RS Aggarwal Solutions: Volumes and Surface Area of Solids- 2 | Mathematics (Maths) Class 9 PDF Download

 Page 1


                      
              
             
      
                                  
                         
         
          
                    
                       
           
    
       
      
 
                     
         
                        
                        
                 
           
      
     
    
  
        
Question:29
The diameter of a cylinder is 28 cm and its height is 40 cm. Find the curved surface area, total surface area and the volume of the cylinder.
Solution:
Here, r = 28/2 = 14 cm; h = 40 cm
Curved surface area of the cylinder = 2 prh                                                             = 2 ×
22
7
×14 ×40  cm
2
                                                            = 2 ×22 ×2 ×40 cm
2
                                           
Total surface area of the cylinder = 2 prh + 2 pr
2
                                                        = 3520 +2 ×
22
7
×14
2
 cm
2
                                                       = 3520 +1232 = 4752 cm
2
? Volume of the cylinder = pr
2
h                                          =
22
7
×14
2
×40 cm
3
                                         = 22 ×14 ×2 ×40 cm
3
                                         = 24640 cm
3
Question:30
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:
Radius of the bowl, r = 
7
2
 cm
Height of soup in the bowl, h = 4 cm
Volume of soup in one bowl = pr
2
h =
22
7
×
7
2
2
×4 = 154 cm
3
? Amount of soup the hospital has to prepare daily to serve 250 patients
= Volume of soup in one bowl × 250
( ) ( )
( )
Page 2


                      
              
             
      
                                  
                         
         
          
                    
                       
           
    
       
      
 
                     
         
                        
                        
                 
           
      
     
    
  
        
Question:29
The diameter of a cylinder is 28 cm and its height is 40 cm. Find the curved surface area, total surface area and the volume of the cylinder.
Solution:
Here, r = 28/2 = 14 cm; h = 40 cm
Curved surface area of the cylinder = 2 prh                                                             = 2 ×
22
7
×14 ×40  cm
2
                                                            = 2 ×22 ×2 ×40 cm
2
                                           
Total surface area of the cylinder = 2 prh + 2 pr
2
                                                        = 3520 +2 ×
22
7
×14
2
 cm
2
                                                       = 3520 +1232 = 4752 cm
2
? Volume of the cylinder = pr
2
h                                          =
22
7
×14
2
×40 cm
3
                                         = 22 ×14 ×2 ×40 cm
3
                                         = 24640 cm
3
Question:30
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:
Radius of the bowl, r = 
7
2
 cm
Height of soup in the bowl, h = 4 cm
Volume of soup in one bowl = pr
2
h =
22
7
×
7
2
2
×4 = 154 cm
3
? Amount of soup the hospital has to prepare daily to serve 250 patients
= Volume of soup in one bowl × 250
( ) ( )
( )
= 154 × 250
= 38500 cm
3
Question:31
The pillars of a temple are cylindrically shaped. Each pillar has a circular base of radius 20 cm and height 10 m. How much concrete mixture would be required to build 14 such pillars?
Solution:
Radius of each pillar, r = 20 cm = 0.2 m   1m = 100cm
Height of each pillar, h = 10 m
Volume of concrete mixture used in each pillar = pr
2
h =
22
7
×(0. 2)
2
×10 m
3
? Amount of concrete mixture required to build 14 such pillars
= Volume of concrete mixture used in each pillar × 14
= 
22
7
×(0. 2)
2
×10 ×14 
= 17.6 m
3
Question:32
A soft drink is available in two packs: i a tin can with a rectangular base of length 5 cm, breadth 4 cm and height 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:
i Length of tin can, l = 5 cm
Breadth of tin can, b = 4 cm
Height of tin can, h = 15 cm
? Volume of soft drink in tin can = l × b × h = 5 × 4 × 15 = 300 cm
3
ii Radius of plastic cylinder, r = 
7
2
 cm
Height of plastic cylinder, h = 10 cm
? Volume of soft drink in plastic cylinder = pr
2
h =
22
7
×
7
2
2
×10 = 385 cm
3
So, the capacity of the plastic cylinder pack is greater than the capacity of the tin can pack.
Difference in the capacities of the two packs = 385 - 300 = 85 cm
3
Thus, the capacity of the plastic cylinder pack is 85 cm
3
 more than the capacity of the tin can pack.
Question:33
There are 20 cylindrical pillars in a building, each having a diameter of 50 cm and height 4 m. Find the cost of cleaning them at  14 per m
2
.
Solution:
Radius of each pillar, r = 
50
2
 = 25 cm = 0.25 m        1m = 100cm
Height of each pillar, h = 4 m
? Surface area of each pillar = 2 prh = 2 ×
22
7
×0. 25 ×4 cm
2
Surface area of 20 pillars = Surface area of each pillar × 20 = 2 ×
22
7
×0. 25 ×4 ×20 cm
2
Rate of cleaning =  14 per m
2
? Total cost of cleaning the 20 pillars
= Surface area of 20 pillars × Rate of cleaning 
= 2 ×
22
7
×0. 25 ×4 ×20 ×14
=  1760
Question:34
The curved surface area of a right circular cylinder is 4.4 m
2
. If the radius of its base is 0.7 m, find its i height and ii volume.
Solution:
Radius of the cylinder, r = 0.7 m
Curved surface area of cylinder = 4.4 m
2
i Let the height of the cylinder be h m.
? 2 prh = 4. 4 ? 2 ×
22
7
×0. 7 ×h = 4. 4 ? h =
4.4×7
2×22×0.7
= 1 m
Thus, the height of the cylinder is 1 m.
( )
Page 3


                      
              
             
      
                                  
                         
         
          
                    
                       
           
    
       
      
 
                     
         
                        
                        
                 
           
      
     
    
  
        
Question:29
The diameter of a cylinder is 28 cm and its height is 40 cm. Find the curved surface area, total surface area and the volume of the cylinder.
Solution:
Here, r = 28/2 = 14 cm; h = 40 cm
Curved surface area of the cylinder = 2 prh                                                             = 2 ×
22
7
×14 ×40  cm
2
                                                            = 2 ×22 ×2 ×40 cm
2
                                           
Total surface area of the cylinder = 2 prh + 2 pr
2
                                                        = 3520 +2 ×
22
7
×14
2
 cm
2
                                                       = 3520 +1232 = 4752 cm
2
? Volume of the cylinder = pr
2
h                                          =
22
7
×14
2
×40 cm
3
                                         = 22 ×14 ×2 ×40 cm
3
                                         = 24640 cm
3
Question:30
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:
Radius of the bowl, r = 
7
2
 cm
Height of soup in the bowl, h = 4 cm
Volume of soup in one bowl = pr
2
h =
22
7
×
7
2
2
×4 = 154 cm
3
? Amount of soup the hospital has to prepare daily to serve 250 patients
= Volume of soup in one bowl × 250
( ) ( )
( )
= 154 × 250
= 38500 cm
3
Question:31
The pillars of a temple are cylindrically shaped. Each pillar has a circular base of radius 20 cm and height 10 m. How much concrete mixture would be required to build 14 such pillars?
Solution:
Radius of each pillar, r = 20 cm = 0.2 m   1m = 100cm
Height of each pillar, h = 10 m
Volume of concrete mixture used in each pillar = pr
2
h =
22
7
×(0. 2)
2
×10 m
3
? Amount of concrete mixture required to build 14 such pillars
= Volume of concrete mixture used in each pillar × 14
= 
22
7
×(0. 2)
2
×10 ×14 
= 17.6 m
3
Question:32
A soft drink is available in two packs: i a tin can with a rectangular base of length 5 cm, breadth 4 cm and height 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:
i Length of tin can, l = 5 cm
Breadth of tin can, b = 4 cm
Height of tin can, h = 15 cm
? Volume of soft drink in tin can = l × b × h = 5 × 4 × 15 = 300 cm
3
ii Radius of plastic cylinder, r = 
7
2
 cm
Height of plastic cylinder, h = 10 cm
? Volume of soft drink in plastic cylinder = pr
2
h =
22
7
×
7
2
2
×10 = 385 cm
3
So, the capacity of the plastic cylinder pack is greater than the capacity of the tin can pack.
Difference in the capacities of the two packs = 385 - 300 = 85 cm
3
Thus, the capacity of the plastic cylinder pack is 85 cm
3
 more than the capacity of the tin can pack.
Question:33
There are 20 cylindrical pillars in a building, each having a diameter of 50 cm and height 4 m. Find the cost of cleaning them at  14 per m
2
.
Solution:
Radius of each pillar, r = 
50
2
 = 25 cm = 0.25 m        1m = 100cm
Height of each pillar, h = 4 m
? Surface area of each pillar = 2 prh = 2 ×
22
7
×0. 25 ×4 cm
2
Surface area of 20 pillars = Surface area of each pillar × 20 = 2 ×
22
7
×0. 25 ×4 ×20 cm
2
Rate of cleaning =  14 per m
2
? Total cost of cleaning the 20 pillars
= Surface area of 20 pillars × Rate of cleaning 
= 2 ×
22
7
×0. 25 ×4 ×20 ×14
=  1760
Question:34
The curved surface area of a right circular cylinder is 4.4 m
2
. If the radius of its base is 0.7 m, find its i height and ii volume.
Solution:
Radius of the cylinder, r = 0.7 m
Curved surface area of cylinder = 4.4 m
2
i Let the height of the cylinder be h m.
? 2 prh = 4. 4 ? 2 ×
22
7
×0. 7 ×h = 4. 4 ? h =
4.4×7
2×22×0.7
= 1 m
Thus, the height of the cylinder is 1 m.
( )
ii Volume of the cylinder = pr
2
h =
22
7
×(0. 7)
2
×1 = 1.54 m
3
Thus, the volume of the cylinder is 1.54 m
3
.
Question:35
The lateral surface area of a cylinder is 94.2 cm
2
 and its height is 5 cm. Find i the radius of its base and ii its volume. Take p = 3.14.
Solution:
Height of the cylinder, h = 5 cm
Lateral orcurved surface area of cylinder = 94.2 cm
2
i Let the radius of the cylinder be r cm.
? 2 prh = 94. 2 cm
2
? 2 ×3. 14 ×r×5 = 94. 2 ? r =
94.2
2×3.14×5
= 3 cm
Thus, the radius of the cylinder is 3 cm.
ii Volume of the cylinder = pr
2
h = 3. 14 ×(3)
2
×5 = 141.3 cm
3
Thus, the volume of the cylinder is 141.3 cm
3
.
Question:36
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. Find the area of the metal sheet needed to make it.
Solution:
Height of the cylinder, h = 1 m
Capacity of the cylinder = 15.4 L = 15.4 × 0.001 m
3
 = 0.0154 m
3
? pr
2
h = 0. 0154 m
3
?
22
7
×r
2
×1 = 0. 0154 ? r =
0.0154×7
22
= 0. 07 m
? Area of the metal sheet needed to make the cylinder 
= Total surface area of the cylinder
= 2 prh(r+h) = 2 ×
22
7
×0. 07 ×1 ×(0. 07 +1) = 2 ×
22
7
×0. 07 ×1 ×1. 07 = 0. 4708 m
2
Thus, the area of the metal sheet needed to make the cylinder is 0.4708 m
2
.
Question:37
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 cm
3 
of wood has a mass of 0.6
g.
Solution:
Inner radius of the wooden pipe, r = 
24
2
 = 12 cm
Outer radius of the wooden pipe, R = 
28
2
 = 14 cm
Length of the wooden pipe, h = 35 cm
? Volume of wood in the pipe = p R
2
-r
2
h =
22
7
× 14
2
-12
2
×35 = 5720 cm
3
It is given that 1 cm
3 
of wood has a mass of 0.6 g.
? Mass of the pipe = Volume of wood in the pipe × 0.6 = 5720 × 0.6 = 3432 g = 
3432
1000
 = 3.432 kg
Thus, the mass of the pipe is 3.432 kg.
Question:38
In a 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:
Length of the cylindrical pipe, h = 28 m
Radius of the cylindrical pipe, r = 
5
2
 = 2.5 cm = 0.025 m         1m = 100cm
? Total radiating surface in the system
= Curved surface area of the cylindrical pipe
= 2 prh = 2 ×
22
7
×0. 025 ×28 = 4. 4 m
2
Thus, the total radiating surface in the system is 4.4 m
2
.
Question:39
Find the weight of a solid cylinder of radius 10.5 cm and height 60 cm if the material of the cylinder weighs 5 g per cm
2
.
Solution:
Here, r = 10.5 cm; h = 60 cm
Now, volume of the cylinder = pr
2
h                                       =
22
7
× 10. 5)
2
×60 cm
3
                                      = 22 ×10. 5 ×1. 5 ×60 cm
3
= 20790 cm
3
v
( ) ( )
(
Page 4


                      
              
             
      
                                  
                         
         
          
                    
                       
           
    
       
      
 
                     
         
                        
                        
                 
           
      
     
    
  
        
Question:29
The diameter of a cylinder is 28 cm and its height is 40 cm. Find the curved surface area, total surface area and the volume of the cylinder.
Solution:
Here, r = 28/2 = 14 cm; h = 40 cm
Curved surface area of the cylinder = 2 prh                                                             = 2 ×
22
7
×14 ×40  cm
2
                                                            = 2 ×22 ×2 ×40 cm
2
                                           
Total surface area of the cylinder = 2 prh + 2 pr
2
                                                        = 3520 +2 ×
22
7
×14
2
 cm
2
                                                       = 3520 +1232 = 4752 cm
2
? Volume of the cylinder = pr
2
h                                          =
22
7
×14
2
×40 cm
3
                                         = 22 ×14 ×2 ×40 cm
3
                                         = 24640 cm
3
Question:30
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:
Radius of the bowl, r = 
7
2
 cm
Height of soup in the bowl, h = 4 cm
Volume of soup in one bowl = pr
2
h =
22
7
×
7
2
2
×4 = 154 cm
3
? Amount of soup the hospital has to prepare daily to serve 250 patients
= Volume of soup in one bowl × 250
( ) ( )
( )
= 154 × 250
= 38500 cm
3
Question:31
The pillars of a temple are cylindrically shaped. Each pillar has a circular base of radius 20 cm and height 10 m. How much concrete mixture would be required to build 14 such pillars?
Solution:
Radius of each pillar, r = 20 cm = 0.2 m   1m = 100cm
Height of each pillar, h = 10 m
Volume of concrete mixture used in each pillar = pr
2
h =
22
7
×(0. 2)
2
×10 m
3
? Amount of concrete mixture required to build 14 such pillars
= Volume of concrete mixture used in each pillar × 14
= 
22
7
×(0. 2)
2
×10 ×14 
= 17.6 m
3
Question:32
A soft drink is available in two packs: i a tin can with a rectangular base of length 5 cm, breadth 4 cm and height 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:
i Length of tin can, l = 5 cm
Breadth of tin can, b = 4 cm
Height of tin can, h = 15 cm
? Volume of soft drink in tin can = l × b × h = 5 × 4 × 15 = 300 cm
3
ii Radius of plastic cylinder, r = 
7
2
 cm
Height of plastic cylinder, h = 10 cm
? Volume of soft drink in plastic cylinder = pr
2
h =
22
7
×
7
2
2
×10 = 385 cm
3
So, the capacity of the plastic cylinder pack is greater than the capacity of the tin can pack.
Difference in the capacities of the two packs = 385 - 300 = 85 cm
3
Thus, the capacity of the plastic cylinder pack is 85 cm
3
 more than the capacity of the tin can pack.
Question:33
There are 20 cylindrical pillars in a building, each having a diameter of 50 cm and height 4 m. Find the cost of cleaning them at  14 per m
2
.
Solution:
Radius of each pillar, r = 
50
2
 = 25 cm = 0.25 m        1m = 100cm
Height of each pillar, h = 4 m
? Surface area of each pillar = 2 prh = 2 ×
22
7
×0. 25 ×4 cm
2
Surface area of 20 pillars = Surface area of each pillar × 20 = 2 ×
22
7
×0. 25 ×4 ×20 cm
2
Rate of cleaning =  14 per m
2
? Total cost of cleaning the 20 pillars
= Surface area of 20 pillars × Rate of cleaning 
= 2 ×
22
7
×0. 25 ×4 ×20 ×14
=  1760
Question:34
The curved surface area of a right circular cylinder is 4.4 m
2
. If the radius of its base is 0.7 m, find its i height and ii volume.
Solution:
Radius of the cylinder, r = 0.7 m
Curved surface area of cylinder = 4.4 m
2
i Let the height of the cylinder be h m.
? 2 prh = 4. 4 ? 2 ×
22
7
×0. 7 ×h = 4. 4 ? h =
4.4×7
2×22×0.7
= 1 m
Thus, the height of the cylinder is 1 m.
( )
ii Volume of the cylinder = pr
2
h =
22
7
×(0. 7)
2
×1 = 1.54 m
3
Thus, the volume of the cylinder is 1.54 m
3
.
Question:35
The lateral surface area of a cylinder is 94.2 cm
2
 and its height is 5 cm. Find i the radius of its base and ii its volume. Take p = 3.14.
Solution:
Height of the cylinder, h = 5 cm
Lateral orcurved surface area of cylinder = 94.2 cm
2
i Let the radius of the cylinder be r cm.
? 2 prh = 94. 2 cm
2
? 2 ×3. 14 ×r×5 = 94. 2 ? r =
94.2
2×3.14×5
= 3 cm
Thus, the radius of the cylinder is 3 cm.
ii Volume of the cylinder = pr
2
h = 3. 14 ×(3)
2
×5 = 141.3 cm
3
Thus, the volume of the cylinder is 141.3 cm
3
.
Question:36
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. Find the area of the metal sheet needed to make it.
Solution:
Height of the cylinder, h = 1 m
Capacity of the cylinder = 15.4 L = 15.4 × 0.001 m
3
 = 0.0154 m
3
? pr
2
h = 0. 0154 m
3
?
22
7
×r
2
×1 = 0. 0154 ? r =
0.0154×7
22
= 0. 07 m
? Area of the metal sheet needed to make the cylinder 
= Total surface area of the cylinder
= 2 prh(r+h) = 2 ×
22
7
×0. 07 ×1 ×(0. 07 +1) = 2 ×
22
7
×0. 07 ×1 ×1. 07 = 0. 4708 m
2
Thus, the area of the metal sheet needed to make the cylinder is 0.4708 m
2
.
Question:37
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 cm
3 
of wood has a mass of 0.6
g.
Solution:
Inner radius of the wooden pipe, r = 
24
2
 = 12 cm
Outer radius of the wooden pipe, R = 
28
2
 = 14 cm
Length of the wooden pipe, h = 35 cm
? Volume of wood in the pipe = p R
2
-r
2
h =
22
7
× 14
2
-12
2
×35 = 5720 cm
3
It is given that 1 cm
3 
of wood has a mass of 0.6 g.
? Mass of the pipe = Volume of wood in the pipe × 0.6 = 5720 × 0.6 = 3432 g = 
3432
1000
 = 3.432 kg
Thus, the mass of the pipe is 3.432 kg.
Question:38
In a 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:
Length of the cylindrical pipe, h = 28 m
Radius of the cylindrical pipe, r = 
5
2
 = 2.5 cm = 0.025 m         1m = 100cm
? Total radiating surface in the system
= Curved surface area of the cylindrical pipe
= 2 prh = 2 ×
22
7
×0. 025 ×28 = 4. 4 m
2
Thus, the total radiating surface in the system is 4.4 m
2
.
Question:39
Find the weight of a solid cylinder of radius 10.5 cm and height 60 cm if the material of the cylinder weighs 5 g per cm
2
.
Solution:
Here, r = 10.5 cm; h = 60 cm
Now, volume of the cylinder = pr
2
h                                       =
22
7
× 10. 5)
2
×60 cm
3
                                      = 22 ×10. 5 ×1. 5 ×60 cm
3
= 20790 cm
3
v
( ) ( )
(
? Weight of cylinder = volume of cylinder ×weight of cylinder per gram
                              = 20790 ×5  g = 103950 g = 103. 95 kg 
Question:40
The curved surface area of a cylinder is 1210 cm
2
 and its diameter is 20 cm. Find its height and volume.
Solution:
Curved surface area = 1210 cm
2
Suppose that the height of cylinder is h cm.
We have r = 10 cm
Now, 1210 = 2 prh ? 1210 = 2 ×
22
7
×10 ×h ? h =
1210×7
2×22×10
=
11×7
2×2
= 19. 25 cm
? Volume of the cylinder = pr
2
h                                            =
22
7
×10
2
×19. 25 cm
3
                                           = 2200 ×2. 75 cm
3
 = 6050 cm
3
Question:41
The curved surface area of a cylinder is 4400 cm
2
 and the circumference of its base is 110 cm. Find the height and the volume of the cylinder.
Solution:
Let r be the radius and h be the height of the cylinder.
Circumference of its basecircle = 110 cm.
? 2 pr = 110 ? r =
110
2 p
? r =
110
2×
22
7
? r =
110×7
2×22
? r =
35
2
cm
Curved surface area of a cylinder = 4400 cm
2
.
? 2 prh = 4400 ? h =
4400
2 p r
? h =
4400
2×
22
7
×
35
2
? h =
4400×7×2
2×22×35
? h = 40 cm
Also, Volume of the cylinder = pr
2
h =
22
7
×
35
2
2
×40 =
22×35×35×40
7×2×2
= 38500 cm
3
.
Question:42
The radius of the base and the height of a cylinder are in the ratio 2 : 3. If its volume is 1617 cm
3
, find the total surface area of the cylinder.
Solution:
Suppose that the radius of the base and the height of the cylinder are 2x cm and 3x cm, respectively.
Then 1617 = pr
2
h =
22
7
× 2x)
2
×3x                   = 
22
7
×12x
3
? x
3
=
1617×7
22×12
= 42. 875 ? x =
3
v 42. 875 = 3. 5 cm
Hence, radius = 7 cm; height of the cylinder = 10.5 cm
? Total surface area of the cylinder = 2 prh +2 pr
2
                                                         = 2 ×
22
7
7 ×10. 5 +7 ×7 cm
2
                                                       =
44
7
× 73. 5 +49 cm
2
 = 770  cm
2
Question:43
The total surface area of a cylinder is 462 cm
2
. Its curved surface area is one-third of its total surface area. Find the volume of the cylinder.
Solution:
Total surface area = 462 cm
2
Given: Curved surface area =
1
3
×total surface area = 
1
3
×462 = 154 cm
2
Now, total surface area - curved surface area = 2 prh +2 pr
2
 -2 prh
? ? 462 -154 = 2 pr
2
? 308 = 2 ×
22
7
×r
2
? r
2
=
308×7
44
= 49 ? r = 7 cm
Now, curved surface area = 154 cm
2
? 2 prh = 154 ? 2 ×
22
7
×7 ×h = 154 ? h =
154
44
= 3. 5 cm
 ? Volume of the cylinder = pr
2
h                                              =
22
7
×7
2
×3. 5                                           = 539 cm
3
Question:44
The total surface area of a solid cylinder is 231 cm
2
 and its curved surface area is 
2
3
 of the total surface area. Find the volume of the cylinder.
Solution:
Curved surface area = 
2
3
×total surface area = 
2
3
×231 = 2 ×77 = 154 cm
2
Now, total surface area - curved surface area = 2 prh +2 pr
2
-2 prh
Then 231 -154 = 2 pr
2
? 2 ×
22
7
×r
2
= 77 ? r
2
=
77×7
44
= 12. 25 ? r = 3. 5 cmAlso, curved surface area = 154 cm
2
? 2 prh = 154 ? 2 ×
22
7
×3. 5 ×h = 154 ? h =
154×7
44×3.5
= 7 cm
? Volume of the cylinder = pr
2
h                                           =
22
7
× 3. 5)
2
×7 = 269. 5 cm
3
Question:45
The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder if its total surface area is 616 cm
2
.
Solution:
Suppose that the curved surface area and the total surface area of the right circular cylinder are x cm
2
 and 2x cm
2
.
Then we have:
2x = 616
x = 308 sq cm
Hence, the curved surface area of the cylinder is 308 sq cm.
Let  r cm and h cm be the radius and height of the cylinder, respectively.
(
( )
(
( ) ( )
(
Page 5


                      
              
             
      
                                  
                         
         
          
                    
                       
           
    
       
      
 
                     
         
                        
                        
                 
           
      
     
    
  
        
Question:29
The diameter of a cylinder is 28 cm and its height is 40 cm. Find the curved surface area, total surface area and the volume of the cylinder.
Solution:
Here, r = 28/2 = 14 cm; h = 40 cm
Curved surface area of the cylinder = 2 prh                                                             = 2 ×
22
7
×14 ×40  cm
2
                                                            = 2 ×22 ×2 ×40 cm
2
                                           
Total surface area of the cylinder = 2 prh + 2 pr
2
                                                        = 3520 +2 ×
22
7
×14
2
 cm
2
                                                       = 3520 +1232 = 4752 cm
2
? Volume of the cylinder = pr
2
h                                          =
22
7
×14
2
×40 cm
3
                                         = 22 ×14 ×2 ×40 cm
3
                                         = 24640 cm
3
Question:30
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:
Radius of the bowl, r = 
7
2
 cm
Height of soup in the bowl, h = 4 cm
Volume of soup in one bowl = pr
2
h =
22
7
×
7
2
2
×4 = 154 cm
3
? Amount of soup the hospital has to prepare daily to serve 250 patients
= Volume of soup in one bowl × 250
( ) ( )
( )
= 154 × 250
= 38500 cm
3
Question:31
The pillars of a temple are cylindrically shaped. Each pillar has a circular base of radius 20 cm and height 10 m. How much concrete mixture would be required to build 14 such pillars?
Solution:
Radius of each pillar, r = 20 cm = 0.2 m   1m = 100cm
Height of each pillar, h = 10 m
Volume of concrete mixture used in each pillar = pr
2
h =
22
7
×(0. 2)
2
×10 m
3
? Amount of concrete mixture required to build 14 such pillars
= Volume of concrete mixture used in each pillar × 14
= 
22
7
×(0. 2)
2
×10 ×14 
= 17.6 m
3
Question:32
A soft drink is available in two packs: i a tin can with a rectangular base of length 5 cm, breadth 4 cm and height 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:
i Length of tin can, l = 5 cm
Breadth of tin can, b = 4 cm
Height of tin can, h = 15 cm
? Volume of soft drink in tin can = l × b × h = 5 × 4 × 15 = 300 cm
3
ii Radius of plastic cylinder, r = 
7
2
 cm
Height of plastic cylinder, h = 10 cm
? Volume of soft drink in plastic cylinder = pr
2
h =
22
7
×
7
2
2
×10 = 385 cm
3
So, the capacity of the plastic cylinder pack is greater than the capacity of the tin can pack.
Difference in the capacities of the two packs = 385 - 300 = 85 cm
3
Thus, the capacity of the plastic cylinder pack is 85 cm
3
 more than the capacity of the tin can pack.
Question:33
There are 20 cylindrical pillars in a building, each having a diameter of 50 cm and height 4 m. Find the cost of cleaning them at  14 per m
2
.
Solution:
Radius of each pillar, r = 
50
2
 = 25 cm = 0.25 m        1m = 100cm
Height of each pillar, h = 4 m
? Surface area of each pillar = 2 prh = 2 ×
22
7
×0. 25 ×4 cm
2
Surface area of 20 pillars = Surface area of each pillar × 20 = 2 ×
22
7
×0. 25 ×4 ×20 cm
2
Rate of cleaning =  14 per m
2
? Total cost of cleaning the 20 pillars
= Surface area of 20 pillars × Rate of cleaning 
= 2 ×
22
7
×0. 25 ×4 ×20 ×14
=  1760
Question:34
The curved surface area of a right circular cylinder is 4.4 m
2
. If the radius of its base is 0.7 m, find its i height and ii volume.
Solution:
Radius of the cylinder, r = 0.7 m
Curved surface area of cylinder = 4.4 m
2
i Let the height of the cylinder be h m.
? 2 prh = 4. 4 ? 2 ×
22
7
×0. 7 ×h = 4. 4 ? h =
4.4×7
2×22×0.7
= 1 m
Thus, the height of the cylinder is 1 m.
( )
ii Volume of the cylinder = pr
2
h =
22
7
×(0. 7)
2
×1 = 1.54 m
3
Thus, the volume of the cylinder is 1.54 m
3
.
Question:35
The lateral surface area of a cylinder is 94.2 cm
2
 and its height is 5 cm. Find i the radius of its base and ii its volume. Take p = 3.14.
Solution:
Height of the cylinder, h = 5 cm
Lateral orcurved surface area of cylinder = 94.2 cm
2
i Let the radius of the cylinder be r cm.
? 2 prh = 94. 2 cm
2
? 2 ×3. 14 ×r×5 = 94. 2 ? r =
94.2
2×3.14×5
= 3 cm
Thus, the radius of the cylinder is 3 cm.
ii Volume of the cylinder = pr
2
h = 3. 14 ×(3)
2
×5 = 141.3 cm
3
Thus, the volume of the cylinder is 141.3 cm
3
.
Question:36
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. Find the area of the metal sheet needed to make it.
Solution:
Height of the cylinder, h = 1 m
Capacity of the cylinder = 15.4 L = 15.4 × 0.001 m
3
 = 0.0154 m
3
? pr
2
h = 0. 0154 m
3
?
22
7
×r
2
×1 = 0. 0154 ? r =
0.0154×7
22
= 0. 07 m
? Area of the metal sheet needed to make the cylinder 
= Total surface area of the cylinder
= 2 prh(r+h) = 2 ×
22
7
×0. 07 ×1 ×(0. 07 +1) = 2 ×
22
7
×0. 07 ×1 ×1. 07 = 0. 4708 m
2
Thus, the area of the metal sheet needed to make the cylinder is 0.4708 m
2
.
Question:37
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 cm
3 
of wood has a mass of 0.6
g.
Solution:
Inner radius of the wooden pipe, r = 
24
2
 = 12 cm
Outer radius of the wooden pipe, R = 
28
2
 = 14 cm
Length of the wooden pipe, h = 35 cm
? Volume of wood in the pipe = p R
2
-r
2
h =
22
7
× 14
2
-12
2
×35 = 5720 cm
3
It is given that 1 cm
3 
of wood has a mass of 0.6 g.
? Mass of the pipe = Volume of wood in the pipe × 0.6 = 5720 × 0.6 = 3432 g = 
3432
1000
 = 3.432 kg
Thus, the mass of the pipe is 3.432 kg.
Question:38
In a 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:
Length of the cylindrical pipe, h = 28 m
Radius of the cylindrical pipe, r = 
5
2
 = 2.5 cm = 0.025 m         1m = 100cm
? Total radiating surface in the system
= Curved surface area of the cylindrical pipe
= 2 prh = 2 ×
22
7
×0. 025 ×28 = 4. 4 m
2
Thus, the total radiating surface in the system is 4.4 m
2
.
Question:39
Find the weight of a solid cylinder of radius 10.5 cm and height 60 cm if the material of the cylinder weighs 5 g per cm
2
.
Solution:
Here, r = 10.5 cm; h = 60 cm
Now, volume of the cylinder = pr
2
h                                       =
22
7
× 10. 5)
2
×60 cm
3
                                      = 22 ×10. 5 ×1. 5 ×60 cm
3
= 20790 cm
3
v
( ) ( )
(
? Weight of cylinder = volume of cylinder ×weight of cylinder per gram
                              = 20790 ×5  g = 103950 g = 103. 95 kg 
Question:40
The curved surface area of a cylinder is 1210 cm
2
 and its diameter is 20 cm. Find its height and volume.
Solution:
Curved surface area = 1210 cm
2
Suppose that the height of cylinder is h cm.
We have r = 10 cm
Now, 1210 = 2 prh ? 1210 = 2 ×
22
7
×10 ×h ? h =
1210×7
2×22×10
=
11×7
2×2
= 19. 25 cm
? Volume of the cylinder = pr
2
h                                            =
22
7
×10
2
×19. 25 cm
3
                                           = 2200 ×2. 75 cm
3
 = 6050 cm
3
Question:41
The curved surface area of a cylinder is 4400 cm
2
 and the circumference of its base is 110 cm. Find the height and the volume of the cylinder.
Solution:
Let r be the radius and h be the height of the cylinder.
Circumference of its basecircle = 110 cm.
? 2 pr = 110 ? r =
110
2 p
? r =
110
2×
22
7
? r =
110×7
2×22
? r =
35
2
cm
Curved surface area of a cylinder = 4400 cm
2
.
? 2 prh = 4400 ? h =
4400
2 p r
? h =
4400
2×
22
7
×
35
2
? h =
4400×7×2
2×22×35
? h = 40 cm
Also, Volume of the cylinder = pr
2
h =
22
7
×
35
2
2
×40 =
22×35×35×40
7×2×2
= 38500 cm
3
.
Question:42
The radius of the base and the height of a cylinder are in the ratio 2 : 3. If its volume is 1617 cm
3
, find the total surface area of the cylinder.
Solution:
Suppose that the radius of the base and the height of the cylinder are 2x cm and 3x cm, respectively.
Then 1617 = pr
2
h =
22
7
× 2x)
2
×3x                   = 
22
7
×12x
3
? x
3
=
1617×7
22×12
= 42. 875 ? x =
3
v 42. 875 = 3. 5 cm
Hence, radius = 7 cm; height of the cylinder = 10.5 cm
? Total surface area of the cylinder = 2 prh +2 pr
2
                                                         = 2 ×
22
7
7 ×10. 5 +7 ×7 cm
2
                                                       =
44
7
× 73. 5 +49 cm
2
 = 770  cm
2
Question:43
The total surface area of a cylinder is 462 cm
2
. Its curved surface area is one-third of its total surface area. Find the volume of the cylinder.
Solution:
Total surface area = 462 cm
2
Given: Curved surface area =
1
3
×total surface area = 
1
3
×462 = 154 cm
2
Now, total surface area - curved surface area = 2 prh +2 pr
2
 -2 prh
? ? 462 -154 = 2 pr
2
? 308 = 2 ×
22
7
×r
2
? r
2
=
308×7
44
= 49 ? r = 7 cm
Now, curved surface area = 154 cm
2
? 2 prh = 154 ? 2 ×
22
7
×7 ×h = 154 ? h =
154
44
= 3. 5 cm
 ? Volume of the cylinder = pr
2
h                                              =
22
7
×7
2
×3. 5                                           = 539 cm
3
Question:44
The total surface area of a solid cylinder is 231 cm
2
 and its curved surface area is 
2
3
 of the total surface area. Find the volume of the cylinder.
Solution:
Curved surface area = 
2
3
×total surface area = 
2
3
×231 = 2 ×77 = 154 cm
2
Now, total surface area - curved surface area = 2 prh +2 pr
2
-2 prh
Then 231 -154 = 2 pr
2
? 2 ×
22
7
×r
2
= 77 ? r
2
=
77×7
44
= 12. 25 ? r = 3. 5 cmAlso, curved surface area = 154 cm
2
? 2 prh = 154 ? 2 ×
22
7
×3. 5 ×h = 154 ? h =
154×7
44×3.5
= 7 cm
? Volume of the cylinder = pr
2
h                                           =
22
7
× 3. 5)
2
×7 = 269. 5 cm
3
Question:45
The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder if its total surface area is 616 cm
2
.
Solution:
Suppose that the curved surface area and the total surface area of the right circular cylinder are x cm
2
 and 2x cm
2
.
Then we have:
2x = 616
x = 308 sq cm
Hence, the curved surface area of the cylinder is 308 sq cm.
Let  r cm and h cm be the radius and height of the cylinder, respectively.
(
( )
(
( ) ( )
(
Then 2 prh +2 pr
2
 = 616 cm
2
  and 2 prh = 308 cm
2
? 2 prh +2 pr
2
 - 2 prh = 616 -308 ? 2 pr
2
= 308 ? 2 ×
22
7
×r
2
= 308 ? r
2
=
308×7
44
= 49 ? r = 7 cmNow, 2 prh = 308 cm
2
? 2 ×
22
7
×7 ×h = 308
? Volume of the cylinder = pr
2
h  cubic cm                                          =
22
7
×7
2
×7                                           = 1078 cm
3
Question:46
A cylindrical bucket, 28 cm in diameter and 72 cm and high, is full of water. The water is emptied into a rectangular tank, 66 cm long and 28 cm wide. Find the height of the water level in
the tank.
Solution:
 Given: Diameter of the cylindrical bucket = 28 cm
i.e., radius = 14 cm
Height of the cylindrical bucket, h
1
 = 72 cm
Length of the rectangular tank, l = 66 cm
Breadth of the rectangular tank, b = 28 cm
Let the height of the rectangular tank be h cm.
The water from the cylindrical bucket is emptied into the rectangular tank.
i.e., volume of the bucket = volume of the tank
? pr
2
h
1
 = l ×b ×h ?
22
7
×14
2
×72 = 66 ×28 ×h ? h =
22×14×2×72
66×28
= 24 cm
? Height of the rectangular tank = 24 cm
Question:47
The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up on writing 330 words on an average. How many words
would use up a bottle of ink containing one-fifth of a litre?
Solution:
Height of the barrel = h = 7 cm
Radius of the barrel = r = 2.5 mm = 0.25 cm
Volume of the barrel = pr
2
h 
                               =
22
7
×0. 25 ×0. 25 ×7 = 22 ×0. 25 ×0. 25 = 1. 375  cm
3
i.e., 1.375 cm
3
 of ink is used for writing 330 words
Now, number of words that could be written with one-fifth of a litre, i.e., 
1
5
×1000 cm
3
 = 330 ×
1
1.375
×
1
5
×1000 = 48000   
? 48000 words would use up a bottle of ink containing one-fifith of a litre.  
Question:48
1 cm
3
 of gold is drawn into a wire 0.1 mm in diameter. Find the length of the wire.
Solution:
Let r cm be the radius of the wire and h cm be the height of the wire.
Volume of the gold = 1 cm
3
1 cm
3 
of gold is drawn into a wire of diameter 0.1 mm.
Here, r =  0. 1 mm =
0.1
20
cm =
1
200
cm
? pr
2
h = 1 ?
22
7
×
1
200
×
1
200
×h = 1 ? h =
40000×7
22
= 12727. 27 cm
Hence, length of the wire = 127.27 m
Question:49
If 1 cm
3
 of case iron weighs 21 g, find the weight of a cast iron pipe of length 1 m with a bore of 3 cm in which the thickness of the metal is 1 cm.
Solution:
Internal radius of the pipe = 1.5 cm
External radius of the pipe = 1.5 +1 cm = 2.5 cm
Height of the pipe = 1 m = 100 cm
Volume of the cast iron = total volume of the pipe - internal volume of the pipe
                                   = p × 2. 5)
2
×100 - p ×(1. 5)
2
×100 =
22
7
×100 ×[6. 25 -2. 25] =
2200
7
×4 =
8800
7
cm
3
1 cm
3
 of cast iron weighs 21 g.
? Weight of 
8800
7
cm
3
cast iron = 
8800
7
×
21
1000
 kg = 88 ×0. 3 = 26. 4 kg
Question:50
A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate
the volume of the metal.
Solution:
Inner radius of the cylindrical tube, r = 5.2 cm
Height of the cylindrical tube, h = 25 cm
Outer radius of the cylindrical tube, R = 5.2 +0.8 cm = 6 cm
? Volume of the metal = external volume - internal volume                                         = pR
2
h - pr
2
h       where R and r are the outer and inner radii, respectively                                     
Question:51
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 metres of the sheet are required for the same?
Solution:
Height of the cylindrical tank, h = 1 m
Radius of the cylindrical tank, r = 
140
2
 = 70 cm = 0.7 m       1m = 100cm
? Area of the metal sheet required to make the cylindrical tank
= Total surface area of the cylindrical tank
(
( )