CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set 19

Download CBSE MCQs for Class 10 Mathematics: Chapter 12 Surface Areas and Volumes

Review structured MCQ sets for Class 10 Mathematics Chapter 12 Surface Areas and Volumes. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Chapter-wise Objective Questions: Chapter 12 Surface Areas and Volumes

View or download the dedicated Chapter 12 Surface Areas and Volumes MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: A metallic cuboid 100 cm × 80 cm × 64 cm is melted and recast into a cube. The surface area of the cube is:
(a) 38400 cm²
(b) 19200 cm²
(c) 32000 cm²
(d) 51200 cm²
Show Answer & Explanation

Answer: (a) 38400 cm²

Explanation:
1. Volume = 100 × 80 × 64 = 512000 cm³, so the edge is the cube root, 80 cm. Surface area = 6 × 80² = 38400 cm².

Teacher's Note:
1. Volume stays the same on melting.
2. Surface area of a cube = 6a².

Question 2: Water in a canal 6 m wide and 1.5 m deep flows at 10 km/h. The area it can irrigate in 30 minutes if 8 cm of standing water is needed is:
(a) 5.625 hectares
(b) 562.5 hectares
(c) 28.125 hectares
(d) 56.25 hectares
Show Answer & Explanation

Answer: (d) 56.25 hectares

Explanation:
1. Length of water in 30 min = 5000 m. Volume = 6 × 1.5 × 5000 = 45000 m³. Area = 45000/0.08 = 562500 m² = 56.25 hectares (1 hectare = 10000 m²).

Teacher's Note:
1. Volume of the water column = cross-section × distance.
2. Area = volume / depth.

Question 3: Water flows at 3 km/h through a circular pipe of internal diameter 20 cm into a cylindrical cistern of diameter 10 m and depth 2 m. The time taken to fill the cistern is:
(a) 1 hour 20 minutes
(b) 2 hours
(c) 1 hour 40 minutes
(d) 50 minutes
Show Answer & Explanation

Answer: (c) 1 hour 40 minutes

Explanation:
1. Cistern volume = π × 5² × 2 = 50π m³. The pipe delivers π × 0.1² × 3000 = 30π m³ per hour. Time = 50/30 = 5/3 h = 1 h 40 min.

Teacher's Note:
1. Compare cistern volume with the volume delivered per hour.

Question 4: A bucket is 32 cm in diameter at the top and 20 cm at the bottom and is 21 cm deep. Its capacity (π = 22/7) is:
(a) 10.5 litres
(b) 11.352 litres
(c) 22.7 litres
(d) 5.676 litres
Show Answer & Explanation

Answer: (b) 11.352 litres

Explanation:
1. R = 16, r = 10, h = 21. V = (1/3)π h (R² + r² + Rr) = (1/3)(22/7)(21)(256 + 100 + 160) = 22 × 516 = 11352 cm³ = 11.352 L.

Teacher's Note:
1. Frustum volume formula.
2. 1000 cm³ = 1 litre.

Question 5: A metallic sphere of diameter 6 cm is melted and drawn into a wire of uniform cross-section and length 36 m. The radius of the wire is:
(a) 1 cm
(b) 1 mm
(c) 0.5 mm
(d) 2 mm
Show Answer & Explanation

Answer: (b) 1 mm

Explanation:
1. Sphere volume = (4/3)π × 27 = 36π cm³. Wire length = 3600 cm. π r² × 3600 = 36π gives r² = 0.01, r = 0.1 cm = 1 mm.

Teacher's Note:
1. Equate volumes; use the same unit.

Question 6: A cylindrical tube open at both ends is made of iron 1 cm thick. Its external diameter is 22 cm and its length is 50 cm. The volume of iron (π = 22/7) is:
(a) 3300 cm³
(b) 3850 cm³
(c) 2750 cm³
(d) 6600 cm³
Show Answer & Explanation

Answer: (a) 3300 cm³

Explanation:
1. R = 11, r = 10. Volume = π(R² − r²)h = (22/7)(121 − 100)(50) = (22/7)(1050) = 3300 cm³.

Teacher's Note:
1. Material = outer volume minus inner volume.

Question 7: A well of diameter 7 m is dug 22.5 m deep and the earth is used to form a circular embankment around it, 1.5 m high. The width of the embankment is:
(a) 7 m
(b) 14 m
(c) 12 m
(d) 10.5 m
Show Answer & Explanation

Answer: (d) 10.5 m

Explanation:
1. Earth dug = π(3.5)²(22.5). Embankment: π(R² − 3.5²)(1.5), where R is the outer radius. So R² − 12.25 = 183.75, R² = 196, R = 14. Width = 14 − 3.5 = 10.5 m.

Teacher's Note:
1. The embankment is a hollow cylinder around the well.

Question 8: The cost of painting the whole outside of a closed cylindrical tank at 60 paise per dm² is Rs. 237.60 and the height is 6 times the radius. The volume of the tank (π = 22/7) is about:
(a) 318.86 dm³
(b) 254.57 dm³
(c) 509.14 dm³
(d) 1018.29 dm³
Show Answer & Explanation

Answer: (c) 509.14 dm³

Explanation:
1. Area = 237.60/0.60 = 396 dm². With h = 6r: 2πr(7r) = 14πr² = 396, so r² = 9, r = 3 dm, h = 18 dm. Volume = π × 9 × 18 = 162π ≈ 509.14 dm³.

Teacher's Note:
1. Find the area from the cost.

Question 9: The volume of the largest right circular cone that can be cut from a cube of edge 9 cm is about:
(a) 243 cm³
(b) 382 cm³
(c) 191 cm³
(d) 127 cm³
Show Answer & Explanation

Answer: (c) 191 cm³

Explanation:
1. The cone has base radius 4.5 cm and height 9 cm. Volume = (1/3)π(4.5)²(9) = 60.75π ≈ 190.9 cm³.

Teacher's Note:
1. Base circle fits in a face of the cube.

Question 10: A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical ball of radius 9 cm is dropped in. The water level rises by:
(a) 4.5 cm
(b) 6.75 cm
(c) 9 cm
(d) 3.375 cm
Show Answer & Explanation

Answer: (b) 6.75 cm

Explanation:
1. Ball volume = (4/3)π × 729 = 972π. This equals π × 144 × h, so h = 6.75 cm.

Teacher's Note:
1. Rise = volume of ball / base area of tub.

Question 11: The number of coins of diameter 1.75 cm and thickness 2 mm that must be melted to form a cuboid 11 cm × 10 cm × 7 cm is:
(a) 1600
(b) 400
(c) 800
(d) 3200
Show Answer & Explanation

Answer: (a) 1600

Explanation:
1. Coin volume = π(0.875)² × 0.2 = 0.48125 cm³ (π = 22/7). Cuboid volume = 770 cm³. Number = 770/0.48125 = 1600.

Teacher's Note:
1. Convert 2 mm to 0.2 cm.

Question 12: The interior of a building is a cylinder of diameter 4.3 m and height 3.8 m surmounted by a cone whose vertical angle is 90°. The surface area of the interior is about:
(a) 51.3 m²
(b) 92.4 m²
(c) 35.9 m²
(d) 71.9 m²
Show Answer & Explanation

Answer: (d) 71.9 m²

Explanation:
1. r = 2.15 m. A 90° vertical angle gives slant height l = r√2 = 3.04 m. Area = 2π(2.15)(3.8) + π(2.15)(3.04) = 51.3 + 20.5 ≈ 71.9 m².

Teacher's Note:
1. Half the vertical angle is 45°, so l = r√2.

Question 13: A solid is a cylinder with a cone at one end and a hemisphere at the other, with common diameter 7 cm and total length 20 cm. The cylindrical part has height 4.5 cm. The total surface area (π = 22/7) is:
(a) 285 cm²
(b) 342 cm²
(c) 220 cm²
(d) 313.5 cm²
Show Answer & Explanation

Answer: (d) 313.5 cm²

Explanation:
1. r = 3.5. Cone height = 20 − 4.5 − 3.5 = 12 cm, slant height = √(144 + 12.25) = 12.5 cm. Area = πr(l + 2h + 2r) = π(3.5)(12.5 + 9 + 7) = (22/7)(3.5)(28.5) = 313.5 cm².

Teacher's Note:
1. Cone height = total − cylinder − hemisphere radius.

Question 14: A solid cylinder is melted into cones. Three times the radius of a cone equals twice the radius of the cylinder, and the heights of cylinder and cone are in the ratio 4 : 3. The number of cones made is:
(a) 6
(b) 12
(c) 9
(d) 8
Show Answer & Explanation

Answer: (c) 9

Explanation:
1. Let the cylinder radius R. Cone radius = 2R/3. Let the cone height be 3k and the cylinder height 4k. Cylinder volume = πR²(4k). Cone volume = (1/3)π(4R²/9)(3k) = 4πR²k/9. Number = 4/(4/9) = 9.

Teacher's Note:
1. Divide the cylinder volume by one cone's volume.

Question 15: A cylindrical boiler 2 m high and 3.5 m in diameter has a hemispherical lid. The volume of its interior, including the part under the lid, is (π = 22/7):
(a) 19.25 m³
(b) 30.48 m³
(c) 38.5 m³
(d) 25.7 m³
Show Answer & Explanation

Answer: (b) 30.48 m³

Explanation:
1. r = 1.75 m. Cylinder = (22/7)(3.0625)(2) = 19.25. Hemisphere = (2/3)(22/7)(5.359) = 11.23. Total ≈ 30.48 m³.

Teacher's Note:
1. Cylinder plus hemisphere.

Question 16: A godown has a cuboidal part 10 m × 7 m × 3 m (inner measures) with a half-cylinder of radius 3.5 m along its length on top. Its volume is (π = 22/7):
(a) 402.5 m³
(b) 210 m³
(c) 192.5 m³
(d) 595 m³
Show Answer & Explanation

Answer: (a) 402.5 m³

Explanation:
1. Cuboid = 210. Half cylinder = (1/2)(22/7)(3.5²)(10) = 192.5. Total = 402.5 m³.

Teacher's Note:
1. Add cuboid and half-cylinder.

Question 17: For the same godown (cuboid 10 m × 7 m × 3 m under a half-cylinder of radius 3.5 m and length 10 m), the total internal surface area excluding the floor is:
(a) 250.5 m²
(b) 220.5 m²
(c) 210 m²
(d) 302.5 m²
Show Answer & Explanation

Answer: (a) 250.5 m²

Explanation:
1. Long walls: 2 × 10 × 3 = 60. End walls: 2 × 7 × 3 = 42. Two semicircular ends: 2 × (1/2)π(3.5²) = 38.5. Curved roof: π(3.5)(10) = 110. Total = 250.5 m².

Teacher's Note:
1. Do not count the floor.

Question 18: A cone is 30 cm high. A small cone is cut off at the top by a plane parallel to the base and its volume is 1/27 of the given cone. The height above the base at which the section is made is:
(a) 10 cm
(b) 27 cm
(c) 15 cm
(d) 20 cm
Show Answer & Explanation

Answer: (d) 20 cm

Explanation:
1. Volumes in ratio 1/27 means heights in ratio 1/3, so the small cone has height 10 cm. The section is 30 − 10 = 20 cm above the base.

Teacher's Note:
1. Volume ratio is the cube of the height ratio.

Question 19: A bucket 8 cm high is a frustum of a cone with lower and upper radii 3 cm and 9 cm. The height of the cone of which the bucket is a part is:
(a) 16 cm
(b) 10 cm
(c) 12 cm
(d) 24 cm
Show Answer & Explanation

Answer: (c) 12 cm

Explanation:
1. The height H of the full cone satisfies (H − 8)/H = 3/9, so H = 12 cm.

Teacher's Note:
1. Similar triangles.

Question 20: A bucket 8 cm high is a frustum with lower radius 3 cm and upper radius 9 cm (closed at the lower end, open at the top). The area of copper sheet needed is:
(a) \( 120\pi \) cm²
(b) \( 129\pi \) cm²
(c) \( 210\pi \) cm²
(d) \( 147\pi \) cm²
Show Answer & Explanation

Answer: (b) \( 129\pi \) cm²

Explanation:
1. Slant height = √(64 + 36) = 10 cm. Curved surface = π(3 + 9)(10) = 120π. Bottom = π × 9 = 9π. Total = 129π cm².

Teacher's Note:
1. Open top: add only the bottom disc.

Question 21: A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid, a cone on a hemisphere of radius 3.5 cm with cone height outside the hemisphere 5 cm, is immersed. The volume of water left (π = 22/7) is:
(a) 245 cm³
(b) 616 cm³
(c) 49 cm³
(d) 770 cm³
Show Answer & Explanation

Answer: (b) 616 cm³

Explanation:
1. Tub = π(25)(9.8) = 245π. Solid = (2/3)π(3.5)³ + (1/3)π(3.5)²(5) = 28.58π + 20.42π = 49π. Left = 196π = 616 cm³.

Teacher's Note:
1. Volume left = tub minus solid.

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Multiple Choice Questions (MCQs) for Class 10 Mathematics Chapter 12 Surface Areas and Volumes

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