Class 10 Maths Chapter 13 Surface Areas and Volumes Practice Paper 1

PRACTICE PAPER 01 (2025-26) CHAPTER 12: SURFACE AREAS AND VOLUMES SUBJECT: MATHEMATICS STANDARD CLASS: X MAX. MARKS: 40 DURATION: 1½ hrs

General Instructions:

(i) All questions are compulsory. (ii) This question paper contains 20 questions divided into five Sections A, B, C, D and E. (iii) Section A comprises of 10 MCQs of 1 mark each. Section B comprises of 4 questions of 2 marks each. Section C comprises of 3 questions of 3 marks each. Section D comprises of 1 question of 5 marks and Section E comprises of 2 Case Study Based Questions of 4 marks each. (iv) There is no overall choice. (v) Use of Calculators is not permitted. SECTION – A Questions 1 to 10 carry 1 mark each.

1. A cylindrical pillar has diameter 56 cm and is 3.5 m high. The cost of painting the curved surface at the rate of ₹12.50 per m² is:
(a) ₹77
(b) ₹154
(c) ₹231
(d) ₹308

2. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. The height of the cylinder is:
(a) 2.744 cm
(b) 3.5 cm
(c) 4.116 cm
(d) 2.058 cm

3. The total surface area of a solid hemisphere of radius r is:
(a) 2πr²
(b) 3πr²
(c) 4πr²
(d) πr²

4. A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is:
(a) 1 : 2 : 3
(b) 2 : 1 : 3
(c) 2 : 3 : 1
(d) 3 : 2 : 1

5. The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is:
(a) 1 : 4
(b) 1 : 3
(c) 2 : 3
(d) 1 : 2

6. A solid cylinder of radius r and height h is placed over another cylinder of same height and radius. The total surface area of the shape so formed is:
(a) 4πrh + 4πr²
(b) 2πrh + 2πr²
(c) 4πrh + 2πr²
(d) 2πrh + 4πr²

7. If the radius of base of a right circular cylinder is halved, keeping the height same, the ratio of the volume of the reduced cylinder to that of the original cylinder is:
(a) 1 : 2
(b) 1 : 4
(c) 1 : 8
(d) 8 : 1

8. Water flows at the rate of 10 m/minute through a cylindrical pipe having diameter 5 mm. How much time would it take to fill a conical vessel whose diameter of the base is 40 cm and depth 24 cm?
(a) 48 minutes 15 seconds
(b) 51 minutes 12 seconds
(c) 52 minutes 1 second
(d) 55 minutes In the following questions 9 and 10, a statement of assertion
(a) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both assertion
(a) and reason (R) are true and reason (R) is the correct explanation of assertion
(a) .

(b) Both assertion
(a) and reason (R) are true but reason (R) is not the correct explanation of assertion
(a) .
(c) Assertion
(a) is true but reason (R) is false.
(d) Assertion
(a) is false but reason (R) is true.

9. Assertion
(a) : A solid ball is exactly fitted inside a cubical box of side 'a'. The volume of the ball is πa³/6. Reason (R): Diameter of the ball will be equal to the side of the cube.

10. Assertion
(a) : If the height of a cone is doubled and radius remains same, then the volume becomes double. Reason (R): Volume of cone is directly proportional to its height when radius is constant. SECTION – B Questions 11 to 14 carry 2 marks each.

11. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

12. A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base of the cone. [Use π = 22/7]

13. 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 of the toy. [Use π = 22/7] OR A cylindrical container of radius 6 cm and height 15 cm is filled with ice-cream. The whole ice-cream is to be distributed to 10 children in equal cones with hemispherical tops. If the height of the conical portion is 4 times the radius of its base, find the radius of the ice-cream cone.

14. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment. [Use π = 22/7] SECTION – C Questions 15 to 17 carry 3 marks each.

15. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 6 cm and its height is 4 cm. Find the cost of painting the toy at the rate of ₹5 per 100 cm². [Use π = 22/7]

16. A milk container is made of metal sheet in the shape of frustum of a cone whose volume is 10459 3/7 cm³. The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of ₹1.40 per cm². [Use π = 22/7]

17. 504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area. [Use π = 22/7] SECTION – D Question 18 carries 5 marks.

18. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 4 cm and the diameter of the base is 8 cm. Determine the volume of the toy. If a cube circumscribes the toy, then find the difference of the volumes of cube and the toy. Also, find the total surface area of the toy. [Use π = 3.14] OR A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find:

(a) The inner surface area of the vessel
(b) The volume of water the vessel can hold [Use π = 22/7] SECTION – E (Case Study Based Questions) Questions 19 to 20 carry 4 marks each.

19. Water Tank Design A water tank for a building is designed to store water. The tank consists of a cylindrical portion and a hemispherical top. The total height of the tank is 4.5 m and the diameter is 4 m. [Diagram: A cylinder with hemispherical top, Total height = 4.5 m, Diameter = 4 m] Based on the above, answer the following questions: (i) Find the height of the cylindrical portion. (1 mark) (ii) Find the curved surface area of the hemispherical portion. (1 mark) (iii)
(a) Find the total capacity of the tank (volume). (2 marks) OR
(b) Find the cost of painting the outer surface of the tank (excluding the base) at the rate of ₹25 per m².

(2 marks)

20. Ice-cream Factory An ice-cream factory makes ice-cream in two shapes: Shape A: A cone with hemispherical top (both having radius 3.5 cm). The height of the cone is 10 cm. Shape B: A cylinder with two hemispherical ends (all having radius 3.5 cm). The height of the cylinder is 7 cm. Based on the above, answer the following questions:
(a) Which ice-cream shape has more quantity? Find the difference in volumes. (2 marks)
(b) If the factory wants to pack the ice-creams in wrappers, which shape requires more wrapper material? Find the difference in surface areas. (2 marks) DETAILED ANSWER KEY PRACTICE PAPER 01 SECTION – A (Answers) Answer 1:
(a) ₹77

Solution:

Diameter = 56 cm = 0.56 m, so radius r = 0.28 m Height h = 3.5 m Curved surface area = 2πrh = 2 × (22/7) × 0.28 × 3.5 = 2 × 22 × 0.04 × 3.5 = 6.16 m² Cost = 6.16 × 12.50 = ₹77 Answer 2:
(a) 2.744 cm

Solution:

Volume of sphere = (4/3)πr³ = (4/3)π(4.2)³ Volume of cylinder = πR²h = π(6)²h Since volumes are equal: (4/3)π(4.2)³ = π(6)²h h = (4/3) × (4.2)³ / 36 = (4/3) × 74.088 / 36 = 2.744 cm Answer 3:
(b) 3πr²

Solution:

Total surface area of solid hemisphere = Curved surface area + Base area = 2πr² + πr² = 3πr² Answer 4:
(a) 1 : 2 : 3

Solution:

Let radius = r and height = h = r (for hemisphere) Volume of cone = (1/3)πr²h = (1/3)πr³ Volume of hemisphere = (2/3)πr³ Volume of cylinder = πr²h = πr³ Ratio = (1/3)πr³ : (2/3)πr³ : πr³ = 1 : 2 : 3 Answer 5:
(a) 1 : 4

Solution:

Surface area of hemisphere = 3πr² Initial surface area = 3π(6)² = 108π Final surface area = 3π(12)² = 432π Ratio = 108π : 432π = 1 : 4 Answer 6:
(c) 4πrh + 2πr²

Solution:

Total surface area = 2 curved surfaces + 2 circular ends = 2πrh + 2πrh + 2πr² = 4πrh + 2πr² Answer 7:
(b) 1 : 4

Solution:

Original volume = πr²h New volume = π(r/2)²h = πr²h/4 Ratio = πr²h/4 : πr²h = 1 : 4 Answer 8:
(b) 51 minutes 12 seconds

Solution:

Volume of cone = (1/3)πr²h = (1/3) × π × (20)² × 24 = 3200π cm³ Pipe radius = 2.5 mm = 0.25 cm Volume flowing per minute = π(0.25)² × 1000 = 62.5π cm³ Time = 3200π / 62.5π = 51.2 minutes = 51 minutes 12 seconds Answer 9:
(a)

Solution:

Both assertion and reason are true and reason correctly explains the assertion. If diameter = a, then radius = a/2 Volume = (4/3)π(a/2)³ = (4/3)π(a³/8) = πa³/6 Answer 10:
(a)

Solution:

Both assertion and reason are true and reason correctly explains the assertion. V = (1/3)πr²h. When h is doubled and r is constant, volume doubles. SECTION – B (Answers) Answer 11:

Solution:

Volume of cone = (1/3)πr²h = (1/3)π(5)²(8) = 200π/3 cm³ Water flown out = (1/4) × 200π/3 = 50π/3 cm³ Volume of one lead shot = (4/3)π(0.5)³ = π/6 cm³ Number of shots = (50π/3) ÷ (π/6) = (50π/3) × (6/π) = 100 Therefore, 100 lead shots were dropped. Answer 12:

Solution:

Volume of hemisphere = (2/3)πr³ = (2/3)π(7)³ = 686π/3 cm³ Volume of cone = (1/3)πR²h = (1/3)πR²(49) Equating volumes: 686π/3 = (1/3)πR²(49) R² = 686/49 = 14 R = √14 cm Radius of the cone = √14 cm ≈ 3.74 cm Answer 13:

Solution:

Radius r = 3.5 cm Height of cone = 15.5 - 3.5 = 12 cm Slant height l = √(r² + h²) = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm TSA = Curved surface of cone + Curved surface of hemisphere = πrl + 2πr² = (22/7) × 3.5 × 12.5 + 2 × (22/7) × 3.5 × 3.5 = 137.5 + 77 = 214.5 cm² Total surface area = 214.5 cm² OR Volume of cylinder = π(6)²(15) = 540π cm³ For each cone: let radius = r, height = 4r Volume = (1/3)πr²(4r) + (2/3)πr³ = (4πr³/3) + (2πr³/3) = 2πr³ 10 cones: 10 × 2πr³ = 20πr³ 20πr³ = 540π r³ = 27, r = 3 cm Radius of ice-cream cone = 3 cm Answer 14:

Solution:

Volume of earth dug out = π(1.5)²(14) = 31.5π m³ Inner radius of embankment = 1.5 m Outer radius = 1.5 + 4 = 5.5 m Volume of embankment = π(R² - r²)h = π(5.5² - 1.5²)h 31.5π = π(30.25 - 2.25)h 31.5 = 28h h = 31.5/28 = 1.125 m = 1.125 m Height of embankment = 1.125 m or 1 m 12.5 cm SECTION – C (Answers) Answer 15:

Solution:

Radius = 3 cm, Height of cone = 4 cm Slant height l = √(3² + 4²) = 5 cm Surface area to be painted = πrl + 2πr² = πr(l + 2r) = (22/7) × 3 × (5 + 6) = (22/7) × 3 × 11 = 103.71 cm² Cost = (103.71/100) × 5 = ₹5.19 ≈ ₹5.20 Cost of painting = ₹5.19 Answer 16:

Solution:

Volume = 10459 3/7 = 73216/7 cm³ r₁ = 8 cm, r₂ = 20 cm Volume of frustum = (1/3)πh(r₁² + r₂² + r₁r₂) 73216/7 = (1/3) × (22/7) × h × (64 + 400 + 160) 73216/7 = (22/21) × h × 624 h = 73216 × 21/(7 × 22 × 624) = 16 cm Slant height l = √[(r₂-r₁)² + h²] = √[(12)² + (16)²] = 20 cm Surface area = π(r₁ + r₂)l + πr₁² + πr₂² = π(28)(20) + π(64) + π(400) = π(560 + 64 + 400) = 1024π = 3217.14 cm² Cost = 3217.14 × 1.40 = ₹4504 Cost of metal sheet = ₹4504 Answer 17:

Solution:

Volume of one cone = (1/3)πr²h = (1/3)π(1.75)²(3) cm³ Volume of 504 cones = 504 × (1/3)π(1.75)²(3) = 1617π cm³ Volume of sphere = (4/3)πR³ (4/3)πR³ = 1617π R³ = 1617 × 3/4 = 1212.75 R = 10.66 cm (approximately) Diameter = 2R = 21.32 cm Surface area = 4πR² = 4π(10.66)² = 1424.7 cm² Diameter = 21 cm, Surface area ≈ 1386 cm² SECTION – D (Answers) Answer 18:

Solution:

Radius = 4 cm, Height of cone = 4 cm Volume of toy = Volume of cone + Volume of hemisphere = (1/3)πr²h + (2/3)πr³ = (1/3)π(4)²(4) + (2/3)π(4)³ = 64π/3 + 128π/3 = 192π/3 = 64π = 201.06 cm³ Side of cube = 8 cm (diameter) Volume of cube = 8³ = 512 cm³ Difference = 512 - 201.06 = 310.94 cm³ TSA = πrl + 2πr² l = √(16 + 16) = 4√2 cm TSA = π(4)(4√2) + 2π(16) = 16π√2 + 32π = 103.58 cm² Volume of toy = 201.06 cm³, Difference = 310.94 cm³, TSA = 103.58 cm² OR Radius = 7 cm, Height of cylinder = 13 - 7 = 6 cm
(a) Inner surface area = 2πrh + 2πr² = 2π(7)(6) + 2π(7)² = 84π + 98π = 182π = 572 cm²
(b) Volume = πr²h + (2/3)πr³ = π(49)(6) + (2/3)π(343) = 294π + 228.67π = 522.67π = 1641.14 cm³
(a) Inner surface area = 572 cm²,
(b) Volume = 1641.14 cm³ SECTION – E (Answers) Answer 19:

(i) Radius = 2 m Height of cylindrical portion = 4.5 - 2 = 2.5 m (ii) Curved surface area of hemisphere = 2πr² = 2π(2)² = 8π = 25.13 m² (iii)
(a) Volume = πr²h + (2/3)πr³ = π(4)(2.5) + (2/3)π(8) = 10π + 16π/3 = 46π/3 = 48.17 m³ (i) 2.5 m, (ii) 25.13 m², (iii)
(a) 48.17 m³ OR (iii)
(b) Surface area = 2πrh + 2πr² = 2π(2)(2.5) + 2π(4) = 10π + 8π = 18π = 56.55 m² Cost = 56.55 × 25 = ₹1413.75 (iii)
(b) Cost = ₹1413.75 Answer 20:
(a) Shape A volume = (1/3)πr²h + (2/3)πr³ = (1/3)π(3.5)²(10) + (2/3)π(3.5)³ = 128.33π + 28.58π = 156.91π = 493.04 cm³ Shape B volume = πr²h + 2 × (2/3)πr³ = π(3.5)²(7) + (4/3)π(3.5)³ = 269.5π + 57.17π = 326.67π = 1026.43 cm³ Shape B has more quantity.

Difference = 1026.43 - 493.04 = 533.39 cm³
(b) Shape A surface area = πrl + 2πr² (cone + hemisphere) l = √(3.5² + 10²) = 10.59 cm = π(3.5)(10.59) + 2π(3.5)² = 116.49π + 24.5π = 140.99π = 442.91 cm² Shape B surface area = 2πrh + 2(2πr²) = 2π(3.5)(7) + 4π(3.5)² = 154π + 49π = 203π = 637.74 cm² Shape B requires more wrapper material. Difference = 637.74 - 442.91 = 194.83 cm²
(a) Shape B, Difference = 533.39 cm³,
(b) Shape B, Difference = 194.83 cm²