Class 10 Mathematics: Surface Areas and Volumes important questions

Surface Areas and Volumes is a chapter of Class 10 Mathematics for the Andhra Pradesh State Board. These 16 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. Find the volume of a cylinder whose base radius is 3 cm and height is 7 cm.

   V = πr²h = (22/7) × 9 × 7 = 198 cm³.

2. Important · 2 marks

   2. While building a kitchen shed at school, a truck unloaded sand in the shape of a cone of base radius 2.7 m and height 7 m. Find the volume of sand.

   V = (1/3)πr²h = (1/3) × (22/7) × 2.7 × 2.7 × 7 = (22 × 7.29)/3 = 53.46 m³.

3. Very Important · 8 marks

   3. Two cubes, each of volume 64 cm³, are joined end to end. Find the lateral surface area and total surface area of the resulting cuboid.

   Side of each cube = ∛64 = 4 cm. The cuboid is 8 cm × 4 cm × 4 cm. Lateral surface area = 2h(l + b) = 2 × 4 × (8 + 4) = 96 cm². Total surface area = 2(lb + bh + hl) = 2(32 + 16 + 32) = 160 cm².

4. Practice · 1 marks

   4. Draw a rough figure of a cylinder with height h cm and base radius r cm.

   Draw two equal ellipses one above the other (the circular top and base), and join their left and right ends with two vertical lines. Mark the radius r from the centre of the base to its edge, and the height h as the vertical distance between the two circles.

5. Important · 2 marks

   5. Find the volume of a cylinder with base radius 6 cm and height 7 cm.

   V = πr²h = (22/7) × 6 × 6 × 7 = 22 × 36 = 792 cm³.

6. Practice · 1 marks

   6. The total surface area of a solid hemisphere of radius r is A) πr² B) 2πr² C) 3πr² D) 4πr²

   C) 3πr² Curved surface 2πr² plus the flat circular face πr². (The curved surface area alone is 2πr².)

7. Important · 2 marks

   7. Find the volume of the largest right circular cone that can be cut out of a cube of edge 7 cm.

   The largest cone has base diameter 7 cm (r = 3.5 cm) and height 7 cm. Volume = (1/3)πr²h = (1/3) × (22/7) × 3.5 × 3.5 × 7 = 269.5/3 ≈ 89.83 cm³.

8. Important · 4 marks

   8. A solid is a cone standing on a hemisphere, both of radius 1 cm, and the height of the cone equals its radius. Find the volume of the solid in terms of π.

   r = 1 cm, cone height h = 1 cm. Volume = (1/3)πr²h + (2/3)πr³ = (1/3)π + (2/3)π = π cm³.

9. Practice · 1 marks

   9. Match each item with its formula. i) Curved surface area of a cone ii) Total surface area of a cone iii) Volume of a cone p) πr(l + r) q) (1/3)πr²h r) πrl A) i–r, ii–p, iii–q B) i–p, ii–q, iii–r C) i–p, ii–r, iii–q D) i–q, ii–p, iii–r

   A) i–r (CSA = πrl), ii–p (TSA = πr(l + r)), iii–q (Volume = (1/3)πr²h).

10. Important · 4 marks

    10. A hemispherical depression is cut out of one face of a cubical wooden block so that the diameter l of the hemisphere equals the edge of the cube. Find the surface area of the remaining solid.

    Edge of cube = l, radius of hemisphere = l/2. Surface area = 6l² − (area of circular opening) + (curved surface of hemisphere) = 6l² − π(l/2)² + 2π(l/2)² = 6l² + πl²/4 = (l²/4)(24 + π).

11. Practice · 1 marks

    11. The total surface area of a cylinder of radius r and height h is A) πr(r + h) B) 2πr(r + h) C) 2πr²(r + h) D) πr²(r + h)

    B) 2πr(r + h) TSA = curved surface 2πrh + two circular ends 2πr² = 2πr(h + r).

12. Important · 4 marks

    12. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

    The greatest diameter equals the side of the cube: 7 cm, so r = 3.5 cm. Surface area = surface of cube − base area of hemisphere + curved surface of hemisphere = 6a² − πr² + 2πr² = 6a² + πr² = 6 × 49 + (22/7) × 3.5 × 3.5 = 294 + 38.5 = 332.5 cm².

13. Very Important · 8 marks

    13. A solid cylinder has height 2.4 cm and radius 0.7 cm. A conical cavity of the same height and radius is hollowed out of it. Find the total surface area of the remaining solid to the nearest cm².

    Slant height of the cone l = √(0.7² + 2.4²) = √(0.49 + 5.76) = √6.25 = 2.5 cm. TSA = curved surface of cylinder + area of the top circle + curved surface of the cone = 2πrh + πr² + πrl = πr(2h + r + l) = (22/7) × 0.7 × (4.8 + 0.7 + 2.5) = 2.2 × 8 = 17.6 cm². To the nearest cm², the total surface area is 18 cm².

14. Practice · 1 marks

    14. The volume of a sphere of radius r cm is A) (4/3)πr² B) (4/3)πr³ C) (3/4)πr³ D) (3/4)πr²

    B) (4/3)πr³

15. Important · 4 marks

    15. A solid toy is a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Find the volume of the toy. (Take π = 3.14.)

    Radius r = 2 cm (for both the cone and the hemisphere), cone height h = 2 cm. Volume = (2/3)πr³ + (1/3)πr²h = (2/3)π(8) + (1/3)π(4)(2) = 16π/3 + 8π/3 = 8π. Volume = 8 × 3.14 = 25.12 cm³.

16. Very Important · 8 marks

    16. A solid made of a right circular cone (height 120 cm, radius 60 cm) standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water so that it touches the bottom. The cylinder has radius 60 cm and height 180 cm. Find the volume of water left in the cylinder.

    Volume of cylinder = πr²h = π × 3600 × 180 = 648000π cm³. Volume of cone = (1/3)πr²h = (1/3)π × 3600 × 120 = 144000π cm³. Volume of hemisphere = (2/3)πr³ = (2/3)π × 216000 = 144000π cm³. Water left = 648000π − 288000π = 360000π = 360000 × 22/7 ≈ 1131428.6 cm³ ≈ 1.131 m³.