NCERT Solutions Class 9 Mathematics Chapter 13 Surface Area and Volume

Get the most accurate NCERT Solutions for Class 9 Mathematics Chapter 11 Surface Area and Volume here. Updated for the 2026-27 academic session, these solutions are based on the latest NCERT textbooks for Class 9 Mathematics. Our expert-created answers for Class 9 Mathematics are available for free download in PDF format.

Detailed Chapter 11 Surface Area and Volume NCERT Solutions for Class 9 Mathematics

For Class 9 students, solving NCERT textbook questions is the most effective way to build a strong conceptual foundation. Our Class 9 Mathematics solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 11 Surface Area and Volume solutions will improve your exam performance.

Class 9 Mathematics Chapter 11 Surface Area and Volume NCERT Solutions PDF

Exercise 13.1

Q.1) A plastic box 1.5 π‘š long, 1.25 π‘š wide and 65 π‘π‘š deep is to be made. It is opened 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 sheet for it, if a sheet measuring 1π‘š2 costs 𝑅𝑠 20.
Sol.1) Length of plastic box = 1.5 π‘š
Width of plastic box = 1.25 π‘š
Depth of plastic box = 1.25 π‘š
(i) The area of sheet required to make the box is equal to the surface area of the box excluding the top. 
Surface area of the box = Lateral surface area + Area of the base
= 2(𝑙 + 𝑏) Γ— β„Ž + (𝑙 Γ— 𝑏)
= 2[(1.5 + 1.25) Γ— 1.25] + (1.5 Γ— 1.25) π‘š2
= (3.575 + 1.875) π‘š2
= 5.45 π‘š2
The sheet required required to make the box is 5.45 π‘š2
(ii) Cost of 1 π‘š2 of sheet = 𝑅𝑠 20
∴ Cost of 5.45 π‘š2 of sheet = 𝑅𝑠 (20 Γ— 5.45) = 𝑅𝑠 109

Q.2) The length, breadth and height of a room are 5π‘š, 4π‘š and 3π‘š respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of 𝑅𝑠. 7.50 π‘π‘’π‘Ÿ π‘š2.
Sol.2) length of the room = 5π‘š
breadth of the room = 4π‘š
height of the room = 3π‘š
Area of four walls including the ceiling = 2(𝑙 + 𝑏) Γ— β„Ž + (𝑙 Γ— 𝑏)
= 2(5 + 4) Γ— 3 + (5 Γ— 4) π‘š2
= (54 + 20) π‘š2
= 74 π‘š2 Cost of white washing = 𝑅𝑠. 7.50 π‘π‘’π‘Ÿ π‘š2
Total cost = 𝑅𝑠. (74 Γ— 7.50) = 𝑅𝑠. 555

Q.3) The floor of a rectangular hall has a perimeter 250 π‘š. If the cost of painting the four walls at the rate of 𝑅𝑠. 10 π‘π‘’π‘Ÿ π‘š2 is Rs.15000, find the height of the hall.
[Hint : Area of the four walls = Lateral surface area.]
Sol.3) Perimeter of rectangular hall = 2(𝑙 + 𝑏) = 250 π‘š
Total cost of painting = 𝑅𝑠. 15000
Rate π‘π‘’π‘Ÿ π‘š2 = 𝑅𝑠. 10
Area of four walls = 2(𝑙 + 𝑏)β„Ž π‘š2 = (250 Γ— β„Ž) π‘š2
A/q
β‡’ β„Ž = 15000/2500π‘š
β‡’ β„Ž = 6 π‘š
Thus the height of the hall is 6 π‘š.

Q.4) The paint in a certain container is sufficient to paint an area equal to 9.375 π‘π‘š2. How many bricks of dimensions 22.5 π‘π‘š Γ— 10 π‘π‘š Γ— 7.5 π‘π‘š can be painted out of this container?
Sol.4) Volume of paint = 9.375 π‘š2 = 93750 π‘π‘š2
Dimension of brick = 22.5 π‘π‘š Γ— 10 π‘π‘š Γ— 7.5 π‘π‘š
Total surface area of a brick = 2(𝑙𝑏 + π‘β„Ž + π‘™β„Ž) π‘π‘š2
= 2(22.5 Γ— 10 + 10 Γ— 7.5 + 22.5 Γ— 7.5) π‘π‘š2
= 2(225 + 75 + 168.75) π‘π‘š2
= 2 Γ— 468.75 π‘π‘š2 = 937.5 π‘π‘š2
Number of bricks can be painted = 93750/937.5 = 100

Q.5) A cubical box has each edge 10 π‘π‘š and another cuboidal box is 12.5 π‘π‘š long, 10 π‘π‘š wide and 8 π‘π‘š 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?
Sol.5) (i) Lateral surface area of cubical box of edge 10π‘π‘š = 4 Γ— 102 π‘π‘š2 = 400 π‘π‘š2
Lateral surface area of cuboid box = 2(𝑙 + 𝑏) Γ— β„Ž
= 2 Γ— (12.5 + 10) Γ— 8 π‘π‘š2
= 2 Γ— 22.5 Γ— 8 π‘π‘š2 = 360 π‘π‘š2
Thus, lateral surface area of the cubical box is greater by (400 – 360) π‘π‘š2 = 40 π‘π‘š2
(ii) Total surface area of cubical box of edge 10 π‘π‘š = 6 Γ— 102π‘π‘š2 = 600π‘π‘š2
Total surface area of cuboidal box = 2(𝑙𝑏 + π‘β„Ž + π‘™β„Ž)
= 2(12.5 Γ— 10 + 10 Γ— 8 + 8 Γ— 12.5)π‘π‘š2
= 2(125 + 80 + 100) π‘π‘š2
= (2 Γ— 305) π‘π‘š2 = 610 π‘π‘š2
Thus, total surface area of cubical box is smaller by 10 π‘π‘š2

Q.6) A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?
Sol.6) i) Dimensions of greenhouse:
𝑙 = 30 π‘π‘š, 𝑏 = 25 π‘π‘š, β„Ž = 25 π‘π‘š
Total surface area of greenhouse = 2(𝑙𝑏 + π‘β„Ž + π‘™β„Ž)
= 2(30 Γ— 25 + 25 Γ— 25 + 25 Γ— 30) π‘π‘š2
= 2(750 + 625 + 750) π‘π‘š2
= 4250 π‘π‘š2
(ii) Length of the tape needed = 4(𝑙 + 𝑏 + β„Ž)
= 4(30 + 25 + 25) π‘π‘š
= 4 Γ— 80 π‘π‘š = 320 π‘π‘š

Q.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 π‘π‘š Γ— 20 π‘π‘š Γ— 5 π‘π‘š and the smaller of dimensions 15 π‘π‘š Γ— 12 π‘π‘š Γ— 5 π‘π‘š. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs.4 for 1000 π‘π‘š2 , find the cost of cardboard required for supplying 250 boxes of each kind.
Sol.7) Dimension of bigger box = 25 π‘π‘š Γ— 20 π‘π‘š Γ— 5 π‘π‘š
Total surface area of bigger box = 2(𝑙𝑏 + π‘β„Ž + π‘™β„Ž)
= 2(25 Γ— 20 + 20 Γ— 5 + 25 Γ— 5) π‘π‘š2
= 2(500 + 100 + 125) π‘π‘š2
= 1450 π‘π‘š2
Dimension of smaller box = 15 π‘π‘š Γ— 12 π‘π‘š Γ— 5 π‘π‘š
Total surface area of smaller box = 2(𝑙𝑏 + π‘β„Ž + π‘™β„Ž)
= 2(15 Γ— 12 + 12 Γ— 5 + 15 Γ— 5) π‘π‘š2
= 2(180 + 60 + 75) π‘π‘š2
= 630 π‘π‘š2
Total surface area of 250 boxes of each type = 250(1450 + 630)π‘π‘š2
= 250 Γ— 2080 π‘π‘š2 = 520000 π‘π‘š2
Extra area required = 5/100(1450 + 630) Γ— 250 π‘π‘š2 = 26000 π‘π‘š2
Total Cardboard required = 520000 + 26000 π‘π‘š2 = 546000 π‘π‘š2
Total cost of cardboard sheet = 𝑅𝑠.
(546000 Γ— 4) / 1000 = 𝑅𝑠. 2184

Q.8) Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the 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 much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4π‘š Γ— 3π‘š?
Sol.8) Dimensions of the box- like structure = 4π‘š Γ— 3π‘š Γ— 2.5
Tarpaulin only required for all the four sides and top.
Thus, Tarpaulin required = 2(𝑙 + 𝑏) Γ— β„Ž + 𝑙𝑏
= [2(4 + 3) Γ— 2.5 + 4 Γ— 3] π‘š2
= (35 Γ— 12) π‘š2
= 47 π‘š2

Exercise 13.2

Q.1) The curved surface area of a right circular cylinder of height 14 cm is 88 π‘π‘š2. Find the diameter of the base of the cylinder.
Sol.1) Let π‘Ÿ be the radius of the base and β„Ž = 14 π‘π‘š be the height of the cylinder.
Curved surface area of cylinder = 2πœ‹π‘Ÿβ„Ž = 88 π‘π‘š2

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume

β‡’ π‘Ÿ = 1 π‘π‘š
Thus, the diameter of the base = 2π‘Ÿ = 2 Γ— 1 = 2π‘π‘š

Q.2) It is required to make a closed cylindrical tank of height 1 π‘š and base diameter 140 π‘π‘š from a metal sheet. How many square metres of the sheet are required for the same?
Sol.2) Let r be the radius of the base and h be the height of the cylinder.
Base diameter = 140 π‘π‘š and Height (β„Ž) = 1π‘š
Radius of base (π‘Ÿ) = 140/2 = 70 π‘π‘š = 0.7 π‘š
Metal sheet required to make a closed cylindrical tank = 2πœ‹π‘Ÿ(β„Ž + π‘Ÿ)
= 7.48 π‘š2

Q.3) A metal pipe is 77 π‘π‘š long. The inner diameter of a cross section is 4 π‘π‘š, the outer diameter being 4.4 π‘π‘š (see Fig.). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-1

Sol.3) Let 𝑅 be external radius and π‘Ÿ be the internal radius β„Ž be the length of the pipe.

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-2

Q.4) The diameter of a roller is 84 π‘π‘š and its length is 120 π‘π‘š. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in π‘š2.
Sol.4) Length of the roller (β„Ž) = 120 π‘π‘š = 1.2 π‘š
Radius of the cylinder = 84/2 π‘π‘š = 42 π‘π‘š = 0.42 π‘š
Total no. of revolutions = 500
Distance covered by roller in one revolution = Curved surface area = 2πœ‹π‘Ÿβ„Ž
= (2 Γ— 22/7 Γ— 0.42 Γ— 1.2) π‘š2
= 3.168 π‘š2
Area of the playground = (500 Γ— 3.168) π‘š2 = 1584 π‘š2

Q.5) A cylindrical pillar is 50 π‘π‘š in diameter and 3.5 π‘š in height. Find the cost of painting the curved surface of the pillar at the rate of 𝑅𝑠. 12.50 π‘π‘’π‘Ÿ π‘š2.
Sol.5) Radius of the pillar (π‘Ÿ) = 50/2 π‘π‘š = 25 π‘π‘š = 0.25 π‘š
Height of the pillar (h) = 3.5 m.
Rate of painting = 𝑅𝑠. 12.50 π‘π‘’π‘Ÿ π‘š2
Curved surface = 2πœ‹π‘Ÿβ„Ž
= (2 Γ— 22 / 7 Γ— 0.25 Γ— 3.5) π‘š2
= 5.5 π‘š2
Total cost of painting = (5.5 Γ— 12.5) = 𝑅𝑠. 68.75

Q.6) Curved surface area of a right circular cylinder is 4.4 π‘š2. If the radius of the base of the cylinder is 0.7 π‘š, find its height.
Sol.6) Let r be the radius of the base and h be the height of the cylinder.
Curved surface area = 2πœ‹π‘Ÿβ„Ž = 4.4 π‘š2
β‡’ 2 Γ— 22 / 7 Γ— 0.7 Γ— β„Ž = 4.4
β‡’ β„Ž = 4.4/(2 Γ— 22 / 7 Γ— 0.7) = 1π‘š
β‡’ β„Ž = 1π‘š

Q.7) The inner diameter of a circular well is 3.5 π‘š. It is 10 π‘š deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of 𝑅𝑠. 40 π‘π‘’π‘Ÿ π‘š2.
Sol.7) Radius of circular well (π‘Ÿ) = 3.5/2 π‘š = 1.75 π‘š
Depth of the well (β„Ž) = 10 π‘š
Rate of plastering = 𝑅𝑠. 40 π‘π‘’π‘Ÿ π‘š2
(i) Curved surface = 2πœ‹π‘Ÿβ„Ž
= (2 Γ— 22 / 7 Γ— 1.75 Γ— 10) π‘š2 = 110 π‘š2
(ii) Cost of plastering = 𝑅𝑠. (110 Γ— 40) = 𝑅𝑠. 4400

Q.8) In a hot water heating system, there is a cylindrical pipe of length 28 π‘š and diameter 5 π‘π‘š. Find the total radiating surface in the system.
Sol.8) Radius of the pipe (π‘Ÿ) = 5/2 π‘π‘š = 2.5 π‘π‘š = 0.025 π‘š
Length of the pipe (β„Ž) = 28/2 π‘š = 14 π‘š
Total radiating surface = Curved surface area of the pipe = 2πœ‹π‘Ÿβ„Ž
= (2 Γ— 22 / 7 Γ— 0.025 Γ— 28) π‘š2
= 4.4 π‘š2

Q.9) Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 π‘š in diameter and 4.5 π‘š high.
(ii) how much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank.
Sol.9) (i) Radius of the tank (π‘Ÿ) = 4.2/2 π‘š = 2.1 π‘š
Height of the tank (β„Ž) = 4.5 π‘š
Curved surface area = 2πœ‹π‘Ÿβ„Ž π‘š2

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-3

Q.10) In Fig., you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 π‘π‘š and height of 30 π‘π‘š. A margin of 2.5 π‘π‘š 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.

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-4

Sol.10) Radius of the frame (π‘Ÿ) = 20/2 π‘π‘š = 10 π‘π‘š
Height of the frame (β„Ž) = 30 π‘π‘š + 2 Γ— 2.5 π‘π‘š = 35 π‘π‘š
2.5 π‘π‘š of margin will be added both side in the height.
Cloth required for covering the lampshade = curved surface area = 2πœ‹π‘Ÿβ„Ž 
= (2 Γ— 22 / 7 Γ— 10 Γ— 35) π‘π‘š2
= 2200 π‘π‘š2

Q.11) The students of a 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 π‘π‘š and height 10.5 π‘π‘š. 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?
Sol.11) Radius of the penholder (π‘Ÿ) = 3π‘π‘š
Height of the penholder (β„Ž) = 10.5π‘π‘š
Cardboard required by 1 competitor = CSA of one penholder + area of the base
= 2πœ‹π‘Ÿβ„Ž + πœ‹π‘Ÿ2

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-5

Exercise 13.3

Q.1) Diameter of the base of a cone is 10.5 π‘π‘š and its slant height is 10 π‘π‘š. Find its curved surface area.
Sol.1) Radius (π‘Ÿ) = 10.5/2 π‘π‘š = 5.25 π‘π‘š
Slant height (𝑙) = 10 π‘π‘š
Curved surface area of the cone = (πœ‹π‘Ÿπ‘™)π‘π‘š2
= (22 / 7 Γ— 5.25 Γ— 10) π‘π‘š2
= 165 π‘π‘š2

Q.2) Find the total surface area of a cone, if its slant height is 21 π‘š and diameter of its base is 24 π‘š.
Sol.2) Radius (π‘Ÿ) = 24/2 π‘š = 12 π‘š
Slant height (𝑙) = 21 π‘š
Total surface area of the cone = πœ‹π‘Ÿ (𝑙 + π‘Ÿ) π‘š2
= 22/7 Γ— 12 Γ— (21 + 12) π‘š2
= (22/7 Γ— 12 Γ— 33) π‘š2
= 1244.57 π‘š2

Q.3) Curved surface area of a cone is 308 π‘π‘š2 and its slant height is 14 π‘π‘š. Find
(i) radius of the base and (ii) total surface area of the cone.
Sol.3) (i) Curved surface of a cone = 308 π‘π‘š2
Slant height (𝑙) = 14π‘π‘š
Let r be the radius of the base
∴ πœ‹π‘Ÿπ‘™ = 308

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-6

Q.4) A conical tent is 10 π‘š high and the radius of its base is 24 π‘š. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 π‘š2 canvas is Rs.70
Sol.4) Radius of the base (π‘Ÿ) = 24 π‘š
Height of the conical tent (β„Ž) = 10 π‘š
Let l be the slant height of the cone.
∴ 𝑙2 = β„Ž2 + π‘Ÿ2
β‡’ 𝑙 = βˆšβ„Ž2 + βˆšπ‘Ÿ2
β‡’ 𝑙 = √102 + 242
β‡’ 𝑙 = √100 + 576
β‡’ 𝑙 = 26 π‘š
(ii) Canvas required to make the conical tent = Curved surface of the cone Cost of 1 π‘š2 canvas = β‚Ή70

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-7

Q.5) What length of tarpaulin 3π‘š wide will be required to make conical tent of height 8m and base radius 6π‘š? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20π‘π‘š (π‘ˆπ‘ π‘’ πœ‹ = 3.14).
Sol.5) Radius of the base (π‘Ÿ) = 6 π‘š
Height of the conical tent (β„Ž) = 8 π‘š
Let l be the slant height of the cone.
β‡’ 𝑙 = √100
β‡’ 𝑙 = 10 π‘š
CSA of conical tent = πœ‹π‘Ÿπ‘™
= (3.14 Γ— 6 Γ— 10) π‘š2 = 188.4 π‘š2
Breadth of tarpaulin = 3 m
Let length of tarpaulin sheet required be π‘₯.
20 cm will be wasted in cutting.
So, the length will be (π‘₯ – 0.2 π‘š)
Breadth of tarpaulin = 3 π‘š
Area of sheet = CSA of tent
[(π‘₯ – 0.2 π‘š) Γ— 3] π‘š = 188.4 π‘š2
β‡’ π‘₯ – 0.2 π‘š = 62.8 π‘š
β‡’ π‘₯ = 63 π‘š

Q.6) The slant height and base diameter of a conical tomb are 25 π‘š and 14 π‘š respectively.
Find the cost of white-washing its curved surface at the rate of β‚Ή210 per 100 π‘š2.
Sol.6) Radius (r) = 14/2 m = 7 m
Slant height tomb (l) = 25 m
Curved surface area = Ο€rl π‘š2
= (227Γ—25Γ—7) π‘š2
= 550 π‘š2
Rate of white- washing = β‚Ή210 per 100 π‘š2
Total cost of white-washing the tomb = β‚Ή(550 Γ— 210/100) = β‚Ή1155

Q.7) A joker’s cap is in the form of a right circular cone of base radius 7 π‘π‘š and height 24 π‘π‘š.
Find the area of the sheet required to make 10 such caps.
Sol.7) Radius of the cone (π‘Ÿ) = 7 π‘π‘š
Height of the cone (β„Ž) = 24 π‘π‘š
Let l be the slant height
∴ 𝑙 = βˆšβ„Ž2 + π‘Ÿ2
β‡’ 𝑙 = √242 + 72
β‡’ 𝑙 = √625
β‡’ 𝑙 = 25 π‘š
Sheet required for one cap = Curved surface of the cone
= πœ‹π‘Ÿπ‘™ π‘π‘š2
= (22 / 7 Γ— 7 Γ— 25) π‘π‘š2 = 550 π‘π‘š2
Sheet required for 10 π‘π‘Žπ‘π‘  = 550 Γ— 10 π‘π‘š2 = 5500 π‘π‘š2

Q.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 π‘š2, what will be the cost of painting all these cones? (π‘ˆπ‘ π‘’ πœ‹ = 3.14 and take √1.04 = 1.02)
Sol.8) Radius of the cone (π‘Ÿ) = 40/2 π‘π‘š = 20 π‘π‘š = 0.2 π‘š
Height of the cone (β„Ž) = 1 π‘š
Let l be the slant height of a cone.
∴ 𝑙 = βˆšβ„Ž2 + π‘Ÿ2
β‡’ 𝑙 = √12 + 0.22
β‡’ 𝑙 = √1.04
β‡’ 𝑙 = 1.02 π‘š
Rate of painting = 𝑅𝑠. 12 π‘π‘’π‘Ÿ π‘š2
Curved surface of 1 π‘π‘œπ‘›π‘’ = πœ‹π‘Ÿπ‘™ π‘š2
= (3.14 Γ— 0.2 Γ— 1.02) π‘š2
= 0.64056 π‘š2
Curved surface of such 50 cones = (50 Γ— 0.64056) π‘š2
= 32.028 π‘š2
Cost of painting all these cones = 𝑅𝑠. (32.028 Γ— 12)
= 𝑅𝑠. 384.34

Exercise 13.4

Q.1) Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
Sol.1) (i) Radius of the sphere (π‘Ÿ) = 10.5 π‘π‘š
Surface area = 4πœ‹π‘Ÿ2

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-18

Q.3) Find the total surface area of a hemisphere of radius 10 π‘π‘š. (π‘ˆπ‘ π‘’ πœ‹ = 3.14)
Sol.3) π‘Ÿ = 10 π‘π‘š
Total surface area of hemisphere = 3πœ‹π‘Ÿ2
= (3 Γ— 3.14 Γ— 10 Γ— 10)π‘π‘š2
= 942 π‘π‘š2

Q.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.
Sol.4) Let r be the initial radius and R be the increased radius of balloons.
π‘Ÿ = 7π‘π‘š and 𝑅 = 14π‘π‘š
Ratio of the surface area = 4πœ‹π‘Ÿ2/4πœ‹π‘…2
= π‘Ÿ2/𝑅2
= 7Γ—7/14Γ—14 = 1/4
Thus, the ratio of surface areas = 1 : 4

Q.5) A hemispherical bowl made of brass has inner diameter 10.5 π‘π‘š. Find the cost of tinplating it on the inside at the rate of Rs.16 per 100 π‘π‘š2.
Sol.5) Radius of the bowl (π‘Ÿ) = 10.5/2 π‘π‘š = 5.25 π‘π‘š
Curved surface area of the hemispherical bowl = 2πœ‹π‘Ÿ2
= (2 Γ— 22/7 Γ— 5.25 Γ— 5.25) π‘π‘š2
= 173.25 π‘π‘š2
Rate of tin - plating is = 𝑅𝑠. 16 π‘π‘’π‘Ÿ 100 π‘π‘š2
Therefore, cost of 1 π‘π‘š2 = 𝑅𝑠. 16/100
Total cost of tin-plating the hemisphere bowl = 173.25 Γ— 16 / 100
= 𝑅𝑠. 27.72

Q.6) Find the radius of a sphere whose surface area is 154 π‘π‘š2.
Sol.6) Let r be the radius of the sphere.
Surface area = 154 π‘π‘š2
β‡’ 4πœ‹π‘Ÿ2 = 154

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-19

Q.7) The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Sol.7) Let the diameter of earth be r and that of the moon will be r/4
Radius of the earth = π‘Ÿ/2
Radius of the moon = π‘Ÿ/8

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-20

Q.8) A hemispherical bowl is made of steel, 0.25 π‘π‘š thick. The inner radius of the bowl is 5π‘π‘š. Find the outer curved surface area of the bowl.
Sol.8) Inner radius of the bowl (π‘Ÿ) = 5 π‘π‘š
Thickness of the steel = 0.25 π‘π‘š
∴ outer radius (𝑅) = (π‘Ÿ + 0.25) π‘π‘š
= (5 + 0.25) π‘π‘š = 5.25 π‘π‘š
Outer curved surface = 2πœ‹π‘Ÿ2
= (2 Γ— 22/7 Γ— 5.25 Γ— 5.25) π‘π‘š2
= 173.25 π‘π‘š2

Q.9) A right circular cylinder just encloses a sphere of radius r (see Fig.). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
Sol.9) (i) The surface area of the sphere with raius π‘Ÿ = 4πœ‹π‘Ÿ2
(ii) The right circular cylinder just encloses a sphere of radius r.
∴ the radius of the cylinder = π‘Ÿ and its height = 2π‘Ÿ
∴ Curved surface of cylinder = 2πœ‹π‘Ÿβ„Ž
= 2πœ‹ Γ— π‘Ÿ Γ— 2π‘Ÿ = 4πœ‹π‘Ÿ2
(iii) Ratio of the areas = 4πœ‹π‘Ÿ2: 4πœ‹π‘Ÿ2 = 1: 1

Exercise 13.5

Q.1) A matchbox measures 4π‘π‘š Γ— 2.5π‘π‘š Γ— 1.5π‘π‘š. What will be the volume of a packet containing 12 such boxes?
Sol.1) Dimension of matchbox = 4π‘π‘š Γ— 2.5π‘π‘š Γ— 1.5π‘π‘š
𝑙 = 4 π‘π‘š, 𝑏 = 2.5 π‘π‘š and β„Ž = 1.5 π‘π‘š
Volume of one matchbox = (𝑙 Γ— 𝑏 Γ— β„Ž)
= (4 Γ— 2.5 Γ— 1.5) π‘π‘š3 = 15 π‘π‘š3
Volume of a packet containing 12 such boxes = (12 Γ— 15) π‘π‘š3 = 180 π‘π‘š33

Q.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? (1π‘š3 = 1000 𝑙)
Sol.2) Dimensions of water tank = 6π‘š Γ— 5π‘š Γ— 4.5π‘š
𝑙 = 6π‘š , 𝑏 = 5π‘š and β„Ž = 4.5π‘š
Therefore Volume of the tank = π‘™π‘β„Ž π‘š3
= (6 Γ— 5 Γ— 4.5)π‘š3 = 135 π‘š3
[𝑆𝑖𝑛𝑐𝑒 1π‘š3 = 1000π‘™π‘–π‘‘π‘Ÿπ‘’π‘ ] = 135000 π‘™π‘–π‘‘π‘Ÿπ‘’π‘  of water.
Therefore , the tank can hold = 135 Γ— 1000 π‘™π‘–π‘‘π‘Ÿπ‘’π‘ 

Q.3) A cuboidal vessel is 10 π‘š long and 8 π‘š wide. How high must it be made to hold 380 cubic metres of a liquid?
Sol.3) Length = 10 π‘š , Breadth = 8 π‘š and Volume = 380 π‘š3
Volume of cuboid = πΏπ‘’π‘›π‘”π‘‘β„Ž Γ— π΅π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž Γ— π»π‘’π‘–π‘”β„Žπ‘‘
β‡’ Height = π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘π‘’π‘π‘œπ‘–π‘‘/πΏπ‘’π‘›π‘”π‘‘β„Ž Γ— π΅π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž
= 380/10 Γ— 8 π‘š
= 4.75π‘š

Q.4) Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of π‘…𝑠. 30 π‘π‘’π‘Ÿ π‘š3.
Sol.4) 𝑙 = 8 π‘š, 𝑏 = 6 π‘š and β„Ž = 3 π‘š
Volume of the pit = π‘™π‘β„Ž π‘š3
= (8 Γ— 6 Γ— 3)π‘š3
= 144 π‘š3
Rate of digging = 𝑅𝑠. 30 π‘π‘’π‘Ÿ π‘š3
Total cost of digging the pit = 𝑅𝑠. (144 Γ— 30) = 𝑅𝑠. 4320

Q.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 π‘š and 10 π‘š.
Sol.5) length = 2.5 π‘š, depth = 10 π‘š and volume = 50000 π‘™π‘–π‘‘π‘Ÿπ‘’π‘ 
1π‘š3 = 1000 π‘™π‘–π‘‘π‘Ÿπ‘’π‘ 
∴ 50000 π‘™π‘–π‘‘π‘Ÿπ‘’π‘  = (50000/1000)π‘š3 = 50 π‘š3
Breadth = π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘π‘’π‘π‘œπ‘–π‘‘ / πΏπ‘’π‘›π‘”π‘‘β„Ž Γ— π΅π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž
= (50/2.5 Γ— 10) π‘š = 2 π‘š

Q.6) A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20π‘š Γ— 15π‘š Γ— 6π‘š. For how many days will the water of this tank last?
Sol.6) Dimension of tank = 20π‘š Γ— 15π‘š Γ— 6π‘š
𝑙 = 20 π‘š , 𝑏 = 15 π‘š and β„Ž = 6 π‘š
Capacity of the tank = π‘™π‘β„Ž π‘š3
= (20 Γ— 15 Γ— 6)π‘š3
= 1800 π‘š3
Water requirement per person per day = 150 π‘™π‘–π‘‘π‘Ÿπ‘’π‘ 
Water required for 4000 person per day = (4000 Γ— 150) 𝑙
= 4000 Γ— 150/1000
= 600 π‘š3
Number of days the water will last = Capacity of tank Total water required per day
= (1800/600) = 3
The water will last for 3 days.

Q.7) A godown measures 40π‘š Γ— 25π‘š Γ— 15π‘š. Find the maximum number of wooden crates each measuring 1.5π‘š Γ— 1.25π‘š Γ— 0.5π‘š that can be stored in the godown.
Sol.7) Dimension of godown = 40 π‘š Γ— 25 π‘š Γ— 15 π‘š
Volume of the godown = (40 Γ— 25 Γ— 15)π‘š3 = 10000 π‘š3
Dimension of crates = 1.5π‘š Γ— 1.25π‘š Γ— 0.5π‘š
Volume of 1 crates = (1.5 Γ— 1.25 Γ— 0.5)π‘š3 = 0.9375 π‘š3
Number of crates that can be stored =Volume of the godown/Volume of 1 crate
= 10000/0.9375
= 10666.66 = 10666

Q.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.
Sol.8) Edge of the cube = 12 π‘π‘š.
Volume of the cube = (𝑒𝑑𝑔𝑒)3 π‘π‘š3
= (12 Γ— 12 Γ— 12)π‘π‘š3
= 1728 π‘π‘š3
Number of smaller cube = 8
Volume of the 1 smaller cube = (1728/8) π‘π‘š3 = 216 π‘π‘š3
Side of the smaller cube = π‘Ž
π‘Ž3 = 216
β‡’ π‘Ž = 6 π‘π‘š
Surface area of the cube = 6 (𝑠𝑖𝑑𝑒)2
Ratio of their surface area = 6 Γ— 12 Γ— 12/6 Γ— 6 Γ— 6 
= 4/1 = 4: 1

Q.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?
Sol.9) Depth of river (β„Ž) = 3 π‘š
Width of river (𝑏) = 40 π‘š
Rate of flow of water (𝑙) = 2 π‘˜π‘š π‘π‘’π‘Ÿ β„Žπ‘œπ‘’π‘Ÿ = (2000/60) π‘š π‘π‘’π‘Ÿ π‘šπ‘–π‘›π‘’π‘‘π‘’
= (100/3)π‘š π‘π‘’π‘Ÿ π‘šπ‘–π‘›π‘’π‘‘π‘’
Volume of water flowing into the sea in a minute = π‘™π‘β„Ž π‘š3 = (100/3 Γ— 40 Γ— 3)π‘š3
= 4000 π‘š3

Exercise 13.6

Q.1) The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm.
How many litres of water can it hold? (1000 π‘π‘š3 = 1𝑙)
Sol.1) Here, β„Ž = 25 π‘π‘š, 2πœ‹π‘Ÿ = 132 π‘π‘š.
2πœ‹π‘Ÿ = 132

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-30

Q.2) The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm.
The length of the pipe is 35 cm. Find the mass of the pipe, if 1 π‘π‘š3 of wood has a mass of 0.6 g.
Sol.2) 

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-31

Mass of 1 π‘π‘š3 of wood = 0.6 𝑔
∴ Mass of 5720 π‘π‘š3 of wood = 0.6 Γ— 5720 𝑔 = 3432 𝑔 = 3.432 π‘˜π‘”

Q.3) A soft drink is available in two packs β€”
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm.
Which container has greater capacity and by how much?
Sol.3) For tin can with rectangular 6 base.
𝑙 = 5 π‘π‘š, 𝑏 = 4 π‘π‘š, β„Ž = 15 π‘π‘š
Volume of the tin can = π‘™π‘β„Ž = 5 Γ— 4 Γ— 15 π‘π‘š3 = 300 π‘π‘š3
For plastic cylinder with circular base.

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-32

Difference in the capacities of the two containers= (385 – 300)π‘π‘š3 = 85 π‘π‘š3
Hence, the plastic cylinder with circular base has greater capacity by 85 π‘π‘š3

Q.4) If the lateral surface of a cylinder is 94.2 π‘π‘š2 and its height is 5 cm, then find
(i) radius of its base (ii) its volume (π‘ˆπ‘ π‘’ πœ‹ = 3.14)
Sol.4) Here, β„Ž = 5 π‘π‘š, 2πœ‹π‘Ÿβ„Ž = 94.2 π‘π‘š2.
(i) 2πœ‹π‘Ÿβ„Ž = 94.2
β‡’ 2 Γ— 3.14 Γ— π‘Ÿ Γ— 5 = 94.2

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-33

Hence, base radius of the cylinder = 3 π‘π‘š
(ii) Volume of the cylinder = πœ‹π‘Ÿ2β„Ž
= 3.14 Γ— 3 Γ— 3 Γ— 5 π‘π‘š3 = 141.3 π‘π‘š3

Q.5) It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel 10π‘š deep. If the cost of painting is at the rate of 𝑅𝑠 20 π‘π‘’π‘Ÿ π‘š2, find
(i) Inner curved surface area of the vessel, (ii) radius of the base,
(iii) capacity of the vessel.
Sol.5) Here, β„Ž = 10 π‘š
(i) Inner curved surface area = Total cost / Cost of painting per sq. metre

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-29

Q.6) The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?
Sol.6) Here, β„Ž = 1 π‘š, volume = 15.4 π‘™π‘–π‘‘π‘Ÿπ‘’π‘ 

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-28

= 2 Γ— 22/7 Γ— 0.07 (1 + 0.07)π‘š2
= 44 Γ— 0.01 Γ— 1.07 π‘š2 = 0.4708 π‘š2
Hence, 0.4708 π‘š2 of metal sheet would be needed

Q.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 dimeter 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.
Sol.7) Here, β„Ž = 14 π‘π‘š.
Radius of the pencil (𝑅) = 7/2 π‘šπ‘š = 0.35 π‘π‘š.
Radius of the graphite (π‘Ÿ) = 1/2 π‘šπ‘š = 0.05 π‘π‘š.
Volume of the the graphite = πœ‹π‘Ÿ2β„Ž
 = 22/7 Γ— 0.05 Γ— 0.05 Γ— 14 π‘π‘š3 = 0.11 π‘π‘š3
Volume of the the wood = πœ‹ (𝑅2 – π‘Ÿ2)β„Ž 
= 22/7 Γ— [(0.35)2 – (0.05)2] Γ— 14 π‘π‘š3
= 22/7 Γ— 0.4 Γ— 0.3 Γ— 14 π‘π‘š3 = 5.28 π‘π‘š3
Hence, volume of the wood = 5.28 π‘π‘š3 and
volume of the graphite = 0.11 π‘π‘š3

Q.8) A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Sol.8) Here, π‘Ÿ = 7/2 π‘π‘š = 3.5 π‘π‘š, β„Ž = 4 π‘π‘š
Capacity of 1 cylindrical bowl = πœ‹π‘Ÿ2β„Ž
= 22/7 Γ— 3.5 Γ— 3.5 Γ— 4 π‘π‘š3 = 154 π‘π‘š3
Hence, soup consumed by 250 patients per day = 250 Γ— 154 π‘π‘š3 = 38500 π‘π‘š3

Exercise 13.7

Q.1) Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm
Sol.1) (i) Radius (π‘Ÿ) = 6 π‘π‘š Height (β„Ž) = 7 π‘π‘š

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-27

Q.2) Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm
Sol.2) (i) Radius (π‘Ÿ) = 7 π‘π‘š
Slant height (𝑙) = 25 π‘π‘š
Let h be the height of the conical vessel.
∴ β„Ž = βˆšπ‘™2 βˆ’ π‘Ÿ2

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-26

Q.3) . The height of a cone is 15 π‘π‘š. If its volume is 1570 π‘π‘š3, find the radius of the base.
(Use πœ‹ = 3.14)
Sol.3) Height (β„Ž) = 15 π‘π‘š
Volume = 1570 π‘π‘š3
Let the radius of the base of cone be π‘Ÿ π‘π‘š
∴ Volume = 1570 π‘π‘š3
β‡’ (1/3)πœ‹π‘Ÿ2β„Ž = 1570
β‡’ 13 Γ— 3.14 Γ— π‘Ÿ2 Γ— 15 = 1570
β‡’ π‘Ÿ2 = 1570/3.14 Γ— 5 = 100
β‡’ π‘Ÿ = 10

Q.4) If the volume of a right circular cone of height 9 cm is 48πœ‹ π‘π‘š3 , find the diameter of its base
Sol.4) Height (β„Ž) = 9 π‘π‘š
Volume = 48πœ‹ π‘π‘š3
Let the radius of the base of the cone be π‘Ÿ π‘π‘š
∴ Volume = 48πœ‹ π‘π‘š3
β‡’ (1/3)πœ‹π‘Ÿ2β„Ž = 48πœ‹
β‡’ 13 Γ— π‘Ÿ2 Γ— 9 = 48
β‡’ 3π‘Ÿ2 = 48
β‡’ π‘Ÿ2 = 48/3 = 16
β‡’ π‘Ÿ = 4

Q.5) A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Sol.5) Diameter of the top of the conical pit = 3.5 π‘š
Radius (π‘Ÿ) = (3.5/2) π‘š = 1.75 π‘š
Depth of the pit (β„Ž) = 12 π‘š
Volume = (1/3)πœ‹π‘Ÿ2β„Ž
= (13 Γ— 22/7 Γ— 1.75 Γ— 1.75 Γ— 12) π‘š3
= 38.5 π‘š3 ( 1 π‘š3 = 1 π‘˜π‘–π‘™π‘œπ‘™π‘–π‘‘π‘Ÿπ‘’ )
Capacity of pit = 38.5 π‘˜π‘–π‘™π‘œπ‘™π‘–π‘‘π‘Ÿπ‘’π‘ 

Q.6) The volume of a right circular cone is 9856 π‘π‘š3 . If the diameter of the base is 28 cm,
find: (i) height of the cone (ii) slant height of the cone
(iii) curved surface area of the cone
Sol.6) (i) Diameter of the base of the cone = 28 π‘π‘š

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-25

Q.7) A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm.
Find the volume of the solid so obtained
Sol.7) On revolving the Δ𝐴𝐡𝐢 along the side 12 π‘π‘š,
a right circular cone of height(β„Ž) 12 π‘π‘š,
radius(r) 5 cm and slant height(𝑙) 13 π‘π‘š will be formed.
Volume of solid so obtained = 1/3 πœ‹π‘Ÿ 2β„Ž
= (1/3 Γ— πœ‹ Γ— 5 Γ— 5 Γ— 12) π‘π‘š3 = 100πœ‹ π‘π‘š3

Q.8) If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Sol.8) On revolving the Δ𝐴𝐡𝐢 along the side 12 π‘π‘š,
a cone of radius(π‘Ÿ) 12 π‘π‘š,
height(β„Ž) 5 π‘π‘š, and
slant height(𝑙) 13 π‘π‘š will be formed.
Volume of solid so obtained = (1/3)πœ‹π‘Ÿ2β„Ž
= (1/3 Γ— πœ‹ Γ— 12 Γ— 12 Γ— 5) π‘π‘š3 = 240πœ‹ π‘π‘š3
Ratio of the volumes = 100πœ‹/240πœ‹ = 5/12 = 5: 12

Q.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 required
Sol.9) Diameter of the base of the cone = 10.5 π‘š

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-24

Exercise 13.8

Q.1) Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m
Sol.1) (i) Radius of the sphere(π‘Ÿ) = 7 π‘π‘š
Therefore, Volume of the sphere = 4/3 πœ‹π‘Ÿ2

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-23

Q.2) Find the amount of water displaced by a solid spherical ball of diameter.
(i) 28 cm (ii) 0.21 m
Sol.2) (i) Diameter of the spherical ball = 28 π‘π‘š

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-22

Q.3) The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 𝑔 π‘π‘’π‘Ÿ π‘π‘š3?
Sol.3) Diameter of the ball = 4.2 π‘π‘š
Radius = (4.2/2) π‘π‘š = 2.1 π‘π‘š
Volume of the ball = 4/3 πœ‹π‘Ÿ3
= (4/3 Γ— 22/7 Γ— 2.1 Γ— 2.1 Γ— 2.1) π‘π‘š3
= 38.808 π‘π‘š3
Density of the metal is 8.9𝑔 π‘π‘’π‘Ÿ π‘π‘š3
Mass of the ball = (38.808 Γ— 8.9) 𝑔 = 345.3912 𝑔

Q.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?
Sol.4) Let the diameter of the moon be r Radius of the moon = π‘Ÿ/2
A/q,
Diameter of the earth = 4π‘Ÿ
Radius(π‘Ÿ) = 4π‘Ÿ/2 = 2π‘Ÿ

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-21

Q.5) How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Sol.5) Diameter of a hemispherical bowl = 10.5 π‘π‘š
Radius(π‘Ÿ) = (10.5/2) π‘π‘š = 5.25π‘π‘š

= 0.3031875 π‘™π‘–π‘‘π‘Ÿπ‘’π‘  (approx.)

Q.6) A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Sol.6) Internal radius = π‘Ÿ = 1π‘š
External radius = 𝑅 = (1 + 0.1) π‘π‘š = 1.01 π‘π‘š
Volume of iron used = External volume – Internal volume

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-16

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-15

Q.7) Find the volume of a sphere whose surface area is 154 π‘π‘š2.
Sol.7) Let π‘Ÿ π‘π‘š be the radius of the sphere So,
surface area = 154π‘π‘š2
β‡’ 4πœ‹π‘Ÿ2 = 154
β‡’ 4 Γ— 22/7 Γ— π‘Ÿ2 = 154

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-14

Q.8) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs.498.96. If the cost of white-washing is Rs.2.00 per square metre, find the
(i) inside surface area of the dome, (ii) volume of the air inside the dome.
Sol.8) (i) Inside surface area of the dome =Total cost of white washing/Rate of white washing
= ( 498.96/2.00) π‘š2 = 249.48 π‘š2
(ii) Let π‘Ÿ be the radius of the dome.
Surface area = 2πœ‹π‘Ÿ2

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-13

Q.9) Twenty -seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area Sβ€². Find the
(i) radius rβ€² of the new sphere, (ii) ratio of S and S
Sol.9)

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-12

(ii) Required ratio = π‘†/𝑆′ = 4πœ‹π‘Ÿ2/4πœ‹π‘Ÿβ€²2
= π‘Ÿ 2 /(3π‘Ÿ) 2 = π‘Ÿ 2 /9π‘Ÿ 2
= 1/9 = 1: 9

Q.10) A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (𝑖𝑛 π‘šπ‘š3 ) is needed to fill this capsule?
Sol.10) Diameter of the spherical capsule = 3.5 π‘šπ‘š
Radius(π‘Ÿ) = 3.52π‘šπ‘š = 1.75π‘šπ‘š
Medicine needed for its filling = Volume of spherical capsule
= 4/3 πœ‹π‘Ÿ3
= (4/3 Γ— 22 7 Γ— 1.75 Γ— 1.75 Γ— 1.75) π‘šπ‘š3
= 22.46 π‘šπ‘š3(π‘Žπ‘π‘π‘Ÿπ‘œπ‘₯. )

Exercise 13.9

Q.1) A wooden bookshelf has external dimensions as follows:
Height = 110 cm, Depth = 25 cm, Breadth = 85 cm [See fig.].
The thickness of the planks is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 π‘π‘Žπ‘–π‘ π‘’ π‘π‘’π‘Ÿ π‘π‘š2 and the rate of painting is 10 π‘π‘Žπ‘–π‘ π‘’ π‘π‘’π‘Ÿ π‘π‘š2, find the total expenses required for polishing and painting the surface of the bookshelf.

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-11

Sol.1) External faces to be polished = Area of six faces of cuboidal bookshelf – 3 (Area of open portion ABCD)
= 2 (110 Γ— 25 + 25 Γ— 85 + 85 Γ— 110) – 3 (75 Γ— 30)
[𝐴𝐡 = 85 – 5 – 5 = 75 π‘π‘š and 𝐴𝐷
= 1/3 Γ— 110 – 5 – 5 – 5 – 5 = 30 π‘π‘š]
= 2 (2750 + 2125 + 9350) – 3 Γ— 2250
= 2 Γ— 14225 – 6750
= 28450 – 6750
= 21700 π‘π‘š2
Now, cost of painting outer faces of wodden bookshelf at the rate of 20 paise.
= 𝑅𝑠. 0.20 π‘π‘’π‘Ÿ π‘π‘š2 = 𝑅𝑠. 0.20 Γ— 21700 = 𝑅𝑠. 4340
Here, three equal five sides inner faces.
Therefore, total surface area
= 3 [ 2 (30 + 75) 20 + 30 Γ— 75] [ π·π‘’π‘π‘‘β„Ž = 25 – 5 = 20 π‘π‘š]
= 3 [ 2 Γ— 105 Γ— 20 + 2250] = 3
[ 4200 + 2250]
= 3 Γ— 6450 = 19350 π‘π‘š2
Now, cost of painting inner faces at the rate of 10 π‘π‘Žπ‘–π‘ π‘’ i.e. 𝑅𝑠. 0.10 π‘π‘’π‘Ÿ π‘π‘š2.
= 𝑅𝑠. 0.10 Γ— 19350 = 𝑅𝑠. 1935
Total expenses required = 𝑅𝑠. 4340 + 𝑅𝑠. 1935 = 𝑅𝑠. 6275

Q.2) The front compound wall of a house is decorated by wooden spheres of diameter 21 π‘π‘š, placed on small supports as shown in figure. Eight such spheres are used for this purpose and are to be painted silver. Each support is a cylinder of radius 1.5 π‘π‘š and height 7 π‘π‘š and is to be painted black. Find he cost of paint required if silver paint costs 25 π‘π‘Žπ‘–π‘ π‘’ π‘π‘’π‘Ÿ π‘π‘š2 and black paint costs 5 π‘π‘Žπ‘–π‘ π‘’ π‘π‘’π‘Ÿ π‘π‘š2.
Sol.2) Diameter of a wooden sphere = 21 π‘π‘š.
∴ Radius of wooden sphere(𝑅) = 21/2 π‘π‘š
And Radius of the cylinder (π‘Ÿ) = 1.5 π‘π‘š
Surface area of silver painted part = Surface area of sphere – Upper part of cylinder for support

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-8

Surface area of such type of 8 spherical part = 8 Γ— 1378.928
= 11031.424 π‘π‘š2
Cost of silver paint over 1 π‘π‘š2 = 𝑅𝑠. 0.25
Cost of silver paint over 11031.928 π‘π‘š2 = 0.25 Γ— 11031.928
= 𝑅𝑠. 2757.85
Now, curved surface area of a cylindrical support = 2πœ‹π‘Ÿβ„Ž
= 2 Γ— 22 / 7 Γ— 15/10 Γ— 7 = 66π‘π‘š2
Curved surface area of 8 such cylindrical supports = 66 Γ— 8 = 528 π‘π‘š2
Cost of black paint over 1 cπ‘š2 of cylindrical support = 𝑅𝑠. 0.50
Cost of black paint over 528 cπ‘š2 of cylindrical support = 0.50 Γ— 528
= 𝑅𝑠. 26.40
Total cost of paint required = 𝑅𝑠. 2757.85 + 𝑅𝑠. 26.4
= 𝑅𝑠. 2784.25

Q.3) If diameter of a sphere is decreased by 25% then what percent does its curved surface area decrease?
Sol.3) Diameter of original sphere = 𝐷 = 2𝑅
𝑅 = π·/2
Curved surface area of original sphere = 4πœ‹π‘…2 = 4πœ‹ (𝐷/2)2 = πœ‹π·2
According to the question,
Decreased diameter = 25% π‘œπ‘“ 𝐷

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-9

Q.4) Sameera wants to celebrate the fifth birthday of her daughter with a party. She bought thick paper to make the conical party caps. Each cap is to have a base diameter of 10 cm and height 12 cm. A sheet of the paper is 25 cm by 40 cm and approximately 82% of the sheet can be effectively used for making the caps after cutting. What is the minimum number of sheets of paper that Sameera would need to buy, if there are to be 15 children at the party? (Use Ο€ = 3.14)

""NCERT-Solutions-Class-9-Mathematics-Chapter-13-Surface-Area-and-Volume-10

Free study material for Mathematics

NCERT Solutions Class 9 Mathematics Chapter 11 Surface Area and Volume

Students can now access the NCERT Solutions for Chapter 11 Surface Area and Volume prepared by teachers on our website. These solutions cover all questions in exercise in your Class 9 Mathematics textbook. Each answer is updated based on the current academic session as per the latest NCERT syllabus.

Detailed Explanations for Chapter 11 Surface Area and Volume

Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 9 Mathematics chapter. Along with the final answers, we have also explained the concept behind it to help you build stronger understanding of each topic. This will be really helpful for Class 9 students who want to understand both theoretical and practical questions. By studying these NCERT Questions and Answers your basic concepts will improve a lot.

Benefits of using Mathematics Class 9 Solved Papers

Using our Mathematics solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 9 solutions are a guide for self-study and homework assistance. Along with the chapter-wise solutions, you should also refer to our Revision Notes and Sample Papers for Chapter 11 Surface Area and Volume to get a complete preparation experience.

FAQs

Where can I find the latest [current-page:node:field_title] for the 2026-27 session?

The complete and updated is available for free on StudiesToday.com. These solutions for Class 9 Mathematics are as per latest NCERT curriculum.

Are the Mathematics NCERT solutions for Class 9 updated for the new 50% competency-based exam pattern?

Yes, our experts have revised the as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Mathematics concepts are applied in case-study and assertion-reasoning questions.

How do these Class 9 NCERT solutions help in scoring 90% plus marks?

Toppers recommend using NCERT language because NCERT marking schemes are strictly based on textbook definitions. Our will help students to get full marks in the theory paper.

Do you offer [current-page:node:field_title] in multiple languages like Hindi and English?

Yes, we provide bilingual support for Class 9 Mathematics. You can access in both English and Hindi medium.

Is it possible to download the Mathematics NCERT solutions for Class 9 as a PDF?

Yes, you can download the entire in printable PDF format for offline study on any device.