CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set 24

Answer: (a) 512

Explanation:
1. Volume of the big sphere = (4/3)π(8)³ and of one ball = (4/3)π(1)³. Number = 8³/1³ = 512.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (d) 0.1 cm

Explanation:
1. Rod volume = π(2)²(100) = 400π cm³. Each wire: πr²(20), so 2000 x 20 x r² = 400 and r² = 0.01, giving r = 0.1 cm.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (c) 9

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

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (b) 180

Explanation:
1. Cone volume = (1/3)π(3)²(10) = 30π. Ball volume = (4/3)π(1/2)³ = π/6. Number = 30π / (π/6) = 180.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (b) 18

Explanation:
1. Cone = (1/3)π(81)(72) = 1944π. Can = π(9)(12) = 108π. Number = 1944/108 = 18.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (a) \( \frac13 \)

Explanation:
1. Let the radius be r, so the height is 2r. Cylinder = π r²(2r) = 2πr³. Sphere = (4/3)πr³. Unused = 2πr³ - (4/3)πr³ = (2/3)πr³, which is (2/3)/2 = 1/3 of the cylinder.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (d) 51.2 minutes

Explanation:
1. Pipe radius = 0.25 cm, so the water per minute = π(0.25)²(1000) = 62.5π cm³. Cone = (1/3)π(20)²(24) = 3200π cm³. Time = 3200/62.5 = 51.2 minutes.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (c) 26.74 minutes

Explanation:
1. Volume = (2/3)(22/7)(175)³ cm³ = 11,239,583.3 cm³ = 11,239.58 litres. Time = 11,239.58/7 = 1605.65 s = 26.76 minutes.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (c) 0.4 m

Explanation:
1. Earth dug = π(1)²(14) = 14π m³. The ring has inner radius 1 m and outer radius 6 m, so its volume = π(36 - 1)h = 35πh. Then 35h = 14 and h = 0.4 m.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (b) 225000 m²

Explanation:
1. Cross-section = 3 m x 1.2 m = 3.6 m². In 30 minutes the water flowing = 3.6 x 5000 = 18000 m³. Area = 18000 / 0.08 = 225000 m².

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (a) 19.272 litres

Explanation:
1. R = 20, r = 14, h = 21. Volume = (1/3)(22/7)(21)(400 + 196 + 280) = 22 x 876 = 19272 cm³ = 19.272 litres.

Teacher's Note:
1. For a frustum with radii R and r, height h and slant height l: volume = (1/3)π h (R² + r² + R r), curved surface area = π (R + r) l and l = √(h² + (R - r)²).
2. A frustum is the part of a cone left after a smaller cone is cut off by a plane parallel to the base.

Answer: (d) 44.25 Rs.

Explanation:
1. Slant height = √(21² + 6²) = √477 = 21.84 cm. Area = π(20 + 14)(21.84) + π(14)² = 2333.8 + 616 = 2949.8 cm² = 29.498 dm². Cost = 29.498 x 1.50 = Rs. 44.25.

Teacher's Note:
1. For a frustum with radii R and r, height h and slant height l: volume = (1/3)π h (R² + r² + R r), curved surface area = π (R + r) l and l = √(h² + (R - r)²).
2. A frustum is the part of a cone left after a smaller cone is cut off by a plane parallel to the base.

Answer: (d) 1 : 26

Explanation:
1. The small cone has 1/3 of the height, so its volume is (1/3)³ = 1/27 of the whole cone. The frustum is 26/27 of the whole. Ratio = 1 : 26.

Teacher's Note:
1. For a frustum with radii R and r, height h and slant height l: volume = (1/3)π h (R² + r² + R r), curved surface area = π (R + r) l and l = √(h² + (R - r)²).
2. A frustum is the part of a cone left after a smaller cone is cut off by a plane parallel to the base.

Answer: (c) 156 π m²

Explanation:
1. r = √(13² - 5²) = 12 m. CSA = π r l = π x 12 x 13 = 156π m².

Teacher's Note:
1. Curved surface area of a cylinder = 2π r h, of a cone = π r l, of a sphere = 4π r² and of a hemisphere = 2π r², where l = √(r² + h²).
2. Always check which faces are actually present (open top, closed base) before adding areas.

Answer: (b) 2 : 3

Explanation:
1. Take radius r. The cylinder has height r, so its volume is πr³. The hemisphere is (2/3)πr³. Ratio = 2 : 3.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (a) 2.25 cm

Explanation:
1. Sphere = (4/3)π(27) = 36π. Rise = 36π / (π x 16) = 2.25 cm.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (a) 3388 cm³

Explanation:
1. The length 44 cm becomes the circumference, so 2πr = 44 and r = 7 cm. The height is the other side, 22 cm. Volume = (22/7)(49)(22) = 3388 cm³.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (d) 3 : 1

Explanation:
1. Volume is proportional to r²h. Ratio = (3² x 1) : (1² x 3) = 9 : 3 = 3 : 1.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (c) \( \sqrt{\pi} : \sqrt{6} \)

Explanation:
1. 6a² = 4πr² gives a = r√(2π/3). Cube/sphere = a³ / ((4/3)πr³) = (2π/3)^(3/2) / (4π/3) = √(π/6). So the ratio is √π : √6.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (b) 1 : 7

Explanation:
1. The small cone has half the height, so its volume is (1/2)³ = 1/8 of the whole. The frustum has 7/8. Ratio = 1 : 7.

Teacher's Note:
1. For a frustum with radii R and r, height h and slant height l: volume = (1/3)π h (R² + r² + R r), curved surface area = π (R + r) l and l = √(h² + (R - r)²).
2. A frustum is the part of a cone left after a smaller cone is cut off by a plane parallel to the base.

Answer: (b) 3768 cm³

Explanation:
1. Hypotenuse = 25 cm. The common radius = (15 x 20)/25 = 12 cm. The two cones have heights adding to 25, so volume = (1/3)π(144)(25) = 3.14 x 1200 = 3768 cm³.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (a) 1318.8 cm²

Explanation:
1. The radius is 12 cm and the slant heights are 15 cm and 20 cm. Area = π(12)(15) + π(12)(20) = π x 12 x 35 = 3.14 x 420 = 1318.8 cm².

Teacher's Note:
1. Curved surface area of a cylinder = 2π r h, of a cone = π r l, of a sphere = 4π r² and of a hemisphere = 2π r², where l = √(r² + h²).
2. Always check which faces are actually present (open top, closed base) before adding areas.

Answer: (d) 0.05 cm

Explanation:
1. Water displaced = 50 x 0.04 = 2 m³. Base area = 80 x 50 = 4000 m². Rise = 2/4000 = 0.0005 m = 0.05 cm.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (c) 2 hours

Explanation:
1. Water needed = 50 x 44 x 0.07 = 154 m³. Pipe area = (22/7)(0.07)² = 0.0154 m². Flow per hour = 0.0154 x 5000 = 77 m³. Time = 154/77 = 2 hours.

Teacher's Note:
1. Volume of a cuboid = l x b x h, of a cylinder = π r² h, of a cone = (1/3) π r² h, of a sphere = (4/3) π r³ and of a hemisphere = (2/3) π r³.
2. When a solid is melted and recast, its volume stays the same, so the number of pieces = volume of the big solid / volume of one piece.

Answer: (c) 9735 m²

Explanation:
1. Cylinder part: 2π(52.5)(3) = 315π. Cone part: π(52.5)(53) = 2782.5π. Total = 3097.5π = 3097.5 x 22/7 = 9735 m².

Teacher's Note:
1. Curved surface area of a cylinder = 2π r h, of a cone = π r l, of a sphere = 4π r² and of a hemisphere = 2π r², where l = √(r² + h²).
2. Always check which faces are actually present (open top, closed base) before adding areas.