Inner radius of cylindrical pipe, say r1 = diameter1/ 2 = 24/2 cm = 12cm Outer radius of cylindrical pipe, say r2 = diameter2/ 2 = 28/2 cm = 14 cm Height of pipe, h = Length of pipe = 35cm Now, the Volume of pipe = π(r22-r12)h cm3 Substitute the values. Volume of pipe = 110×52 cm3 = 5720 cm3 Since Mass of 1 cm3 wood = 0.6 g Mass of 5720 cm3 wood = (5720×0.6) g = 3432 g or 3.432 kg. 3. A soft drink is available in two packs - (i) a tin can with a rectangular base of length 5cm and width 4cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7cm and height 10cm. Which container has greater capacity and by how much? (Assume π=22/7) Solution: (i) Tin can will be cuboidal in shape  The dimensions of the tin can are Length, l = 5 cm Breadth, b = 4 cm Height, h = 15 cm Capacity of tin can = l×b×h= (5×4×15) cm3 = 300 cm3 (ii) Plastic cylinder will be cylindrical in shape.  Dimensions of the plastic can are: Radius of the circular end of the plastic cylinder, r = 3.5cm Height , H = 10 cm Capacity of the plastic cylinder = πr2H Capacity of the plastic cylinder = (22/7)×(3.5)2×10 = 385 Capacity of the plastic cylinder is 385 cm3 From results (i) and (ii), the plastic cylinder has more capacity. Difference in capacity = (385-300) cm3 = 85cm3 4. If the lateral surface of a cylinder is 94.2cm2and its height is 5cm, then find (i) radius of its base (ii) its volume.[Use π= 3.14] Solution: CSA of cylinder = 94.2 cm2 Height of cylinder, h = 5cm (i) Let the radius of the cylinder be r. Using the CSA of the cylinder, we get 2πrh = 94.2 2×3.14×r×5 = 94.2 r = 3 radius is 3 cm (ii) Volume of cylinder Formula for volume of cylinder = πr2h Now, πr2h = (3.14×(3)2×5) (using the value of r from (i)) = 141.3 Volume is 141.3 cm3 5. It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10m deep. If the cost of painting is at the rate of Rs 20 per m2, find (i) inner curved surface area of the vessel (ii) radius of the base (iii) capacity of the vessel (Assume π = 22/7) Solution: (i) Rs 20 is the cost of painting 1 m2 area. Rs 1 is the cost to paint 1/20 m2 area So, Rs 2200 is the cost of painting = (1/20×2200) m2 = 110 m2 area The inner surface area of the vessel is 110m2. (ii) Radius of the base of the vessel, let us say r. Height (h) = 10 m and Surface area formula = 2πrh Using the result of (i) 2πrh = 110 m2 2×22/7×r×10 = 110 r = 1.75 Radius is 1.75 m. (iii) Volume of vessel formula = πr2h Here r = 1.75 and h = 10 Volume = (22/7)×(1.75)2×10 = 96.25 Volume of the vessel is 96.25 m3 Therefore, the capacity of the vessel is 96.25 m3or 96250 litres. 6. The capacity of a closed cylindrical vessel of height 1m is15.4 liters. How many square meters of metal sheet would be needed to make it? (Assume π = 22/7) Solution: Height of the cylindrical vessel, h = 1 m Capacity of the cylindrical vessel = 15.4 litres = 0.0154 m3 Let r be the radius of the circular end. Now, Capacity of the vessel = (22/7)×r2×1 = 0.0154 After simplifying, we get r = 0.07 m Again, the total surface area of vessel = 2πr(r+h) = 2×22/7×0.07(0.07+1) = 0.44×1.07 = 0.4708 Total surface area of the vessel is 0.4708 m2 Therefore, 0.4708 m2 of the metal sheet would be required to make the cylindrical vessel. 7. A lead pencil consists of a cylinder of wood with solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite. (Assume π = 22/7) Solution:  Radius of pencil, r1 = 7/2 mm = 0.7/2 cm = 0.35 cm Radius of graphite, r2 = 1/2 mm = 0.1/2 cm = 0.05 cm Height of pencil, h = 14 cm Formula to find the volume of wood in pencil = (r12-r22)h cubic units Substituting the values we have = [(22/7)×(0.352-0.052)×14] = 44×0.12 = 5.28 This implies the volume of wood in pencil = 5.28 cm3 Again, The volume of graphite = r22h cubic units Substituting the values, we have = (22/7)×0.052×14 = 44×0.0025 = 0.11 So, the volume of graphite is 0.11 cm3. 8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7cm. If the bowl is filled with soup to a height of 4cm, how much soup the hospital has to prepare daily to serve 250 patients? (Assume π = 22/7) Solution: Diameter of the cylindrical bowl = 7 cm Radius of cylindrical bowl, r = 7/2 cm = 3.5 cm Bowl is filled with soup to a height of4cm, so h = 4 cm  Volume of soup in one bowl= πr2h (22/7)×3.52×4 = 154 Volume of soup in one bowl is 154 cm3 Therefore, Volume of soup given to 250 patients = (250×154) cm3= 38500 cm3 = 38.5litres. We hope this information on “NCERT Solution for Class 9 Maths Chapter 13 - Surface Areas and Volumes is useful for students preparing for exams. Stay tuned for further updates on CBSE and other competitive exams. To access interactive Maths and Science Videos, download the BYJU’S App and subscribe to RUclips Channel. Key Features of NCERT Solutions for Class 9 Maths Chapter 13 - Surface Areas and Volume Exercise 13.6 Solving these solutions help students to: Students can self-assess their knowledge of the conceptHelps to solve the cylinder concept problemsIt helps the student to find the inner curved surface area of the vessel, the radius of the base of the vessel and determine the capacity of the vesselImprove efficiency in solving the problemsRemember the formulas in an easy wa
![HIGH RATED GABRU [SLOWED+REVERB]](http://i.ytimg.com/vi/ESS1lLSEs5M/mqdefault.jpg)
![HIGH RATED GABRU [SLOWED+REVERB]](/img/tr.png)
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Inner radius of cylindrical pipe, say r1 = diameter1/ 2 = 24/2 cm = 12cm
Outer radius of cylindrical pipe, say r2 = diameter2/ 2 = 28/2 cm = 14 cm
Height of pipe, h = Length of pipe = 35cm
Now, the Volume of pipe = π(r22-r12)h cm3
Substitute the values.
Volume of pipe = 110×52 cm3 = 5720 cm3
Since Mass of 1 cm3 wood = 0.6 g
Mass of 5720 cm3 wood = (5720×0.6) g = 3432 g or 3.432 kg.
3. A soft drink is available in two packs - (i) a tin can with a rectangular base of length 5cm and width 4cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7cm and height 10cm. Which container has greater capacity and by how much? (Assume π=22/7)
Solution:
(i) Tin can will be cuboidal in shape

The dimensions of the tin can are
Length, l = 5 cm
Breadth, b = 4 cm
Height, h = 15 cm
Capacity of tin can = l×b×h= (5×4×15) cm3 = 300 cm3
(ii) Plastic cylinder will be cylindrical in shape.

Dimensions of the plastic can are:
Radius of the circular end of the plastic cylinder, r = 3.5cm
Height , H = 10 cm
Capacity of the plastic cylinder = πr2H
Capacity of the plastic cylinder = (22/7)×(3.5)2×10 = 385
Capacity of the plastic cylinder is 385 cm3
From results (i) and (ii), the plastic cylinder has more capacity.
Difference in capacity = (385-300) cm3 = 85cm3
4. If the lateral surface of a cylinder is 94.2cm2and its height is 5cm, then find
(i) radius of its base (ii) its volume.[Use π= 3.14]
Solution:
CSA of cylinder = 94.2 cm2
Height of cylinder, h = 5cm
(i) Let the radius of the cylinder be r.
Using the CSA of the cylinder, we get
2πrh = 94.2
2×3.14×r×5 = 94.2
r = 3
radius is 3 cm
(ii) Volume of cylinder
Formula for volume of cylinder = πr2h
Now, πr2h = (3.14×(3)2×5) (using the value of r from (i))
= 141.3
Volume is 141.3 cm3
5. It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10m deep. If the cost of painting is at the rate of Rs 20 per m2, find
(i) inner curved surface area of the vessel
(ii) radius of the base
(iii) capacity of the vessel
(Assume π = 22/7)
Solution:
(i) Rs 20 is the cost of painting 1 m2 area.
Rs 1 is the cost to paint 1/20 m2 area
So, Rs 2200 is the cost of painting = (1/20×2200) m2
= 110 m2 area
The inner surface area of the vessel is 110m2.
(ii) Radius of the base of the vessel, let us say r.
Height (h) = 10 m and
Surface area formula = 2πrh
Using the result of (i)
2πrh = 110 m2
2×22/7×r×10 = 110
r = 1.75
Radius is 1.75 m.
(iii) Volume of vessel formula = πr2h
Here r = 1.75 and h = 10
Volume = (22/7)×(1.75)2×10 = 96.25
Volume of the vessel is 96.25 m3
Therefore, the capacity of the vessel is 96.25 m3or 96250 litres.
6. The capacity of a closed cylindrical vessel of height 1m is15.4 liters. How many square meters of metal sheet would be needed to make it? (Assume π = 22/7)
Solution:
Height of the cylindrical vessel, h = 1 m
Capacity of the cylindrical vessel = 15.4 litres = 0.0154 m3
Let r be the radius of the circular end.
Now,
Capacity of the vessel = (22/7)×r2×1 = 0.0154
After simplifying, we get r = 0.07 m
Again, the total surface area of vessel = 2πr(r+h)
= 2×22/7×0.07(0.07+1)
= 0.44×1.07
= 0.4708
Total surface area of the vessel is 0.4708 m2
Therefore, 0.4708 m2 of the metal sheet would be required to make the cylindrical vessel.
7. A lead pencil consists of a cylinder of wood with solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite. (Assume π = 22/7)
Solution:

Radius of pencil, r1 = 7/2 mm = 0.7/2 cm = 0.35 cm
Radius of graphite, r2 = 1/2 mm = 0.1/2 cm = 0.05 cm
Height of pencil, h = 14 cm
Formula to find the volume of wood in pencil = (r12-r22)h cubic units
Substituting the values we have
= [(22/7)×(0.352-0.052)×14]
= 44×0.12
= 5.28
This implies the volume of wood in pencil = 5.28 cm3
Again,
The volume of graphite = r22h cubic units
Substituting the values, we have
= (22/7)×0.052×14
= 44×0.0025
= 0.11
So, the volume of graphite is 0.11 cm3.
8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7cm. If the bowl is filled with soup to a height of 4cm, how much soup the hospital has to prepare daily to serve 250 patients? (Assume π = 22/7)
Solution:
Diameter of the cylindrical bowl = 7 cm
Radius of cylindrical bowl, r = 7/2 cm = 3.5 cm
Bowl is filled with soup to a height of4cm, so h = 4 cm

Volume of soup in one bowl= πr2h
(22/7)×3.52×4 = 154
Volume of soup in one bowl is 154 cm3
Therefore,
Volume of soup given to 250 patients = (250×154) cm3= 38500 cm3
= 38.5litres.
We hope this information on “NCERT Solution for Class 9 Maths Chapter 13 - Surface Areas and Volumes is useful for students preparing for exams. Stay tuned for further updates on CBSE and other competitive exams. To access interactive Maths and Science Videos, download the BYJU’S App and subscribe to RUclips Channel.
Key Features of NCERT Solutions for Class 9 Maths Chapter 13 - Surface Areas and Volume Exercise 13.6
Solving these solutions help students to:
Students can self-assess their knowledge of the conceptHelps to solve the cylinder concept problemsIt helps the student to find the inner curved surface area of the vessel, the radius of the base of the vessel and determine the capacity of the vesselImprove efficiency in solving the problemsRemember the formulas in an easy wa
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