CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set 14

Answer: (a) 5 cm

Explanation:
1. Volume of the cuboid = 15 × 11 × 10.5 = 1732.5 cm³.
2. \( \frac{22}{7} \times 10.5^2 \times h = 1732.5 \) gives \( 346.5\,h = 1732.5 \), so h = 5 cm.

Teacher's Note:
1. Equal capacity means equal volumes.

Answer: (c) Rs. 924

Explanation:
1. Height of the cylindrical part = 105 − 2 × 14 = 77 cm.
2. Surface area = CSA of cylinder + 2 hemispheres = \( 2 \times \frac{22}{7} \times 14 \times 77 + 4 \times \frac{22}{7} \times 14^2 = 6776 + 2464 = 9240 \) cm².
3. Cost = 9240 × 0.1 = Rs. 924.

Teacher's Note:
1. Subtract both hemisphere radii from the total height.
2. Two hemispheres make one sphere: 4πr².

Answer: (d) \( 126.62 \text{ cm}^3 \)

Explanation:
1. r = 1.75 cm. Hemisphere: \( \frac{2}{3} \times 3.14 \times 1.75^3 = 11.22 \) cm³.
2. Cylinder: \( 3.14 \times 1.75^2 \times 10 = 96.16 \) cm³.
3. Cone: \( \frac{1}{3} \times 3.14 \times 1.75^2 \times 6 = 19.23 \) cm³.
4. Total = 126.62 cm³.

Teacher's Note:
1. Add the three volumes.

Answer: (b) 9 : 4

Explanation:
1. Volume = \( \frac{1}{3}\pi r^2 h \) with the same h, so volume ∝ r².
2. Ratio = 3² : 2² = 9 : 4.

Teacher's Note:
1. Equal heights make volume proportional to r².

Answer: (a) \( 3432 \text{ cm}^2 \)

Explanation:
1. Area painted in one rotation = CSA of the cylinder = \( 2\pi r h = \frac{22}{7} \times 7 \times 26 = 572 \) cm².
2. In 6 rotations = 6 × 572 = 3432 cm².

Teacher's Note:
1. The roller paints its curved surface once per turn.
2. Area is measured in cm², not cm³.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. TSA = \( 2\pi r(h + r) = 2 \times \frac{22}{7} \times 14 \times 44 = 3872 \) cm². So A is true.
2. R is the correct formula used to get this value, so R explains A.

Teacher's Note:
1. Use TSA = 2πr(h + r).

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Earth dug out = \( \pi \times 2^2 \times 21 = 84\pi \) m³.
2. Ring: outer radius 5 m, inner radius 2 m, area = π(25 − 4) = 21π m². Height = 84π ÷ 21π = 4 m. A is true.
3. R is a true formula, but A also needs the ring area, so R alone is not the full explanation.

Teacher's Note:
1. Volume of earth = volume of embankment.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( r_1^3 : r_2^3 = 64 : 125 \), so \( r_1 : r_2 = 4 : 5 \).
2. Surface areas = \( r_1^2 : r_2^2 = 16 : 25 \). A is true, and R gives the formulas that were used.

Teacher's Note:
1. Cube root first, then square.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The cuboid is 20 cm × 10 cm × 10 cm, so TSA = 2(200 + 100 + 200) = 1000 cm². A is true.
2. R is the formula applied here, so R explains A.

Teacher's Note:
1. Joining end to end gives length 20 cm.

Answer: (a) 142296

Explanation:
1. Volume filled = \( \frac{7}{8} \times 22^3 = 9317 \) cm³.
2. Volume of one marble = \( \frac{4}{3} \times \frac{22}{7} \times 0.25^3 = \frac{11}{168} \) cm³.
3. Number = \( 9317 \times \frac{168}{11} = 142296 \).

Teacher's Note:
1. Only 7/8 of the cube is filled.

Answer: (c) 700%

Explanation:
1. New radius = 2r, so the new volume = \( \frac{4}{3}\pi (2r)^3 = 8 \times \) old volume.
2. Increase = 7 × old volume = 700%.

Teacher's Note:
1. Volume scales with the cube of the radius.

Answer: (c) \( 96 \text{ cm}^2 \)

Explanation:
1. \( (l + b + h)^2 = l^2 + b^2 + h^2 + 2(lb + bh + hl) \), that is TSA = \( (l + b + h)^2 - d^2 \).
2. TSA = 108 − 12 = 96 cm².

Teacher's Note:
1. Expand the square of the sum.

Answer: (d) 4 : 1

Explanation:
1. The largest sphere has diameter equal to the radius of the hemisphere, so its radius is 3.5 cm.
2. Ratio = \( \frac{\frac{2}{3}\pi \times 7^3}{\frac{4}{3}\pi \times 3.5^3} = \frac{2 \times 343}{4 \times 42.875} = 4 \).

Teacher's Note:
1. Largest sphere inside a hemisphere has radius r/2.

Answer: (b) \( 360 \text{ cm}^2 \)

Explanation:
1. The cuboid is 12 cm × 6 cm × 6 cm.
2. TSA = 2(12 × 6 + 6 × 6 + 12 × 6) = 2(72 + 36 + 72) = 360 cm².

Teacher's Note:
1. Do not just add 2 × 216 and forget the joined faces.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The two plane (flat) faces are joined, so they are hidden and are not part of the outer surface. Hence TSA = CSA of hemisphere + CSA of cone. A is true.
2. R states exactly why this is so, so R is true and explains A.

Teacher's Note:
1. Joined faces are not counted in surface area.

Answer: (a) \( 392\pi \text{ m}^3 \)

Explanation:
1. Capacity = \( \pi \times 7^2 \times 10 = 490\pi \) m³.
2. 80% of it = 0.8 × 490π = 392π m³.

Teacher's Note:
1. Find the full volume first.

Answer: (c) \( 1386\pi \text{ cm}^2 \)

Explanation:
1. Outer diameter = 4 × 4.5 = 18 cm.
2. Outer CSA = \( \pi d h = \pi \times 18 \times 77 = 1386\pi \text{ cm}^2 \).

Teacher's Note:
1. Use the outer diameter, not the inner one.

Answer: (b) 758.95 m

Explanation:
1. Radius = 3 × 240 = 720 m.
2. \( l = \sqrt{240^2 + 720^2} = 240\sqrt{10} = 758.95 \) m.

Teacher's Note:
1. l² = r² + h².

Answer: (a) \( 21600 \text{ m}^3 \)

Explanation:
1. In 30 minutes the water travels 3.6 × 0.5 = 1.8 km = 1800 m.
2. Volume = 8 × 1.5 × 1800 = 21600 m³.

Teacher's Note:
1. Volume = cross-section × length of flow.

Answer: (c) A is true, but R is false.

Explanation:
1. Volume = \( \frac{1}{3}\pi r^2 h \) is proportional to r². Halving r makes the volume \( \frac{1}{4} \) of the original (h unchanged). A is true.
2. The volume is proportional to r², not inversely proportional to r. R is false.

Teacher's Note:
1. Volume depends on r².

Answer: (c) A is true, but R is false.

Explanation:
1. Surface area = 6s² and volume = s³ both increase with s, so A is true.
2. They are proportional to s² and s³, not to s. R is false.

Teacher's Note:
1. Both grow with the side, but not in direct proportion.

Answer: (b) \( 78.57 \text{ m}^2 \)

Explanation:
1. Cloth needed = 2 × CSA of a hemisphere = \( 2 \times 2\pi r^2 = 4 \times \frac{22}{7} \times 2.5^2 \).
2. = 78.57 m².

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

Answer: (b) \( 123.2 \text{ m}^2 \)

Explanation:
1. CSA of one pillar = \( 2\pi r h = 2 \times \frac{22}{7} \times 1.4 \times 7 = 61.6 \) m².
2. For two pillars = 123.2 m².

Teacher's Note:
1. Lateral area means curved surface only.

Answer: (b) 1 : 8

Explanation:
1. Two hemispheres of radius 1 make one sphere: \( \frac{4}{3}\pi \).
2. Sphere of radius 2: \( \frac{4}{3}\pi \times 8 \).
3. Ratio = 1 : 8.

Teacher's Note:
1. Two hemispheres of equal radius form one sphere.

Answer: (c) \( 38500 \text{ cm}^2 \)

Explanation:
1. r = 35 cm, h = 140 cm.
2. Area = \( 2\pi r h + 2\pi r^2 = 2\pi r(h + r) = 2 \times \frac{22}{7} \times 35 \times 175 = 38500 \text{ cm}^2 \).

Teacher's Note:
1. Convert 1.4 m to 140 cm.

Answer: (b) 6 : 1

Explanation:
1. Ratio = \( \frac{\pi r^2 h}{\frac{2}{3}\pi r^3} = \frac{3h}{2r} = \frac{3 \times 140}{2 \times 35} \).
2. = 6 : 1.

Teacher's Note:
1. Cancel πr² from both volumes.

Answer: (c) \( 6\sqrt{3} \) cm

Explanation:
1. Diagonal of a cube = √3 × edge.
2. = 6√3 cm.

Teacher's Note:
1. Space diagonal = √(l² + b² + h²).

Answer: (c) \( 343 \text{ cm}^3 \)

Explanation:
1. Volume = edge³ = 7³ = 343 cm³.

Teacher's Note:
1. Volume of a cube = a³.

Answer: (a) \( 858 \text{ cm}^2 \)

Explanation:
1. CSA of cone = \( \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \) cm².
2. CSA of hemisphere = \( 2\pi r^2 = 308 \) cm².
3. Total = 858 cm².

Teacher's Note:
1. Add the two curved surfaces.

Answer: (b) 27.5 litres

Explanation:
1. Height of cone = 2r, so cone volume = \( \frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3 \) and cylinder volume = \( 3\pi r^3 \).
2. Cylinder ÷ cone = \( 3 \div \frac{2}{3} = \frac{9}{2} \), so cylinder = 4.5 × 5 = 22.5 litres.
3. Whole container = 22.5 + 5 = 27.5 litres.

Teacher's Note:
1. Compare the two volumes as a ratio.

Answer: (b) \( 572 \text{ cm}^2 \)

Explanation:
1. Radius = 7 cm, so the cylinder height = 13 − 7 = 6 cm.
2. Inner surface = \( 2\pi r^2 + 2\pi r h = 2\pi \times 7(7 + 6) = 182\pi \).
3. = 182 × 22/7 = 572 cm².

Teacher's Note:
1. The hollow vessel has no bottom face inside: hemisphere + cylinder wall.

Answer: (a) 1 : 2

Explanation:
1. \( r^3 : R^3 = 8 : 27 \), so \( r : R = 2 : 3 \) and \( R = \frac{3r}{2} \).
2. \( (R - r) : r = \frac{r}{2} : r = 1 : 2 \).

Teacher's Note:
1. Take the cube root of the ratio.

Answer: (b) 100 minutes

Explanation:
1. Volume of tank = \( \pi \times 5^2 \times 2 = 50\pi \) m³.
2. Water per minute = \( \pi \times 0.1^2 \times 50 = 0.5\pi \) m³ (speed 3 km/h = 50 m/min).
3. Time = 50π ÷ 0.5π = 100 minutes.

Teacher's Note:
1. Volume flowing per minute = area × speed.

Answer: (a) \( 7.4\sqrt{5} \) cm

Explanation:
1. \( \pi r^2 = 770 \) gives \( r^2 = 245 \), so \( r = 7\sqrt{5} \) cm.
2. \( \pi r l = 814 \) gives \( l = \frac{814 \times 7}{22 \times 7\sqrt{5}} = \frac{37}{\sqrt{5}} = 7.4\sqrt{5} \) cm.

Teacher's Note:
1. Divide CSA by base area to get l/r = 37/35.

Answer: (b) \( 583 \text{ cm}^2 \)

Explanation:
1. Surface area of the cuboid = 2(150 + 50 + 75) = 550 cm².
2. Curved surface of the hole = \( 2\pi r h = 2 \times \frac{22}{7} \times 3.5 \times 5 = 110 \) cm²; two circular holes on the faces = \( 2 \times \frac{22}{7} \times 3.5^2 = 77 \) cm².
3. Remaining = 550 + 110 − 77 = 583 cm².

Teacher's Note:
1. The hole removes two circles but adds an inner curved wall.

Answer: (a) 24.5 cm

Explanation:
1. Volume of the tube = \( \frac{22}{7} \times 4^2 \times 28 = 1408 \) cm³, holding 3000 g of food.
2. Food eaten in 5 hours = 375 g, which is \( 1408 \times \frac{375}{3000} = 176 \) cm³.
3. Remaining volume = 1232 cm³, so \( \frac{22}{7} \times 16 \times H = 1232 \), giving H = 24.5 cm.

Teacher's Note:
1. Food mass is proportional to volume.

Answer: (a) 3 : 4

Explanation:
1. \( 2\pi r h : \pi r l = 8 : 5 \), so \( \frac{2h}{l} = \frac{8}{5} \) and \( l = \frac{5h}{4} \).
2. \( r = \sqrt{l^2 - h^2} = \sqrt{\frac{25h^2}{16} - h^2} = \frac{3h}{4} \).
3. So r : h = 3 : 4.

Teacher's Note:
1. Use l² = r² + h².

Answer: (a) 1 : 1

Explanation:
1. Cone: \( \frac{1}{3}\pi r^2 (3h) = \pi r^2 h \). Cylinder: \( \pi r^2 h \).
2. Ratio = 1 : 1.

Teacher's Note:
1. The factor 1/3 cancels with the factor 3.

Answer: (d) \( 1056 \text{ cm}^3 \)

Explanation:
1. Cylinder = \( \frac{22}{7} \times 36 \times 14 = 1584 \) cm³; cone = 528 cm³.
2. Remaining = 1584 − 528 = 1056 cm³.

Teacher's Note:
1. Cone is one third of the cylinder.

Answer: (b) \( 374 \text{ cm}^2 \)

Explanation:
1. The flat ends are removed and replaced by hemispherical hollows, so the surface = CSA of cylinder + 2 × CSA of hemisphere.
2. CSA of cylinder = \( 2 \times \frac{22}{7} \times 3.5 \times 10 = 220 \) cm²; two hemispherical hollows = \( 4\pi r^2 = 154 \) cm².
3. Total = 220 + 154 = 374 cm².

Teacher's Note:
1. Scooping out a hemisphere replaces πr² of flat face by 2πr² of curved face.