Class 9 RD SHARMA Solutions Maths Chapter 19 - Surface Areas and Volume of a Circular Cylinder
Ex. 19.1
Ex. 19.2
19.28
19.29
19.30
Surface Areas and Volume of a Circular Cylinder Exercise Ex. 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.
Radius (r) of circular end of pipe =
cm = 2.5 cm = 0.025 mCSA 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.
Radius of the circular end of the pillar =
cm = 25 cm = 0.25 mCSA of pillar =
= Cost of painting 1
area = Rs 12.50Cost of painting 5.5
area = Rs (5.5 12.50) = Rs 68.75Thus, 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
Base radius (r) of cylindrical tank =
= 70 cm = 0.7 mArea 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
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
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
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 cm2Thus, 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 =
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. Surface Areas and Volume of a Circular Cylinder Exercise Ex. 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
b h = (5 4 15) cm3 = 300 cm3
Breadth (b) of tin can = 4 cm
Height (h) of tin can = 15 cm
Capacity of tin can = l
b h = (5 4 15) cm3 = 300 cm3Radius (R) of circular end of plastic cylinder = 
Height (H) of plastic cylinder = 10 cm
Capacity of plastic cylinder = 
R2H =
=385 cm3
R2H ==385 cm3Thus, the plastic cylinder has greater capacity.
Difference in capacity = (385 - 300) cm3 = 85 cm3
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
2
rh = 94.2 cm2
(2
3.14 r 5) cm = 94.2 cm2
r = 3 cm
Let radius of cylinder be r.
CSA of cylinder = 94.2 cm2
2
rh = 94.2 cm2(2
3.14 r 5) cm = 94.2 cm2r = 3 cm
(ii) Volume of cylinder = r2h = (3.14 (3)2 5) cm3 = 141.3 cm3
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
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)
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
cm = 3.5 cmHeight (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.
r2h= Volume of soup in 250 bowls = (250
154) cm3 = 38500 cm3 = 38.5 litresThus, 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
Surface Areas and Volume of a Circular Cylinder Exercise 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).
Surface Areas and Volume of a Circular Cylinder Exercise 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).
Surface Areas and Volume of a Circular Cylinder Exercise 19.30
Solution 18
Solution 19
Solution 20
