Chapter 19 : Surface Areas and Volume of a Circular Cylinder - Rd Sharma Solutions for Class 9 Maths CBSE

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Chapter 19 - Surface Areas and Volume of a Circular Cylinder Excercise Ex. 19.1

Question 1
Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.

Solution 1
Question 2
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Solution 2
 Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 m
  Radius (r) of circular end of pipe =  cm = 2.5 cm = 0.025 m
  CSA of cylindrical pipe =   = 4.4
    Thus, the area of radiating surface of the system is 4.4 m2 or 44000 cm2.

Question 3
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs.12.50 per m2.
Solution 3
Height of the pillar = 3.5 m
Radius of the circular end of the pillar = cm = 25 cm  = 0.25 m
CSA of pillar =  =
Cost of painting 1  area = Rs 12.50
Cost of painting 5.5  area = Rs (5.5 12.50) = Rs 68.75

Thus, the cost of painting the CSA of pillar is Rs 68.75.

Question 4
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?   
Solution 4
Height (h) of cylindrical tank = 1 m.
    Base radius (r) of cylindrical tank =  = 70 cm = 0.7 m

Area of sheet required = total surface area of tank = 
   

So, it will require 7.48 of metal sheet.

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find

 (i)    Its inner curved surface area,
 (ii)    The cost of plastering this curved surface at the rate of Rs 40 per m2

Solution 8
Inner radius (r) of circular well = 1.75 m
Depth (h) of circular well = 10 m 
(i) Inner curved surface area =
 

                                     
          = (44 x 0.25 x 10)
          = 110 m2
(ii) Cost of plastering 1 m2 area = Rs 40                    
    Cost of plastering 110 m2 area = Rs (110 x 40) = Rs 4400

Question 9
The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?   
Solution 9
Radius of circular end of cylindrical penholder = 3 cm
Height of penholder = 10.5 cm
Surface area of 1 penholder = CSA of penholder + Area of base of     
SA of 1 penholder =  +
                  
Area of cardboard sheet used by 1 competitor =  
    Area of cardboard sheet used by 35 competitors

 = 7920 cm2
    Thus, 7920 cm2 of cardboard sheet will be required for the competition.


Question 10

Solution 10

Question 11

Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m. What will be the cost of cleaning them at the rate of Rs 2.50 per square metre?

Solution 11

Question 12

Solution 12

Question 13

The total surface area of a hollow metal cylinder open at both ends of external radius 8 cm and height 10 cm is 338 cm2. Taking r to be inner radius, obtain an equation in r and use it to obtain the thickness of the metal in the cylinder.

Solution 13

Question 14

Find the lateral or curved surface area of a cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high. How much steel was actually used, if  of the steel actually used was wasted in making the closed tank?

Solution 14
Height (h) cylindrical tank = 4.5 m
Radius (r) of circular end of cylindrical tank =m = 2.1m
    (i)    Lateral or curved surface area of tank =

                                    =                                          
                                    = 59.4 m2            
    
    (ii)    Total surface area of tank = 2 (r + h)

              =                           
              = 87.12 m2

    Let A m2 steel sheet be actually used in making the tank.

    
    
Thus, 95.04  steel was used in actual while making the tank.    

Chapter 19 - Surface Areas and Volume of a Circular Cylinder Excercise Ex. 19.2

Question 1
A soft drink is available in two packs -

(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?   
Solution 1
The tin can will be cuboidal in shape.
Length (l) of tin can = 5 cm
Breadth (b) of tin can = 4 cm
Height (h) of tin can = 15 cm
Capacity of tin can = l  h = (5  15) cm3 = 300 cm3
Radius (R) of circular end of plastic cylinder =
Height (H) of plastic cylinder = 10 cm
Capacity of plastic cylinder = R2H  ==385 cm3
Thus, the plastic cylinder has greater capacity.
Difference in capacity = (385 - 300) cm3 = 85 cm3

Question 2
The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 20 cm and height 10m. How much concrete mixture would be required to build 14 such pillars?

Solution 2
Question 3
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.
Solution 3
Question 4
If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then find
(i) radius of its base    (ii) its volume. (Use  = 3.14)
Solution 4
(i)    Height (h) of cylinder = 5 cm
        Let radius of cylinder be r.
        CSA of cylinder = 94.2 cm2
        2rh = 94.2 cm2
        (2  3.14  5) cm = 94.2 cm2
        r = 3 cm
 
(ii)    Volume of cylinder = r2h = (3.14  (3)2  5) cm3 = 141.3 cm3

Question 5
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?
Solution 5
Let radius of the circular ends of the cylinder be r.
Height (h) of the cylindrical vessel = 1 m
Volume of cylindrical vessel = 15.4 litres = 0.0154 m3
Total  Surface area of vessel = 2 r(r+h)
                                            
Thus, 0.4708 m2 of metal sheet would be needed to make the cylindrical vessel.   
Question 6
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Solution 6
Radius (r) of cylindrical bowl = cm = 3.5 cm
Height (h) up to which the bowl is filled with soup = 4 cm
Volume of soup in 1 bowl = r2h=  
Volume of soup in 250 bowls = (250  154) cm3 = 38500 cm3 = 38.5 litres

Thus, the hospital will have to prepare 38.5 litres of soup daily to serve 250 patients.

Question 7
Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

A cylinderical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm x 22 cm x 14 cm. Find the rise in the level of the water when the solid is completely submerged.

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to around it to a width of 21 m to form an embankment. Find the height of the embankment.

Solution 27

Question 28

Solution 28

Question 29

Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 meters per second into cylindrical tank. The water is collected in a cylindrical vessel radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes?

Solution 29

Question 30

Solution 30

Question 31

The sum of the radius of the base and height of a solid cylinder is 37 m. If the total surface area of the solid cylinder is 1628 m2. Find the volume of the cylinder.

Solution 31

Question 32
A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.
Solution 32

Chapter 19 - Surface Areas and Volume of a Circular Cylinder Excercise 19.28

Question 1

In a cylinder, if radius is doubled and height is halved, curved surface area will be

(a) halved

(b) doubled

(c) same

(d) four times

Solution 1

begin mathsize 12px style Curved space surface space Area space of space straight a space cylinder space of space radius space apostrophe straight r apostrophe space and space height space apostrophe straight h apostrophe space is space given space by
straight A space equals space 2 πrh
Now comma space if space straight r apostrophe space equals space 2 straight r space and space straight h apostrophe space equals space straight h over 2
then space straight A apostrophe space equals space 2 straight pi space cross times space open parentheses up diagonal strike 2 straight r close parentheses cross times fraction numerator straight h over denominator up diagonal strike 2 end fraction equals space 2 πrh space equals space straight A
rightwards double arrow straight C. straight S. straight A. space remains space the space same
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 2

Two cylindrical jars have their diameters in the ratio 3 : 1, but height 1 : 3. Then the ratio of their volumes is

(a) 1 : 4

(b) 1 : 3

(c) 3 : 1

(d) 2 : 5

Solution 2

begin mathsize 12px style Volume space of space any space cylinder space equals space πr squared straight h
straight r space equals space straight d over 2
If space straight d subscript 1 space colon space straight d subscript 2 space equals space 3 space colon space 1 space then comma space straight r subscript 1 space colon space straight r subscript 2 space equals space 3 space colon space 1
straight h subscript 1 space colon space straight h subscript 2 space equals space 1 space colon space 3
Now comma
straight V subscript 1 over straight V subscript 2 equals space fraction numerator straight pi open parentheses straight r subscript 1 close parentheses squared straight h subscript 1 over denominator straight pi open parentheses straight r subscript 2 close parentheses squared straight h subscript 2 end fraction equals open parentheses straight r subscript 1 over straight r subscript 2 close parentheses squared space straight h subscript 1 over straight h subscript 2 equals space open parentheses 3 over 1 close parentheses squared open parentheses 1 third close parentheses equals 3 over 1 equals 3 space colon space 1
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 3

The number of surfaces in right cylinder is

(a) 1

(b) 2

(c) 3

(d) 4

Solution 3

Number of Surfaces In a Right cylinder are 3.

Top surface, bottom surface and curved surface.

Hence, correct option is (c).

Question 4

Vertical cross-section of a right circular cylinder is always a

(a) square

(b) rectangle

(c) rhombus

(d) trapezium

Solution 4

Vertical cross-section of cylinder will always be a Rectangle of sides 'h', and 'r',

where h is the height of a cylinder and r is the radius of a cylinder.

Hence, correct option is (b).

Question 5

If r is the radius and h is height of the cylinder the volume will be

begin mathsize 12px style left parenthesis straight a right parenthesis space 1 third πr squared straight h
left parenthesis straight b right parenthesis space πr squared straight h
left parenthesis straight c right parenthesis space 2 πr open parentheses straight h plus straight r close parentheses
left parenthesis straight d right parenthesis space 2 πrh end style

Solution 5

Volume of cylinder

= Area of Base × Height

= (∏r2) × h

V = ∏r2h

Hence, correct option is (b).

Question 6

The number of surfaces of a hollow cylindrical object is

(a) 1

(b) 2

(c) 3

(d) 4

Solution 6

A Hollow cylinder has only 2 surfaces.

One is outer-curved surface and another is inner-curved surface.

Hence, correct option is (b).

Chapter 19 - Surface Areas and Volume of a Circular Cylinder Excercise 19.29

Question 1

If the radius of a cylinder is doubled and the height remains same, the volume will be

(a) doubled

(b) halved

(c) same

(d) four times

Solution 1

Volume  of a cylinder = V = ∏r2h

If r' = 2r and h' = h, then

V' = ∏(2r)2h   = 4∏r2h

V' = 4V

Hence, correct option is (d).

Question 2

If the height of a cylinder is doubled and radius remains the same, then volume will be

(a) doubled

(b) halved

(c) same

(d) four times 

Solution 2

Volume of cylinder V = ∏r2h

If h' = 2h and r' = r, then

V' = ∏(r)2(2h) = 2∏r2h = 2V

Hence, correct option is (a).

Question 3

In a cylinder, if radius is halved and height is doubled, the volume will be

(a) same

(b) doubled

(c) halved

(d) four times

Solution 3

begin mathsize 12px style Volume space of space cylinder space equals space straight V space equals space πr squared straight h
If space straight r apostrophe space equals straight r over 2 space space and space space straight h apostrophe space equals space 2 straight h
then space straight V apostrophe space equals space straight pi open parentheses straight r over 2 close parentheses squared space 2 straight h space equals πr squared over up diagonal strike 4 subscript 2 cross times up diagonal strike 2 straight h space equals fraction numerator πr squared straight h over denominator 2 end fraction space equals space straight V over 2
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 4

If the height of a cylinder is doubled, by what number must the radius of the base be multiplied so that the resulting cylinder has the same volume as the original cylinder?

begin mathsize 12px style left parenthesis straight a right parenthesis space 4
left parenthesis straight b right parenthesis space fraction numerator 1 over denominator square root of 2 end fraction
left parenthesis straight c right parenthesis space 2
left parenthesis straight d right parenthesis space 1 half end style

Solution 4

begin mathsize 12px style Volume space of space straight a space cylinder space equals space straight V space equals space πr squared straight h
Now comma space if space straight h apostrophe space equals space 2 straight h space and space new space radius space equals space straight r apostrophe comma space then
straight V apostrophe space equals space πr apostrophe squared straight h apostrophe space equals space πr apostrophe squared space left parenthesis 2 straight h right parenthesis space equals space 2 πr apostrophe squared straight h
Now space if space volume space should space remain space same comma space then
straight V apostrophe space equals straight V
rightwards double arrow space 2 up diagonal strike straight pi straight r apostrophe squared up diagonal strike straight h space equals space up diagonal strike straight pi straight r squared up diagonal strike straight h
rightwards double arrow 2 straight r apostrophe squared equals straight r squared
rightwards double arrow straight r apostrophe squared equals straight r squared over 2
rightwards double arrow straight r apostrophe space equals space fraction numerator straight r over denominator square root of 2 end fraction
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 5

The volume of a cylinder of radius r is 1/4 of the volume of a rectangular box with a square base of side length x. If the cylinder and the box have equal heights, what is r in terms of x?

begin mathsize 12px style left parenthesis straight a right parenthesis space fraction numerator straight x squared over denominator 2 straight pi end fraction
left parenthesis straight b right parenthesis space fraction numerator straight x over denominator 2 square root of straight pi end fraction
left parenthesis straight c right parenthesis space fraction numerator square root of 2 straight x end root over denominator straight pi end fraction
left parenthesis straight d right parenthesis space fraction numerator straight pi over denominator 2 square root of straight x end fraction end style

Solution 5

begin mathsize 12px style Area space of space Base space of space cylinder space equals πr squared
Area space of space Base space of space Box space left parenthesis square right parenthesis equals space straight x squared
Let space the space height space of space both space objects space equals space straight h
Then comma space straight V subscript cylinder space equals space πr squared straight h
space space space space space space space space space space space space straight V subscript box space equals space straight x squared straight h
Now comma space straight V subscript cylinder equals 1 fourth space straight V subscript box
rightwards double arrow πr squared up diagonal strike straight h space equals space 1 fourth straight x squared up diagonal strike straight h
rightwards double arrow straight r squared space equals space fraction numerator straight x squared over denominator 4 straight pi end fraction
rightwards double arrow straight r space equals space fraction numerator straight x over denominator 2 square root of straight pi end fraction
hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 6

The height h of a cylinder equals the circumference of the cylinder. In terms of h, what is the volume of the cylinder?

begin mathsize 12px style left parenthesis straight a right parenthesis space fraction numerator straight h cubed over denominator 4 straight pi end fraction
left parenthesis straight b right parenthesis space fraction numerator straight h squared over denominator 2 straight pi end fraction
left parenthesis straight c right parenthesis space straight h cubed over 2
left parenthesis straight d right parenthesis space πh cubed end style

Solution 6

begin mathsize 12px style Circumference space of space cylinder space equals space 2 πr
Height space equals space straight h
If space straight h space equals space 2 πr comma space then space straight r space equals space fraction numerator straight h over denominator 2 straight pi end fraction
rightwards double arrow Volume space equals space πr squared straight h space equals space straight pi open parentheses fraction numerator straight h squared over denominator 4 straight pi squared end fraction close parentheses straight h equals fraction numerator straight h cubed over denominator 4 straight pi end fraction
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 7

If the diameter of the base of a closed right circular cylinder be equal to its height h, then its whole surface area is

begin mathsize 12px style left parenthesis straight a right parenthesis space 2 πh squared
left parenthesis straight b right parenthesis space 3 over 2 πh squared
left parenthesis straight c right parenthesis space 4 over 3 πh squared
left parenthesis straight d right parenthesis space πh squared end style

Solution 7

begin mathsize 12px style Diameter space equals space 2 straight r space equals space straight h space space space left parenthesis Given right parenthesis
rightwards double arrow straight r equals straight h over 2
Cylinder space is space closed.
rightwards double arrow Its space surface space area space equals space 2 πr left parenthesis straight h space plus space straight r right parenthesis
straight S space equals space 2 πr space left parenthesis straight h space plus space straight r right parenthesis space equals space up diagonal strike 2 straight pi space fraction numerator straight h over denominator up diagonal strike 2 end fraction open parentheses straight h plus straight h over 2 close parentheses equals πh open parentheses 3 over 2 straight h close parentheses equals 3 over 2 πh squared
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 8

A right circular cylindrical tunnel of diameter 2 m and length 40 m is to be constructed from a sheet of iron. The area of the iron sheet required in m2, is

(a) 40∏

(b) 80∏

(c) 160∏

(d) 200∏

Solution 8

begin mathsize 12px style Cylindrical space tunnel space will space be space hollow space cylinder space of space radius space equals space 1 space straight m space open curly brackets straight r equals straight d over 2 equals 2 over 2 equals 1 space straight m close curly brackets
Length space equals space 40 space straight m
Area space of space iron space sheet space equals space Curved space Surface space Area space of space cylinder
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 2 πrh
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 2 straight pi left parenthesis 1 right parenthesis 40
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 80 straight pi
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 9

Two circular cylinders of equal volume have their heights in the ratio 1 : 2. Ratio of their radii is

begin mathsize 12px style left parenthesis straight a right parenthesis space 1 space colon space square root of 2
left parenthesis straight b right parenthesis space square root of 2 space colon space 1
left parenthesis straight c right parenthesis space 1 space colon space 2
left parenthesis straight d right parenthesis space 1 space colon space 4 end style

Solution 9

begin mathsize 12px style Volume space of space cylinder space 1 comma space straight v subscript 1 space equals space πr subscript 1 squared straight h subscript 1
Volume space of space cylinder space 2 comma space straight v subscript 2 space equals space πr subscript 2 squared straight h subscript 2
straight v subscript 1 over straight v subscript 2 equals straight r subscript 1 squared over straight r subscript 2 squared space space straight h subscript 1 over straight h subscript 2 space space space space space space.... left parenthesis 1 right parenthesis
Now comma space space straight v subscript 1 equals straight v subscript 2 space and space straight h subscript 1 over straight h subscript 2 equals 1 half
Hence comma space equation space left parenthesis 1 right parenthesis space reduces space to space
1 equals straight r subscript 1 squared over straight r subscript 2 squared space 1 half
rightwards double arrow straight r subscript 2 squared over straight r subscript 1 squared space equals space 1 half
rightwards double arrow straight r subscript 1 squared over straight r subscript 2 squared equals 2
rightwards double arrow straight r subscript 1 over straight r subscript 2 equals fraction numerator square root of 2 over denominator 1 end fraction
space rightwards double arrow straight r subscript 1 space colon space straight r subscript 2 space equals space square root of 2 space colon space 1
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 10

The radius of a wire is decreased to one-third. If volume remains the same, the length will become

(a) 3 times

(b) 6 times

(c) 9 times

(d) 27 times

Solution 10

begin mathsize 12px style Volume space of space wire space equals space πr squared
If space volume space remains space same comma space then
πr subscript 1 squared straight l subscript 1 space equals space πr subscript 2 squared straight l subscript 2 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
If space straight r subscript 2 space equals space straight r subscript 1 over 3 comma space then
up diagonal strike straight pi straight r subscript 1 squared straight l subscript 1 space equals space up diagonal strike straight pi open parentheses straight r subscript 1 over 3 close parentheses squared space straight l subscript 2
rightwards double arrow straight l subscript 2 space equals space 9 straight l subscript 1
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 11

A cylinder with radius r and height h is closed on the top and bottom. Which of the following expressions represents the total surface area of this cylinder?

(a) 2∏r (r + h)

(b) ∏r (r + 2h)

(c) ∏r (2r + h)

(d) 2∏r+ h

Solution 11

Total surface Area = Area of Top + Area of bottom + Curved Surface Area

T.S.A. = ∏r2 + ∏r2 + 2∏rh = 2∏r2 + 2∏rh = 2∏r (r + h)

Hence, correct option is (a).

Chapter 19 - Surface Areas and Volume of a Circular Cylinder Excercise 19.30

Question 1

The height of sand in a cylindrical-shaped can drops 3 inches when 1 cubic foot of sand is poured out. What is the diameter, in inches, of the cylinder?

begin mathsize 12px style left parenthesis straight a right parenthesis space fraction numerator 24 over denominator square root of straight pi end fraction
left parenthesis straight b right parenthesis space fraction numerator 48 over denominator square root of straight pi end fraction
left parenthesis straight c right parenthesis space fraction numerator 32 over denominator square root of straight pi end fraction
left parenthesis straight d right parenthesis space 48 over straight pi end style

Solution 1

when space sand space is space poured space out comma space height space dropped space equals space 3 space inches
Volume space vacant space equals πr squared space cross times space 3 space inches space space
Now comma space volume space vacant space equals space volume space of space sand space poured space out space equals space 1 space cubic space foot
1 space foot space equals space 12 space inches
rightwards double arrow 1 space cubic space foot space equals space 12 cross times 12 cross times 12 space inches space equals space 1728 space inches
Thus comma space we space have
3 πr squared equals 1728
rightwards double arrow πr squared equals 576
rightwards double arrow straight r squared equals 576 over straight pi
rightwards double arrow straight r equals fraction numerator 24 over denominator square root of straight pi end fraction
rightwards double arrow Diameter equals 2 straight r equals 2 cross times fraction numerator 24 over denominator square root of straight pi end fraction equals fraction numerator 48 over denominator square root of straight pi end fraction
Hence comma space correct space option space is space left parenthesis straight b right parenthesis.

Question 2

Two steel sheets each of length a1 and breadth a2 are used to prepare the surfaces of two right circular cylinders - one having volume v1 and height a2 and other having volume v2 and height a1. Then,

(a) v1 = v2

(b) a1v1 = a2v2

(c) a2v1 = a1v2

(d) begin mathsize 12px style straight v subscript 1 over straight a subscript 1 equals straight v subscript 2 over straight a subscript 2 end style 

Solution 2

begin mathsize 12px style Surface space Area space of space both space cylinders space is space going space to space be space same space because space same space sheet space is space used.
rightwards double arrow straight S. straight A. space equals space straight a subscript 1 straight a subscript 2
Surface space area space of space cylinders space is space same
straight S subscript 1 space equals space left parenthesis 2 πr subscript 1 right parenthesis straight a subscript 2 equals space straight a subscript 1 straight a subscript 2 space space space space space.... left parenthesis 1 right parenthesis
straight S subscript 2 space equals left parenthesis 2 πr subscript 2 right parenthesis straight a subscript 1 equals space straight a subscript 1 straight a subscript 2 space space space space space space.... open parentheses 2 close parentheses
From space equations space open parentheses 1 close parentheses space & space open parentheses 2 close parentheses
2 πr subscript 1 space equals space straight a subscript 1 space and space 2 πr subscript 2 space equals space straight a subscript 2
Volume space of space cylinder space 1 comma space straight v subscript 1 equals space left parenthesis πr subscript 1 squared right parenthesis space straight a subscript 2 space space space space space space.... open parentheses 3 close parentheses
Volume space of space cylinder space 2 comma space straight v subscript 2 space equals space left parenthesis πr subscript 2 squared right parenthesis space straight a subscript 1 space space space space.... open parentheses 4 close parentheses

Dividing space equation space open parentheses 3 close parentheses space by space equation space open parentheses 4 close parentheses
straight v subscript 1 over straight v subscript 2 equals straight r subscript 1 squared over straight r subscript 2 squared space straight a subscript 2 over straight a subscript 1 space equals space open parentheses begin display style fraction numerator straight a subscript 1 over denominator up diagonal strike 2 straight pi end strike end fraction end style close parentheses squared over open parentheses begin display style fraction numerator straight a subscript 2 over denominator up diagonal strike 2 straight pi end strike end fraction end style close parentheses squared straight a subscript 2 over straight a subscript 1 equals straight a subscript 1 to the power of up diagonal strike 2 end exponent over straight a subscript 2 to the power of up diagonal strike 2 end exponent cross times space fraction numerator up diagonal strike straight a subscript 2 squared end strike over denominator up diagonal strike straight a subscript 1 end strike end fraction space equals straight a subscript 1 over straight a subscript 2
rightwards double arrow straight a subscript 2 straight v subscript 1 equals straight a subscript 1 straight v subscript 2
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 3

The altitude of a circular cylinder is increased six times and the base area is decreased one-ninth of its value. The factor by which the lateral surface of the cylinder increases, is

begin mathsize 12px style left parenthesis straight a right parenthesis space 2 over 3
left parenthesis straight b right parenthesis space 1 half
left parenthesis straight c right parenthesis space 3 over 2
left parenthesis straight d right parenthesis space 2 end style

Solution 3

begin mathsize 12px style If space straight h space is space initial space altitude comma space then space straight h apostrophe space equals space 6 straight h
Initial space Base space Area space equals space πr squared space equals space straight B
New space Base space Area space equals space straight B apostrophe space equals πr apostrophe squared
Now comma space straight B apostrophe equals space straight B over 9
rightwards double arrow πr apostrophe squared equals πr squared over 9
rightwards double arrow straight r apostrophe squared equals straight r squared over 9
rightwards double arrow straight r apostrophe space equals space straight r over 3
Intial space Lateral space surface space Area equals space 2 πrh
New space Lateral space Surface space Area
equals 2 πr apostrophe straight h apostrophe
equals 2 straight pi space open parentheses straight r over 3 close parentheses 6 straight h
equals 2 space open parentheses 2 πrh close parentheses
equals 2 open parentheses Initial space Lateral space Surface space Area close parentheses
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

CBSE Class 9 Maths Homework Help

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