NCERT Solutions for Class 9 Maths Chapter 13:
NCERT Solutions for Class 9 Maths Chapter 13, Surface Areas and Volumes provide detailed guidance for understanding and solving problems related to geometric shapes and their dimensions.
This chapter focuses on calculating the surface areas and volumes of various solids such as cubes, cuboids, cylinders, cones, and spheres. The solutions provided here are designed to help students grasp the formulas and methods for finding surface areas and volumes effectively.
By working through these solutions, students can develop a solid understanding of three-dimensional geometry, enhance their problem-solving skills, and prepare effectively for their exams.
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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Overview
The NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volume are created by experts from Physics Wallah. These solutions explain how to find the surface areas and volumes of different shapes like cubes, cuboids, cylinders, cones, and spheres.
The clear instructions help students understand how to use formulas and solve problems correctly. By using these solutions students can get a better grasp of 3D shapes and improve their skills, which will help them do well in their exams.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes PDF
The PDF for NCERT Solutions for Class 9 Maths Chapter 13, "Surface Areas and Volumes," is available below. This resource provides detailed solutions and explanations for problems related to calculating the surface areas and volumes of various geometric shapes.
By referring to this PDF, students can find step-by-step guidance on solving exercises, which will help them better understand the concepts and improve their problem-solving skills.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes PDF
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes
Below we have provided NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes:
NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.1 Page No: 213
1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine
(I) The area of the sheet required for making the box.
(ii) The cost of the sheet for it, if a sheet measuring 1m
2
costs Rs. 20.
Solution:
Given: length (l) of box = 1.5m
Breadth (b) of box = 1.25 m
Depth (h) of box = 0.65m
(i) Box is to be open at the top.
Area of sheet required.
= 2lh+2bh+lb
= [2×1.5×0.65+2×1.25×0.65+1.5×1.25]m
2
= (1.95+1.625+1.875) m
2
= 5.45 m
2
(ii) Cost of sheet per m
2
area = Rs.20
Cost of sheet of 5.45 m
2
area = Rs (5.45×20)
= Rs.109.
2. The length, breadth and height of a room are 5 m, 4 m and 3 m, respectively. Find the cost of whitewashing the walls of the room and ceiling at the rate of Rs 7.50 per m
2
.
Solution:
Length (l) of room = 5m
Breadth (b) of room = 4m
Height (h) of room = 3m
It can be observed that four walls and the ceiling of the room are to be whitewashed.
Total area to be whitewashed = Area of walls + Area of the ceiling of the room
= 2lh+2bh+lb
= [2×5×3+2×4×3+5×4]
= (30+24+20)
= 74
Area = 74 m
2
Also,
Cost of whitewash per m
2
area = Rs.7.50 (Given)
Cost of whitewashing 74 m
2
area = Rs. (74×7.50)
= Rs. 555
3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs.10 per m
2
is Rs.15,000, find the height of the hall.
[Hint: Area of the four walls = Lateral surface area.]
Solution:
Let the length, breadth, and height of the rectangular hall be l, b, and h, respectively.
Area of four walls = 2lh+2bh
= 2(l+b)h
Perimeter of the floor of hall = 2(l+b)
= 250 m
Area of four walls = 2(l+b) h = 250h m
2
Cost of painting per square metre area = Rs.10
Cost of painting 250h square metre area = Rs (250h×10) = Rs.2500h
However, it is given that the cost of painting the walls is Rs. 15,000.
15000 = 2500h
Or h = 6
Therefore, the height of the hall is 6 m.
4. The paint in a certain container is sufficient to paint an area equal to 9.375 m
2
. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container?
Solution:
Total surface area of one brick = 2(lb +bh+lb)
= [2(22.5×10+10×7.5+22.5×7.5)] cm
2
= 2(225+75+168.75) cm
2
= (2×468.75) cm
2
= 937.5 cm
2
Let n bricks can be painted out by the paint of the container.
Area of n bricks = (n×937.5) cm
2
= 937.5n cm
2
As per the given instructions, the area that can be painted by the paint of the container = 9.375 m
2
= 93750 cm
2
So, we have 93750 = 937.5n
n = 100
Therefore, 100 bricks can be painted out by the paint of the container.
5. A cubical box has each edge 10 cm, and another cuboidal box is 12.5cm long, 10 cm wide, and 8 cm high.
(i) Which box has the greater lateral surface area, and by how much?
(ii) Which box has the smaller total surface area, and by how much?
Solution:
From the question statement, we have
Edge of a cube = 10cm
Length, l = 12.5 cm
Breadth, b = 10cm
Height, h = 8 cm
(i) Find the lateral surface area for both figures.
Lateral surface area of cubical box = 4 (edge)
2
= 4(10)
2
= 400 cm
2
…(1)
Lateral surface area of cuboidal box = 2[lh+bh]
= [2(12.5×8+10×8)]
= (2×180) = 360
Therefore, the lateral surface area of the cuboidal box is 360 cm
2
. …(2)
From (1) and (2), the lateral surface area of the cubical box is more than the lateral surface area of the cuboidal box. The difference between both lateral surfaces is 40 cm
2
.
(Lateral surface area of the cubical box – Lateral surface area of cuboidal box=400cm
2
–360cm
2
= 40 cm
2
)
(ii) Find the total surface area for both figures.
The total surface area of the cubical box = 6(edge)
2
= 6(10 cm)
2
= 600 cm
2
…(3)
The total surface area of the cuboidal box
= 2[lh+bh+lb]
= [2(12.5×8+10×8+12.5×100)]
= 610
This implies that the total surface area of the cuboidal box is 610 cm
2
..(4)
From (3) and (4), the total surface area of the cubical box is smaller than that of the cuboidal box. And their difference is 10cm
2
.
Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm
2
6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including the base) held together with tape. It is 30cm long, 25 cm wide, and 25 cm high.
(i) What is the area of the glass?
(ii)How much tape is needed for all 12 edges?
Solution:
Length of the greenhouse, say l = 30cm
Breadth of the greenhouse, say b = 25 cm
Height of greenhouse, say h = 25 cm
(i) Total surface area of greenhouse = Area of the glass = 2[lb+lh+bh]
= [2(30×25+30×25+25×25)]
= [2(750+750+625)]
= (2×2125) = 4250
The total surface area of the glass is 4250 cm
2
(ii)
From the figure, the tape is required along sides AB, BC, CD, DA, EF, FG, GH, HE AH, BE, DG, and CF.
Total length of tape = 4(l+b+h)
= [4(30+25+25)] (after substituting the values)
= 320
Therefore, 320 cm tape is required for all 12 edges.
7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm×20cm×5cm, and the smaller of dimension 15cm×12cm×5cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000 cm
2
, find the cost of cardboard required for supplying 250 boxes of each kind.
Solution:
Let l, b and h be the length, breadth and height of the box.
Bigger Box
l = 25cm
b = 20 cm
h = 5 cm
Total surface area of bigger box = 2(lb+lh+bh)
= [2(25×20+25×5+20×5)]
= [2(500+125+100)]
= 1450 cm
2
Extra area required for overlapping 1450×5/100 cm
2
= 72.5 cm
2
While considering all overlaps, the total surface area of the bigger box.
= (1450+72.5) cm
2
= 1522.5 cm
2
Area of cardboard sheet required for 250 such bigger boxes
= (1522.5×250) cm
2
= 380625 cm
2
Smaller Box
Similarly, total surface area of the smaller box = [2(15×12+15×5+12×5)] cm
2
= [2(180+75+60)] cm
2
= (2×315) cm
2
= 630 cm
2
Therefore, the extra area required for overlapping 630×5/100 cm
2
= 31.5 cm
2
The total surface area of 1 smaller box while considering all overlaps
= (630+31.5) cm
2
= 661.5 cm
2
Area of cardboard sheet required for 250 smaller boxes = (250×661.5) cm
2
= 165375 cm
2
In Short
|
Box
|
Dimensions (in cm)
|
Total surface area (in cm
2
)
|
Extra area required for overlapping (in cm
2
)
|
Total surface area for all overlaps (in cm
2
)
|
Area for 250 such boxes (in cm
2
)
|
|
Bigger Box
|
l = 25
b = 20
c = 5
|
1450
|
1450×5/100
= 72.5
|
(1450+72.5) = 1522.5
|
(1522.5×250) = 380625
|
|
Smaller Box
|
l = 15
b = 12
h =5
|
630
|
630×5/100 = 31.5
|
(630+31.5) = 661.5
|
( 250×661.5) = 165375
|
Now, total cardboard sheet required = (380625+165375) cm
2
= 546000 cm
2
Given, cost of 1000 cm
2
cardboard sheet = Rs. 4
Therefore, the cost of 546000 cm
2
cardboard sheet =Rs. (546000×4)/1000 = Rs. 2184
Therefore, the cost of cardboard required for supplying 250 boxes of each kind will be Rs. 2,184.
8. Praveen wanted to make a temporary shelter for her car by making a box-like structure with a tarpaulin that covers all four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small and therefore negligible, how many tarpaulins would be required to make the shelter of height 2.5m, with base dimensions 4m×3m?
Solution:
Let l, b and h be the length, breadth and height of the shelter.
Given:
l = 4m
b = 3m
h = 2.5m
Tarpaulins will be required for the top and four wall sides of the shelter.
Using formula, area of tarpaulin required = 2(lh+bh)+lb
On putting the values of l, b and h, we get
= [2(4×2.5+3×2.5)+4×3] m
2
= [2(10+7.5)+12]m
2
= 47m
2
Therefore, 47 m
2
of tarpaulin will be required.
NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.2 Page No: 216
1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm
2
. Find the diameter of the base of the cylinder (Assume π =22/7 ).
Solution:
Height of cylinder, h = 14cm
Let the diameter of the cylinder be d.
The curved surface area of cylinder = 88 cm
2
We know that the formula to find the Curved surface area of the cylinder is 2πrh.
So 2πrh =88 cm
2
(r is the radius of the base of the cylinder)
2×(22/7)×r×14 = 88 cm
2
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 metres 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 metres
Therefore, 7.48 square metres 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 r
1
and r
2
inner and outer radii of the cylindrical pipe.
r
1
= 4/2 cm = 2 cm
r
2
= 4.4/2 cm = 2.2 cm
Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm
(i) Curved surface area of the outer surface of pipe = 2πr
1
h
= 2×(22/7)×2×77 cm
2
= 968 cm
2
(ii) Curved surface area of the outer surface of pipe = 2πr
2
h
= 2×(22/7)×2.2×77 cm
2
= (22×22×2.2) cm
2
= 1064.8 cm
2
(iii) Total surface area of pipe = inner curved surface area+ outer curved surface area+ area of both circular ends of pipe
= 2r
1
h+2r
2
h+2π(r
1
2
-r
2
2
)
= 9668+1064.8+2×(22/7)×(2.2
2
-2
2
)
= 2031.8+5.28
= 2038.08 cm
2
Therefore, the total surface area of the cylindrical pipe is 2038.08 cm
2
.
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 m
2
(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 cm
2
Area of field = 500×CSA of roller
= (500×31680) cm
2
= 15840000 cm
2
= 1584 m
2
.
Therefore, the area of the playground is 1584 m
2
.
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 m
2
.
(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 = 25cm = 0.25m
CSA of pillar = 2πrh
= 2×(22/7)×0.25×3.5
= 5.5 m
2
Cost of painting 1 m
2
area = Rs. 12.50
Cost of painting 5.5 m
2
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 m
2
is Rs 68.75.
6. Curved surface area of a right circular cylinder is 4.4 m
2
. If the radius 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 = 0.7m
CSA of cylinder = 2πrh
CSA of cylinder = 4.4m
2
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 m
2
.
(Assume π = 22/7)
Solution:
Inner radius of circular well, r = 3.5/2m = 1.75m
Depth of circular well, say h = 10m
(i) Inner curved surface area = 2πrh
= (2×(22/7 )×1.75×10)
= 110 m
2
Therefore, the inner curved surface area of the circular well is 110 m
2
.
(ii) Cost of plastering 1 m
2
area = Rs.40
Cost of plastering 110 m
2
area = Rs (110×40)
= Rs.4400
Therefore, the cost of plastering the curved surface of the well is Rs. 4,400.
8. In a hot water heating system, there is a 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 m
2
= 4.4m
2
The area of the radiating surface of the system is 4.4m
2
.
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 m
2
= (44×0.3×4.5) m
2
= 59.4 m
2
Therefore, the CSA of the tank is 59.4 m
2
.
(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 m
2
Now, Let S m
2
steel sheet be actually used in making the tank.
S(1 -1/12) = 87.12 m
2
This implies, S = 95.04 m
2
Therefore, 95.04m
2
steel was used in actuality 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 a 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 to cover the lampshade. (Assume π = 22/7)
Solution:
Say h = height of the frame of the lampshade, which looks like 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) cm
2
= 2200 cm
2
Hence, 2200 cm
2
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+πr
2
= 2×(22/7)×3×10.5+(22/7)×3
2
= 1584/7
Therefore, the Area of cardboard sheet used by one competitor is 1584/7 cm
2
So, the Area of cardboard sheet used by 35 competitors = 35×1584/7 = 7920 cm
2
Therefore, a 7920 cm
2
cardboard sheet will be needed for the competition.
NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.3 Page No: 221
1. Diameter of the base of a cone is 10.5 cm, and its slant height is 10 cm. Find its curved surface area. (Assume π=22/7)
Solution:
Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm
The slant height of the cone, say l = 10 cm
CSA of the cone is = πrl
= (22/7)×5.25×10 = 165 cm
2
Therefore, the curved surface area of the cone is 165 cm
2
.
2. Find the total surface area of a cone, if its slant height is 21 m and the diameter of its base is 24 m. (Assume π = 22/7)
Solution:
Radius of cone, r = 24/2 m = 12m
Slant height, l = 21 m
Formula: Total Surface area of the cone = πr(l+r)
Total Surface area of the cone = (22/7)×12×(21+12) m
2
= 1244.57m
2
3.
Curved surface area of a cone is 308 cm
2,
and its slant height is 14 cm. Find
(i) radius of the base and (ii) total surface area of the cone.
(Assume π = 22/7)
Solution:
The slant height of the cone, l = 14 cm
Let the radius of the cone be r.
(i) We know the CSA of cone = πrl
Given: Curved surface area of a cone is 308 cm
2
(308 ) = (22/7)×r×14
308 = 44 r
r = 308/44 = 7 cm
The radius of a cone base is 7 cm.
(ii) Total surface area of cone = CSA of cone + Area of base (πr
2
)
Total surface area of cone = 308+(22/7)×7
2
= 308+154 = 462 cm
2
Therefore, the total surface area of the cone is 462 cm
2
.
4. A conical tent is 10 m high, and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m
2
canvas is Rs 70.
(Assume π=22/7)
Solution:
Let ABC be a conical tent.
Height of conical tent, h = 10 m
Radius of conical tent, r = 24m
Let the slant height of the tent be l.
(i) In the right triangle ABO, we have
AB
2
= AO
2
+BO
2
(using Pythagoras’ theorem)
l
2
= h
2
+r
2
= (10)
2
+(24)
2
= 676
l = 26 m
Therefore, the slant height of the tent is 26 m.
(ii) CSA of tent = πrl
= (22/7)×24×26 m
2
Cost of 1 m
2
canvas = Rs 70
Cost of (13728/7)m
2
canvas is equal to Rs (13728/7)×70 = Rs 137280
Therefore, the cost of the canvas required to make such a tent is Rs 137280.
5. What length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. [Use π=3.14]
Solution:
Height of the conical tent, h = 8m
Radius of the base of the tent, r = 6m
Slant height of the tent, l
2
= (r
2
+h
2
)
l
2
= (6
2
+8
2
) = (36+64) = (100)
or l = 10 m
Again, CSA of conical tent = πrl
= (3.14×6×10) m
2
= 188.4m
2
Let the length of the tarpaulin sheet required be L.
As 20 cm will be wasted,
The effective length will be (L-0.2m).
The breadth of tarpaulin = 3m (given)
Area of sheet = CSA of the tent
[(L–0.2)×3] = 188.4
L-0.2 = 62.8
L = 63 m
Therefore, the length of the required tarpaulin sheet will be 63 m.
6. The slant height and base diameter of the conical tomb are 25m and 14 m, respectively. Find the cost of whitewashing its curved surface at the rate of Rs. 210 per 100 m
2
. (Assume π = 22/7)
Solution:
Slant height of the conical tomb, l = 25m
Base radius, r = diameter/2 = 14/2 m = 7m
CSA of the conical tomb = πrl
= (22/7)×7×25 = 550
CSA of the conical tomb= 550m
2
Cost of whitewashing 550 m
2
area, which is Rs (210×550)/100
= Rs. 1155
Therefore, the cost will be Rs. 1155 while whitewashing the tomb.
7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. (Assume π =22/7)
Solution:
Radius of the conical cap, r = 7 cm
Height of the conical cap, h = 24cm
Slant height, l
2
= (r
2
+h
2
)
= (7
2
+24
2
)
= (49+576)
= (625)
Or l = 25 cm
CSA of 1 conical cap = πrl
= (22/7)×7×25
= 550 cm
2
CSA of 10 caps = (10×550) cm
2
= 5500 cm
2
Therefore, the area of the sheet required to make 10 such caps is 5500 cm
2
.
8. A bus stop is barricaded from the remaining part of the road by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m
2
, what will be the cost of painting all these cones? (Use π = 3.14 and take √(1.04) =1.02)
Solution
:
Given:
Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m
Height of cone, h = 1m
Slant height of cone is l, and l
2
= (r
2
+h
2
)
Using given values, l
2
= (0.2
2
+1
2
)
= (1.04)
Or l = 1.02 m
Slant height of the cone is 1.02 m.
Now,
CSA of each cone = πrl
= (3.14×0.2×1.02)
= 0.64056 m
CSA of 50 such cones = (50×0.64056) = 32.028
CSA of 50 such cones = 32.028 m
2
Again,
Cost of painting 1 m
2
area = Rs 12 (given)
Cost of painting 32.028 m
2
area = Rs (32.028×12)
= Rs.384.336
= Rs.384.34 (approximately)
Therefore, the cost of painting all these cones is Rs. 384.34.
NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4 Page No: 225
1. Find the surface area of a sphere of radius
(i) 10.5cm (ii) 5.6cm (iii) 14cm
(Assume π=22/7)
Solution
:
Formula: Surface area of a sphere (SA) = 4πr
2
(i) Radius of a sphere, r = 10.5 cm
SA = 4×(22/7)×10.5
2
= 1386
Surface area of a sphere is 1386 cm
2
(ii) Radius of a sphere, r = 5.6cm
Using formula, SA = 4×(22/ 7)×5.6
2
= 394.24
Surface area of a sphere is 394.24 cm
2
(iii) Radius of a sphere, r = 14cm
SA = 4πr
2
= 4×(22/7)×(14)
2
= 2464
Surface area of a sphere is 2464 cm
2
2. Find the surface area of a sphere of diameter
(i) 14cm (ii) 21cm (iii) 3.5cm
(Assume π = 22/7)
Solution:
(i) Radius of sphere, r = diameter/2 = 14/2 cm = 7 cm
Formula for the surface area of sphere = 4πr
2
= 4×(22/7)×7
2
= 616
Surface area of a sphere is 616 cm
2
(ii) Radius (r) of sphere = 21/2 = 10.5 cm
Surface area of a sphere = 4πr
2
= 4×(22/7)×10.5
2
= 1386
Surface area of a sphere is 1386 cm
2
Therefore, the surface area of a sphere having a diameter 21 cm is 1386 cm
2
(iii) Radius(r) of a sphere = 3.5/2 = 1.75 cm
Surface area of a sphere = 4πr
2
= 4×(22/7)×1.75
2
= 38.5
Surface area of a sphere is 38.5 cm
2
3. Find the total surface area of a hemisphere of radius 10 cm. [Use π=3.14]
Solution:
Radius of the hemisphere, r = 10cm
Formula: Total surface area of the hemisphere = 3πr
2
= 3×3.14×10
2
= 942
The total surface area of the given hemisphere is 942 cm
2
.
4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
Let r
1
and r
2
be the radii of the spherical balloon and spherical balloon when air is pumped into it, respectively. So,
r
1
= 7cm
r
2
= 14 cm
Now, Required ratio = (initial surface area)/(Surface area after pumping air into balloon)
= 4πr
1
2
/4πr
2
2
= (r
1
/r
2
)
2
= (7/14)
2
= (1/2)
2
= ¼
Therefore, the ratio between the surface areas is 1:4.
5. A hemispherical bowl made of brass has an inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm
2
. (Assume π = 22/7)
Solution:
Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm
Formula for the surface area of hemispherical bowl = 2πr
2
= 2×(22/7)×(5.25)
2
= 173.25
Surface area of the hemispherical bowl is 173.25 cm
2
Cost of tin-plating 100 cm
2
area = Rs 16
Cost of tin-plating 1 cm
2
area = Rs 16 /100
Cost of tin-plating 173.25 cm
2
area = Rs. (16×173.25)/100 = Rs 27.72
Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm
2
is Rs
27.72.
6. Find the radius of a sphere whose surface area is 154 cm
2
. (Assume π = 22/7)
Solution:
Let the radius of the sphere be r.
Surface area of sphere = 154 (given)
Now,
4πr
2
= 154
r
2
= (154×7)/(4×22) = (49/4)
r = (7/2) = 3.5
The radius of the sphere is 3.5 cm.
7. The diameter of the moon is approximately one-fourth of the diameter of the earth.
Find the ratio of their surface areas.
Solution:
If the diameter of the earth is said d, then the diameter of the moon will be d/4 (as per the given statement).
Radius of earth = d/2
Radius of moon = ½×d/4 = d/8
Surface area of moon = 4π(d/8)
2
Surface area of earth = 4π(d/2)
2
The ratio between their surface areas is 1:16.
8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. (Assume π =22/7)
Solution:
Given:
Inner radius of the hemispherical bowl = 5cm
Thickness of the bowl = 0.25 cm
Outer radius of the hemispherical bowl = (5+0.25) cm = 5.25 cm
Formula for outer CSA of the hemispherical bowl = 2πr
2
, where r is the radius of the hemisphere.
= 2×(22/7)×(5.25)
2
= 173.25 cm
2
Therefore, the outer curved surface area of the bowl is 173.25 cm
2
.
9. A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in(i) and (ii).
Solution:
(i) Surface area of the sphere = 4πr
2
, where r is the radius of sphere
(ii) Height of the cylinder, h = r+r =2r
The radius of the cylinder = r
CSA of the cylinder formula = 2πrh = 2πr(2r) (using value of h)
= 4πr
2
(iii) Ratio between areas = (Surface area of sphere)/(CSA of Cylinder)
= 4πr
2
/4πr
2
= 1/1
The ratio of the areas obtained in (i) and (ii) is 1:1.
NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.5 Page No: 228
1. A matchbox measures 4 cm×2.5cm×1.5cm. What will be the volume of a packet containing 12 such boxes?
Solution:
Dimensions of a matchbox (a cuboid) are l×b×h = 4 cm×2.5 cm×1.5 cm, respectively
Formula to find the volume of matchbox = l×b×h = (4×2.5×1.5) = 15
Volume of matchbox = 15 cm
3
Now, volume of 12 such matchboxes = (15×12) cm
3
= 180 cm
3
Therefore, the volume of 12 matchboxes is 180 cm
3
.
2. A cuboidal water tank is 6m long, 5m wide and 4.5m deep. How many litres of water can it hold? (1 m
3
= 1000 l)
Solution:
Dimensions of a 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
Putting the values, we get
V = (6×5×4.5) = 135
The volume of the water tank is 135 m
3
Again,
We are given that the amount of water that 1m
3
volume can hold = 1000 l
Amount of water, 135 m
3
volume hold = (135×1000) litres = 135000 litres
Therefore, given cuboidal water tank can hold up to 135000 litres of water.
3. A cuboidal vessel is 10m long and 8m 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 m
3
Let the height of the given vessel be h.
Formula for volume of a cuboid, V = l×b×h
Using the formula, we have
l×b×h = 380
10×8×h= 380
Or h = 4.75
Therefore, the height of the vessels is 4.75 m.
4. Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m
3
.
Solution:
The given pit has its length(l) as 8m, width (b)as 6m and depth (h)as 3 m.
Volume of cuboidal pit = l×b×h = (8×6×3) = 144 (using formula)
The required Volume is 144 m
3
Now,
Cost of digging per m
3
volume = Rs 30
Cost of digging 144 m
3
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 are 2.5 m and 10 m, respectively.
To find: The value of breadth, say b.
Formula to find the volume of a tank = l×b×h = (2.5× b×10) m
3
= 25b m
3
The capacity of tank= 25b m
3
, 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 a 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)
Formula to find the capacity of the tank, C = l×b×h
Using the given data, we have
C = (20×15×6) m
3
= 1800 m
3
Or C = 1800000 litres
Let 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×25m×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
The godown and wooden crate are in cuboidal shape. Find the volume of each using the formula V = lbh
Now,
Volume of godown = (40×25×15) m
3
= 15000 m
3
Volume of a wooden crate = (1.5×1.25×0.5) m
3
= 0.9375 m
3
Let us consider that, n wooden crates can be stored in the godown, then
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 a cube = 12cm (Given)
Find the volume of the cube.
Volume of cube = (Side)
3
= (12)
3
cm
3
= 1728cm
3
Surface area of a cube with side 12 cm = 6a
2
= 6(12)
2
cm
2
…(1)
The cube is cut into eight small cubes of equal volume; say, the side of each cube is p.
Volume of a small cube = p
3
Surface area = 6p
2
…(2)
Volume of each small cube = (1728/8) cm
3
= 216 cm
3
Or (p)
3
= 216 cm
3
Or p = 6 cm
Now, Surface areas of the cubes ratios = (Surface area of the bigger cube)/(Surface area of the smaller cubes)
From equations (1) and (2), we get
Surface areas of the cubes ratios = (6a
2
)/(6p
2
) = a
2
/p
2
= 12
2
/6
2
= 4
Therefore, the required ratio is 4 : 1.
9. A river 3m deep and 40m wide is flowing at the rate of 2km per hour. How much water will fall into the sea in a minute?
Solution:
Given:
Depth of river, h = 3 m
Width of river, b = 40 m
Rate of water flow = 2km per hour = 2000m/60min = 100/3 m/min
Now, volume of water flowed in 1 min = (100/3) × 40 × 3 = 4000m
3
Therefore, 4000 m
3
water will fall into the sea in a minute.
NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.6 Page No: 230
1. The circumference of the base of the cylindrical vessel is 132cm, and its height is 25cm.
How many litres of water can it hold? (1000 cm
3
= 1L) (Assume π = 22/7)
Solution
:
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 the vessel.
We know that the circumference of the base = 2πr, so
2πr = 132 (given)
r = (132/(2 π))
r = 66×7/22 = 21
The radius is 21 cm.
Step 2: Find the volume of the vessel.
Formula: Volume of cylindrical vessel = πr
2
h
= (22/7)×21
2
×25
= 34650
Therefore, the volume is 34650 cm
3
Since
1000 cm
3
= 1L
So, Volume = 34650/1000 L= 34.65L
Therefore, the 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 1cm
3
of wood has a mass of 0.6g. (Assume π = 22/7)
Solution:
Inner radius of cylindrical pipe, say r
1
= diameter
1
/ 2 = 24/2 cm = 12cm
Outer radius of cylindrical pipe, say r
2
= diameter
2
/ 2 = 28/2 cm = 14 cm
Height of pipe, h = Length of pipe = 35cm
Now, the Volume of pipe = π(r
2
2
-r
1
2
)h cm
3
Substitute the values.
Volume of pipe = 110×52 cm
3
= 5720 cm
3
Since
Mass of 1 cm
3
wood = 0.6 g
Mass of 5720 cm
3
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 a 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.
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) cm
3
= 300 cm
3
(ii) Plastic cylinder will be cylindrical in shape.
Dimensions of the plastic can are
Radius of the circular end of plastic cylinder, r = 3.5cm
Height , H = 10 cm
Capacity of the plastic cylinder = πr
2
H
Capacity of the plastic cylinder = (22/7)×(3.5)
2
×10 = 385
Capacity of the plastic cylinder is 385 cm
3
From the results of (i) and (ii), the plastic cylinder has more capacity.
Difference in capacity = (385-300) cm
3
= 85cm
3
4. If the lateral surface of a cylinder is 94.2cm
2
and its height is 5cm, then find
(i) radius of its base (ii) its volume.[Use π= 3.14]
Solution:
CSA of cylinder = 94.2 cm
2
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
The radius is 3 cm.
(ii) Volume of cylinder
The formula for the volume of the cylinder = πr
2
h
Now, πr
2
h = (3.14×(3)
2
×5) (using the value of r from (i))
= 141.3
Volume is 141.3 cm
3
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 m
2
, 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 m
2
area.
Rs 1 is the cost to paint 1/20 m
2
area.
So, Rs 2200 is the cost of painting = (1/20×2200) m
2
= 110 m
2
area
The inner surface area of the vessel is 110m
2
.
(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 m
2
2×22/7×r×10 = 110
r = 1.75
The radius is 1.75 m.
(iii) Volume of vessel formula = πr
2
h
Here r = 1.75 and h = 10
Volume = (22/7)×(1.75)
2
×10 = 96.25
The volume of vessel is 96.25 m
3
Therefore, the capacity of the vessel is 96.25 m
3
or 96250 litres.
6. The capacity of a closed cylindrical vessel of height 1m is 15.4 litres. How many square metres of the metal sheet would be needed to make it? (Assume π = 22/7)
Solution:
Height of cylindrical vessel, h = 1 m
Capacity of cylindrical vessel = 15.4 litres = 0.0154 m
3
Let r be the radius of the circular end.
Now,
Capacity of cylindrical vessel = (22/7)×r
2
×1 = 0.0154
After simplifying, we get r = 0.07 m
Again, the total surface area of the 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 m
2
Therefore, 0.4708 m
2
of the metal sheet would be required to make the cylindrical vessel.
7. A lead pencil consists of a cylinder of wood with a 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, r
1
= 7/2 mm = 0.7/2 cm = 0.35 cm
Radius of graphite, r
2
= 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 = (r
1
2
-r
2
2
)h cubic units
Substituting values, we have,
= [(22/7)×(0.35
2
-0.05
2
)×14]
= 44×0.12
= 5.28
This implies that the volume of wood in pencil = 5.28 cm
3
Again,
Volume of graphite = r
2
2
h cubic units
Substituting the values, we have,
= (22/7)×0.05
2
×14
= 44×0.0025
= 0.11
So, the volume of graphite is 0.11 cm
3
.
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 the cylindrical bowl, r = 7/2 cm = 3.5 cm
Bowl is filled with soup to a height of4cm, so h = 4 cm
Volume of the soup in one bowl= πr
2
h
(22/7)×3.5
2
×4 = 154
Volume of the soup in one bowl is 154 cm
3
Therefore,
Volume of the soup given to 250 patients = (250×154) cm
3
= 38500 cm
3
= 38.5litres.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.7 Page No: 233
1. Find the volume of the right circular cone with
(i) radius 6cm, height 7 cm (ii) radius 3.5 cm, height 12 cm (Assume π = 22/7)
Solution:
Volume of cone = (1/3) πr
2
h cube units
Where r be radius and h be the height of the cone
(i) Radius of the cone, r = 6 cm
Height of the cone, h = 7cm
Ley V be the volume of the cone, so we have
V = (1/3)×(22/7)×36×7
= (12×22)
= 264
The volume of the cone is 264 cm
3
.
(ii) Radius of the cone, r = 3.5cm
Height of the cone, h = 12cm
Volume of the cone = (1/3)×(22/7)×3.5
2
×7 = 154
Hence,
The volume of the cone is 154 cm
3
.
2. Find the capacity in litres of a conical vessel with
(i) radius 7cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm
(Assume π = 22/7)
Solution:
(i) Radius of the cone, r =7 cm
Slant height of the cone, l = 25 cm
or h = 24
Height of the cone is 24 cm
Now,
Volume of the cone, V = (1/3) πr
2
h (formula)
V = (1/3)×(22/7) ×7
2
×24
= (154×8)
= 1232
So, the volume of the vessel is 1232 cm
3
Therefore, the capacity of the conical vessel = (1232/1000) liters (because 1L = 1000 cm
3
)
= 1.232 Liters.
(ii) Height of the cone, h = 12 cm
Slant height of the cone, l = 13 cm
r = 5
Hence, the radius of the cone is 5 cm.
Now, Volume of the cone, V = (1/3)πr
2
h
V = (1/3)×(22/7)×52×12 cm
3
= 2200/7
Volume of the cone is 2200/7 cm
3
Now, Capacity of the conical vessel= 2200/7000 litres (1L = 1000 cm
3
)
= 11/35 litres
3. The height of a cone is 15cm. If its volume is 1570cm
3
, find the diameter of its base. (Use π = 3.14)
Solution:
Height of the cone, h = 15 cm
Volume of cone =1570 cm
3
Let r be the radius of the cone
As we know, volume of the cone, V = (1/3) πr
2
h
So, (1/3) πr
2
h = 1570
(1/3)×3.14×r
2
×15 = 1570
r
2
= 100
r = 10
Radius of the base of the cone 10 cm.
4.
If the volume of a right circular cone of height 9cm is 48πcm
3
, find the diameter of its base.
Solution:
Height of cone, h = 9cm
Volume of cone =48π cm
3
Let r be the radius of the cone.
As we know, volume of the cone, V = (1/3) πr
2
h
So, 1/3 π r
2
(9) = 48 π
r
2
= 16
r = 4
Radius of the cone is 4 cm.
So, diameter = 2×Radius = 8
Thus, diameter of the base is 8cm.
5. A conical pit of a top diameter 3.5m is 12m deep. What is its capacity in kilolitres?
(Assume π = 22/7)
Solution:
Diameter of conical pit = 3.5 m
Radius of conical pit, r = diameter/ 2 = (3.5/2)m = 1.75m
Height of pit, h = Depth of pit = 12m
Volume of cone, V = (1/3) πr
2
h
V = (1/3)×(22/7) ×(1.75)
2
×12 = 38.5
Volume of the cone is 38.5 m
3
Hence, capacity of the pit = (38.5×1) kiloliters = 38.5 kiloliters.
6. The volume of a right circular cone is 9856cm
3
. If the diameter of the base is 28cm, find
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone
(Assume π = 22/7)
Solution:
Volume of a right circular cone = 9856 cm
3
Diameter of the base = 28 cm
(i) Radius of cone, r = (28/2) cm = 14 cm
Let the height of the cone be h
Volume of cone, V = (1/3) πr
2
h
(1/3) πr
2
h = 9856
(1/3)×(22/7) ×14×14×h = 9856
h = 48
The height of the cone is 48 cm.
Slant height of the cone is 50 cm.
(iii) curved surface area of cone = πrl
= (22/7)×14×50
= 2200
Curved surface area of the cone is 2200 cm
2
.
7. A right triangle ABC with sides 5cm, 12cm and 13cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Solution:
Height (h)= 12 cm
Radius (r) = 5 cm, and
Slant height (l) = 13 cm
Volume of cone, V = (1/3) πr
2
h
V = (1/3)×π×5
2
×12
= 100π
Volume of the cone so formed is 100π cm
3
.
8. If the triangle ABC in Question 7 is revolved about the side 5cm, then find the volume of the solids so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Solution:
A right-angled ΔABC is revolved about its side 5cm, a cone will be formed of radius as 12 cm, height as 5 cm, and slant height as 13 cm.
Volume of cone = (1/3) πr
2
h, where r is the radius and h is the height of the cone.
= (1/3)×π×12×12×5
= 240 π
The volume of the cones formed is 240π cm
3
.
So, the required ratio = (the result of question 7) / (the result of question 8) = (100π)/(240π) = 5/12 = 5:12.
9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas.
(Assume π = 22/7)
Solution:
Radius (r) of heap = (10.5/2) m = 5.25
Height (h) of heap = 3m
Volume of heap = (1/3)πr
2
h
= (1/3)×(22/7)×5.25×5.25×3
= 86.625
The volume of the heap of wheat is 86.625 m
3
.
Again,
= (22/7)×5.25×6.05
= 99.825
Therefore, the area of the canvas is 99.825 m
2
.
NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 Page No: 236
1. Find the volume of a sphere whose radius is
(i) 7 cm (ii) 0.63 m
(Assume π =22/7)
Solution:
(i) Radius of the sphere, r = 7 cm
Using, Volume of the sphere = (4/3) πr
3
= (4/3)×(22/7)×7
3
= 4312/3
Hence, volume of the sphere is 4312/3 cm
3
(ii) Radius of the sphere, r = 0.63 m
Using, volume of sphere = (4/3) πr
3
= (4/3)×(22/7)×0.63
3
= 1.0478
Hence, volume of the sphere is 1.05 m
3
(approx).
2. Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm (ii) 0.21 m
(Assume π =22/7)
Solution:
(i) Diameter = 28 cm
Radius, r = 28/2 cm = 14cm
Volume of the solid spherical ball = (4/3) πr
3
Volume of the ball = (4/3)×(22/7)×14
3
= 34496/3
Hence, volume of the ball is 34496/3 cm
3
(ii) Diameter = 0.21 m
Radius of the ball =0.21/2 m= 0.105 m
Volume of the ball = (4/3 )πr
3
Volume of the ball = (4/3)× (22/7)×0.105
3
m
3
Hence, volume of the ball = 0.004851 m
3
3. The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm
3
? (Assume π=22/7)
Solution:
Given,
Diameter of a metallic ball = 4.2 cm
Radius(r) of the metallic ball, r = 4.2/2 cm = 2.1 cm
Volume formula = 4/3 πr
3
Volume of the metallic ball = (4/3)×(22/7)×2.1 cm
3
Volume of the metallic ball = 38.808 cm
3
Now, using the relationship between density, mass and volume,
Density = Mass/Volume
Mass = Density × volume
= (8.9×38.808) g
= 345.3912 g
Mass of the ball is 345.39 g (approx).
4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Solution:
Let the diameter of the earth be “d”. Therefore, the radius of the earth will be d/2.
Diameter of the moon will be d/4, and the radius of the moon will be d/8.
Find the volume of the moon.
Volume of the moon = (4/3) πr
3
= (4/3) π (d/8)
3
= 4/3π(d
3
/512)
Find the volume of the earth
Volume of the earth = (4/3) πr
3
= (4/3) π (d/2)
3
= 4/3π(d
3
/8)
Fraction of the volume of the earth is the volume of the moon
Answer: Volume of the moon is of the 1/64 volume of the earth.
5. How many litres of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume π = 22/7)
Solution:
Diameter of the hemispherical bowl = 10.5 cm
Radius of the hemispherical bowl, r = 10.5/2 cm = 5.25 cm
Formula for volume of the hemispherical bowl = (2/3) πr
3
Volume of the hemispherical bowl = (2/3)×(22/7)×5.25
3
= 303.1875
Volume of the hemispherical bowl is 303.1875 cm
3
Capacity of the bowl = (303.1875)/1000 L = 0.303 litres(approx.)
Therefore, the hemispherical bowl can hold 0.303 litres of milk.
6. A hemispherical tank is made up of an iron sheet 1cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. (Assume π = 22/7)
Solution:
Inner Radius of the tank, (r ) = 1m
Outer Radius (R ) = 1.01m
Volume of the iron used in the tank = (2/3) π(R
3
– r
3
)
Put values,
Volume of the iron used in the hemispherical tank = (2/3)×(22/7)×(1.01
3
– 1
3
) = 0.06348
So, volume of the iron used in the hemispherical tank is 0.06348 m
3
.
7. Find the volume of a sphere whose surface area is 154 cm
2
. (Assume π = 22/7)
Solution:
Let r be the radius of a sphere.
Surface area of the sphere = 4πr
2
4πr
2
= 154 cm
2
(given)
r
2
= (154×7)/(4 ×22)
r = 7/2
The radius is 7/2 cm.
Now,
Volume of the sphere = (4/3) πr
3
8. A dome of a building is in the form of a hemisphere. From inside, it was whitewashed at the cost of Rs. 4989.60. If the cost of white-washing is 20 per square metre, find the
(i) inside surface area of the dome (ii) volume of the air inside the dome
(Assume π = 22/7)
Solution:
(i) Cost of whitewashing the dome from inside = Rs 4989.60
Cost of whitewashing 1m
2
area = Rs 20
CSA of the inner side of dome = 498.96/2 m
2
= 249.48 m
2
(ii) Let the inner radius of the hemispherical dome be r.
CSA of the inner side of dome = 249.48 m
2
(from (i))
Formula to find CSA of a hemisphere = 2πr
2
2πr
2
= 249.48
2×(22/7)×r
2
= 249.48
r
2
= (249.48×7)/(2×22)
r
2
= 39.69
r = 6.3
So, the radius is 6.3 m.
Volume of air inside the dome = Volume of hemispherical dome
Using the formula, the volume of the hemisphere = 2/3 πr
3
= (2/3)×(22/7)×6.3×6.3×6.3
= 523.908
= 523.9(approx.)
Answer: The volume of air inside the dome is 523.9 m
3
.
9. Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the
(i) radius r’ of the new sphere,
(ii) ratio of Sand S’.
Solution:
Volume of the solid sphere = (4/3)πr
3
Volume of twenty seven solid sphere = 27×(4/3)πr
3
= 36 π r
3
(i) New solid iron sphere radius = r’
Volume of this new sphere = (4/3)π(r’)
3
(4/3)π(r’)
3
= 36 π r
3
(r’)
3
= 27r
3
r’= 3r
Radius of the new sphere will be 3r (thrice the radius of the original sphere)
(ii) Surface area of the iron sphere of radius r, S =4πr
2
Surface area of the iron sphere of radius r’= 4π (r’)
2
Now
S/S’ = (4πr
2
)/( 4π (r’)
2
)
S/S’ = r
2
/(3r’)
2
= 1/9
The ratio of S and S’ is 1: 9.
10. A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm
3
) is needed to fill this capsule? (Assume π = 22/7)
Solution:
Diameter of the capsule = 3.5 mm
Radius of the capsule, say r = diameter/ 2 = (3.5/2) mm = 1.75mm
Volume of the spherical capsule = 4/3 πr
3
Volume of the spherical capsule = (4/3)×(22/7)×(1.75)
3
= 22.458
Answer: The volume of the spherical capsule is 22.46 mm
3
.
NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.9 Page No: 236
1. A wooden bookshelf has external dimensions as follows: Height = 110cm, Depth = 25cm,
Breadth = 85cm (see fig. 13.31). The thickness of the plank is 5cm everywhere. The external faces are to be polished, and the inner faces are to be painted. If the rate of polishing is 20 paise per cm
2
and the rate of painting is 10 paise per cm
2
, find the total expenses required for polishing and painting the surface of the bookshelf.
Solution
:
External dimensions of book self:
Length, l = 85cm
Breadth, b = 25 cm
Height, h = 110 cm
External surface area of the shelf while leaving out the front face of the shelf.
= lh+2(lb+bh)
= [85×110+2(85×25+25×110)] = (9350+9750) = 19100
External surface area of the shelf is 19100 cm
2
Area of front face = [85×110-75×100+2(75×5)] = 1850+750
So, the area is 2600 cm
2
Area to be polished = (19100+2600) cm
2
= 21700 cm
2
.
Cost of polishing 1 cm
2
area = Rs 0.20
Cost of polishing 21700 cm
2
area Rs. (21700×0.20) = Rs 4340
Dimensions of the row of the bookshelf
Length(l) = 75 cm
Breadth (b) = 20 cm and
Height(h) = 30 cm
Area to be painted in one row= 2(l+h)b+lh = [2(75+30)× 20+75×30] = (4200+2250) = 6450
So, the area is 6450 cm
2
.
Area to be painted in 3 rows = (3×6450)cm
2
= 19350 cm
2
.
Cost of painting 1 cm
2
area = Rs. 0.10
Cost of painting 19350 cm
2
area = Rs (19350 x 0.1) = Rs 1935
Total expense required for polishing and painting = Rs. (4340+1935) = Rs. 6275
Answer:
The cost for polishing and painting the surface of the bookshelf is Rs. 6275.
2. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in fig. 13.32. Eight such spheres are used forth is the purpose and are to be painted silver. Each support is a cylinder of radius 1.5cm and height 7cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm
2,
and black paint costs 5 paise per cm
2
.
Solution:
Diameter of the wooden sphere = 21 cm
Radius of the wooden sphere, r = diameter/ 2 = (21/2) cm = 10.5 cm
Formula: Surface area of the wooden sphere = 4πr
2
= 4×(22/7)×(10.5)
2
= 1386
So, the surface area is 1386 cm
3
Radius of the circular end of cylindrical support = 1.5 cm
Height of the cylindrical support = 7 cm
Curved surface area = 2πrh
= 2×(22/7)×1.5×7 = 66
So, CSA is 66 cm
2
Now,
Area of the circular end of cylindrical support = πr
2
= (22/7)×1.5
2
= 7.07
Area of the circular end is 7.07 cm
2
Again,
Area to be painted silver = [8 ×(1386-7.07)] = 8×1378.93 = 11031.44
Area to be painted is 11031.44 cm
2
Cost for painting with silver colour = Rs(11031.44×0.25) =Rs 2757.86
Area to be painted black = (8×66) cm
2
= 528 cm
2
Cost for painting with black colour =Rs (528×0.05) = Rs26.40
Therefore, the total painting cost is
= Rs(2757.86 +26.40)
= Rs 2784.26
3. The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?
Solution:
Let the diameter of the sphere be “d”.
Radius of the sphere, r
1
= d/2
New radius of the sphere, say r
2
= (d/2)×(1-25/100) = 3d/8
Curved surface area of the sphere, (CSA)
1
= 4πr
1
2
= 4π×(d/2)
2
= πd
2
…(1)
Curved surface area of the sphere when the radius is decreased (CSA)
2
= 4πr
2
2
= 4π×(3d/8)
2
= (9/16)πd
2
…(2)
From equations (1) and (2), we have
Decrease in surface area of sphere = (CSA)
1
– (CSA)
2
= πd
2
– (9/16)πd
2
= (7/16)πd
2
= (7d
2
/16d
2
)×100 = 700/16 = 43.75% .
Therefore, the percentage decrease in the surface area of the sphere is 43.75%.
Benefits of NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes
-
Comprehensive Understanding
: The solutions provide clear step-by-step explanations for calculating surface areas and volumes of different geometric shapes helping students grasp these essential concepts thoroughly.
-
Problem-Solving Skills
: By working through the exercises and solutions students develop strong problem-solving skills and learn various techniques for tackling complex questions.
-
Concept Reinforcement
: The detailed solutions reinforce key concepts and help students practice and apply their knowledge, leading to better retention and understanding.
-
Clarity on Difficult Topics
: Students can clarify doubts and understand difficult topics more easily, thanks to the well-explained solutions provided.
-
Enhanced Exam Preparation
: Regular practice using these solutions improves exam readiness and boosts confidence in handling surface area and volume problems during tests.
