CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set 22

Practice MCQs for Class 10 Mathematics Chapter 12 Surface Areas and Volumes

Explore reliable objective questions for Chapter 12 Surface Areas and Volumes tailored for Class 10 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

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Multiple Choice Questions

Question 1: A cube of edge 9 cm is completely immersed in a rectangular vessel containing water. The base of the vessel is 15 cm by 12 cm. The rise in the water level is:
(a) 4.05 cm
(b) 3.60 cm
(c) 4.50 cm
(d) 5.40 cm
Show Answer & Explanation

Answer: (a) 4.05 cm

Explanation:
1. The volume of water displaced equals the cube's volume = 9 x 9 x 9 = 729 cm³. Rise = 729/(15 x 12) = 729/180 = 4.05 cm.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 2: Three cubes of edges 3 cm, 4 cm and 5 cm are melted to form a single cube. The surface area of the new cube is:
(a) 150 cm²
(b) 180 cm²
(c) 196 cm²
(d) 216 cm²
Show Answer & Explanation

Answer: (d) 216 cm²

Explanation:
1. The total volume = 27 + 64 + 125 = 216 cm³, so the edge = 6 cm. Surface area = 6 x 6² = 216 cm².

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 3: Water flows into a tank 150 m x 100 m at its base through a pipe of cross-section 2 dm by 1.5 dm at 15 km per hour. The time in which the water level in the tank will be 3 m deep is:
(a) 50 hours
(b) 150 hours
(c) 100 hours
(d) 200 hours
Show Answer & Explanation

Answer: (c) 100 hours

Explanation:
1. Volume needed = 150 x 100 x 3 = 45000 m³. Cross-section = 0.2 m x 0.15 m = 0.03 m², so the flow per hour = 0.03 x 15000 = 450 m³. Time = 45000/450 = 100 hours.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 4: A right circular cone of height 3.6 cm and base radius 1.6 cm is melted and recast into a right circular cone with base radius 1.2 cm. The height of the new cone is:
(a) 4.8 cm
(b) 6.4 cm
(c) 5.6 cm
(d) 8.1 cm
Show Answer & Explanation

Answer: (b) 6.4 cm

Explanation:
1. Equating volumes: (1/3)π(1.6)²(3.6) = (1/3)π(1.2)²h, so h = 2.56 x 3.6/1.44 = 6.4 cm.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 5: A solid cube of side 7 cm is melted to make a cone of height 5 cm. The radius of the base of the cone, to two decimal places, is:
(a) 6.54 cm
(b) 8.09 cm
(c) 7.64 cm
(d) 9.12 cm
Show Answer & Explanation

Answer: (b) 8.09 cm

Explanation:
1. (1/3)πr²(5) = 343 gives r² = 1029/(5π) = 65.5, so r = 8.09 cm.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 6: A toy is a right circular cylinder (height 13 cm, radius 5 cm) with a hemisphere on one end and a cone of height 12 cm on the other, both with the radius of the cylinder. The total surface area of the toy (use π = 22/7) is:
(a) 770 cm²
(b) 660 cm²
(c) 880 cm²
(d) 550 cm²
Show Answer & Explanation

Answer: (a) 770 cm²

Explanation:
1. The cone's slant height = √(25 + 144) = 13 cm. Total = cone CSA π(5)(13) + cylinder CSA 2π(5)(13) + hemisphere CSA 2π(5)² = (65 + 130 + 50)π = 245π = 770 cm².

Teacher's Note:
1. Curved surface area of a cylinder = 2 pi r h, of a cone = pi r l, of a sphere = 4 pi r^2 and of a hemisphere = 2 pi r^2, where l = square root of (r^2 + h^2).

Question 7: A solid sphere of radius 3 cm is melted and cast into small spherical balls each of diameter 0.6 cm. The number of balls is:
(a) 100
(b) 900
(c) 3000
(d) 1000
Show Answer & Explanation

Answer: (d) 1000

Explanation:
1. Radius of a ball = 0.3 cm. Number = (3/0.3)³ = 10³ = 1000.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 8: Three solid iron spheres of diameters 2 cm, 12 cm and 16 cm are melted into a single solid sphere. The radius of the new sphere is:
(a) 8 cm
(b) 10 cm
(c) 9 cm
(d) 15 cm
Show Answer & Explanation

Answer: (c) 9 cm

Explanation:
1. The radii are 1, 6, 8, so R³ = 1 + 216 + 512 = 729 and R = 9 cm.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 9: How many spherical bullets, each 4 cm in diameter, can be made out of a solid cube of lead of edge 44 cm? (Use π = 22/7)
(a) 2451
(b) 2514
(c) 2541
(d) 5082
Show Answer & Explanation

Answer: (c) 2541

Explanation:
1. Volume of cube = 44³ = 85184 cm³. Volume of a bullet = (4/3)(22/7)(2)³ = 704/21 cm³. Number = 85184 x 21/704 = 2541.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 10: A sphere of diameter 6 cm is dropped into a cylindrical vessel of diameter 12 cm partly filled with water. When the sphere is completely submerged, the water level rises by:
(a) 2 cm
(b) 1 cm
(c) 3 cm
(d) 4 cm
Show Answer & Explanation

Answer: (b) 1 cm

Explanation:
1. Volume of sphere = (4/3)π(27) = 36π. Rise = 36π/(π x 36) = 1 cm.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 11: A spherical cannon ball 28 cm in diameter is melted and cast into a right circular cone of base diameter 35 cm. The height of the cone, to one decimal place, is:
(a) 35.8 cm
(b) 28.0 cm
(c) 32.0 cm
(d) 41.2 cm
Show Answer & Explanation

Answer: (a) 35.8 cm

Explanation:
1. (4/3)π(14)³ = (1/3)π(17.5)²h, so h = 4 x 2744/306.25 = 35.84, about 35.8 cm.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 12: The length of a classroom is twice its height and its breadth is 1 1/2 times its height. The cost of white-washing the walls at Rs. 1.60 per square metre is Rs. 179.20. The cost of tiling the floor at Rs. 6.75 per square metre, in rupees, is:
(a) 288
(b) 342
(c) 432
(d) 324
Show Answer & Explanation

Answer: (d) 324

Explanation:
1. Walls area = 179.20/1.60 = 112 m². With height h, length 2h and breadth 1.5h, the wall area is 2(3.5h)h = 7h² = 112, so h = 4, l = 8, b = 6. Floor area = 48 m² and cost = 48 x 6.75 = Rs. 324.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 13: A room is half as long again as it is broad. The cost of carpeting the room at Rs. 3.25 per m² is Rs. 175.50, and the cost of papering the walls at Rs. 1.40 per m² is Rs. 240.80, with 8 m² occupied by a door and two windows. The height of the room is:
(a) 4 m
(b) 5 m
(c) 8 m
(d) 6 m
Show Answer & Explanation

Answer: (d) 6 m

Explanation:
1. Carpet area = 175.50/3.25 = 54 m². With l = 1.5b, 1.5b² = 54 gives b = 6 and l = 9. Papered area = 240.80/1.40 = 172 m², so wall area = 172 + 8 = 180 m² = 2(9 + 6)h = 30h, giving h = 6 m.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 14: A rectangular tank is 225 m by 162 m at the base. Water flows into it through an aperture 60 cm by 45 cm so that the level rises by 20 cm in 5 hours. The speed of the water in metres per hour is:
(a) 2700 m/h
(b) 4500 m/h
(c) 5400 m/h
(d) 6750 m/h
Show Answer & Explanation

Answer: (c) 5400 m/h

Explanation:
1. Volume added = 225 x 162 x 0.2 = 7290 m³. Aperture area = 0.6 x 0.45 = 0.27 m². Speed = 7290/(0.27 x 5) = 5400 m/h.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 15: An agricultural field is a rectangle 20 m by 14 m. A pit 6 m long, 3 m wide and 2.5 m deep is dug in a corner and the earth is spread uniformly over the remaining area of the field. The level of the field is raised by about:
(a) 15.00 cm
(b) 17.18 cm
(c) 18.75 cm
(d) 21.50 cm
Show Answer & Explanation

Answer: (b) 17.18 cm

Explanation:
1. Earth dug = 6 x 3 x 2.5 = 45 m³. Remaining area = 280 - 18 = 262 m². Rise = 45/262 m = 0.1718 m, which is 17.18 cm.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 16: A rectangular sheet of paper 44 cm x 18 cm is rolled along its length to form a cylinder. The radius of the cylinder (use π = 22/7) is:
(a) 7 cm
(b) 4 cm
(c) 14 cm
(d) 22 cm
Show Answer & Explanation

Answer: (a) 7 cm

Explanation:
1. The length 44 cm becomes the circumference: 2πr = 44, so r = 7 cm.

Teacher's Note:
1. Curved surface area of a cylinder = 2 pi r h, of a cone = pi r l, of a sphere = 4 pi r^2 and of a hemisphere = 2 pi r^2, where l = square root of (r^2 + h^2).

Question 17: A wooden toy is a cone surmounted on a hemisphere. The base diameter of the cone is 6 cm and its height is 4 cm. The cost of painting the toy at Rs. 5 per 1000 cm², in paise (use π = 22/7), is:
(a) 51.86 paise
(b) 41.49 paise
(c) 62.23 paise
(d) 77.79 paise
Show Answer & Explanation

Answer: (a) 51.86 paise

Explanation:
1. Slant height = √(9 + 16) = 5 cm. Surface = π(3)(5) + 2π(9) = 33π = 103.71 cm². Cost = 103.71 x 5/1000 = Rs. 0.5186 = 51.86 paise.

Teacher's Note:
1. Curved surface area of a cylinder = 2 pi r h, of a cone = pi r l, of a sphere = 4 pi r^2 and of a hemisphere = 2 pi r^2, where l = square root of (r^2 + h^2).

Question 18: A cylindrical container of radius 6 cm and height 15 cm is filled with ice-cream to be distributed equally among 10 children in cones with hemispherical tops. The height of each conical part is four times the radius of its base. The radius of the ice-cream cone is:
(a) 2 cm
(b) 4 cm
(c) 6 cm
(d) 3 cm
Show Answer & Explanation

Answer: (d) 3 cm

Explanation:
1. Each child gets 540π/10 = 54π cm³. Cone + hemisphere: (1/3)πr²(4r) + (2/3)πr³ = 2πr³ = 54π, so r³ = 27 and r = 3 cm.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 19: A solid wooden toy is a right circular cone mounted on a hemisphere. The radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm. The volume of the toy (use π = 22/7) is:
(a) 177.41 cm³
(b) 199.58 cm³
(c) 266.11 cm³
(d) 332.64 cm³
Show Answer & Explanation

Answer: (c) 266.11 cm³

Explanation:
1. The cone height = 10.2 - 4.2 = 6 cm. Volume = (1/3)π(4.2)²(6) + (2/3)π(4.2)³ = π(35.28 + 49.392) = 84.672π = 266.11 cm³.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 20: A vessel is a hemispherical bowl mounted on a hollow cylinder. The diameter of the sphere is 14 cm and the total height of the vessel is 13 cm. Its capacity (use π = 22/7) is:
(a) 1078.00 cm³
(b) 1642.67 cm³
(c) 1232.00 cm³
(d) 2156.00 cm³
Show Answer & Explanation

Answer: (b) 1642.67 cm³

Explanation:
1. The hemisphere radius = 7 cm, so the cylinder height = 13 - 7 = 6 cm. Capacity = (2/3)π(343) + π(49)(6) = (686/3 + 294)π = (1568/3)π = 1642.67 cm³.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

Question 21: A solid is a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Its total surface area (use π = 22/7) is:
(a) 374 cm²
(b) 418 cm²
(c) 462 cm²
(d) 528 cm²
Show Answer & Explanation

Answer: (b) 418 cm²

Explanation:
1. The cylinder height = 19 - 7 = 12 cm and r = 3.5 cm. Surface = 2π(3.5)(12) + 4π(3.5)² = 84π + 49π = 133π = 418 cm².

Teacher's Note:
1. Curved surface area of a cylinder = 2 pi r h, of a cone = pi r l, of a sphere = 4 pi r^2 and of a hemisphere = 2 pi r^2, where l = square root of (r^2 + h^2).

Question 22: A solid is a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Its volume (use π = 22/7) is:
(a) 641.67 cm³
(b) 538.33 cm³
(c) 718.67 cm³
(d) 1026.67 cm³
Show Answer & Explanation

Answer: (a) 641.67 cm³

Explanation:
1. Volume = π(3.5)²(12) + (4/3)π(3.5)³ = π(147 + 57.1667) = 204.1667π = 641.67 cm³.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cube = a^3, of a cylinder = pi r^2 h, of a cone = (1/3) pi r^2 h, of a sphere = (4/3) pi r^3 and of a hemisphere = (2/3) pi r^3.
2. When a solid is melted and recast, its volume stays the same.

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Chapter 12 Surface Areas and Volumes Objective Questions & Solutions for Class 10 Mathematics

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FAQs

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