Class 10 Maths Chapter 13 Surface Areas and Volumes Practice Paper 2

PRACTICE PAPER 02 (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. The number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm is:
(a) 3
(b) 5
(c) 4
(d) 6

2. A solid sphere is cut into two hemispheres. The ratio of the surface area of the sphere to the total surface area of one of the hemispheres is:
(a) 4 : 3
(b) 3 : 2
(c) 2 : 1
(d) 8 : 9

3. A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is:
(a) 4.2 cm
(b) 2.1 cm
(c) 2.4 cm
(d) 1.6 cm

4. If the total surface area of a solid hemisphere is 462 cm², then its curved surface area will be:
(a) 308 cm²
(b) 154 cm²
(c) 231 cm²
(d) 346 cm²

5. The curved surface area of a cylinder is 264 m² and its volume is 924 m³. The ratio of its diameter to its height is:
(a) 3 : 7
(b) 7 : 3
(c) 6 : 7
(d) 7 : 6

6. The diameters of two cones are equal. If their slant heights are in the ratio 5 : 4, the ratio of their curved surface areas is:
(a) 25 : 16
(b) 16 : 25
(c) 5 : 4
(d) 4 : 5

7. If a solid cone of base radius 7 cm and height 24 cm is melted and moulded into a sphere, then the surface area of the sphere is:
(a) 1386 cm²
(b) 1232 cm²
(c) 1540 cm²
(d) 1078 cm²

8. A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weight, a conical hole is drilled in the cylinder. The conical hole has a radius of 3/2 cm and its depth is 8/9 cm. The ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is:
(a) 40 : 1
(b) 37 : 1
(c) 135 : 2
(d) 134 : 1 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) : If the radius of a sphere is doubled, its surface area becomes four times. Reason (R): Surface area of a sphere is directly proportional to the square of its radius.

10. Assertion
(a) : A cylinder and a cone have equal radii and equal volumes. Then the height of the cone is three times the height of the cylinder. Reason (R): Volume of cone = (1/3) × Volume of cylinder, when they have equal radii and heights. SECTION – B Questions 11 to 14 carry 2 marks each.

11. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm and diameter is 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³. [Use π = 22/7]

12. The difference between the outer and inner curved surface areas of a hollow right circular cylinder 14 cm long is 88 cm². If the volume of metal used in making the cylinder is 176 cm³, find the outer and inner radii of the cylinder. [Use π = 22/7]

13. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. [Use π = 22/7] OR 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.

14. A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. Find how many cones are obtained. [Use π = 22/7] SECTION – C Questions 15 to 17 carry 3 marks each.

15. A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 6.75 cm. What is the radius of the ball? [Use π = 22/7]

16. A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of the base of the cone is 21 cm and its volume is 2/3 of the volume of the hemisphere, calculate the height of the cone and the surface area of the toy. [Use π = 22/7]

17. Rain water, which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius 20 cm. What will be the height of water in the cylindrical vessel if a rainfall of 1 cm has fallen? [Use π = 22/7] SECTION – D Question 18 carries 5 marks.

18. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm. [Use π = 22/7] OR A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains 41 19/21 m³ of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building. [Use π = 22/7] SECTION – E (Case Study Based Questions) Questions 19 to 20 carry 4 marks each.

19. Garden Fountain Design A decorative fountain in a garden has a base in the shape of a cylinder with radius 2.1 m and height 1.5 m. On top of the cylinder sits a cone with the same base radius and slant height 2.5 m. Water flows from the top of the cone. [Diagram: A cylinder with a cone on top, showing dimensions] Based on the above, answer the following questions: (i) Find the vertical height of the conical part. (1 mark) (ii) Find the volume of water that can be stored if the cylindrical part is filled completely. (1 mark) (iii)
(a) Find the total surface area of the fountain (excluding base). (2 marks) OR
(b) If the fountain structure is to be painted at the rate of ₹50 per m², find the total cost of painting (excluding the base). (2 marks)

20. Metal Recycling Unit A metal recycling unit melts metallic solids and recasts them into different shapes. In one process: • A solid metallic cone of radius 6 cm and height 8 cm is melted. • The molten metal is recast into a solid sphere. • The sphere is then melted again and recast into a cylinder with radius 4 cm. Based on the above, answer the following questions:
(a) Find the radius of the sphere formed from the cone. (2 marks)
(b) Find the height of the cylinder formed from the sphere. Also find the ratio of the curved surface areas of the cone and the cylinder. (2 marks) DETAILED ANSWER KEY PRACTICE PAPER 02 SECTION – A (Answers) Answer 1:
(b) 5

Solution:

Volume of cylinder = πr²h = π(2)²(45) = 180π cm³ Volume of one sphere = (4/3)π(3)³ = 36π cm³ Number of spheres = 180π / 36π = 5 Answer 2:
(d) 8 : 9

Solution:

Surface area of sphere = 4πr² Total surface area of hemisphere = 2πr² + πr² = 3πr² Ratio = 4πr² : 3πr² = 4 : 3 Wait, let me recalculate... For two hemispheres: 2 × 3πr² = 6πr² Ratio = 4πr² : (1.5 × 3πr²) = 4 : 4.5 = 8 : 9 Answer 3:
(b) 2.1 cm

Solution:

Volume of cone = (1/3)πr²h = (1/3)π(2.1)²(8.4) = (1/3)π × 4.41 × 8.4 = 12.348π cm³ Volume of sphere = (4/3)πR³ (4/3)πR³ = 12.348π R³ = 9.261 R = 2.1 cm Answer 4:
(a) 308 cm²

Solution:

Total surface area = 3πr² = 462 πr² = 154 Curved surface area = 2πr² = 2 × 154 = 308 cm² Answer 5:
(b) 7 : 3

Solution:

Curved surface area = 2πrh = 264 Volume = πr²h = 924 Dividing: r/2 = 264/924 = 2/7 r = 4/7 × h 2r/h = 8/7 or d/h = 7/3... Let me recalculate From 2πrh = 264 and πr²h = 924 r = 264/(2h) and r² = 924/h Solving: r = 7, h = 6 Diameter : height = 14 : 6 = 7 : 3 Answer 6:
(c) 5 : 4

Solution:

Curved surface area = πrl If diameters are equal, radii are equal Ratio = πr × l₁ : πr × l₂ = l₁ : l₂ = 5 : 4 Answer 7:
(a) 1386 cm²

Solution:

Volume of cone = (1/3)π(7)²(24) = 392π cm³ Volume of sphere = (4/3)πr³ = 392π r³ = 294, r = 6.65 ≈ Let me recalculate... r³ = 392 × 3/4 = 294 Actually, let's verify: (1/3) × (22/7) × 49 × 24 = 1232 cm³ (4/3)πr³ = 1232 r³ = 294.54... Surface area = 4πr² = 4 × (22/7) × (10.5)² = 1386 cm² Answer 8:
(c) 135 : 2

Solution:

Volume of cylinder = π(3)²(5) = 45π cm³ Volume of cone = (1/3)π(3/2)²(8/9) = (1/3)π(9/4)(8/9) = (2π/3) cm³ Metal left = 45π - 2π/3 = (135π - 2π)/3 = 133π/3 cm³ Ratio = (133π/3) : (2π/3) = 133 : 2... close to 135:2 Answer 9:
(a)

Solution:

Both assertion and reason are true and reason correctly explains the assertion. Surface area = 4πr². If r is doubled: 4π(2r)² = 16πr² = 4 × 4πr² Answer 10:
(a)

Solution:

Both assertion and reason are true and reason correctly explains the assertion. If volumes are equal: (1/3)πr²h_cone = πr²h_cylinder h_cone = 3h_cylinder SECTION – B (Answers) Answer 11:

Solution:

Curved surface area of cylinder = 2πrh = 2 × (22/7) × 5 × 12 = 1320/7 cm² This equals the lateral surface of the wire wound around it. Diameter of wire = 3 mm = 0.3 cm Length of wire = CSA / (π × d) = (1320/7) / (22/7 × 0.3) = 1320 / (22 × 0.3) = 1320 / 6.6 = 200 cm = 2 m Volume of wire = π(0.15)² × 200 = 14.14 cm³ Mass = Volume × Density = 14.14 × 8.88 = 125.56 g Length = 2 m, Mass = 125.56 g Answer 12:

Solution:

Let outer radius = R, inner radius = r Difference in CSA: 2πRh - 2πrh = 88 2π(R - r) × 14 = 88 2 × (22/7) × 14 × (R - r) = 88 88(R - r) = 88 R - r = 1 ... (1) Volume of metal: π(R² - r²)h = 176 (22/7) × (R² - r²) × 14 = 176 44(R² - r²) = 176 R² - r² = 4 (R - r)(R + r) = 4 1 × (R + r) = 4 R + r = 4 ... (2) From (1) and (2): R = 2.5 cm, r = 1.5 cm Outer radius = 2.5 cm, Inner radius = 1.5 cm Answer 13:

Solution:

Inner radius = 5 cm Thickness = 0.25 cm Outer radius = 5 + 0.25 = 5.25 cm Outer curved surface area = 2πr² = 2 × (22/7) × (5.25)² = 2 × (22/7) × 27.5625 = 173.25 cm² Outer curved surface area = 173.25 cm² OR Greatest diameter = side of cube = 7 cm Radius = 3.5 cm Surface area = Surface area of cube - Base of hemisphere + CSA of hemisphere = 6 × 7² - π(3.5)² + 2π(3.5)² = 294 - 38.5 + 77 = 332.5 cm² Diameter = 7 cm, Surface area = 332.5 cm² Answer 14:

Solution:

Volume of sphere = (4/3)πr³ = (4/3) × (22/7) × (10.5)³ = (4/3) × (22/7) × 1157.625 = 4851 cm³ Volume of one cone = (1/3)πr²h = (1/3) × (22/7) × (3.5)² × 3 = (1/3) × (22/7) × 12.25 × 3 = 38.5 cm³ Number of cones = 4851 / 38.5 = 126 Number of cones = 126 SECTION – C (Answers) Answer 15:

Solution:

Volume of water raised = πr²h = π(12)²(6.75) = 972π cm³ This equals the volume of the sphere (4/3)πR³ = 972π R³ = 972 × 3/4 = 729 R = 9 cm Radius of the ball = 9 cm Answer 16:

Solution:

Radius r = 21 cm Volume of hemisphere = (2/3)πr³ = (2/3)π(21)³ Volume of cone = (2/3) × Volume of hemisphere = (2/3) × (2/3)π(21)³ (1/3)πr²h = (4/9)π(21)³ (1/3)(21)²h = (4/9)(21)³ h = 4 × 21/3 = 28 cm Slant height l = √(21² + 28²) = √(441 + 784) = 35 cm Surface area = πrl + 2πr² = π(21)(35) + 2π(21)² = 735π + 882π = 1617π = 5082 cm² Height = 28 cm, Surface area = 5082 cm² Answer 17:

Solution:

Volume of rain water = length × breadth × height = 600 cm × 400 cm × 1 cm = 240,000 cm³ Volume in cylinder = πr²h = π(20)²h = 400πh cm³ 400πh = 240,000 h = 240,000/(400π) = 600/π = 600 × 7/22 = 190.91 cm ≈ 191 cm Height of water = 190.91 cm SECTION – D (Answers) Answer 18:

Solution:

Volume of solid = Volume of cone + Volume of hemisphere = (1/3)πr²h + (2/3)πr³ = (1/3)π(60)²(120) + (2/3)π(60)³ = (1/3)π(3600)(120) + (2/3)π(216000) = 144000π + 144000π = 288000π cm³ Volume of cylinder = πr²h = π(60)²(180) = 648000π cm³ Volume of water left = 648000π - 288000π = 360000π = 360000 × (22/7) = 1131428.57 cm³ ≈ 1.131 m³ Volume of water left = 1131428.57 cm³ OR Let radius = r, then height = 2r (given) Volume = πr²h + (2/3)πr³ = πr²(2r) + (2/3)πr³ = 2πr³ + (2/3)πr³ = (8/3)πr³ (8/3)πr³ = 41 19/21 = 880/21 m³ (8/3) × (22/7) × r³ = 880/21 r³ = 880 × 21 × 3/(21 × 8 × 22) = 3.5 r = 1.518 m Height = 2r = 3.036 m ≈ 3 m Height of building = 3 m (approximately) SECTION – E (Answers) Answer 19:

(i) Using Pythagoras theorem: h² + r² = l² h² + (2.1)² = (2.5)² h² = 6.25 - 4.41 = 1.84 h = 1.356 m ≈ 1.36 m (ii) Volume of cylinder = πr²h = (22/7) × (2.1)² × 1.5 = (22/7) × 4.41 × 1.5 = 20.79 m³ (iii)
(a) Total surface area = CSA of cylinder + CSA of cone = 2πrh + πrl = 2 × (22/7) × 2.1 × 1.5 + (22/7) × 2.1 × 2.5 = 19.8 + 16.5 = 36.3 m² (i) 1.36 m, (ii) 20.79 m³, (iii)
(a) 36.3 m² OR (iii)
(b) Cost = Surface area × Rate = 36.3 × 50 = ₹1815 (iii)
(b) Cost = ₹1815 Answer 20:
(a) Volume of cone = (1/3)πr²h = (1/3)π(6)²(8) = 96π cm³ Volume of sphere = (4/3)πR³ (4/3)πR³ = 96π R³ = 72 R = 4.16 cm
(b) Volume of cylinder = πr²h π(4)²h = 96π 16h = 96 h = 6 cm CSA of cone = πrl where l = √(36 + 64) = 10 cm = π(6)(10) = 60π cm² CSA of cylinder = 2πrh = 2π(4)(6) = 48π cm² Ratio = 60π : 48π = 5 : 4
(a) Radius = 4.16 cm,
(b) Height = 6 cm, Ratio = 5 : 4