Surface area is the total area covered by the outer boundary of a three-dimensional solid.
Volume is the amount of space enclosed inside a solid, measured in cubic units.
Real objects are rarely perfect cubes or simple cylinders. A water tanker is a cylinder with two hemispherical ends. A toy rocket is a cone mounted on a cylinder. A medicine capsule is a cylinder capped with two hemispheres. Chapter 12 teaches one principle for all these shapes: break any combined solid into basic components, calculate each part separately, then add or subtract as needed. CBSE 2026 tests Important Questions Class 10 Maths Chapter 12 through surface area and volume of combinations only, based on the NCERT Reprint 2026-27.
Key Takeaways
- Surface Areas And Volumes: Chapter 12 focuses on surface area and volume of combinations of solids.
- Formula Choice: Use surface area formulas for visible outer faces and volume formulas for enclosed space.
- Exam Trap: Remove the common face from surface area when two solids are joined.
- CBSE 2026 Update: Frustum of Cone is not included in the current NCERT Reprint 2026-27 scope.
Important Questions Class 10 Maths Chapter 12 Structure 2026
| Concept |
Formula |
Key Variables |
| Cylinder |
CSA = 2πrh, Volume = πr²h |
r, h |
| Cone |
CSA = πrl, Volume = (1/3)πr²h |
r, h, l |
| Hemisphere |
CSA = 2πr², Volume = (2/3)πr³ |
r |
Important Questions Class 10 Maths Chapter 12: Formulas You Need First
Every question in this chapter uses at least one mensuration formula. Write the formula before substituting values because CBSE marking schemes award steps for it.
Q1. What Are The Cylinder Formulas In Surface Areas And Volumes?
A cylinder has two circular bases and one curved surface. Use CSA = 2πrh, TSA = 2πr(r + h), and Volume = πr²h.
- Given Data:
Radius = r
Height = h
- Formula Used:
CSA = 2πrh
TSA = 2πr(r + h)
Volume = πr²h
- Final Result:
Cylinder formulas depend on radius and height
Q2. What Are The Cone Formulas In Surface Areas And Volumes?
A cone has one circular base and one curved surface. Use CSA = πrl, TSA = πr(r + l), and Volume = (1/3)πr²h.
- Given Data:
Radius = r
Height = h
Slant height = l
- Formula Used:
l = √(r² + h²)
CSA = πrl
TSA = πr(r + l)
Volume = (1/3)πr²h
- Final Result:
Cone formulas use radius, height, and slant height
Q3. What Are The Sphere And Hemisphere Formulas In Class 10?
A sphere has a completely curved surface. A hemisphere is half of a sphere.
- Given Data:
Radius = r
- Formula Used:
Sphere surface area = 4πr²
Sphere volume = (4/3)πr³
Hemisphere CSA = 2πr²
Hemisphere TSA = 3πr²
Hemisphere volume = (2/3)πr³
- Final Result:
Hemisphere formulas come from half of sphere formulas
Q4. What Happens To Common Faces In Combination Of Solids?
The common face does not count in surface area. The full volumes of both solids still add.
- Surface Area Rule:
Remove joined or hidden faces.
- Volume Rule:
Add complete volumes of all components.
- Final Result:
Common faces disappear only from surface area calculations
Class 10 Maths Chapter 12 MCQ With Answers
These class 10 maths chapter 12 MCQ questions test formula recall and shape identification. Each question follows the CBSE 2026 board pattern.
Q1. If A Sphere And Cone Have Equal Radii And Equal Volumes, Find The Cone Height.
The height of the cone is 4r. Equate the volume of the sphere with the volume of the cone.
- Given Data:
Radius of sphere = r
Radius of cone = r
Volumes are equal
- Formula Used:
Volume of sphere = (4/3)πr³
Volume of cone = (1/3)πr²h
- Calculation:
(4/3)πr³ = (1/3)πr²h
4r = h
- Final Result:
Answer: (D) 4r
Q2. If Volumes Of Two Cubes Are In Ratio 8 : 125, Find The Ratio Of Their Surface Areas.
The ratio of their surface areas is 4 : 25. Cube surface area varies with the square of its edge.
- Given Data:
Volume ratio = 8 : 125
- Formula Used:
Volume ratio = a³ : b³
Surface area ratio = a² : b²
- Calculation:
a³ : b³ = 8 : 125
a : b = 2 : 5
a² : b² = 4 : 25
- Final Result:
Answer: (B) 4 : 25
Q3. A Medicine Capsule Has Diameter 5 mm And Total Length 14 mm. Find Its Surface Area.
The surface area of the capsule is 220 mm². Treat it as a cylinder with two hemispherical ends.
- Given Data:
Diameter = 5 mm
Radius = 2.5 mm
Total length = 14 mm
Cylinder height = 14 - 5 = 9 mm
- Formula Used:
TSA = CSA of cylinder + 2 × CSA of hemisphere
TSA = 2πrh + 4πr²
- Calculation:
TSA = 2πr(h + 2r)
TSA = 2 × (22/7) × 2.5 × (9 + 5)
TSA = 220 mm²
- Final Result:
Answer: (A) 220 mm²
Q4. A Solid Hemisphere And Cone Have Equal Radii. Find The Total Surface Area Of The Combined Solid.
The total surface area is πr(2r + l). The common circular base does not count.
- Given Data:
Radius = r
Slant height of cone = l
- Formula Used:
TSA = CSA of cone + CSA of hemisphere
- Calculation:
TSA = πrl + 2πr²
TSA = πr(l + 2r)
- Final Result:
Answer: (B) πr(2r + l)
Q5. Volume And Surface Area Of A Solid Hemisphere Are Numerically Equal. Find Its Diameter.
The diameter of the hemisphere is 9 units. Equate volume and total surface area.
- Given Data:
Volume = Surface area
- Formula Used:
Volume of hemisphere = (2/3)πr³
TSA of hemisphere = 3πr²
- Calculation:
(2/3)πr³ = 3πr²
(2/3)r = 3
r = 4.5 units
- Final Result:
Answer: (C) 9 units
Q6. A Cylinder Has Height 2.4 cm And Diameter 1.4 cm. A Conical Cavity Of Same Size Is Hollowed Out. Find TSA.
The total surface area is 17.6 cm², which is closest to 18 cm². Include the curved cylinder, one base, and conical cavity.
- Given Data:
Radius = 0.7 cm
Height = 2.4 cm
- Formula Used:
l = √(r² + h²)
TSA = 2πrh + πr² + πrl
- Calculation:
l = √(0.7² + 2.4²) = 2.5 cm
TSA = πr(2h + r + l)
TSA = (22/7) × 0.7 × (4.8 + 0.7 + 2.5)
TSA = 17.6 cm²
- Final Result:
Answer: (B) 18 cm²
Q7. Two Cubes Each Of Volume 64 cm³ Are Joined End To End. Find Surface Area Of The Cuboid.
The surface area of the resulting cuboid is 160 cm². Two joined faces disappear.
- Given Data:
Volume of each cube = 64 cm³
Edge of each cube = 4 cm
- Formula Used:
TSA of cuboid = 2(lb + bh + hl)
- Calculation:
New dimensions = 8 cm, 4 cm, 4 cm
TSA = 2(8 × 4 + 4 × 4 + 4 × 8)
TSA = 2(32 + 16 + 32)
TSA = 160 cm²
- Final Result:
Answer: (B) 160 cm²
Q8. A Wooden Article Has A Cylinder With A Hemisphere Scooped From Each End. Find The Surface Area.
The total surface area is 374 cm². Add the cylinder’s curved surface and two hemispherical curved surfaces.
- Given Data:
Cylinder height = 10 cm
Radius = 3.5 cm
- Formula Used:
TSA = CSA of cylinder + 2 × CSA of hemisphere
- Calculation:
TSA = 2πrh + 4πr²
TSA = 2πr(h + 2r)
TSA = 2 × (22/7) × 3.5 × (10 + 7)
TSA = 374 cm²
- Final Result:
Answer: (A) 374 cm²
Surface Area Of Combination Of Solids Class 10 Questions With Solutions
The surface area of a combined solid includes only visible outer surfaces. Joined or hidden faces do not count in the final surface area.
Q1. Rasheed’s Birthday Top Is A Cone Surmounted By A Hemisphere. Find The Area To Be Coloured.
The area to be coloured is 39.6 cm² approximately. Add only the curved surface of the cone and hemisphere.
- Given Data:
Total height = 5 cm
Diameter = 3.5 cm
Radius = 1.75 cm
Cone height = 5 - 1.75 = 3.25 cm
- Formula Used:
l = √(r² + h²)
Surface area = CSA of hemisphere + CSA of cone
- Calculation:
l = √(1.75² + 3.25²)
l = √(3.0625 + 10.5625)
l = √13.625 = 3.7 cm approximately
- Surface Area:
CSA of hemisphere = 2πr²
CSA of hemisphere = 2 × (22/7) × 1.75 × 1.75 = 19.25 cm²
CSA of cone = πrl
CSA of cone = (22/7) × 1.75 × 3.7 = 20.35 cm²
- Final Result:
Area to be coloured = 39.6 cm² approximately
Q2. A Cube Of Edge 5 cm Has A Hemisphere Of Diameter 4.2 cm Fixed On Top. Find TSA.
The total surface area is 163.86 cm². Remove the covered circular base and add the curved hemisphere.
- Given Data:
Cube edge = 5 cm
Hemisphere diameter = 4.2 cm
Radius = 2.1 cm
- Formula Used:
TSA = TSA of cube - base area of hemisphere + CSA of hemisphere
- Calculation:
TSA = 6 × 5² - πr² + 2πr²
TSA = 150 + πr²
TSA = 150 + (22/7) × 2.1 × 2.1
TSA = 163.86 cm²
- Final Result:
Total surface area = 163.86 cm²
Q3. A Vessel Is A Hollow Hemisphere Mounted On A Hollow Cylinder. Find Inner Surface Area.
The inner surface area is 572 cm². Add the inner curved surface of cylinder and hemisphere.
- Given Data:
Diameter = 14 cm
Radius = 7 cm
Total height = 13 cm
Cylinder height = 13 - 7 = 6 cm
- Formula Used:
Inner surface area = CSA of cylinder + CSA of hemisphere
- Calculation:
Inner surface area = 2πrh + 2πr²
Inner surface area = 2πr(h + r)
Inner surface area = 2 × (22/7) × 7 × (6 + 7)
Inner surface area = 44 × 13
- Final Result:
Inner surface area = 572 cm²
Q4. A Toy Is A Cone Mounted On A Hemisphere. Find Total Surface Area.
The total surface area is 214.5 cm². Add only the curved surfaces of the cone and hemisphere.
- Given Data:
Radius = 3.5 cm
Total height = 15.5 cm
Cone height = 15.5 - 3.5 = 12 cm
- Formula Used:
l = √(r² + h²)
TSA = CSA of hemisphere + CSA of cone
- Calculation:
l = √(3.5² + 12²)
l = √(12.25 + 144)
l = √156.25 = 12.5 cm
- Surface Area:
TSA = 2πr² + πrl
TSA = πr(2r + l)
TSA = (22/7) × 3.5 × (7 + 12.5)
TSA = 214.5 cm²
- Final Result:
Total surface area = 214.5 cm²
Q5. A Cubical Block Of Side 7 cm Has A Hemisphere On Top. Find Surface Area.
The total surface area is 332.5 cm². The greatest hemisphere diameter equals the cube edge.
- Given Data:
Cube side = 7 cm
Greatest hemisphere diameter = 7 cm
Radius = 3.5 cm
- Formula Used:
Surface area = TSA of cube - base area of hemisphere + CSA of hemisphere
- Calculation:
Surface area = 6 × 7² - πr² + 2πr²
Surface area = 294 + πr²
Surface area = 294 + (22/7) × 3.5 × 3.5
Surface area = 332.5 cm²
- Final Result:
Surface area = 332.5 cm²
Q6. A Medicine Capsule Has Total Length 14 mm And Diameter 5 mm. Find Surface Area.
The capsule surface area is 220 mm². Treat the capsule as a cylinder with two hemispheres.
- Given Data:
Radius = 2.5 mm
Total length = 14 mm
Cylinder height = 14 - 5 = 9 mm
- Formula Used:
TSA = CSA of cylinder + 2 × CSA of hemisphere
- Calculation:
TSA = 2πrh + 4πr²
TSA = 2πr(h + 2r)
TSA = 2 × (22/7) × 2.5 × (9 + 5)
TSA = 220 mm²
- Final Result:
Surface area = 220 mm²
Q7. A Tent Is A Cylinder Surmounted By A Cone. Find Canvas Area And Cost.
The canvas area is 44 m² and the cost is ₹22,000. Add only the curved surface areas.
- Given Data:
Cylindrical height = 2.1 m
Diameter = 4 m
Radius = 2 m
Slant height of cone = 2.8 m
Cost = ₹500 per m²
- Formula Used:
Canvas area = CSA of cylinder + CSA of cone
- Calculation:
Canvas area = 2πrh + πrl
Canvas area = πr(2h + l)
Canvas area = (22/7) × 2 × (4.2 + 2.8)
Canvas area = 44 m²
- Cost:
Cost = 44 × 500
Cost = ₹22,000
- Final Result:
Canvas area = 44 m² and cost = ₹22,000
Q8. A Bird-Bath Is A Cylinder With A Hemispherical Depression. Find Total Surface Area.
The total surface area is 3.3 m². Add the curved surface of cylinder and hemispherical depression.
- Given Data:
Radius = 30 cm
Cylinder height = 145 cm
- Formula Used:
TSA = CSA of cylinder + CSA of hemisphere
- Calculation:
TSA = 2πrh + 2πr²
TSA = 2πr(h + r)
TSA = 2 × (22/7) × 30 × (145 + 30)
TSA = 33000 cm²
- Conversion:
33000 cm² = 3.3 m²
- Final Result:
Total surface area = 3.3 m²
Volume Of Combination Of Solids Class 10 Questions With Solutions
In volume problems, all component volumes are added in full. No volume disappears where two solids join.
Q1. A Toy Is A Hemisphere Surmounted By A Cone. Find Its Volume And Cylinder Difference.
The toy volume is 25.12 cm³, and the difference from the circumscribing cylinder is 25.12 cm³.
- Given Data:
Radius = 2 cm
Cone height = 2 cm
Cylinder height = 4 cm
- Formula Used:
Volume of toy = Volume of hemisphere + Volume of cone
- Calculation:
Volume of toy = (2/3)πr³ + (1/3)πr²h
Volume of toy = (1/3)πr²(2r + h)
Volume of toy = (1/3) × 3.14 × 4 × (4 + 2)
Volume of toy = 25.12 cm³
- Cylinder Volume:
Volume of cylinder = πr²h
Volume of cylinder = 3.14 × 4 × 4
Volume of cylinder = 50.24 cm³
- Final Result:
Difference = 25.12 cm³
Q2. A Solid Has A Cone Standing On A Hemisphere. Both Radii Are 1 cm. Cone Height Is 1 cm. Find Volume.
The volume is π cm³. Add the volume of hemisphere and cone.
- Given Data:
Radius = 1 cm
Cone height = 1 cm
- Formula Used:
Volume = (2/3)πr³ + (1/3)πr²h
- Calculation:
Volume = (2/3)π(1)³ + (1/3)π(1)²(1)
Volume = (2/3)π + (1/3)π
Volume = π cm³
- Final Result:
Volume = π cm³
Q3. A Model Is A Cylinder With Two Cones At Its Ends. Find Volume Of Air.
The volume of air is 66 cm³. Add the cylinder volume and two cone volumes.
- Given Data:
Diameter = 3 cm
Radius = 1.5 cm
Total length = 12 cm
Each cone height = 2 cm
Cylinder height = 12 - 4 = 8 cm
- Formula Used:
Volume = πr²h + 2 × (1/3)πr²h cone
- Calculation:
Volume = πr²[h + (2/3)h cone]
Volume = π × 2.25 × [8 + (4/3)]
Volume = π × 2.25 × (28/3)
Volume = (22/7) × 2.25 × (28/3)
- Final Result:
Volume of air = 66 cm³
Q4. A Gulab Jamun Is A Cylinder With Two Hemispherical Ends. Find Total Syrup In 45 Pieces.
The total syrup is 338.2 cm³ approximately. Syrup equals 30% of the total gulab jamun volume.
- Given Data:
Length = 5 cm
Diameter = 2.8 cm
Radius = 1.4 cm
Cylinder height = 5 - 2.8 = 2.2 cm
- Formula Used:
Volume of one piece = πr²h + (4/3)πr³
- Calculation:
Volume = π × 1.96 × 2.2 + (4/3)π × 2.744
Volume = 4.312π + 3.659π
Volume = 7.971π
Volume = 25.05 cm³ approximately
- Syrup Calculation:
Volume of 45 pieces = 45 × 25.05 = 1127.25 cm³
Syrup = 30% of 1127.25
Syrup = 338.2 cm³ approximately
- Final Result:
Total syrup = 338.2 cm³ approximately
Q5. A Juice Glass Has A Hemispherical Raised Portion At Its Base. Find Apparent And Actual Capacity.
The apparent capacity is 196.25 cm³, and the actual capacity is 163.54 cm³.
- Given Data:
Inner diameter = 5 cm
Radius = 2.5 cm
Height = 10 cm
- Formula Used:
Apparent capacity = Volume of cylinder
Actual capacity = Cylinder volume - Hemisphere volume
- Calculation:
Apparent capacity = πr²h
Apparent capacity = 3.14 × 6.25 × 10
Apparent capacity = 196.25 cm³
- Raised Portion:
Volume of hemisphere = (2/3)πr³
Volume = (2/3) × 3.14 × 15.625
Volume = 32.71 cm³
- Final Result:
Actual capacity = 163.54 cm³
Q6. A Shed Is A Cuboid Surmounted By A Half Cylinder. Find Air Remaining.
The air remaining is 827.15 m³. Subtract the space occupied by machinery and workers.
- Given Data:
Cuboid base = 7 m × 15 m
Cuboid height = 8 m
Half-cylinder radius = 3.5 m
Half-cylinder length = 15 m
- Formula Used:
Total volume = Cuboid volume + Half-cylinder volume
- Calculation:
Cuboid volume = 15 × 7 × 8 = 840 m³
Half-cylinder volume = (1/2)πr²l
Half-cylinder volume = (1/2) × (22/7) × 12.25 × 15
Half-cylinder volume = 288.75 m³
- Air Remaining:
Total volume = 1128.75 m³
Space used = 300 + (20 × 0.08) = 301.6 m³
Air remaining = 1128.75 - 301.6
- Final Result:
Air remaining = 827.15 m³
Surface Areas And Volumes Class 10 Extra Questions With Solutions
These surface areas and volumes class 10 extra questions cover conversion of solids, scooped solids, and hollowed solids. All questions follow the NCERT 2026-27 scope.
Q1. A Metallic Sphere Of Radius 10.5 cm Is Recast Into Cones Of Radius 3.5 cm And Height 3 cm. How Many Cones Form?
The number of cones formed is 126. Use volume conservation because metal volume remains unchanged.
- Given Data:
Sphere radius = 10.5 cm
Cone radius = 3.5 cm
Cone height = 3 cm
- Formula Used:
Number of cones = Volume of sphere / Volume of one cone
- Calculation:
Volume of sphere = (4/3)πr³
Volume of sphere = (4/3) × (22/7) × 10.5³
Volume of sphere = 4851 cm³
- Cone Volume:
Volume of one cone = (1/3)πr²h
Volume = (1/3) × (22/7) × 3.5² × 3
Volume = 38.5 cm³
- Final Result:
Number of cones = 4851/38.5 = 126
Q2. A Solid Iron Pole Has Two Cylinders. Find Its Mass If 1 cm³ Iron Has Mass 8 g.
The mass of the pole is 892.26 kg approximately. Add both cylinder volumes first.
- Given Data:
Lower cylinder height = 220 cm
Lower cylinder radius = 12 cm
Upper cylinder height = 60 cm
Upper cylinder radius = 8 cm
Mass of 1 cm³ iron = 8 g
- Formula Used:
Volume of cylinder = πr²h
- Calculation:
Lower cylinder volume = π × 12² × 220 = 31680π cm³
Upper cylinder volume = π × 8² × 60 = 3840π cm³
Total volume = 35520π
Total volume = 111532.8 cm³
- Mass:
Mass = 111532.8 × 8
Mass = 892262.4 g
Mass = 892.26 kg approximately
- Final Result:
Mass = 892.26 kg approximately
Q3. A Cylinder Has A Conical Cavity Of Same Height And Diameter. Find Total Surface Area.
The total surface area is 17.6 cm². Count the outer curved surface, top base, and inner conical surface.
- Given Data:
Height = 2.4 cm
Diameter = 1.4 cm
Radius = 0.7 cm
- Formula Used:
l = √(r² + h²)
TSA = 2πrh + πr² + πrl
- Calculation:
l = √(0.7² + 2.4²)
l = √(0.49 + 5.76)
l = 2.5 cm
- Surface Area:
TSA = πr(2h + r + l)
TSA = (22/7) × 0.7 × (4.8 + 0.7 + 2.5)
TSA = 17.6 cm²
- Final Result:
Total surface area = 17.6 cm²
Q4. A Hemispherical Depression Is Cut From A Cube. Find Surface Area Of The Remaining Solid.
The remaining surface area is (l²/4)(24 + π). The flat circular base gets removed and the curved depression gets added.
- Given Data:
Edge of cube = l
Diameter of hemisphere = l
Radius = l/2
- Formula Used:
Surface area = TSA of cube - base area of hemisphere + CSA of hemisphere
- Calculation:
Surface area = 6l² - πr² + 2πr²
Surface area = 6l² + πr²
Surface area = 6l² + π(l/2)²
- Final Result:
Surface area = (l²/4)(24 + π)
Q5. A Wooden Article Has A Hemisphere Scooped From Each End Of A Cylinder. Find TSA.
The total surface area is 374 cm². Add the cylinder’s curved surface and two hemispherical depressions.
- Given Data:
Cylinder height = 10 cm
Radius = 3.5 cm
- Formula Used:
TSA = CSA of cylinder + 2 × CSA of hemisphere
- Calculation:
TSA = 2πrh + 4πr²
TSA = 2πr(h + 2r)
TSA = 2 × (22/7) × 3.5 × (10 + 7)
TSA = 374 cm²
- Final Result:
Total surface area = 374 cm²
Surface Area And Volume Class 10 Short Answer Questions
These surface area and volume class 10 short answer questions follow 2-mark and 3-mark CBSE 2026 formats. They test formula choice and visible surface logic.
Q1. What Is The Difference Between CSA And TSA For A Cylinder?
CSA covers only the curved surface, while TSA covers the curved surface and both circular bases. Cylinder questions often ask students to choose between these formulas.
- Given Data:
Radius = r
Height = h
- Formula Used:
CSA = 2πrh
TSA = 2πr(r + h)
- Final Result:
CSA excludes bases, while TSA includes bases
Q2. A Sphere Of Radius 5 cm Is Melted Into Cones Of Height 5 cm And Radius 2 cm. Find Number Of Cones.
The number of cones is 25. Use volume conservation.
- Given Data:
Sphere radius = 5 cm
Cone radius = 2 cm
Cone height = 5 cm
- Formula Used:
Number of cones = Volume of sphere / Volume of one cone
- Calculation:
Volume of sphere = (4/3)π × 5³ = (500/3)π
Volume of one cone = (1/3)π × 2² × 5 = (20/3)π
Number = (500/3)π ÷ (20/3)π
- Final Result:
25 cones
Q3. A Pen Stand Has Four Conical Depressions. Find Volume Of Wood.
The volume of wood is 523.53 cm³ approximately. Subtract the volume of four conical depressions from the cuboid.
- Given Data:
Cuboid dimensions = 15 cm × 10 cm × 3.5 cm
Radius of each cone = 0.5 cm
Depth of each cone = 1.4 cm
- Formula Used:
Volume of wood = Volume of cuboid - Volume of four cones
- Calculation:
Volume of cuboid = 15 × 10 × 3.5 = 525 cm³
Volume of one cone = (1/3) × (22/7) × 0.5² × 1.4
Volume of one cone = 0.3667 cm³
Volume of four cones = 1.4668 cm³
- Final Result:
Volume of wood = 523.53 cm³ approximately
Q4. Lead Shots Are Dropped Into A Cone And One-Fourth Water Flows Out. Find Number Of Shots.
The number of lead shots is 100. The water that flows out equals the total volume of the lead shots.
- Given Data:
Cone height = 8 cm
Cone radius = 5 cm
Lead shot radius = 0.5 cm
- Formula Used:
Volume displaced = One-fourth volume of cone
- Calculation:
Volume of cone = (1/3)π × 5² × 8
Volume of cone = (200/3)π
Volume flowed out = (1/4) × (200/3)π
Volume flowed out = (50/3)π
- Lead Shot Volume:
Volume of one shot = (4/3)π × 0.5³
Volume of one shot = π/6
Number = (50/3)π ÷ (π/6)
- Final Result:
100 lead shots
Surface Area Volume Class 10 Previous Year Questions
These surface area volume class 10 previous year questions follow recurring CBSE-style patterns. They focus on numerical equality, ratios, and conversion of solids.
Q1. Volume And Surface Area Of A Solid Hemisphere Are Numerically Equal. Find Diameter.
The diameter is 9 units. Equate hemisphere volume with hemisphere total surface area.
- Given Data:
Volume = Surface area
- Formula Used:
Volume of hemisphere = (2/3)πr³
TSA of hemisphere = 3πr²
- Calculation:
(2/3)πr³ = 3πr²
(2/3)r = 3
r = 4.5 units
- Final Result:
Diameter = 9 units
Q2. Cylinder And Cone Radii Are In Ratio 3 : 4. Heights Are In Ratio 2 : 3. Find Volume Ratio.
The volume ratio is 9 : 8. Substitute proportional values into both volume formulas.
- Given Data:
Cylinder radius = 3k
Cone radius = 4k
Cylinder height = 2m
Cone height = 3m
- Formula Used:
Cylinder volume = πr²h
Cone volume = (1/3)πr²h
- Calculation:
Cylinder volume = π(3k)²(2m) = 18πk²m
Cone volume = (1/3)π(4k)²(3m) = 16πk²m
- Final Result:
Ratio of volumes = 9 : 8
Q3. A Sphere Of Radius 6 cm Is Recast Into Smaller Spheres Of Radius 2 cm. Find Number Of Spheres.
The number of smaller spheres is 27. Volumes of similar spheres vary as cubes of radii.
- Given Data:
Radius of large sphere = 6 cm
Radius of each small sphere = 2 cm
- Formula Used:
Number of spheres = R³/r³
- Calculation:
Number = 6³/2³
Number = 216/8
Number = 27
- Final Result:
27 smaller spheres
Class 10 Maths Chapter 12 Questions And Answers: Most Repeated Variations
These class 10 maths chapter 12 questions and answers cover recurring CBSE 2026 patterns. Students should practise each variation with formula-first steps.
Q1. What Are The Most Repeated Variations In Surface Areas And Volumes Important Questions?
The most repeated variations include medicine capsule, toy cone-hemisphere, cubical block with hemisphere, melted solids, and hollowed cone or hemisphere.
- Surface Area Pattern:
Add visible curved surfaces.
Remove common faces.
- Volume Pattern:
Add full component volumes.
Subtract hollowed-out volumes.
- Conversion Pattern:
Equate original volume with new total volume.
- Final Result:
These patterns cover most Chapter 12 board-style questions
Q2. How Should Students Solve Important Questions Class 10 Maths Chapter 12 With Solutions?
Students should identify the solid parts first, then choose surface area or volume. Most errors happen when students include hidden faces.
- Identify The Shape:
Cylinder, cone, hemisphere, sphere, cube, or cuboid.
- Choose The Requirement:
Surface area or volume.
- Apply The Rule:
Remove common faces for surface area.
Add full volumes for volume.
- Final Result:
Shape identification decides the solution path
Useful Important Questions Class 10 Maths Links