Class 10 Mathematics: Surface Areas and Volumes important questions

Surface Areas and Volumes is a chapter of Class 10 Mathematics for the CBSE. These 14 questions are marked Very Important, Important or Practice, and each answer has been checked by our team. Use them to revise, not as a prediction of what will come in the exam.

1. Practice · 1 marks

   1. The sum of the length, breadth and height of a cuboid is 6√3 cm and its diagonal is 2√3 cm. Its total surface area is: (a) 48 cm² (b) 72 cm² (c) 96 cm² (d) 108 cm²

   (c) 96 cm² (l + b + h)² = 108 and l² + b² + h² = 12, so 2(lb + bh + hl) = 108 − 12 = 96 cm².

2. Very Important · 5 marks

   2. 50 schools together provide canvas for 1500 tents for flood victims and share the cost equally. Each tent has a cylindrical lower part (radius 2.8 m, height 3.5 m) and a conical top of the same radius and height 2.1 m. If canvas costs ₹120 per m², find each school's share.

   Slant height = √(2.8² + 2.1²) = 3.5 m. Canvas per tent = 2π × 2.8 × 3.5 + π × 2.8 × 3.5 = 61.6 + 30.8 = 92.4 m². Total cost = 1500 × 92.4 × 120 = ₹1,66,32,000. Each school pays 16632000/50 = ₹3,32,640.

3. Very Important · 5 marks

   3. Two identical solid cubes have side 7 cm. From the top face of the first, a hemisphere of diameter 7 cm is scooped out; this hemisphere is placed on top of the second cube as a dome. Find (i) the ratio of the total surface areas of the two new solids, (ii) the volume of each.

   r = 3.5 cm. Flat circle = πr² = 38.5 cm²; hemisphere's curved surface = 2πr² = 77 cm²; cube surface = 294 cm². (i) First solid = 294 − 38.5 + 77 = 332.5 cm²; second solid = 294 − 38.5 + 77 = 332.5 cm². Ratio 1 : 1. (ii) Hemisphere volume = (2/3) × (22/7) × 3.5³ ≈ 89.83 cm³. First solid = 343 − 89.83 = 253.17 cm³; second = 343 + 89.83 = 432.83 cm³.

4. Practice · 1 marks

   4. A top is made of a hemisphere joined to a cone along their flat faces. Assertion (A): The total surface area of the top is the sum of the curved surface area of the hemisphere and the curved surface area of the cone. Reason (R): The top is obtained by joining the plane surfaces of the hemisphere and the cone together. A. Both A and R are true, and R is the correct explanation of A. B. Both A and R are true, and R is not the correct explanation of A. C. A is true but R is false. D. A is false but R is true.

   A. Both are true, and R explains A: the joined flat faces are hidden, so only the two curved surfaces remain.

5. Very Important · 5 marks

   5. Water flows at 15 km/h through a pipe of diameter 14 cm into a cuboidal pond 50 m long and 44 m wide. In what time will the water level rise by 21 cm? What should the speed of the water be for the same rise in 1 hour?

   Volume needed = 50 × 44 × 0.21 = 462 m³. Pipe cross-section = (22/7) × 0.07² = 0.0154 m²; water per hour = 0.0154 × 15000 = 231 m³. Time = 462/231 = 2 hours. To do it in 1 hour the flow must double: 30 km/h.

6. Very Important · 5 marks

   6. A tent is a cylinder of height 3 m and radius 14 m surmounted by a cone; the total height is 13.5 m. Find the canvas needed, allowing 26 m² for stitching and wastage, and its cost at ₹500 per m².

   Cone height = 13.5 − 3 = 10.5 m; slant height = √(14² + 10.5²) = 17.5 m. CSA of cone = (22/7) × 14 × 17.5 = 770 m²; CSA of cylinder = 2 × (22/7) × 14 × 3 = 264 m². Canvas = 770 + 264 + 26 = 1060 m²; cost = 1060 × 500 = ₹5,30,000.

7. Practice · 1 marks

   7. The radius of the base of a right circular cone and the radius of a sphere are each 5 cm. If the volume of the cone equals the volume of the sphere, the height of the cone is: (A) 5 cm (B) 20 cm (C) 10 cm (D) 4 cm

   (B) 20 cm ⅓πr²h = (4/3)πr³ gives h = 4r = 20 cm.

8. Practice · 1 marks

   8. The volume of a solid hemisphere is 396/7 cm³. Its total surface area (in cm²) is: (A) 396/7 (B) 594/7 (C) 549/7 (D) 604/7

   (B) 594/7 (2/3)πr³ = 396/7 gives r³ = 27, r = 3 cm. TSA = 3πr² = 3 × (22/7) × 9 = 594/7 cm².

9. Very Important · 4 marks

   9. A metallic silo is a cylinder mounted by a cone. The cone has base radius 1.5 m and height 2 m; the cylinder has radius 1.5 m and height 7 m. (i) Find the slant height of the cone. (ii) Find the curved surface area of the cone. (iii) (A) Find the cost of the metal sheet for the curved cylindrical part at ₹2000 per m². OR (B) Find the total capacity of the silo.

   (i) l = √(1.5² + 2²) = √6.25 = 2.5 m. (ii) CSA = πrl = (22/7) × 1.5 × 2.5 ≈ 11.79 m². (iii) (A) CSA of cylinder = 2πrh = 2 × (22/7) × 1.5 × 7 = 66 m²; cost = 66 × 2000 = ₹1,32,000. OR (B) Cylinder = πr²h = (22/7) × 2.25 × 7 = 49.5 m³; cone = ⅓ × (22/7) × 2.25 × 2 ≈ 4.71 m³; total ≈ 54.21 m³.

10. Practice · 1 marks

    10. The base of a right circular cone has area 51 cm² and its volume is 85 cm³. Its height is: (A) 6 cm (B) 3 cm (C) 2 cm (D) 5 cm

    (D) 5 cm V = ⅓ × base area × h, so 85 = ⅓ × 51 × h and h = 5 cm.

11. Very Important · 5 marks

    11. From a solid cone of height 6 cm and base radius 12 cm, a cylindrical cavity of height 3 cm and radius 4 cm is hollowed out, with the bases forming concentric circles. Find the surface area of the remaining solid in terms of π.

    Slant height of cone = √(12² + 6²) = √180 = 6√5 cm. CSA of cone = π × 12 × 6√5 = 72√5π cm². Base: the cone's base minus the cavity opening plus the cylinder's top = 144π cm². Inner curved surface of the cavity = 2π × 4 × 3 = 24π cm². Total = 72√5π + 144π + 24π = (72√5 + 168)π cm².

12. Very Important · 5 marks

    12. An empty cone of radius 3 cm and height 12 cm is filled with ice cream so that the lower one-sixth of its volume stays empty, and a hemisphere of ice cream is formed on top. Find the volume of ice cream.

    Volume of cone = ⅓ × π × 9 × 12 = 36π cm³; filled part = (5/6) × 36π = 30π cm³. Hemisphere = ⅔ × π × 27 = 18π cm³. Ice cream = 30π + 18π = 48π cm³ ≈ 150.86 cm³.

13. Practice · 1 marks

    13. Two cubes, each of volume 64 cm³, are joined end to end. The surface area of the resulting cuboid is: (A) 192 cm² (B) 160 cm² (C) 96 cm² (D) 80 cm²

    (B) 160 cm² Side = 4 cm, so the cuboid is 8 × 4 × 4 cm. TSA = 2(8×4 + 4×4 + 8×4) = 160 cm².

14. Very Important · 5 marks

    14. A tent is a cylinder of height 3 m and radius 14 m with a conical top; the total height is 13.5 m. Find the canvas needed, allowing 26 m² extra for stitching and wastage, and its cost at ₹250 per m².

    Height of cone = 13.5 − 3 = 10.5 m. Slant height l = √(14² + 10.5²) = 17.5 m. CSA of cone = πrl = (22/7) × 14 × 17.5 = 770 m². CSA of cylinder = 2πrh = 2 × (22/7) × 14 × 3 = 264 m². Canvas = 770 + 264 + 26 = 1060 m². Cost = 1060 × 250 = ₹2,65,000.