number of surfaces of a cone has, is

(a) 1

(b) 2

(c) 3

(d) 4

Solution 1

A cone has two surfaces as follows: one curved surface and another bottom surface.

Hence, correct option is (b).

Question 2

begin mathsize 12px style The space area space of space the space curved space surface space of space straight a space cone space of space radius space 2 straight r space and space slant space height space 1 half comma space is
left parenthesis straight a right parenthesis space πrl
left parenthesis straight b right parenthesis space 2 πrl
left parenthesis straight c right parenthesis space 1 half πrl
left parenthesis straight d right parenthesis space straight pi left parenthesis straight r space plus space straight l right parenthesis space straight r end style

Solution 2

begin mathsize 12px style Curved space surface space area space of space straight a space cone space of space radius space apostrophe straight r apostrophe space and space slant space height space apostrophe straight l apostrophe space equals space πrl
Now comma space if space straight r apostrophe space equals space 2 straight r space and space straight l apostrophe space equals space straight l over 2 comma space then
straight C. straight S. straight A. space equals space straight pi open parentheses up diagonal strike 2 straight r close parentheses space open parentheses fraction numerator straight l over denominator up diagonal strike 2 end fraction close parentheses equals space πrl
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 3

begin mathsize 12px style The space total space surface space area space of space straight a space cone space of space radius space straight r over 2 space and space length space 2 straight l comma space is
left parenthesis straight a right parenthesis space 2 πr open parentheses straight l space plus space straight r close parentheses
left parenthesis straight b right parenthesis space πr open parentheses straight l plus straight r over 4 close parentheses
left parenthesis straight c right parenthesis space πr open parentheses straight l space plus space straight r close parentheses
left parenthesis straight d right parenthesis space 2 πrl end style

Solution 3

begin mathsize 12px style Total space surface space Area space of space straight a space cone space equals space πR left parenthesis straight L space plus space straight R right parenthesis
Where comma space straight R space equals space Radius space and space straight L space equals space Slant space height
therefore space straight T. straight S. straight A. space equals space straight pi space open parentheses straight r over 2 close parentheses space open parentheses 2 straight l space plus space straight r over 2 close parentheses equals πr open parentheses straight l space plus space straight r over 4 close parentheses
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 4

A solid cylinder is melted and cast into a cone of same radius. The heights of the cone and cylinder are in the ratio

(a) 9 : 1

(b) 1 : 9

(c) 3 : 1

(d) 1 : 3

Solution 4

begin mathsize 12px style Volume space of space cylinder space equals space Volume space of space cone
rightwards double arrow up diagonal strike straight pi up diagonal strike straight r squared end strike straight h subscript 1 equals 1 third up diagonal strike straight pi up diagonal strike straight r squared end strike straight h subscript 2 space space space space space space left parenthesis Let space Radius space be space straight r space for space both right parenthesis
rightwards double arrow fraction numerator straight h 1 over denominator straight h 2 end fraction equals 1 third
rightwards double arrow fraction numerator straight h subscript 2 open parentheses cone close parentheses over denominator straight h subscript 1 open parentheses cylinder close parentheses end fraction equals 3 over 1
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 5

begin mathsize 12px style If space the space radius space of space the space base space of space straight a space right space circular space cone space is space 3 straight r space and space its space height space is space equal space to space the space radius space of space the space base comma space
then space its space volume space is
left parenthesis straight a right parenthesis space 1 third πr cubed
left parenthesis straight b right parenthesis space 2 over 3 πr cubed
left parenthesis straight c right parenthesis space 3 πr cubed
left parenthesis straight d right parenthesis space 9 πr cubed end style

Solution 5

begin mathsize 12px style Volume space of space straight a space cone space equals 1 third πR squared straight H
where comma space straight R space equals space Radius space of space Base comma space straight H space equals space height
If space straight R space equals space 3 straight r comma space straight H space equals space straight R space equals space 3 straight r comma
Then space space straight V space equals space fraction numerator 1 over denominator up diagonal strike 3 end fraction cross times space straight pi space cross times space open parentheses 3 straight r close parentheses squared space cross times space up diagonal strike 3 straight r space equals space 9 πr cubed
Heans comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 6

If the volumes of two cones are in the ratio 1 : 4 and their diameters are in the ratio 4 : 5, then the ratio of their heights, is

(a) 1 : 5

(b) 5 : 4

(c) 5 : 16

(d) 25 : 64

Solution 6

begin mathsize 12px style Let space the space volume space of space 1 to the power of st space cone space equals space 1 third πR subscript 1 superscript 2 straight H subscript 1 space equals space straight V subscript 1
Let space the space volume space of space 2 to the power of nd space cone space equals space 1 third πR subscript 2 superscript 2 straight H subscript 2 space equals space straight V subscript 2
straight V subscript 1 over straight V subscript 2 equals fraction numerator straight R subscript 1 superscript 2 straight H subscript 1 over denominator straight R subscript 2 superscript 2 straight H subscript 2 end fraction equals 1 fourth space space space space space and space space space space space straight d subscript 1 over straight d subscript 2 equals 4 over 5 rightwards double arrow fraction numerator 2 straight R subscript 1 over denominator 2 straight R subscript 2 end fraction equals 4 over 5 rightwards double arrow straight R subscript 1 over straight R subscript 2 equals 4 over 5
rightwards double arrow open parentheses 4 over 5 close parentheses squared space straight H subscript 1 over straight H subscript 2 equals 1 fourth space space space space space space space space space space space space space space space space space space space space space space space
rightwards double arrow straight H subscript 1 over straight H subscript 2 equals 25 over 64
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. space end style

Question 7

The curved surface area of one cone is twice that of the other while the slant height of the latter is twice that of the former. The ratio of their radii is

(a) 2 : 1

(b) 4 : 1

(c) 8 : 1

(d) 1 : 1

Solution 7

begin mathsize 12px style Let space the space Curved space Surface space Area space of space one space cone space equals space πR subscript 1 open parentheses straight L subscript 1 close parentheses
where comma space straight L space equals space Slant space height comma space space straight R space equals space Radius
Curved space Surface space Area space of space other space cone space equals space πR subscript 2 open parentheses straight L subscript 2 close parentheses
Now comma space πR subscript 1 open parentheses straight L subscript 1 close parentheses space equals space 2 space πR subscript 2 open parentheses straight L subscript 2 close parentheses space and space space space straight L subscript 2 equals space 2 straight L subscript 1
rightwards double arrow up diagonal strike straight pi straight R subscript 1 straight L subscript 1 space equals space 2 up diagonal strike straight pi straight R subscript 2 space open parentheses 2 straight L subscript 1 close parentheses
rightwards double arrow straight R subscript 1 over straight R subscript 2 equals 4 over 1
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Chapter 20 - Surface Areas and Volume of A Right Circular Cone Excercise 20.25

Question 1

If the height and radius of a cone of volume V are doubled, then the volume of the cone, is

(a) 3V

(b) 4V

(c) 6V

(d) 8V

Solution 1

begin mathsize 12px style straight V space equals 1 third πR squared straight H
If space straight R apostrophe space equals space 2 straight R space space and space straight H apostrophe space equals space 2 straight H comma space then
straight V apostrophe space equals space 1 third straight pi open parentheses 2 straight R close parentheses squared open parentheses 2 straight H close parentheses
space space space space space equals space 8 open parentheses 1 third πR squared straight H close parentheses space space space
space space space space space equals space 8 straight V
rightwards double arrow straight V apostrophe space equals space 8 straight V
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 2

The ratio of the volume of a right circular cylinder and a right circular cone of the same base and height, is

(a) 1 : 3

(b) 3 : 1

(c) 4 : 3

(d) 3 : 4

Solution 2

begin mathsize 12px style Volume space of space straight a space right space circular space cylinder space of space Height space straight H space and space Radius space straight R space equals space πR squared straight H space equals space straight V subscript 1
Volume space of space straight a space cone space of space height space straight H space and space Radius space straight R space equals space 1 third πR squared straight H space equals space straight V subscript 2
straight V subscript 1 over straight V subscript 2 equals fraction numerator straight pi up diagonal strike straight R squared straight H end strike over denominator 1 third straight pi up diagonal strike straight R squared straight H end strike end fraction space equals 3 over 1 equals space 3 space colon space 1
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 3

A right circular cylinder cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone is

(a) 3 : 5

(b) 2 : 5

(c) 3 : 1

(d) 1 : 3

Solution 3

begin mathsize 12px style Volume space of space straight a space Right space circular space cylinder space equals space πR subscript 1 superscript 2 straight H subscript 1 space equals space straight V subscript 1
Volume space of space straight a space Right space circular space cone space equals space 1 third πR subscript 2 superscript 2 straight H subscript 2 space equals space straight V subscript 2
If space straight V subscript 1 space equals space straight V subscript 2 space space and space straight R subscript 1 space space equals space straight R subscript 2 comma space then
up diagonal strike πR subscript 1 superscript 2 end strike straight H subscript 1 space equals space 1 third up diagonal strike straight pi open parentheses straight R subscript 1 close parentheses squared end strike space straight H subscript 2
rightwards double arrow straight H subscript 1 over straight H subscript 2 equals 1 third
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 4

The  slant height of a cone is increased by 10%. If the radius remains the same, the curved surface area is increased by 

(a) 10%

(b) 12.1%

(c) 20%

(d) 21%

Solution 4

begin mathsize 12px style straight C. straight S. straight A space of space straight a space cone space equals space πrl
If space straight l apostrophe space equals straight l space plus space 10 percent sign space of space straight l space equals space straight l space plus space 10 over 100 cross times straight l space equals space space straight l plus straight l over 10
And comma space straight r apostrophe space equals space straight r
straight C. straight S. straight A. equals space πr open parentheses straight l plus straight l over 10 close parentheses space equals space 11 over 10 πrl
So comma space increase space in space straight C. straight S. straight A. space equals space fraction numerator begin display style 11 over 10 πrl space minus space πrl space end style over denominator πrl end fraction cross times space 100 percent sign space equals space 10 percent sign
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 5

The height of a solid cone is 12 cm and the area of the circular base is 64∏ cm2. A plane parallel to the base of the cone cuts through the cone 9 cm above the vertex of the cone, thee area of the base of the new cone so formed is

(a) 9∏ cm2

(b) 16∏ cm2

(c) 25∏ cm2

(d) 36∏ cm2

Solution 5

begin mathsize 12px style AB space equals space 12 space cm
Area space of space circular space Base space equals space up diagonal strike straight pi straight r squared space equals space 64 up diagonal strike straight pi
rightwards double arrow straight r space equals space 8 space cm
AD space equals space 9 space cm
Consider space triangle ADE space and space triangle ABC comma
angle DAE space equals space angle space BAC space space space space space open parentheses common close parentheses
angle ADE space equals space angle ABC space space space space left parenthesis each space 90 degree right parenthesis
angle AED space equals space angle ACB space space space space space space space space open parentheses third space angle space will space also space be space same close parentheses
Hence space triangle ADE space tilde space triangle ABC
So space AD over AB equals DE over BC
rightwards double arrow 9 over 12 equals DE over 8
rightwards double arrow DE space equals space 6 space cm
Radius space of space Base space of space new space cone space equals space 6 space cm
rightwards double arrow Area space equals space straight pi open parentheses 6 close parentheses squared space equals space 36 straight pi space cm squared
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 6

The diameters of two cones are equal. If their slant heights are in the ratio 5 : 4, the ratio of their curved surface areas, is

(a) 4 : 5

(b) 25 : 16

(c) 16 : 25

(d) 5 : 4

Solution 6

begin mathsize 12px style Curved space Surface space Area space of space cone space equals space πRL
where comma space straight R space equals space Radius space and space straight L space equals space Slant space height
Ratio space of space straight C. straight S. straight A. space of space two space cones comma
straight C. straight S. straight A subscript 1 space colon space straight C. straight S. straight A subscript 2 space equals space πR subscript 1 straight L subscript 1 space colon space πR subscript 2 straight L subscript 2
If space straight L subscript 1 over straight L subscript 2 equals 5 over 4 comma space then space fraction numerator 2 straight R subscript 1 over denominator 2 straight R subscript 2 end fraction equals 1 space space space space left parenthesis because space straight R subscript 1 equals straight R subscript 2 right parenthesis
rightwards double arrow fraction numerator straight C. straight S. straight A subscript 1 over denominator straight C. straight S. straight A subscript 2 end fraction space equals fraction numerator πR subscript 1 straight L subscript 1 over denominator πR subscript 2 straight L subscript 2 end fraction
rightwards double arrow fraction numerator straight C. straight S. straight A subscript 1 over denominator straight C. straight S. straight A subscript 2 end fraction equals straight R subscript 1 over straight R subscript 1 cross times space 5 over 4
rightwards double arrow fraction numerator straight C. straight S. straight A subscript 1 over denominator straight C. straight S. straight A subscript 2 end fraction equals 5 over 4
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 7

If the heights of two cones are in the ratio of 1 : 4 and the radii of their bases are in the ratio 4 : 1, then the ratio of their volumes is

(a) 1 : 2

(b) 2 : 3

(c) 3 : 4

(d) 4 : 1

Solution 7

begin mathsize 12px style Let space the space volume space of space cone space 1 space equals 1 third πR subscript 1 superscript 2 straight H subscript 1 equals straight V subscript 1
Let space the space volume space of space cone space 2 space equals 1 third πR subscript 2 superscript 2 straight H subscript 2 equals straight V subscript 2
straight V subscript 1 over straight V subscript 2 equals fraction numerator up diagonal strike begin display style 1 third straight pi space end style end strike straight R subscript 1 superscript 2 straight H subscript 1 over denominator up diagonal strike 1 third straight pi space end strike straight R subscript 2 superscript 2 straight H subscript 2 end fraction equals fraction numerator straight R subscript 1 superscript 2 over denominator straight R subscript 2 superscript 2 end fraction space space straight H subscript 1 over straight H subscript 1 space space space space space space space open curly brackets straight H subscript 1 over straight H subscript 2 equals 1 fourth comma space space straight R subscript 1 over straight R subscript 2 equals 4 over 1 space space space space left parenthesis given right parenthesis close curly brackets 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 open parentheses 4 over 1 close parentheses squared space open parentheses 1 fourth close parentheses
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 4 over 1
rightwards double arrow straight V subscript 1 space colon space straight V subscript 2 space equals 4 space colon space 1
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 8

If the base radius and the height of a right circular cone are increased by 20%, then the percentage increase in volume is approximately

(a) 60

(b) 68

(c) 73

(d) 78

Solution 8

begin mathsize 12px style Let space the space radius space of space the space cone space equals space straight R space and space height space equals space straight H
Then comma space Volume space equals space 1 third πR squared straight H
Now comma space straight R apostrophe space equals space straight R space plus space 20 percent sign space of space straight R space equals straight R plus straight R over 5 equals fraction numerator 6 straight R over denominator 5 end fraction
space space space space space space space space space space straight H apostrophe space equals space straight H plus 20 percent sign space of space straight H space equals space straight H plus straight H over 5 equals fraction numerator 6 straight H over denominator 5 end fraction
New space volume comma space straight V apostrophe space equals space 1 third πR apostrophe squared straight H apostrophe
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 1 third straight pi open parentheses fraction numerator 6 straight R over denominator 5 end fraction close parentheses squared space open parentheses fraction numerator 6 straight H over denominator 5 end fraction close parentheses
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 216 over 125 open parentheses 1 third πR squared straight H close parentheses
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 216 over 125 straight V
percent sign space increase space in space Volume space equals space fraction numerator straight V apostrophe space minus space straight V over denominator straight V end fraction cross times 100
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 fraction numerator begin display style fraction numerator 216 straight v over denominator 125 end fraction minus straight V space end style over denominator straight v end fraction cross times 100
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 91 over 125 cross times 100
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 72.8 percent sign
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 almost equal to 73 percent sign
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 9

If h, S and V denote respectively the height, curved surface area and volume of a right circular cone, then 3∏Vh3 - S2h+ 9V2 is equal to

(a) 8

(b) 0

(c) 4∏

(d) 32∏2

Solution 9

begin mathsize 12px style For space straight a space cone comma
straight V space equals space 1 third πR squared straight h space
straight S space equals space Curved space Surface space Area space equals space πRL
straight L space equals space square root of straight h squared space plus space straight R squared end root
3 πVh cubed space minus space straight S squared straight h squared space plus space 9 straight V squared space equals space 3 straight pi open parentheses 1 third πR squared straight h close parentheses straight h cubed space minus space straight pi squared straight R squared left parenthesis straight h squared space plus space straight R squared right parenthesis space straight h squared space plus space up diagonal strike 9 cross times fraction numerator 1 over denominator up diagonal strike 9 end fraction straight pi squared straight R to the power of 4 straight h squared
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 up diagonal strike straight pi squared straight R squared straight h to the power of 4 end strike space minus space up diagonal strike straight pi squared straight R squared straight h to the power of 4 end strike space minus space straight pi up diagonal strike blank squared straight R to the power of 4 straight h squared end strike space plus space up diagonal strike straight pi squared straight R to the power of 4 straight h squared end strike
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 0
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 10

If a cone is cut into two parts by a horizontal plane passing through the mid-point of its axis, the ratio of the volumes of upper and lower part is

(a) 1 : 2

(b) 2 : 1

(c) 1 : 7

(d) 1 : 8

Solution 10

begin mathsize 12px style AD over AB equals DF over BC
rightwards double arrow fraction numerator straight h divided by 2 over denominator straight h divided by 2 plus straight h divided by 2 end fraction equals DF over BC rightwards double arrow DF over BC equals 1 half rightwards double arrow DF equals BC over 2 equals straight r over 2
Volume space of space full space cone space equals space 1 third πr squared straight h
volume space of space small space cone space formed space equals 1 third straight pi open parentheses straight r over 2 close parentheses squared straight h over 2 equals 1 third straight pi straight r squared over 4 straight h over 2 space space equals 1 over 8 open parentheses fraction numerator πr squared straight h over denominator 3 end fraction close parentheses
Ratio space of space volume space of space two space parts
equals space fraction numerator Volume space of space small space cone over denominator Volume space of space full space cone space minus space Volume space of space small space cone end fraction
equals fraction numerator begin display style 1 over 8 open parentheses fraction numerator πr squared straight h over denominator 3 end fraction close parentheses end style over denominator begin display style fraction numerator πr squared straight h over denominator 3 end fraction minus fraction numerator πr squared straight h over denominator 8 cross times 3 end fraction end style end fraction equals 1 over 7
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

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