CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set 25

Answer: (b) sides

Explanation:
1. Revolving a rectangle about one of its sides sweeps out a right circular cylinder.

Teacher's Note:
1. The side acts as the axis of the cylinder.
2. The opposite side traces the curved surface.

Answer: (b) right circular cone

Explanation:
1. The side about which it revolves becomes the axis (height) and the other side sweeps the base.
2. The solid is a right circular cone.

Teacher's Note:
1. The hypotenuse becomes the slant height.
2. Think of a set-square spun about a leg.

Answer: (b) frustum of the cone

Explanation:
1. The upper piece is a smaller cone; the lower piece with two circular faces is the frustum of the cone.

Teacher's Note:
1. A frustum has two parallel circular ends.
2. The cut is parallel to the base.

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

Explanation:
1. V = ⅓πr²h, and h is the same for both.
2. V₁ : V₂ = r₁² : r₂².

Teacher's Note:
1. The volume is proportional to the square of the radius at constant height.
2. The common factor ⅓πh cancels.

Answer: (d) 2541

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

Teacher's Note:
1. Divide the volume of the cube by the volume of one bullet.
2. Radius of a bullet = 2 cm.

Answer: (b) 42.66 m

Explanation:
1. Volume of the cylinder = π × 2² × 8 = 32π m³.
2. Cone: ⅓ × π × (1.5)² × H = 32π gives H = 96/2.25.
3. H = 42.67 m (about 42.66 m).

Teacher's Note:
1. Volume is conserved when a solid is melted and recast.
2. Use V = ⅓πr²H for the cone.

Answer: (c) 1.6 cm

Explanation:
1. Volume displaced = 8³ = 512 cm³.
2. Rise = 512/(π × 10²) = 512 × 7/2200.
3. Rise = 1.63 cm, about 1.6 cm.

Teacher's Note:
1. The volume of water displaced equals the volume of the cube.
2. Rise = volume ÷ base area.

Answer: (a) 2.5 m and 2 m

Explanation:
1. 2π(R - r)(14) = 44 gives 88(R - r) = 44, so R - r = 0.5 m.
2. π(R² - r²)(14) = 99 gives 44(R² - r²) = 99, so R² - r² = 2.25 and R + r = 4.5 m.
3. R = 2.5 m and r = 2 m.

Teacher's Note:
1. Use the curved surface area 2πrh for each cylinder.
2. Divide R² - r² by R - r to get R + r.

Answer: (c) 8171.42 cm³

Explanation:
1. V = ⅓πh(R² + Rr + r²) = ⅓ × (22/7) × 24 × (225 + 75 + 25).
2. = (22/7) × 8 × 325 = 57200/7 = 8171.43 cm³.

Teacher's Note:
1. Use the frustum volume formula.
2. R² + Rr + r² = 325.

Answer: (d) none of these

Explanation:
1. Let the common radius = height = r. Slant height of the cone = r√2.
2. Curved surface of the hemisphere = 2πr²; of the cone = πr × r√2 = √2 πr².
3. Ratio = 2 : √2 = √2 : 1, which matches none of the options.

Teacher's Note:
1. Equal heights means the height of the cone equals the radius.
2. Find the slant height of the cone.

Answer: (a) 3 : 1 : 2

Explanation:
1. Same base and same height means h = r.
2. Cylinder: πr³; cone: ⅓πr³; hemisphere: ⅔πr³.
3. Ratio = 3 : 1 : 2.

Teacher's Note:
1. The hemisphere has height equal to its radius.
2. Multiply each volume by 3/(πr³).

Answer: (b) \( \frac{r_1}{r_1 - r_2} \)

Explanation:
1. Let H be the height of the whole cone and h that of the frustum. The small cone removed has height H - h.
2. By similar triangles (H - h)/H = r₂/r₁, so H - h = Hr₂/r₁.
3. h = H(r₁ - r₂)/r₁, so H/h = r₁/(r₁ - r₂).

Teacher's Note:
1. Use similar triangles for the whole cone and the small cone cut off.
2. Express h in terms of H.

Answer: (d) 2

Explanation:
1. Volume of the hemisphere = ⅔πR³ and volume of the cone = ⅓πR²H.
2. ⅓πR²H = ⅔πR³, so H = 2R.
3. H/R = 2.

Teacher's Note:
1. Equate the volumes.
2. Cancel πR².

Answer: (c) 2 : 3

Explanation:
1. Cylinder: height r, volume πr³. Hemisphere: ⅔πr³.
2. Ratio = 2/3 : 1 = 2 : 3.

Teacher's Note:
1. The cylinder has the same radius and height as the hemisphere.
2. Compare ⅔πr³ with πr³.

Answer: (b) \( \sqrt{\frac{\pi}{6}} \)

Explanation:
1. 6a² = 4πr², so a² = 2πr²/3.
2. V_cube/V_sphere = a³/((4/3)πr³) = a² × a × 3/(4πr³).
3. a = r√(2π/3), so the ratio = (2π/3) × √(2π/3) × 3/(4π) = (1/2)√(2π/3) = √(π/6).

Teacher's Note:
1. Express a in terms of r.
2. Simplify the ratio carefully.

Answer: (a) \( \left[x + y + z + \frac{xy + yz + zx}{100} + \frac{xyz}{100^2}\right]\% \)

Explanation:
1. New volume = lbh(1 + x/100)(1 + y/100)(1 + z/100).
2. Expanding: 1 + (x + y + z)/100 + (xy + yz + zx)/100² + xyz/100³.
3. Percentage increase = (x + y + z) + (xy + yz + zx)/100 + xyz/100².

Teacher's Note:
1. Multiply the three factors (1 + x/100) etc.
2. Subtract 1 and multiply by 100.

Answer: (c) 924 cm³

Explanation:
1. R = 6 cm and r = 3 cm.
2. V = ⅓πh(R² + Rr + r²) = ⅓ × (22/7) × 14 × (36 + 18 + 9)
3. = (44/3) × 63 = 924 cm³.

Teacher's Note:
1. Convert diameters to radii.
2. Use the frustum formula.

Answer: (d) 1 cm

Explanation:
1. π(81 - r²) × 14 = 748, so 44(81 - r²) = 748 and 81 - r² = 17.
2. r² = 64, so the internal radius r = 8 cm.
3. Thickness = 9 - 8 = 1 cm.

Teacher's Note:
1. Volume of the metal = π(R² - r²)h.
2. Thickness = R - r.

Answer: (a) 14 cm

Explanation:
1. Volume of the hollow sphere = (4/3)π(4³ - 2³) = (4/3)π × 56 = 224π/3.
2. Cone: ⅓π × 4² × h = 16πh/3.
3. 16h/3 = 224/3 gives h = 14 cm.

Teacher's Note:
1. Equate the volumes.
2. The radii are 2 cm and 4 cm.

Answer: (b) 20 m

Explanation:
1. l = √(12² + 16²) = √400 = 20 m.

Teacher's Note:
1. Radius = 12 m.
2. This is the 3-4-5 triangle scaled by 4.

Answer: (c) 754.29 m²

Explanation:
1. Canvas = curved surface area = πrl = (22/7) × 12 × 20
2. = 5280/7 = 754.29 m².

Teacher's Note:
1. Use CSA = πrl.
2. r = 12 m and l = 20 m.

Answer: (a) 44 persons

Explanation:
1. Volume = ⅓πr²h = ⅓ × (22/7) × 144 × 16 = 2412.57 m³.
2. Number = 2412.57/54 = 44.7, so at most 44 persons.

Teacher's Note:
1. Find the volume of the cone.
2. Round down: a fraction of a person cannot be accommodated.

Answer: (a) \( 122.5\pi \) cm³

Explanation:
1. V = πr²h = π × 3.5² × 10 = 122.5π cm³.

Teacher's Note:
1. Use V = πr²h.
2. 3.5² = 12.25.

Answer: (c) 374 cm²

Explanation:
1. Curved surface of the cylinder = 2πrh = 2 × (22/7) × 3.5 × 10 = 220 cm².
2. Inner surfaces of the two hemispheres = 2 × 2πr² = 4 × (22/7) × 12.25 = 154 cm².
3. Total = 220 + 154 = 374 cm².

Teacher's Note:
1. The flat circular ends disappear after scooping out the hemispheres.
2. Add the curved surface of the cylinder and of both hemispheres.

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

Explanation:
1. Volume of two hemispheres = (4/3)πr³ = (4/3)π(3.5)³ = (4/3)π × 42.875.
2. = 171.5π/3 cm³ (about 179.6 cm³).

Teacher's Note:
1. Two hemispheres make one sphere.
2. 3.5³ = 42.875.

Answer: (b) 77/4 cm²

Explanation:
1. Radius = 1.75 cm. CSA = 2πr² = 2 × (22/7) × 3.0625 = 19.25 cm² = 77/4 cm².

Teacher's Note:
1. The radius is half the diameter.
2. CSA of a hemisphere = 2πr².

Answer: (a) 3.25 cm

Explanation:
1. Height of the cone = total height - radius of the hemisphere = 5 - 1.75 = 3.25 cm.

Teacher's Note:
1. Subtract the hemisphere's radius from the total height.
2. The radius is 1.75 cm.

Answer: (c) 3.7 cm

Explanation:
1. l = √(1.75² + 3.25²) = √(3.0625 + 10.5625) = √13.625 = 3.69 cm, about 3.7 cm.

Teacher's Note:
1. Use Pythagoras with the radius and the height of the cone.
2. 1.75² = 3.0625.

Answer: (d) 39.6 cm²

Explanation:
1. Hemisphere: 19.25 cm². Cone: πrl = (22/7) × 1.75 × 3.69 = 20.3 cm².
2. Total = 19.25 + 20.3 = 39.6 cm² (approximately).

Teacher's Note:
1. Add the curved surfaces of the two parts.
2. The base of the cone is covered by the hemisphere.