IMO Maths Olympiad Class 10 Chapter 13 Surface Areas and Volumes MCQs with Answers

Answer: (c) 377.14 cm²

Explanation:
1. Slant height of the cone l = \( \sqrt{12^2 + 9^2} \) = 15 cm, so its curved surface area = π × 9 × 15 = 135π cm².
2. The water forms a similar cone of height 4 cm, so its radius is 9 × 4/12 = 3 cm and its slant height is 5 cm; area in contact = π × 3 × 5 = 15π cm².
3. Area not in contact = 135π − 15π = 120π = \( 120 \times \dfrac{22}{7} \) ≈ 377.14 cm².

Teacher's Note:
1. Similar cones: radius and slant height scale by the same ratio as the heights.
2. Only the curved surface touches the water here, not the open top.

Answer: (b) 48510 cm³

Explanation:
1. Volume of a frustum = \( \dfrac{1}{3}\pi h\left(r_1^2 + r_2^2 + r_1 r_2\right) \) (taking π = 22/7).
2. \( \dfrac{1}{3} \times \dfrac{22}{7} \times 45 \times \left(784 + 49 + 196\right) \) = \( \dfrac{22}{7} \times 15 \times 1029 \).
3. Capacity = 48510 cm³.

Teacher's Note:
1. 1029 = 7 × 147, so the 7 in the denominator cancels exactly.
2. Do not forget the cross term r₁r₂ in the frustum formula.

Answer: (a) 4 : 9

Explanation:
1. \( \dfrac{r_1^3}{r_2^3} = \dfrac{8}{27} \), so \( \dfrac{r_1}{r_2} = \dfrac{2}{3} \).
2. Surface areas are in the ratio of the squares of the radii: \( \dfrac{4\pi r_1^2}{4\pi r_2^2} = \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9} \).
3. Ratio of surface areas = 4 : 9.

Teacher's Note:
1. Volume goes as the cube of the radius and area as the square.
2. Take the cube root of the volume ratio first.

Answer: (b) 4.2 cm

Explanation:
1. Total volume of the cones = \( \dfrac{1}{3}\pi (2.1)^2 (4.1 + 4.3) = \dfrac{1}{3}\pi (2.1)^2 (8.4) \).
2. Equate with the sphere: \( \dfrac{4}{3}\pi r^3 = \dfrac{1}{3}\pi (2.1)^2 (8.4) \Rightarrow r^3 = \dfrac{4.41 \times 8.4}{4} = 9.261 \).
3. So r = 2.1 cm and the diameter is 4.2 cm.

Teacher's Note:
1. Both cones have the same base, so add the heights first.
2. Remember to double the radius to get the diameter.

Answer: (d) 81 cm³

Explanation:
1. Outer volume = 10 × 9 × 2.5 = 225 cm³.
2. The box is open at the top, so the thickness is removed from both ends of the length and breadth but only from the bottom of the height: inner dimensions 9 cm × 8 cm × 2 cm, volume 144 cm³.
3. Volume of metal = 225 − 144 = 81 cm³.

Teacher's Note:
1. Because the top is open, subtract the thickness only once from the height.
2. Metal volume = outer volume − inner volume.

Answer: (a) 20 cm

Explanation:
1. The small cone is similar to the whole cone, so the ratio of volumes is the cube of the ratio of heights: \( \dfrac{h}{30} = \sqrt[3]{\dfrac{1}{27}} = \dfrac{1}{3} \).
2. Height of the small cone = 10 cm, measured from the top (apex).
3. Height of the section above the base = 30 − 10 = 20 cm.

Teacher's Note:
1. The question asks for the height above the base, not the height of the small cone.
2. Volume ratio of similar solids is the cube of the length ratio.

Answer: (d) Only II and either I or III

Explanation:
1. Capacity = πr²h, so both the radius and the height are needed.
2. The radius can be found from statement I (area of base) or from statement III (circumference), so either one is enough.
3. The height comes only from statement II. So II and either I or III are necessary.

Teacher's Note:
1. Check what is needed first, then see which statements supply it.
2. Two different statements may give the same quantity, so only one of them is required.

Answer: (c) \( \dfrac{1}{3}\pi r^3 \)

Explanation:
1. The largest cone has the same base as the hemisphere (radius r) and height equal to the hemisphere's radius, h = r.
2. Volume = \( \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi r^2 \cdot r \).
3. Maximum volume = \( \dfrac{1}{3}\pi r^3 \).

Teacher's Note:
1. The cone's height cannot exceed the radius of the hemisphere.
2. Its volume is exactly half that of the hemisphere.

Answer: (b) 0.4 cm

Explanation:
1. Volume of water displaced = π × 2² × 0.8 = 3.2π cm³.
2. Volume of one ball = \( \dfrac{3.2\pi}{300} \), so \( \dfrac{4}{3}\pi r^3 = \dfrac{3.2\pi}{300} \Rightarrow r^3 = \dfrac{8}{1000} \).
3. r = 0.2 cm, so the diameter is 0.4 cm.

Teacher's Note:
1. The rise of water level gives the total volume of the balls.
2. The answer asks for the diameter, not the radius.

Answer: (d) ₹ 96.28

Explanation:
1. External radius = 12.5 cm and internal radius = 12 cm. Outer curved area = 2π(12.5)² and inner curved area = 2π(12)².
2. The rim (ring on top) has area π(12.5² − 12²) = π × 12.25 = 38.5 cm² (π = 22/7).
3. Total area = \( \dfrac{6875}{7} + \dfrac{6336}{7} + 38.5 \) ≈ 1925.8 cm², so the cost ≈ 0.05 × 1925.8 ≈ ₹ 96.28.

Teacher's Note:
1. Paint everything: the outer surface, the inner surface and the circular rim.
2. The answer matches option (d) when π = 22/7.

Answer: (a) 12960

Explanation:
1. Volume of the wall = 24 × 0.4 × 6 = 57.6 m³; bricks occupy 9/10 of it = 51.84 m³.
2. Volume of one brick = 25 × 16 × 10 = 4000 cm³ = 0.004 m³.
3. Number of bricks = \( \dfrac{51.84}{0.004} \) = 12960.

Teacher's Note:
1. Use the same unit (m³ or cm³) for the wall and for the brick.
2. Mortar takes 1/10 of the volume, so bricks fill only 9/10.

Answer: (b) 15 m

Explanation:
1. Base area of the tent = 150 × 4 = 600 m², so πr² = 600.
2. Volume of air = 150 × 20 = 3000 m³, so \( \dfrac{1}{3}\pi r^2 h = 3000 \).
3. \( \dfrac{1}{3} \times 600 \times h = 3000 \) gives 200h = 3000, so h = 15 m.

Teacher's Note:
1. You never need r itself, since the base area π r² is already known.
2. Total space and air are for all 150 persons.

Answer: (d) 163.86 cm²

Explanation:
1. Surface area of the cube = 6 × 5² = 150 cm². The hemisphere (radius 2.1 cm) covers a circle of the top face and adds its curved surface.
2. Total = 150 − π(2.1)² + 2π(2.1)² = 150 + π(2.1)² = 150 + \( \dfrac{22}{7} \) × 4.41.
3. = 150 + 13.86 = 163.86 cm².

Teacher's Note:
1. Subtract the covered base circle, add the curved surface: the net change is one πr².
2. Use the radius 2.1 cm, not the diameter.

Answer: (d) All of these.

Explanation:
1. (a) is incorrect: cutting a hemisphere out of the top removes a circle πr² but exposes a curved surface 2πr², so the surface area increases.
2. (b) is incorrect: melting keeps the volume, not the surface area, the same. (c) is incorrect: a length cannot equal an area.
3. All three statements are wrong, so the answer is All of these.

Teacher's Note:
1. Melting and recasting preserves volume only, never surface area.
2. A length is measured in cm and an area in cm², so they cannot be equal.

Answer: (a) 65.56 m³, 71.83 m²

Explanation:
1. Radius = 2.15 m. A vertical angle of 90° means the half-angle is 45°, so the cone's height = r = 2.15 m and its slant height = 2.15√2 ≈ 3.04 m.
2. Volume = \( \pi r^2 h + \dfrac{1}{3}\pi r^2 \cdot r = 3.14 \times 4.6225 \times (3.8 + 0.7167) \) ≈ 65.56 m³.
3. Curved area = 2πrh + πrl = 51.31 + 20.53 ≈ 71.8 m², so the answer is 65.56 m³, 71.83 m².

Teacher's Note:
1. A right-angled vertex means the cone's height equals its base radius.
2. Add the curved surfaces of the cylinder and the cone, and leave out the joining circle.