Official Class 10 Mathematics Revision Material: CBSE Class 10 Mathematics Surface Areas and Volumes Notes Set 08
Access comprehensive revision notes for Chapter 12 Surface Areas and Volumes using the CBSE Class 10 Mathematics Surface Areas and Volumes Notes Set 08. Designed to align with the 2026-27 academic syllabus for Class 10 Mathematics, these concept summaries help students streamline their exam preparation and review complex topics efficiently.
Chapter-wise Concept Summaries: Chapter 12 Surface Areas and Volumes
View or download the dedicated Chapter 12 Surface Areas and Volumes revision resource below. Engaging with these structured concept notes daily ensures continuous academic progress and mastery of the 2026-27 curriculum for Mathematics.
Conversion of Solids
In this section, we shall discuss problems pertaining to conversion of a solid (discussed in the previous classes) into another solid of different shape. For example, a metallic sphere is melted and recast into a cylindrical wire, or the earth taken out by digging a well and spreading it uniformly around the well to form an embankment in the form of a cylindrical shell from its original shape of right circular cylinder, etc. The volume of the solid does not change when it is melted and recast.
Solved Examples
Example 10. By melting a solid cylindrical metal, a few conical materials are to be made. If three times the radius of the cone is equal to twice the radius of the cylinder and the ratio of the height of the cylinder and the height of the cone is 4 : 3, find the number of cones which can be made.
Answer:
Let R be the radius and H be the height of the cylinder and let r and h be the radius and height of the cone respectively. Then 3r = 2R ...(i) and H : h = 4 : 3, so 3H = 4h ...(ii).
Let n be the number of cones which can be made from the material of the cylinder. Then the volume of the cylinder is equal to the sum of the volumes of n cones. Hence, \( \pi R^2 H = n \cdot \dfrac{1}{3}\pi r^2 h \), so 3R²H = n r²h.
From (i) and (ii), R = \( \dfrac{3r}{2} \) and H = \( \dfrac{4h}{3} \). Therefore n = \( \dfrac{3R^2H}{r^2h} = \dfrac{3 \times \frac{9r^2}{4} \times \frac{4h}{3}}{r^2h} \) = 9.
Hence, the required number of cones is 9.
Example 11. An iron rod of length 1 m and diameter 4 cm is melted and cast into thin wires of length 20 cm each. If the number of such wires is 2000, find the radius of each thin wire.
Answer:
Let the radius of each thin wire be r cm. The sum of the volumes of the 2000 thin wires will be equal to the volume of the iron rod. The iron rod and each thin wire are cylindrical.
Volume of the iron rod of radius 2 cm = π × 2² × 100 cm³. Volume of each thin wire = πr² × 20.
Hence, π × 2² × 100 = 2000 × πr² × 20, so 400 = 40000 r², r² = \( \dfrac{1}{100} \) and r = \( \dfrac{1}{10} \) [taking the positive square root only].
Hence, the required radius of each thin wire is \( \dfrac{1}{10} \) cm or 0.1 cm.
Example 12. The base diameter of a solid in the form of a cone is 6 cm and the height of the cone is 10 cm. It is melted and recast into spherical balls of diameter 1 cm. Find the number of balls thus obtained.
Answer:
Let the number of spherical balls be n. The volume of the cone will be equal to the sum of the volumes of the spherical balls. The radius of the base of the cone = 3 cm and the radius of the sphere = \( \dfrac{1}{2} \) cm.
Volume of the cone = \( \dfrac{1}{3}\pi \times 3^2 \times 10 \) = 30π cm³ and volume of each sphere = \( \dfrac{4}{3}\pi \left( \dfrac{1}{2} \right)^3 = \dfrac{\pi}{6} \) cm³.
Hence, n × \( \dfrac{\pi}{6} \) = 30π, so n = 6 × 30 = 180. Hence, the required number of balls = 180.
Example 13. A conical empty vessel is to be filled up completely by pouring water into it successively with the help of a cylindrical can of diameter 6 cm and height 12 cm. The radius of the conical vessel is 9 cm and its height is 72 cm. How many times will it require to pour water into the conical vessel to fill it completely, if, in each time, the cylindrical can is filled with water completely?
Answer:
Let n be the required number of times. Then the volume of the conical vessel will be equal to n times the volume of the cylindrical can.
Volume of the conical vessel = \( \dfrac{1}{3}\pi \times 9^2 \times 72 \) = 24 × 81π cm³ and the volume of the cylindrical can = π × 3² × 12 = 9 × 12π cm³.
Hence, 24 × 81π = 9 × 12π × n, so n = \( \dfrac{24 \times 81}{9 \times 12} \) = 18. Hence, the required number of times = 18.
Example 14. The height of a right circular cylinder is equal to its diameter. If it is melted and recast into a sphere of radius equal to the radius of the cylinder, find the part of the material that remained unused.
Answer:
Let h be the height of the cylinder. Then its diameter is h and so its radius is \( \dfrac{h}{2} \). Hence, its volume is V₁ = \( \pi \left( \dfrac{h}{2} \right)^2 h = \dfrac{\pi h^3}{4} \).
The radius of the sphere = \( \dfrac{h}{2} \). Hence, the volume of the sphere is V₂ = \( \dfrac{4}{3}\pi \left( \dfrac{h}{2} \right)^3 = \dfrac{\pi h^3}{6} \).
The volume of the unused material = V₁ − V₂ = \( \dfrac{\pi h^3}{4} - \dfrac{\pi h^3}{6} = \dfrac{\pi h^3}{12} = \dfrac{1}{3} \times \dfrac{\pi h^3}{4} = \dfrac{1}{3}V_1 \).
Hence, the volume of the unused material is \( \dfrac{1}{3} \)rd of the volume of the cylinder.
Example 15. Water flows at the rate of 10 m per minute through a cylindrical pipe having its diameter as 5 mm. How much time will it take to fill a conical vessel whose diameter of the base is 40 cm and depth 24 cm?
Answer:
Diameter of the pipe = 5 mm = \( \dfrac{1}{2} \) cm, so the radius of the pipe = \( \dfrac{1}{4} \) cm.
In 1 minute, the length of the water column in the cylindrical pipe = 10 m = 1000 cm. Volume of water that flows out of the pipe in 1 minute = π × \( \dfrac{1}{4} \) × \( \dfrac{1}{4} \) × 1000 cm³.
Also, volume of the cone = \( \dfrac{1}{3} \times \pi \times 20 \times 20 \times 24 \) cm³.
Hence, the time needed to fill up this conical vessel = \( \dfrac{\frac{1}{3} \times 20 \times 20 \times 24}{\frac{1}{16} \times 1000} \) minutes = \( \dfrac{256}{5} \) minutes = 51.2 minutes.
Hence, the required time is 51.2 minutes.
Example 16. A hemispherical tank of radius \( 1\dfrac{3}{4} \) m is full of water. It is connected with a pipe which empties it at the rate of 7 litres per second. How much time will it take to empty the tank completely?
Answer:
Radius of the hemisphere = \( \dfrac{7}{4} \) m = 175 cm. Volume of the hemisphere = \( \dfrac{2}{3}\pi \times 175^3 \) cm³.
The pipe empties the tank at the rate of 7 litres, that is 7000 cm³ of water per second.
Required time = \( \dfrac{2}{3} \times \dfrac{22}{7} \times 175 \times 175 \times 175 \div 7000 \) s = \( \dfrac{1925}{72} \) min = 26.74 min (nearly).
Answer:
Let h be the required height of the embankment. The shape of the embankment will be like the shape of a cylinder of internal radius 1 m and external radius (5 + 1) m = 6 m.
The volume of the embankment is equal to the volume of the earth dug out from the well, that is π × 1² × 14 = 14π m³. Also, the volume of the embankment = π (6² − 1²) h = 35πh m³.
Hence, 35πh = 14π and h = \( \dfrac{14}{35} = \dfrac{2}{5} \) = 0.4.
Hence, the required height of the embankment = 0.4 m.
Example 18. Water in a canal, 30 dm wide and 12 dm deep, is flowing with a speed of 10 km/hr. How much area will it irrigate in 30 minutes, if 8 cm of standing water is required for irrigation?
Answer:
Speed of water in the canal = 10 km/h = 10000 m / 60 min = \( \dfrac{500}{3} \) m/min.
The volume of the water flowing out of the canal in 1 minute = \( \dfrac{500}{3} \times \dfrac{30}{10} \times \dfrac{12}{10} \) m³ = 600 m³. In 30 minutes, the amount of water flowing out of the canal = 600 × 30 = 18000 m³.
If the required area of the irrigated land is x m², then the volume of water needed to irrigate the land = x × \( \dfrac{8}{100} \) = \( \dfrac{2x}{25} \) m³.
Hence, \( \dfrac{2x}{25} \) = 18000 and x = 18000 × \( \dfrac{25}{2} \) = 225000. Hence, the required area is 225000 m².
Check Your Progress
1. The radius of a metal sphere is 9 cm. It is melted and drawn into a wire of diameter 2 m. Find the length of the wire in metres.
2. A metallic sphere of radius 21 cm is melted and then recast into small cones, each of radius 7 cm and height 6 cm. Find the number of cones made.
3. A vessel is in the form of a cone. Its height is 16 cm and the radius of the open top is 10 cm. It is filled with water completely. Some lead shots which have 1 cm radius are dropped into the vessel. One fourth of the water flows out of the vessel. Find the number of shots dropped into the vessel.
4. A well of diameter 6 m is dug 28 m deep. The earth taken out has been spread evenly all around it in the shape of a circular ring of width 5 m to form an embankment. Find the height of the embankment.
5. Two cylindrical tanks are connected by a tube with a tap. The radii of the tanks are 7 m and 7/2 m. The water stands to a height 6 m and 5 m respectively as shown in the figure. Find the height to which the water will stand in both the tanks, if the tap is opened.
6. A rectangular swimming bath 60 m long and 40 m broad can be filled by a supply pipe in 5 days, but if 6000 cubic metres of water is put in first, the rest can be filled in 3 days and 18 hours. Find the depth of the bath.
Answers
1. 972 × 10⁻⁶ metres
2. 126
3. 100
4. 4.58 m
5. 5.8 m
6. 10 m
Surface Area and Volume of a Combination of Solids
Consider a solid which can be divided into two hemispheres and a cylinder. The surface of the solid is the sum of the curved surface areas of the two hemispheres and the curved surface area of the cylinder. The volume of the solid is the sum of the volumes of the two hemispheres and the cylinder.
Similarly a playing top (lattu) is divided into a hemisphere and a cone. Surface area of the playing top = curved surface area of hemisphere + curved surface area of cone. Volume of the playing top = volume of hemisphere + volume of the cone.
The following solved examples will illustrate the method of finding the total surface area and volume of a combination of solids.
Solved Examples
Answer:
Radius of cone, r₁ = 5 cm. Radius of hemisphere, r = 21 cm. Height of cone, h = 21 − 9 = 12 cm.
Slant height of cone, ℓ = \( \sqrt{r_1^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{169} \) = 13 cm.
Surface area of the solid = T.S.A. of hemisphere + C.S.A. of cone − area of base of cone
= 3πr² + πr₁ℓ − πr₁² = 3 × \( \dfrac{22}{7} \) × 21 × 21 + \( \dfrac{22}{7} \) × 5 × 13 − \( \dfrac{22}{7} \) × 5 × 5
= \( \dfrac{22}{7} \) (1323 + 65 − 25) = \( \dfrac{22}{7} \) × 1363 = 4283.7 cm²
Answer:
Let r be the common radius of the dome and cylinder and h be the height of the building. Then base diameter of the dome = 2r m. According to the question, 2r = \( \dfrac{2}{3} \)h, so r = \( \dfrac{h}{3} \) m.
Height of the cylinder, h₁ = h − r.
Volume of the solid = volume of cylinder + volume of hemisphere. Since the volume of air = volume of solid = \( 67\dfrac{1}{21} = \dfrac{1408}{21} \) m³:
\( \dfrac{1408}{21} = \pi r^2 (h - r) + \dfrac{2}{3}\pi r^3 = \pi \dfrac{h^2}{9}\cdot\dfrac{2h}{3} + \dfrac{2}{3}\pi \dfrac{h^3}{27} = \dfrac{\pi h^3}{27}\left( 2 + \dfrac{2}{3} \right) = \dfrac{\pi h^3}{27} \times \dfrac{8}{3} \)
h³ = \( \dfrac{1408 \times 27 \times 3 \times 7}{21 \times 8 \times 22} \) = 27 × 8, so h = 6. Hence, the height of the building = 6 m.
Answer:
Radius of bigger cylinder, r = 10 cm; radius of smaller cylinder, r₁ = 5 cm. Height of larger cylinder, h = 40 cm; height of each smaller cylinder, h₁ = 10 cm.
T.S.A. of the roller pin = T.S.A. of bigger cylinder + 2 × C.S.A. of smaller cylinder − 2 × area of base of smaller cylinder + 2 × area of top of smaller cylinder = 2πr(h + r) + 2 × 2πr₁h₁ − 2 × πr₁² + 2 × πr₁² = 2πr (h + r) + 4πr₁h₁.
= 2 × \( \dfrac{22}{7} \) × 10 × (40 + 10) + 4 × \( \dfrac{22}{7} \) × 5 × 10 = 2 × \( \dfrac{22}{7} \) × (500 + 100) = 3771.43 cm².
Cost of painting = 0.14 × 3771.43 = Rs. 528.
Answer:
Radius of hemisphere and cone, r = 4 cm. Height of cone, h = 4 cm.
Volume of the toy = volume of hemisphere + volume of cone = \( \dfrac{2}{3}\pi r^3 + \dfrac{1}{3}\pi r^2 h = \dfrac{\pi}{3} r^2 (2r + h) = \dfrac{22}{7} \times \dfrac{1}{3} \times 4 \times 4 (2 \times 4 + 4) \) = \( \dfrac{22}{21} \times 16 \times 12 \) = 201.14 cm³.
The cube circumscribes the toy, so the edge of the cube = diameter of hemisphere = 8 cm. Volume of cube = 8³ = 512 cm³. Difference of the volumes of the cube and the toy = 512 − 201.14 = 310.86 cm³.
Slant height of cone, ℓ = \( \sqrt{r^2 + h^2} = \sqrt{4^2 + 4^2} = 4\sqrt{2} \) cm.
Surface area of the toy = C.S.A. of cone + C.S.A. of hemisphere = πrℓ + 2πr² = πr (ℓ + 2r) = \( \dfrac{22}{7} \times 4 \times (4\sqrt{2} + 2 \times 4) = \dfrac{88}{7} \times (5.656 + 8) = \dfrac{88}{7} \times 13.656 \) = 171.67 cm².
Check Your Progress
1. A biggest sphere is made from a cube of side 14 cm. Find the volume of the solid in cubic centimetre.
2. A solid 21 cm long, 20 cm broad and 10 cm high has a hemisphere fixed on the top. The radius of the hemisphere is 7 cm. Find the total surface area of the combined object.
3. A top is of the form of a hemisphere mounted on a cone. The diameter of the conical portion is 12 cm and height 8 cm. Find the surface area and volume of the top. (Take π = 3.14)
4. 6 cylindrical milk powder tins are packed in a rectangular box, 3 on the longer side and 2 on the breadth side. Find the volume of the empty space if the radius of the tin is 7 cm and height is 10 cm.
5. A cylinder has a hemisphere on one end and a cone on the other end with the same diameter. The height of the cylindrical part is 10 cm. Common diameter is 8 cm. The height of the conical part is 3 cm. Find the total surface area of the solid formed.
6. A circus tent is cylindrical to a height of 8 m surmounted by a conical part. If the total height is 13 m and the diameter of the tent is 24 m, calculate (i) the total surface area of the tent, (ii) the volume of the air contained in it.
7. A cone is surmounted on a cylinder of diameter 10 cm. Total height is 19.5 cm and height of the cylindrical part is 7.5 cm. Find the curved surface area.
8. A cylinder has diameter 7 cm and 8 cm vertical height. From the cylinder a cone of height 3 cm with the same diameter of the cylinder is carved out. Find the volume of the remaining solid.
Answers
1. 1437.33 cubic cm
2. 1814 cm²
3. 414.48 sq. cm, 753.6 cubic cm
4. 2520 cc
5. 415 cm² (approx)
6. (i) \( 1093\dfrac{5}{7} \) m², (ii) \( 4374\dfrac{6}{7} \) cubic metre
7. 440 sq. cm
8. 269.5 cubic cm
Free study material for Mathematics
Download CBSE Revision Notes: Class 10 Mathematics Chapter 12 Surface Areas and Volumes
Download Chapter Summary: Class 10 Mathematics Chapter 12 Surface Areas and Volumes
Access structured revision notes for Chapter 12 Surface Areas and Volumes designed in alignment with the latest CBSE curriculum for Class 10 Mathematics. These summaries help clarify core themes and support effective daily study.
Concept Clarification for Chapter 12 Surface Areas and Volumes
Built using the official NCERT book for Class 10 Mathematics, these notes provide reliable academic guidance. Pair your summary review with our recommended NCERT solutions to master optimal problem-solving approaches.
Additional Study Resources for Class 10 Mathematics
Follow up your revision by attempting the interactive online Mathematics MCQ test and practice worksheets for this chapter to verify your mastery. All platform resources are free to access.
FAQs
You can download the teacher prepared revision notes for CBSE Class 10 Mathematics Surface Areas and Volumes Notes Set 08 from StudiesToday.com. These notes are designed as per 2026-27 academic session to help Class 10 students get the best study material for Mathematics.
Yes, our CBSE Class 10 Mathematics Surface Areas and Volumes Notes Set 08 include 50% competency-based questions with focus on core logic, keyword definitions, and the practical application of Mathematics principles which is important for getting more marks in 2026 CBSE exams.
Yes, our CBSE Class 10 Mathematics Surface Areas and Volumes Notes Set 08 provide a detailed, topic wise breakdown of the chapter. Fundamental definitions, complex numerical formulas and all topics of CBSE syllabus in Class 10 is covered.
These notes for Mathematics are organized into bullet points and easy-to-read charts. By using CBSE Class 10 Mathematics Surface Areas and Volumes Notes Set 08, Class 10 students fast revise formulas, key definitions before the exams.
No, all study resources on StudiesToday, including CBSE Class 10 Mathematics Surface Areas and Volumes Notes Set 08, are available for immediate free download. Class 10 Mathematics study material is available in PDF and can be downloaded on mobile.