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

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.

Exercise/Page

## 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

Solution 2

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

Solution 14

Solution 15

Solution 16

Solution 10

Solution 11

Solution 12

Solution 13

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).

Solution 18

Solution 19

Solution 20

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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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