CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set 21

Mathematics Objective Questions and Answers: 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.

Download Chapter 12 Surface Areas and Volumes MCQs with Answers

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: The internal and external radii of a hollow sphere are 3 cm and 5 cm. It is melted into a solid cylinder of height \( 2\frac23 \) cm. The diameter of the cylinder is:
(a) 14 cm
(b) 7 cm
(c) 28 cm
(d) 10 cm
Show Answer & Explanation

Answer: (a) 14 cm

Explanation:
1. Shell volume = (4/3)π(125 − 27) = 392π/3. Cylinder: π r²(8/3) = 392π/3 gives r² = 49, r = 7, diameter 14 cm.

Teacher's Note:
1. Equate volumes.

Question 2: For the same cylinder (radius 7 cm, height 8/3 cm), the curved surface area (π = 22/7) is about:
(a) 58.67 cm²
(b) 176 cm²
(c) 154 cm²
(d) 117.33 cm²
Show Answer & Explanation

Answer: (d) 117.33 cm²

Explanation:
1. 2πrh = 2(22/7)(7)(8/3) = 117.33 cm².

Teacher's Note:
1. Curved surface of a cylinder.

Question 3: The ratio of the volumes of two cones with the same height is:
(a) \( r_1 : r_2 \)
(b) \( r_1^3 : r_2^3 \)
(c) \( r_1^2 : r_2^2 \)
(d) \( r_2^2 : r_1^2 \)
Show Answer & Explanation

Answer: (c) \( r_1^2 : r_2^2 \)

Explanation:
1. Volume = (1/3)π r² h with equal h, so the ratio is r₁² : r₂².

Teacher's Note:
1. Only r changes.

Question 4: The number of spherical lead bullets, each of diameter 4 cm, that can be made from a lead cube of edge 44 cm is (π = 22/7):
(a) 2462
(b) 2000
(c) 2541
(d) 1682
Show Answer & Explanation

Answer: (c) 2541

Explanation:
1. Cube = 44³ = 85184 cm³. One bullet = (4/3)(22/7)(8) = 704/21 cm³. Number = 85184 × 21/704 = 2541.

Teacher's Note:
1. Divide the cube volume by one bullet's volume.

Question 5: A glass cylinder of diameter 20 cm contains water to a height of 9 cm. A metal cube of edge 8 cm is completely immersed. The water level rises by about (π = 3.14):
(a) 4.3 cm
(b) 1.6 cm
(c) 2.4 cm
(d) 3 cm
Show Answer & Explanation

Answer: (b) 1.6 cm

Explanation:
1. Rise = 512/(π × 100) = 512/314 ≈ 1.63 cm.

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

Question 6: The difference between the outer and inner curved surface areas of a cylindrical metal pipe 14 m long is 44 m², and the pipe is made of 99 m³ of metal. The outer and inner radii are (π = 22/7):
(a) 2.5 m and 2 m
(b) 5 m and 7 m
(c) 9 m and 7 m
(d) 3 m and 2 m
Show Answer & Explanation

Answer: (a) 2.5 m and 2 m

Explanation:
1. 2π(R − r)(14) = 44 gives R − r = 0.5. π(R² − r²)(14) = 99 gives R + r = 99/(44 × 0.5) = 4.5. So R = 2.5 and r = 2.

Teacher's Note:
1. Use difference of areas and the volume of material.

Question 7: A hemisphere and a cone have equal bases and equal heights. The ratio of their curved surface areas is:
(a) 1 : 3
(b) 2 : 3
(c) 1 : 4
(d) \( \sqrt2 : 1 \)
Show Answer & Explanation

Answer: (d) \( \sqrt2 : 1 \)

Explanation:
1. Equal heights and bases means h = r, so the cone's slant height is r√2. Hemisphere CSA = 2πr²; cone CSA = πr(r√2) = √2 πr². The ratio is 2 : √2 = √2 : 1.

Teacher's Note:
1. Hemisphere height equals its radius.

Question 8: A cylinder, a cone and a hemisphere have equal bases and the same height. The ratio of their volumes is:
(a) 1 : 2 : 3
(b) 4 : 5 : 6
(c) 2 : 3 : 1
(d) 3 : 1 : 2
Show Answer & Explanation

Answer: (d) 3 : 1 : 2

Explanation:
1. The height equals the base radius r, so the volumes are πr³, (1/3)πr³ and (2/3)πr³, in the ratio 3 : 1 : 2.

Teacher's Note:
1. Same height as radius.

Question 9: A metallic hemisphere is melted and recast as a cone with the same base radius R. If H is the cone's height, H/R equals:
(a) 1 / 2
(b) 4 / 6
(c) 2 / 1
(d) 1 / 3
Show Answer & Explanation

Answer: (c) 2 / 1

Explanation:
1. (2/3)πR³ = (1/3)πR²H gives H = 2R, so H/R = 2.

Teacher's Note:
1. Equate volumes.

Question 10: A toy is a cone surmounted by a hemisphere of the same radius. The common diameter is 3.5 cm and the total height is 5 cm. The height of the conical part is:
(a) 3.5 cm
(b) 3.25 cm
(c) 2.5 cm
(d) 3.75 cm
Show Answer & Explanation

Answer: (b) 3.25 cm

Explanation:
1. Hemisphere radius = 1.75 cm. Cone height = 5 − 1.75 = 3.25 cm.

Teacher's Note:
1. Total height minus the radius.

Question 11: For the same toy, the curved surface area of the hemisphere is (π = 22/7):
(a) \( \frac{77}{4} \) cm² (19.25)
(b) 38.5 cm²
(c) 9.625 cm²
(d) 77 cm²
Show Answer & Explanation

Answer: (a) \( \frac{77}{4} \) cm² (19.25)

Explanation:
1. 2π r² = 2(22/7)(1.75²) = 19.25 = 77/4 cm².

Teacher's Note:
1. Curved surface of a hemisphere = 2πr².

Question 12: For the same toy, the total surface area of the top is about (π = 22/7):
(a) 39.6 cm²
(b) 19.25 cm²
(c) 58.9 cm²
(d) 20.3 cm²
Show Answer & Explanation

Answer: (a) 39.6 cm²

Explanation:
1. Hemisphere = 19.25. Cone = π(1.75)(3.69) = 20.3. Total ≈ 39.6 cm².

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

Question 13: The total surface area of a cube equals numerically the surface area of a sphere. The ratio of the volume of the cube to the volume of the sphere is:
(a) \( \frac{\pi}{6} \) : 1
(b) \( \frac{\pi}{216} \) : 1
(c) \( \frac{6}{\pi} \) : 1
(d) \( \sqrt{\frac{\pi}{6}} \) : 1
Show Answer & Explanation

Answer: (d) \( \sqrt{\frac{\pi}{6}} \) : 1

Explanation:
1. 6a² = 4πr², so a = r√(2π/3). Ratio = a³/((4/3)πr³) = (2π/3)^(3/2)(3/(4π)) = √(π/6) ≈ 0.724.

Teacher's Note:
1. Express a in terms of r.

Question 14: The length, breadth and height of a cuboid are increased by x%, y% and z%. The percentage increase in volume is:
(a) \( x+y+z \)
(b) \( x+y+z+\frac{xy+yz+zx}{100} \)
(c) \( x+y+z+\frac{xy+yz+zx}{100}+\frac{xyz}{100^2} \)
(d) \( xyz \)
Show Answer & Explanation

Answer: (c) \( x+y+z+\frac{xy+yz+zx}{100}+\frac{xyz}{100^2} \)

Explanation:
1. New volume = lbh(1 + x/100)(1 + y/100)(1 + z/100). Expanding gives the increase x + y + z + (xy + yz + zx)/100 + xyz/100² in percent.

Teacher's Note:
1. Multiply the three factors.

Question 15: 748 cm³ of metal is used to make a cylindrical pipe of length 14 cm and external radius 9 cm. The thickness of the pipe is (π = 22/7):
(a) 2 cm
(b) 1 cm
(c) 2.5 cm
(d) 1.5 cm
Show Answer & Explanation

Answer: (b) 1 cm

Explanation:
1. π(81 − r²)(14) = 748 gives 81 − r² = 17, r² = 64, r = 8 cm. Thickness = 9 − 8 = 1 cm.

Teacher's Note:
1. Material = π(R² − r²)h.

Question 16: A hollow sphere of internal and external diameters 4 cm and 8 cm is melted into a cone of base diameter 8 cm. The height of the cone is:
(a) 12 cm
(b) 14 cm
(c) 16 cm
(d) 10 cm
Show Answer & Explanation

Answer: (b) 14 cm

Explanation:
1. Shell = (4/3)π(64 − 8) = 224π/3. Cone: (1/3)π(16)h. So h = 224/16 = 14 cm.

Teacher's Note:
1. Equate volumes.

Question 17: For the same tent (diameter 24 m, height 16 m), the canvas required (π = 22/7) is:
(a) 754.29 m²
(b) 745.29 m²
(c) 452.92 m²
(d) 755.87 m²
Show Answer & Explanation

Answer: (a) 754.29 m²

Explanation:
1. CSA = π r l = (22/7)(12)(20) = 754.29 m².

Teacher's Note:
1. Curved surface of the cone.

Question 18: For the same tent, if each person needs 54 m³ of air, the largest number of persons it can hold is (π = 22/7):
(a) 45
(b) 49
(c) 48
(d) 44
Show Answer & Explanation

Answer: (d) 44

Explanation:
1. Volume = (1/3)(22/7)(144)(16) = 2413.7 m³. 2413.7/54 = 44.7, so at most 44 persons.

Teacher's Note:
1. Divide volume by 54 and round down.

Question 19: For the same article (cylinder height 10 cm, radius 3.5 cm, hemisphere scooped from each end), the total surface area (π = 22/7) is:
(a) 344 cm²
(b) 379 cm²
(c) 374 cm²
(d) 342 cm²
Show Answer & Explanation

Answer: (c) 374 cm²

Explanation:
1. Curved surface of cylinder = 2π(3.5)(10) = 70π. Two inner hemispherical surfaces = 2 × 2π(3.5²) = 49π. Total = 119π = 374 cm².

Teacher's Note:
1. The flat ends are removed.

Question 20: For the same article, the total volume of the two scooped-out hemispheres is:
(a) \( \frac{162.5\pi}{3} \) cm³
(b) \( \frac{143.9\pi}{3} \) cm³
(c) \( \frac{171.5\pi}{3} \) cm³
(d) \( \frac{176.8\pi}{3} \) cm³
Show Answer & Explanation

Answer: (c) \( \frac{171.5\pi}{3} \) cm³

Explanation:
1. Each hemisphere = (2/3)π(3.5³) = (2/3)π(42.875). Two = (4/3)π(42.875) = 171.5π/3 cm³.

Teacher's Note:
1. Two hemispheres make a sphere.

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

Class 10 Mathematics Chapter 12 Surface Areas and Volumes Objective Test Questions

Test your conceptual understanding of Chapter 12 Surface Areas and Volumes with these targeted multiple-choice questions. Designed in alignment with the latest CBSE curriculum for Class 10 Mathematics, these problem sets build accuracy and prepare students for objective exams.

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FAQs

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