NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.6 contains only 8 questions which are based on the formula of the geometrical objects. NCERT Solutions help students in their studies, here we have provided the solutions to all the questions in PDF format with detailed explanations. Students can download the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.6 PDF, through the links provided further below in this page.
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Access Other Exercise Solutions of Class 9 Maths Chapter 13 – Surface Areas and Volumes
Exercise 13.1 solution (9 questions)
Exercise 13.2 solution (8 questions)
Exercise 13.3 solution (9 questions)
Exercise 13.4 solution (5 questions)
Exercise 13.5 solution (5 questions)
Exercise 13.7 solution (9 questions)
Exercise 13.8 solution (10 questions)
Exercise 13.9 solution (3 questions)
Access Answers to NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.6
1. The circumference of the base of cylindrical vessel is 132cm and its height is 25cm.
How many litres of water can it hold? (1000 cm3= 1L) (Assume π = 22/7)
Circumference of the base of cylindrical vessel = 132 cm
Height of vessel, h = 25 cm
Let r be the radius of the cylindrical vessel.
Step 1: Find the radius of vessel
We know that, circumference of base = 2πr, so
2πr = 132 (given)
r = (132/(2 π))
r = 66×7/22 = 21
Radius is 21 cm
Step 2: Find the volume of vessel
Formula: Volume of cylindrical vessel = πr2h
Therefore, volume is 34650 cm3
Since, 1000 cm3 = 1L
So, Volume = 34650/1000 L= 34.65L
Therefore, vessel can hold 34.65 litres of water.
2. The inner diameter of a cylindrical wooden pipe is 24cm and its outer diameter is 28 cm. The length of the pipe is 35cm.Find the mass of the pipe, if 1cm3 of wood has a mass of 0.6g. (Assume π = 22/7)
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. Answer!
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)
- tin can will be cuboidal in shape
Dimensions of 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
- Plastic cylinder will be cylindrical in shape.
Dimensions of plastic can are:
Radius of circular end of plastic cylinder, r = 3.5cm
Height , H = 10 cm
Capacity of plastic cylinder = πr2H
Capacity of plastic cylinder = (22/7)×(3.5)2×10 = 385
Capacity of plastic cylinder is 385 cm3
From results of (i) and (ii), plastic cylinder has more capacity.
Difference in capacity = (385-300) cm3 = 85cm3
4. If the lateral surface of a cylinder is 94.2cm2 and its height is 5cm, then find
(i) radius of its base (ii) its volume.[Use π= 3.14]
CSA of cylinder = 94.2 cm2
Height of cylinder, h = 5cm
(i) Let radius of cylinder be r.
Using CSA of 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 value of r from (i))
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)
(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 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 vessel is 96.25 m3
Therefore, the capacity of the vessel is 96.25 m3 or 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)
Height of cylindrical vessel, h = 1 m
Capacity of cylindrical vessel = 15.4 litres = 0.0154 m3
Let r be the radius of the circular end.
Capacity of cylindrical vessel = (22/7)×r2×1 = 0.0154
After simplifying, we get, r = 0.07 m
Again, total surface area of vessel = 2πr(r+h)
Total surface area of 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)
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, volume of wood in pencil = (r12-r22)h cubic units
Substitute the values, we have
This implies, volume of wood in pencil = 5.28 cm3
Volume of graphite = r22h cubic units
Substitute the values, we have
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 does the hospital have to prepare daily to serve 250 patients? (Assume π = 22/7)
Diameter of 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
Volume of soup given to 250 patients = (250×154) cm3= 38500 cm3
= 38.5litres. Answer!
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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 about the concept
- Helps to solve the cylinder concept problems
- It 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 vessel
- Improve efficiency in solving the problems
- Remember the formulas in an easy way