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RD Sharma Solution for Class 9 Mathematics Chapter 19 - Surface Areas and Volume of a Circular Cylinder

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Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 19 - Surface Areas and Volume of a Circular Cylinder.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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RD Sharma Solution for Class 9 Mathematics Chapter 19 - Surface Areas and Volume of a Circular Cylinder Page/Excercise 19.1

Solution 1

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.

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.

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.

Solution 5

Solution 6

Solution 7

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

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.


Solution 10

Solution 11

Solution 12

Solution 13

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.    

RD Sharma Solution for Class 9 Mathematics Chapter 19 - Surface Areas and Volume of a Circular Cylinder Page/Excercise 19.2

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

Solution 2

Solution 3

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

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.   

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.

Solution 7


Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

RD Sharma Solution for Class 9 Mathematics Chapter 19 - Surface Areas and Volume of a Circular Cylinder Page/Excercise 19.28

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

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

Solution 3

Number of Surfaces In a Right cylinder are 3.

Top surface, bottom surface and curved surface.

Hence, correct option is (c).

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).

Solution 5

Volume of cylinder

= Area of Base × Height

= (∏r2) × h

V = ∏r2h

Hence, correct option is (b).

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).

RD Sharma Solution for Class 9 Mathematics Chapter 19 - Surface Areas and Volume of a Circular Cylinder Page/Excercise 19.29

Solution 7

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).

Solution 8

Volume of cylinder V = ∏r2h

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

V' = ∏(r)2(2h) = 2∏r2h = 2V

Hence, correct option is (a).

Solution 9

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

Solution 14

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

Solution 15

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

Solution 16

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

Solution 10

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

Solution 11

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

Solution 12

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

Solution 13

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

Solution 17

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).

RD Sharma Solution for Class 9 Mathematics Chapter 19 - Surface Areas and Volume of a Circular Cylinder Page/Excercise 19.30

Solution 18

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.

Solution 19

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

Solution 20

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

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 9. Access the CBSE Class 9 Mathematics Chapter 19 - Surface Areas and Volume of a Circular Cylinder for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 19 - Surface Areas and Volume of a Circular Cylinder.

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