Dimensions of the matchbox (a cuboid) are l, b, h = 4 cm, 2.5 cm, 1.5 cm, respectively Formula to find the volume of the matchbox = l×b×h = (4×2.5×1.5) = 15 Volume of matchbox = 15 cm3 Now, volume of 12 such matchboxes = (15×12) cm3 = 180 cm3 Therefore, the volume of the packet containing 12 matchboxes is 180 cm3. 2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m3 = 1000 l) Solution: Dimensions of the cuboidal water tank are: l = 6 m and b = 5 m and h = 4.5 m Formula to find the volume of the tank, V = l×b×h Put the values, we get V = (6×5×4.5) = 135 The volume of the water tank is 135 m3 Again, The amount of water that 1 m3 volume can hold = 1000 l The amount of water that 135 m3 volume can hold = (135×1000) litres = 135000 litres Therefore, the given cuboidal water tank can hold up to 135000 litres of water. 3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid? Solution: Given: Length of the cuboidal vessel, l = 10 m Width of the cuboidal vessel, b = 8m Volume of the cuboidal vessel, V = 380 m3 Let the height of the given vessel be h. Formula to find the volume of a cuboid, V = l×b×h Using the formula, we get l×b×h = 380 10×8×h = 380 Or h = 4.75 Therefore, the height of the vessels must be 4.75 m. 4. Find the cost of digging a cuboidal pit of 8 m long, 6 m broad and 3 m deep at the rate of Rs. 30 per m3. Solution: The given pit has its length (l) as 8m, width (b) as 6m and depth (h) as 3 m. Volume of the cuboidal pit = l×b×h = (8×6×3) = 144 The required Volume is 144 m3 Now, The cost of digging per m3 volume = Rs. 30 Therefore, the cost of digging 144 m3 volume = Rs. (144×30) = Rs. 4320. 5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m. Solution: The length (l) and depth (h) of the tank is 2.5 m and 10 m, respectively. To find the value of breadth, say b, The formula to find the volume of the tank = l×b×h = (2.5× b×10) m3 = 25b m3 The capacity of tank = 25b m3, which is equal to 25000b litres Also, the capacity of a cuboidal tank is 50000 litres of water (Given) Therefore, 25000 b = 50000 This implies that b = 2 Therefore, the breadth of the tank is 2 m. 6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m×15 m×6 m. For how many days will the water in this tank last? Solution: Length of the tank = l = 20 m Breadth of the tank = b = 15 m Height of the tank = h = 6 m Total population of the village = 4000 Consumption of water per head per day = 150 litres Water consumed by the people in 1 day = (4000×150) litres = 600000 litres …(1) The formula to find the capacity of a tank, C = l×b×h Using the given data, we have C = (20×15×6) m3 = 1800 m3 Or C = 1800000 litres Let the water in this tank last for d days. Water consumed by all people in d days = Capacity of the tank (using equation (1)) 600000 d = 1800000 d = 3 Therefore, the water in this tank will last for 3 days. 7. A godown measures 40 m×25 m×15 m. Find the maximum number of wooden crates, each measuring 1.5m×1.25 m×0.5 m, that can be stored in the godown. Solution: From the statement, we have Length of the godown = 40 m Breadth = 25 m Height = 15 m Whereas, Length of the wooden crate = 1.5 m Breadth = 1.25 m Height = 0.5 m Since the godown and wooden crate are in cuboidal shape, we can find the volume of each using the formula, V = lbh. Now, Volume of the godown = (40×25×15) m3 = 15000 m3 Volume of the wooden crate = (1.5×1.25×0.5) m3 = 0.9375 m3 Let us consider that n wooden crates can be stored in the godown, then The volume of n wooden crates = Volume of godown 0.9375×n = 15000 Or n = 15000/0.9375 = 16000 Hence, the number of wooden crates that can be stored in the godown is 16,000. 8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas. Solution: Side of the cube = 12 cm (Given) To find the volume of the cube: Volume of cube = (Side)3 = (12)3 cm3= 1728 cm3 Surface area of a cube with side 12 cm = 6a2 = 6(12) 2 cm2 …(1) The cube is cut into eight small cubes of equal volume; say the side of each cube is p. The volume of the small cube = p3 Surface area = 6p2 …(2) Volume of each small cube = (1728/8) cm3 = 216 cm3 Or (p)3 = 216 cm3 Or p = 6 cm Now, the surface areas of the cubes ratios = (Surface area of the bigger cube)/(Surface area of smaller cubes) From equations (1) and (2), we get Surface areas of the cubes ratios = (6a2)/(6p2) = a2/p2 = 122/62 = 4 Therefore, the required ratio is 4:1. 9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute? Solution: Given: Depth of the river, h = 3 m Width of the river, b = 40 m Rate of water flow = 2 km per hour = 2000 m/60 min = 100/3 m/min So, the volume of water flowed in 1 min = (100/3) × 40 × 3 = 4000m3 Therefore, 4000 m3 of water will fall into the sea in a minute. By practising problems in NCERT Solutions for Class 9 Maths Chapter 13, students can learn the easy way to solve them and can score well in the annual examination. All the 9 questions included in this exercise will help students analyse the situation and apply the formula. It includes easy-to-solve questions, which can be expected in examinations. Solve these NCERT solutions, where problems are explained in a detailed way, following each and every step. NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes exercise consists of application-level questions that help to determine the volume of the cuboid. It also helps in finding the length and breadth of the cuboid by applying the formula. Key Features of NCERT Solutions for Class 9 Maths Chapter 13 - Surface Areas and Volume Exercise 13.5 Solving the NCERT solutions for Class 9 Maths Chapter 13 helps students in multiple ways, as explained below: Students can self-assess their learning abilities and preparation levelStudents can analyse the type of questions that appear for the examsImprove their efficiency and speed in solving the problemsRemember the formulas in an easy way and help them apply relevantly
Height of cylinder, h = 14 cm Let the diameter of the cylinder be d Curved surface area of cylinder = 88 cm2 We know that the formula to find the curved surface area of a cylinder is 2πrh. So 2πrh =88 cm2 (r is the radius of the base of the cylinder) 2×(22/7)×r×14 = 88 cm2 2r = 2 cm d =2 cm Therefore, the diameter of the base of the cylinder is 2 cm. 2. It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? Assume π = 22/7 Solution: Let h be the height and r be the radius of a cylindrical tank. Height of cylindrical tank, h = 1m Radius = half of diameter = (140/2) cm = 70cm = 0.7m Area of sheet required = Total surface area of tank = 2πr(r+h) unit square = [2×(22/7)×0.7(0.7+1)] = 7.48 square meters Therefore, 7.48 square meters of the sheet are required. 3. A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4cm. (see fig. 13.11). Find its  (i) inner curved surface area, (ii) outer curved surface area (iii) total surface area (Assume π=22/7) Solution: Let r1 and r2 be the inner and outer radii of the cylindrical pipe r1 = 4/2 cm = 2 cm r2 = 4.4/2 cm = 2.2 cm Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm (i) curved surface area of outer surface of pipe = 2πr1h = 2×(22/7)×2×77 cm2 = 968 cm2 (ii) curved surface area of outer surface of pipe = 2πr2h = 2×(22/7)×2.2×77 cm2 = (22×22×2.2) cm2 = 1064.8 cm2 (iii) Total surface area of pipe = inner curved surface area+ outer curved surface area+ Area of both circular ends of pipe. = 2r1h+2r2h+2π(r12-r22) = 9668+1064.8+2×(22/7)×(2.22-22) = 2031.8+5.28 = 2038.08 cm2 Therefore, the total surface area of the cylindrical pipe is 2038.08 cm2. 4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? (Assume π = 22/7) Solution: A roller is shaped like a cylinder. Let h be the height of the roller and r be the radius. h = Length of roller = 120 cm Radius of the circular end of roller = r = (84/2) cm = 42 cm Now, CSA of roller = 2πrh = 2×(22/7)×42×120 = 31680 cm2 Area of field = 500×CSA of roller = (500×31680) cm2 = 15840000 cm2 = 1584 m2. Therefore, the area of the playground is 1584 m2. 5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2. (Assume π = 22/7) Solution: Let h be the height of a cylindrical pillar and r be the radius. Given: Height cylindrical pillar = h = 3.5 m Radius of the circular end of pillar = r = diameter/2 = 50/2 = 25 cm = 0.25 m CSA of pillar = 2πrh = 2×(22/7)×0.25×3.5 = 5.5 m2 Cost of painting 1 m2 area = Rs. 12.50 Cost of painting 5.5 m2 area = Rs (5.5×12.50) = Rs. 68.75 Therefore, the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2 is Rs 68.75. 6. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the base of the cylinder is 0.7 m, find its height. (Assume π = 22/7) Solution: Let h be the height of the circular cylinder and r be the radius. The radius of the base of the cylinder, r = 0.7 m CSA of cylinder = 2πrh CSA of cylinder = 4.4 m2 Equating both equations, we have 2×(22/7)×0.7×h = 4.4 Or h = 1 Therefore, the height of the cylinder is 1 m. 7. The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. (Assume π = 22/7) Solution: Inner radius of circular well, r = 3.5/2 m = 1.75 m Depth of circular well, say h = 10 m (i) Inner curved surface area = 2πrh = (2×(22/7 )×1.75×10) = 110 m2 Therefore, the inner curved surface area of the circular well is 110 m2. (ii) Cost of plastering 1 m2 area = Rs. 40 Cost of plastering 110 m2 area = Rs (110×40) = Rs. 4400 Therefore, the cost of plastering the curved surface of the well is Rs. 4400. 8. In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. (Assume π = 22/7) Solution: Height of cylindrical pipe = Length of cylindrical pipe = 28m Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of the cylinder = 2×(22/7)×0.025×28 m2 = 4.4m2 The area of the radiating surface of the system is 4.4m2. 9. Find (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high. (ii) How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank. (Assume π = 22/7) Solution: Height of cylindrical tank, h = 4.5m Radius of the circular end , r = (4.2/2)m = 2.1m (i) the lateral or curved surface area of the cylindrical tank is 2πrh = 2×(22/7)×2.1×4.5 m2 = (44×0.3×4.5) m2 = 59.4 m2 Therefore, the CSA of the tank is 59.4 m2. (ii) Total surface area of tank = 2πr(r+h) = 2×(22/7)×2.1×(2.1+4.5) = 44×0.3×6.6 = 87.12 m2 Now, let us assume that S m2 steel sheet was used in making the tank. S(1 -1/12) = 87.12 m2 This implies, S = 95.04 m2 Therefore, 95.04m2 steel was actually used while making such a tank. 10. In fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. (Assume π = 22/7)  Solution: Say h = height of the frame of the lampshade, which has a cylindrical shape r = radius Total height is h = (2.5+30+2.5) cm = 35cm and r = (20/2) cm = 10cm Use the curved surface area formula to find the cloth required for covering the lampshade, which is 2πrh = (2×(22/7)×10×35) cm2 = 2200 cm2 Hence, 2200 cm2 cloth is required to cover the lampshade. 11. The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Assume π =22/7) Solution: The radius of the circular end of the cylindrical penholder, r = 3cm Height of penholder, h = 10.5cm Surface area of a penholder = CSA of pen holder + Area of base of penholder = 2πrh+πr2 = 2×(22/7)×3×10.5+(22/7)×32= 1584/7 Therefore, the area of cardboard sheet used by one competitor is 1584/7 cm2 So, the area of cardboard sheets used by 35 competitors = 35×1584/7 = 7920 cm2 Therefore, 7920 cm2 cardboard sheet will be needed for the competition. NCERT Solutions for Class 9 Maths Chapter 13helps to find the surface area of various geometrical objects in a simplified and easy way. NCERT Solutions for Class 9 Maths Exercise 13.2 explains how to find the surface area of the cylinder and helps to solve the problems based on it. Key Features of NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2 These NCERT Solutions help you understand and solve all questions of Chapter 13.It helps to secure the best score in Maths board exams.It contains all the important questions from the examination point of view.Good explanations help students to remember the concept effectively.    NCERT Solutions For Class 9 Maths Chapter 13 Surface Area And Volume Exercises NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.1NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.3NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.4NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.5NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.6NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.7NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.8NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.9 NCERT Solutions For Class 9 Maths Chapters Chapter 1 Number SystemsChapter 2 PolynomialChapter 3 Coordinate GeometryChapter 4 Linear Equation In Two VariablesChapter 5 Introduction To Euclids GeometryChapter 6 Lines And AnglesChapter 7 Geometry Of TrianglesChapter 8 QuadrilateralsChapter 9 Areas Of Parallelograms And TriangleChapter 10 CirclesChapter 11 ConstructionChapter 12 Herons FormulaChapter 13 Surface Area And VolumeChapter 14 StatisticsChapter 15 Introduction To Probability Join BYJU'S Learning Program
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Dimensions of the matchbox (a cuboid) are l, b, h = 4 cm, 2.5 cm, 1.5 cm, respectively
Formula to find the volume of the matchbox = l×b×h = (4×2.5×1.5) = 15
Volume of matchbox = 15 cm3
Now, volume of 12 such matchboxes = (15×12) cm3 = 180 cm3
Therefore, the volume of the packet containing 12 matchboxes is 180 cm3.
2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m3 = 1000 l)
Solution:
Dimensions of the cuboidal water tank are: l = 6 m and b = 5 m and h = 4.5 m
Formula to find the volume of the tank, V = l×b×h
Put the values, we get
V = (6×5×4.5) = 135
The volume of the water tank is 135 m3
Again,
The amount of water that 1 m3 volume can hold = 1000 l
The amount of water that 135 m3 volume can hold = (135×1000) litres = 135000 litres
Therefore, the given cuboidal water tank can hold up to 135000 litres of water.
3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?
Solution:
Given:
Length of the cuboidal vessel, l = 10 m
Width of the cuboidal vessel, b = 8m
Volume of the cuboidal vessel, V = 380 m3
Let the height of the given vessel be h.
Formula to find the volume of a cuboid, V = l×b×h
Using the formula, we get
l×b×h = 380
10×8×h = 380
Or h = 4.75
Therefore, the height of the vessels must be 4.75 m.
4. Find the cost of digging a cuboidal pit of 8 m long, 6 m broad and 3 m deep at the rate of Rs. 30 per m3.
Solution:
The given pit has its length (l) as 8m, width (b) as 6m and depth (h) as 3 m.
Volume of the cuboidal pit = l×b×h = (8×6×3) = 144
The required Volume is 144 m3
Now,
The cost of digging per m3 volume = Rs. 30
Therefore, the cost of digging 144 m3 volume = Rs. (144×30) = Rs. 4320.
5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Solution:
The length (l) and depth (h) of the tank is 2.5 m and 10 m, respectively.
To find the value of breadth, say b,
The formula to find the volume of the tank = l×b×h = (2.5× b×10) m3 = 25b m3
The capacity of tank = 25b m3, which is equal to 25000b litres
Also, the capacity of a cuboidal tank is 50000 litres of water (Given)
Therefore, 25000 b = 50000
This implies that b = 2
Therefore, the breadth of the tank is 2 m.
6. A village, having a population of 4000, requires 150 litres of water per head per day.
It has a tank measuring 20 m×15 m×6 m. For how many days will the water in this tank last?
Solution:
Length of the tank = l = 20 m
Breadth of the tank = b = 15 m
Height of the tank = h = 6 m
Total population of the village = 4000
Consumption of water per head per day = 150 litres
Water consumed by the people in 1 day = (4000×150) litres = 600000 litres …(1)
The formula to find the capacity of a tank, C = l×b×h
Using the given data, we have
C = (20×15×6) m3 = 1800 m3
Or C = 1800000 litres
Let the water in this tank last for d days.
Water consumed by all people in d days = Capacity of the tank (using equation (1))
600000 d = 1800000
d = 3
Therefore, the water in this tank will last for 3 days.
7. A godown measures 40 m×25 m×15 m. Find the maximum number of wooden crates, each
measuring 1.5m×1.25 m×0.5 m, that can be stored in the godown.
Solution:
From the statement, we have
Length of the godown = 40 m
Breadth = 25 m
Height = 15 m
Whereas,
Length of the wooden crate = 1.5 m
Breadth = 1.25 m
Height = 0.5 m
Since the godown and wooden crate are in cuboidal shape, we can find the volume of each using the formula, V = lbh.
Now,
Volume of the godown = (40×25×15) m3 = 15000 m3
Volume of the wooden crate = (1.5×1.25×0.5) m3 = 0.9375 m3
Let us consider that n wooden crates can be stored in the godown, then
The volume of n wooden crates = Volume of godown
0.9375×n = 15000
Or n = 15000/0.9375 = 16000
Hence, the number of wooden crates that can be stored in the godown is 16,000.
8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Solution:
Side of the cube = 12 cm (Given)
To find the volume of the cube:
Volume of cube = (Side)3 = (12)3 cm3= 1728 cm3
Surface area of a cube with side 12 cm = 6a2 = 6(12) 2 cm2 …(1)
The cube is cut into eight small cubes of equal volume; say the side of each cube is p.
The volume of the small cube = p3
Surface area = 6p2 …(2)
Volume of each small cube = (1728/8) cm3 = 216 cm3
Or (p)3 = 216 cm3
Or p = 6 cm
Now, the surface areas of the cubes ratios = (Surface area of the bigger cube)/(Surface area of smaller cubes)
From equations (1) and (2), we get
Surface areas of the cubes ratios = (6a2)/(6p2) = a2/p2 = 122/62 = 4
Therefore, the required ratio is 4:1.
9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Solution:
Given:
Depth of the river, h = 3 m
Width of the river, b = 40 m
Rate of water flow = 2 km per hour = 2000 m/60 min = 100/3 m/min
So, the volume of water flowed in 1 min = (100/3) × 40 × 3 = 4000m3
Therefore, 4000 m3 of water will fall into the sea in a minute.
By practising problems in NCERT Solutions for Class 9 Maths Chapter 13, students can learn the easy way to solve them and can score well in the annual examination. All the 9 questions included in this exercise will help students analyse the situation and apply the formula. It includes easy-to-solve questions, which can be expected in examinations. Solve these NCERT solutions, where problems are explained in a detailed way, following each and every step.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes exercise consists of application-level questions that help to determine the volume of the cuboid. It also helps in finding the length and breadth of the cuboid by applying the formula.
Key Features of NCERT Solutions for Class 9 Maths Chapter 13 - Surface Areas and Volume Exercise 13.5
Solving the NCERT solutions for Class 9 Maths Chapter 13 helps students in multiple ways, as explained below:
Students can self-assess their learning abilities and preparation levelStudents can analyse the type of questions that appear for the examsImprove their efficiency and speed in solving the problemsRemember the formulas in an easy way and help them apply relevantly
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[13/01, 7:20 PM] ABHIJIT: We know that,
Speed=
time
Distance
Time =5s
Distance =3625 m
Total distance to be travelled by sound =2×3625, as it is the distance taken by sonar wave to transmit and receive =7250 m
So, Speed =
5
7250
=1450m/s
[13/01, 7:21 PM] ABHIJIT: Ultrasound is passed through one end of the metal and a detector is installed at the other end. When the metal is perfect, the ultrasonic waves travel as expected and hence are detected on the other end. For a metal with a defect, the waves are reflected back and hence don't reach the other end. When no detection is received, it can be concluded that the metal has a defect.
[13/01, 7:23 PM] ABHIJIT: The outer ear is called ‘pinna’. It collects the sound from the surroundings. The collected sound passes through the auditory canal. At the end of the auditory canal there is a thin membrane called the ear drum or tympanic membrane. When a compression of the medium reaches the eardrum the pressure on the outside of the membrane increases and forces the eardrum inward. Similarly, the eardrum moves outward when a rarefaction reaches it. In this way the eardrum vibrates. The vibrations are amplified several times by three bones (the hammer, anvil and stirrup) in the middle ear. The middle ear transmits the amplified pressure variations received from the sound wave to the inner ear. In the inner ear, the pressure variations are turned into electrical signals by the cochlea. These electrical signals are sent to the brain via the auditory nerve, and the brain interprets them as sound.
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1:43
Height of cylinder, h = 14 cm
Let the diameter of the cylinder be d
Curved surface area of cylinder = 88 cm2
We know that the formula to find the curved surface area of a cylinder is 2πrh.
So 2πrh =88 cm2 (r is the radius of the base of the cylinder)
2×(22/7)×r×14 = 88 cm2
2r = 2 cm
d =2 cm
Therefore, the diameter of the base of the cylinder is 2 cm.
2. It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? Assume π = 22/7
Solution:
Let h be the height and r be the radius of a cylindrical tank.
Height of cylindrical tank, h = 1m
Radius = half of diameter = (140/2) cm = 70cm = 0.7m
Area of sheet required = Total surface area of tank = 2πr(r+h) unit square
= [2×(22/7)×0.7(0.7+1)]
= 7.48 square meters
Therefore, 7.48 square meters of the sheet are required.
3. A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4cm. (see fig. 13.11). Find its

(i) inner curved surface area,
(ii) outer curved surface area
(iii) total surface area
(Assume π=22/7)
Solution:
Let r1 and r2 be the inner and outer radii of the cylindrical pipe
r1 = 4/2 cm = 2 cm
r2 = 4.4/2 cm = 2.2 cm
Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm
(i) curved surface area of outer surface of pipe = 2πr1h
= 2×(22/7)×2×77 cm2
= 968 cm2
(ii) curved surface area of outer surface of pipe = 2πr2h
= 2×(22/7)×2.2×77 cm2
= (22×22×2.2) cm2
= 1064.8 cm2
(iii) Total surface area of pipe = inner curved surface area+ outer curved surface area+ Area of both circular ends of pipe.
= 2r1h+2r2h+2π(r12-r22)
= 9668+1064.8+2×(22/7)×(2.22-22)
= 2031.8+5.28
= 2038.08 cm2
Therefore, the total surface area of the cylindrical pipe is 2038.08 cm2.
4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to
move once over to level a playground. Find the area of the playground in m2? (Assume π = 22/7)
Solution:
A roller is shaped like a cylinder.
Let h be the height of the roller and r be the radius.
h = Length of roller = 120 cm
Radius of the circular end of roller = r = (84/2) cm = 42 cm
Now, CSA of roller = 2πrh
= 2×(22/7)×42×120
= 31680 cm2
Area of field = 500×CSA of roller
= (500×31680) cm2
= 15840000 cm2
= 1584 m2.
Therefore, the area of the playground is 1584 m2.
5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2.
(Assume π = 22/7)
Solution:
Let h be the height of a cylindrical pillar and r be the radius.
Given:
Height cylindrical pillar = h = 3.5 m
Radius of the circular end of pillar = r = diameter/2 = 50/2 = 25 cm = 0.25 m
CSA of pillar = 2πrh
= 2×(22/7)×0.25×3.5
= 5.5 m2
Cost of painting 1 m2 area = Rs. 12.50
Cost of painting 5.5 m2 area = Rs (5.5×12.50)
= Rs. 68.75
Therefore, the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2 is Rs 68.75.
6. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the base of the cylinder is 0.7 m, find its height. (Assume π = 22/7)
Solution:
Let h be the height of the circular cylinder and r be the radius.
The radius of the base of the cylinder, r = 0.7 m
CSA of cylinder = 2πrh
CSA of cylinder = 4.4 m2
Equating both equations, we have
2×(22/7)×0.7×h = 4.4
Or h = 1
Therefore, the height of the cylinder is 1 m.
7. The inner diameter of a circular well is 3.5m. It is 10m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2.
(Assume π = 22/7)
Solution:
Inner radius of circular well, r = 3.5/2 m = 1.75 m
Depth of circular well, say h = 10 m
(i) Inner curved surface area = 2πrh
= (2×(22/7 )×1.75×10)
= 110 m2
Therefore, the inner curved surface area of the circular well is 110 m2.
(ii) Cost of plastering 1 m2 area = Rs. 40
Cost of plastering 110 m2 area = Rs (110×40)
= Rs. 4400
Therefore, the cost of plastering the curved surface of the well is Rs. 4400.
8. In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find
the total radiating surface in the system. (Assume π = 22/7)
Solution:
Height of cylindrical pipe = Length of cylindrical pipe = 28m
Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m
Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of the cylinder
= 2×(22/7)×0.025×28 m2
= 4.4m2
The area of the radiating surface of the system is 4.4m2.
9. Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in
diameter and 4.5m high.
(ii) How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank. (Assume π = 22/7)
Solution:
Height of cylindrical tank, h = 4.5m
Radius of the circular end , r = (4.2/2)m = 2.1m
(i) the lateral or curved surface area of the cylindrical tank is 2πrh
= 2×(22/7)×2.1×4.5 m2
= (44×0.3×4.5) m2
= 59.4 m2
Therefore, the CSA of the tank is 59.4 m2.
(ii) Total surface area of tank = 2πr(r+h)
= 2×(22/7)×2.1×(2.1+4.5)
= 44×0.3×6.6
= 87.12 m2
Now, let us assume that S m2 steel sheet was used in making the tank.
S(1 -1/12) = 87.12 m2
This implies, S = 95.04 m2
Therefore, 95.04m2 steel was actually used while making such a tank.
10. In fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth.
The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. (Assume π = 22/7)

Solution:
Say h = height of the frame of the lampshade, which has a cylindrical shape
r = radius
Total height is h = (2.5+30+2.5) cm = 35cm and
r = (20/2) cm = 10cm
Use the curved surface area formula to find the cloth required for covering the lampshade, which is 2πrh
= (2×(22/7)×10×35) cm2
= 2200 cm2
Hence, 2200 cm2 cloth is required to cover the lampshade.
11. The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Assume π =22/7)
Solution:
The radius of the circular end of the cylindrical penholder, r = 3cm
Height of penholder, h = 10.5cm
Surface area of a penholder = CSA of pen holder + Area of base of penholder
= 2πrh+πr2
= 2×(22/7)×3×10.5+(22/7)×32= 1584/7
Therefore, the area of cardboard sheet used by one competitor is 1584/7 cm2
So, the area of cardboard sheets used by 35 competitors = 35×1584/7 = 7920 cm2
Therefore, 7920 cm2 cardboard sheet will be needed for the competition.
NCERT Solutions for Class 9 Maths Chapter 13helps to find the surface area of various geometrical objects in a simplified and easy way. NCERT Solutions for Class 9 Maths Exercise 13.2 explains how to find the surface area of the cylinder and helps to solve the problems based on it.
Key Features of NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2
These NCERT Solutions help you understand and solve all questions of Chapter 13.It helps to secure the best score in Maths board exams.It contains all the important questions from the examination point of view.Good explanations help students to remember the concept effectively.



NCERT Solutions For Class 9 Maths Chapter 13 Surface Area And Volume Exercises
NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.1NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.3NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.4NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.5NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.6NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.7NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.8NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.9
NCERT Solutions For Class 9 Maths Chapters
Chapter 1 Number SystemsChapter 2 PolynomialChapter 3 Coordinate GeometryChapter 4 Linear Equation In Two VariablesChapter 5 Introduction To Euclids GeometryChapter 6 Lines And AnglesChapter 7 Geometry Of TrianglesChapter 8 QuadrilateralsChapter 9 Areas Of Parallelograms And TriangleChapter 10 CirclesChapter 11 ConstructionChapter 12 Herons FormulaChapter 13 Surface Area And VolumeChapter 14 StatisticsChapter 15 Introduction To Probability
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