CBSE Class 10 Maths Surface Area and Volume Important Questions 2026
Mastering CBSE Class 10 Mathematics requires a thorough grasp of spatial mensuration, and preparing CBSE Class 10 Maths Surface Area and Volume Important Questions 2026 is essential for scoring a perfect 100 in your upcoming board examination. In the Class 10 CBSE Mathematics curriculum, the chapter on Surface Areas and Volumes carries significant weightage (typically 6 to 8 marks across 2-mark, 3-mark, 5-mark, or 4-mark case study questions). Unlike lower grades where you calculated metrics for standalone figures, the Class 10 syllabus focuses on combinations of solids (such as a cone surmounted on a hemisphere or a cylinder with hemispherical ends) and conversion of shapes. This comprehensive guide covers the core formulas, high-yield NCERT and CBSE previous year questions with step-by-step solutions, and exam strategies tailored for the 2026 board exam 10 candidates.
Key Concepts and Essential Formulas
To solve combination-of-solids problems accurately, you must understand how individual geometric surfaces interact when joined together. A common error made by students in CBSE Class 10 board exams is simply adding the Total Surface Areas (TSA) of two joined figures. Remember the golden rule: The Total Surface Area of a combined solid is the sum of the exposed Curved Surface Areas (CSA) of its individual components, not the sum of their individual TSAs. Conversely, the total volume of a combined solid is always the algebraic sum of the individual volumes of its parts.
Formula Quick Reference Table
| Solid Shape | Curved / Lateral Surface Area (CSA/LSA) | Total Surface Area (TSA) | Volume |
|---|---|---|---|
| Cuboid (length l, breadth b, height h) | 2h(l + b) | 2(lb + bh + hl) | l × b × h |
| Cube (side a) | 4a2 | 6a2 | a3 |
| Right Circular Cylinder (radius r, height h) | 2πrh | 2πr(r + h) | πr2h |
| Right Circular Cone (radius r, height h, slant height l = √(r2 + h2)) | πrl | πr(l + r) | (1/3)πr2h |
| Sphere (radius r) | 4πr2 | 4πr2 | (4/3)πr3 |
| Hemisphere (radius r) | 2πr2 | 3πr2 | (2/3)πr3 |
Important CBSE Questions with Answers
Below are selected high-probability questions structured according to the latest CBSE Class 10 Mathematics examination pattern, covering 1-mark MCQs, 2-mark Short Answer, 3-mark Short Answer, 5-mark Long Answer, and 4-mark Case Study questions.
Section A: 1-Mark Multiple Choice Questions
Question 1: If the ratio of the surface areas of two spheres is 16 : 9, then the ratio of their volumes is:
(a) 4 : 3
(b) 64 : 27
(c) 16 : 9
(d) 256 : 81
Answer: (b) 64 : 27
Solution:
Let the radii of the two spheres be r1 and r2.
Ratio of Surface Areas = (4πr12) / (4πr22) = 16 / 9 ⇒ (r1 / r2)2 = (4 / 3)2 ⇒ r1 / r2 = 4 / 3.
Ratio of Volumes = [(4/3)πr13] / [(4/3)πr23] = (r1 / r2)3 = (4 / 3)3 = 64 / 27.
Question 2: Two identical solid cubes each of volume 64 cm3 are joined end to end. The total surface area of the resulting cuboid is:
(a) 128 cm2
(b) 160 cm2
(c) 192 cm2
(d) 216 cm2
Answer: (b) 160 cm2
Solution:
Let side of each cube be a.
Volume = a3 = 64 cm3 ⇒ a = 4 cm.
When joined end to end, the dimensions of the resulting cuboid are:
Length (l) = 4 + 4 = 8 cm, Breadth (b) = 4 cm, Height (h) = 4 cm.
TSA of cuboid = 2(lb + bh + hl) = 2(8 × 4 + 4 × 4 + 4 × 8) = 2(32 + 16 + 32) = 2(80) = 160 cm2.
Section B: 2-Mark Short Answer Question
Question 3: A decorative vessel is in the form of a hollow cylinder mounted on a hollow hemisphere. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. (Take π = 22/7)
Solution:
1. Radius of hemisphere (r) = Diameter / 2 = 14 / 2 = 7 cm.
2. Radius of cylinder (r) = 7 cm.
3. Height of cylinder (h) = Total height − Radius of hemisphere = 13 − 7 = 6 cm.
4. Inner surface area of vessel = Inner CSA of cylinder + Inner CSA of hemisphere
= 2πrh + 2πr2 = 2πr(h + r)
= 2 × (22/7) × 7 × (6 + 7)
= 44 × 13 = 572 cm2.
Section C: 3-Mark Short Answer Question
Question 4: A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area and volume of the toy. (Take π = 22/7)
Solution:
1. Dimensions:
Radius (r) = 3.5 cm = 7/2 cm.
Height of the hemispherical part = r = 3.5 cm.
Height of the conical part (h) = Total height − 3.5 cm = 15.5 − 3.5 = 12 cm.
Slant height of the cone (l) = √(r2 + h2) = √((3.5)2 + (12)2) = √(12.25 + 144) = √156.25 = 12.5 cm.
2. Total Surface Area of the Toy:
TSA = CSA of cone + CSA of hemisphere
TSA = πrl + 2πr2 = πr(l + 2r)
TSA = (22/7) × 3.5 × [12.5 + 2(3.5)]
TSA = 11 × (12.5 + 7.0) = 11 × 19.5 = 214.5 cm2.
3. Volume of the Toy:
Total Volume = Volume of cone + Volume of hemisphere
Volume = (1/3)πr2h + (2/3)πr3 = (1/3)πr2(h + 2r)
Volume = (1/3) × (22/7) × (3.5)2 × [12 + 2(3.5)]
Volume = (1/3) × (22/7) × 12.25 × (12 + 7)
Volume = (1/3) × 38.5 × 19 = 731.5 / 3 = 243.83 cm3.
Section D: 5-Mark Long Answer Question
Question 5: From a solid right circular cylinder of height 2.4 cm and base diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find:
(i) The total surface area of the remaining solid to the nearest cm2.
(ii) The volume of the remaining solid. (Take π = 22/7)
Solution:
1. Given parameters:
Diameter = 1.4 cm ⇒ Radius (r) = 0.7 cm.
Height (h) = 2.4 cm.
Slant height of conical cavity (l) = √(r2 + h2) = √((0.7)2 + (2.4)2) = √(0.49 + 5.76) = √6.25 = 2.5 cm.
2. (i) Total Surface Area of Remaining Solid:
When the cone is scooped out, the remaining surface comprises:
- The outer curved surface area of the cylinder
- The area of the top circular base of the cylinder
- The inner curved surface area of the conical cavity
TSA = CSA of cylinder + Area of top base + CSA of cone
TSA = 2πrh + πr2 + πrl = πr(2h + r + l)
TSA = (22/7) × 0.7 × [2(2.4) + 0.7 + 2.5]
TSA = 2.2 × [4.8 + 0.7 + 2.5] = 2.2 × 8.0 = 17.6 cm2 (or approximately 18 cm2 to the nearest cm2).
3. (ii) Volume of Remaining Solid:
Remaining Volume = Volume of cylinder − Volume of conical cavity
Volume = πr2h − (1/3)πr2h = (2/3)πr2h
Volume = (2/3) × (22/7) × (0.7)2 × 2.4
Volume = (2/3) × (22/7) × 0.49 × 2.4
Volume = (2/3) × 1.54 × 2.4 = 2 × 1.54 × 0.8 = 2.464 cm3.
Section E: 4-Mark Case Study Based Question
Question 6 (Case Study): An agricultural cooperative society builds a metal storage silo to store wheat during the harvest season. The silo consists of a cylindrical central tank of diameter 6 m and height 7 m, surmounted by a conical roof of the same diameter and vertical height 4 m.
Based on the above information, answer the following:
- Find the slant height of the conical roof. (1 Mark)
- Calculate the total internal storage capacity (volume) of the silo. (2 Marks)
- Find the total area of the metal sheet required to build the outer surface of the silo (excluding the bottom base on the ground). (1 Mark)
Solution:
1. Slant Height of Conical Roof:
Radius (r) = 6 / 2 = 3 m, Conical height (hcone) = 4 m.
Slant height (l) = √(r2 + hcone2) = √(32 + 42) = √(9 + 16) = √25 = 5 m.
2. Total Internal Volume:
Height of cylinder (hcyl) = 7 m.
Total Volume = Volume of cylinder + Volume of cone
Total Volume = πr2hcyl + (1/3)πr2hcone = πr2[hcyl + (1/3)hcone]
Total Volume = (22/7) × (3)2 × [7 + (4/3)] = (22/7) × 9 × (25/3)
Total Volume = (22/7) × 3 × 25 = 1650 / 7 ≈ 235.71 m3.
3. Outer Metal Sheet Area (Excluding Base):
Area = CSA of cylinder + CSA of cone
Area = 2πrhcyl + πrl = πr(2hcyl + l)
Area = (22/7) × 3 × [2(7) + 5] = (66/7) × 19 = 1254 / 7 ≈ 179.14 m2.
How to Prepare for This Topic
Achieving full marks in Surface Areas and Volumes in the 2026 CBSE Class 10 board exam requires precision, visualization, and calculation discipline. Follow these teacher-recommended preparation strategies:
- Draw Clear Figures First: Always sketch the composite solid before beginning calculations. Clearly annotate radii, vertical heights, and slant heights on your diagram. Examiners award step marks for accurate representation.
- Factor Out Common Terms: Avoid evaluating individual terms separately. For instance, write 2πrh + 2πr2 as 2πr(h + r) before substituting numbers. Factoring reduces calculation time and eliminates arithmetic errors.
- Check Unit Consistency: Ensure all dimensions are converted to the same unit (e.g., converting cm to m or litres to m3) before applying formulas. Note that 1 m3 = 1000 litres and 1000 cm3 = 1 litre.
- Watch for Value of π: Use π = 3.14 only when explicitly stated in the question; otherwise, default to π = 22/7.
- Master Case Study Formats: Solve at least 10 real-world situational problems (tents, containers, ice cream cones, embankments) to build confidence for Section E of the board paper.
Where to Practice More
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