CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set 15

Answer: (b) 352 cm²

Explanation:
1. V = πr²h gives 448π = πr² × 7, so r² = 64 and r = 8 cm.
2. Curved surface area = 2πrh = 2 × (22/7) × 8 × 7 = 352 cm².

Teacher's Note:
1. Find the radius from the volume first.
2. Lateral surface area = 2πrh.

Answer: (a) 44 m²

Explanation:
1. r = 0.7 m and h = 2 m. Area in one revolution = 2πrh = 2 × (22/7) × 0.7 × 2 = 8.8 m².
2. In 5 revolutions the area is 5 × 8.8 = 44 m².

Teacher's Note:
1. One revolution covers the curved surface area of the roller.
2. Multiply by the number of revolutions.

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

Explanation:
1. Let the diameter be d and the edge be a. 4π(d/2)² = 6a², so πd² = 6a².
2. d²/a² = 6/π, so d : a = √6 : √π.

Teacher's Note:
1. Surface area of a sphere is 4πr² and that of a cube is 6a².
2. Write the radius as d/2.

Answer: (a) 5.27 litres, 3253.25 cm²

Explanation:
1. Internal radius r = 21 cm and external radius R = 22.75 cm.
2. Volume of material = (2/3)π(R³ − r³) = (2/3) × (22/7) × [(22.75)³ − (21)³] = 5266.46 cm³ = 5.27 litres (1000 cm³ = 1 litre).
3. Outer curved surface area = 2πR² = 2 × (22/7) × (22.75)² = 3253.25 cm².

Teacher's Note:
1. Here the volume asked is that of the material between the two hemispheres.
2. Outer curved surface area uses the external radius.

Answer: (c) 0

Explanation:
1. v = (1/3)πr²h and c = πrl with l² = r² + h².
2. 3πvh³ = π²r²h⁴, c²h² = π²r²(h² + r²)h² = π²r²h⁴ + π²r⁴h², and 9v² = π²r⁴h².
3. So the value is π²r²h⁴ − π²r²h⁴ − π²r⁴h² + π²r⁴h² = 0.

Teacher's Note:
1. Substitute v and c in terms of r, h and l.
2. Replace l² by r² + h².

Answer: (b) \( \frac{50}{\pi} \)

Explanation:
1. The rectangle is the curved surface: its width 2 cm is the height and its length 10 cm is the circumference of the base.
2. 2πr = 10 gives r = 5/π.
3. Volume = πr²h = π × 25/π² × 2 = 50/π cm³.

Teacher's Note:
1. The curved surface opened out is a rectangle.
2. One side is the height and the other is the circumference.

Answer: (a) \( \frac{1}{3}\pi h^3\tan^2\alpha \)

Explanation:
1. In the right triangle formed by the height, radius and slant height, tan α = r/h, so r = h tan α.
2. Volume = (1/3)πr²h = (1/3)π(h tan α)²h = (1/3)πh³tan²α.

Teacher's Note:
1. The semi-vertical angle is between the axis and the slant side.
2. tan α = radius / height.

Answer: (d) 36

Explanation:
1. Volume of D = 18³ + 24³ + 30³ = 5832 + 13824 + 27000 = 46656.
2. Edge of D = ∛46656 = 36 cm.

Teacher's Note:
1. Melting does not change the total volume.
2. 46656 = 36³.

Answer: (b) 525 m

Explanation:
1. πr² = 346.5 gives r² = 110.25 and r = 10.5 m.
2. Slant height l = √(10.5² + 14²) = 17.5 m.
3. Canvas area = πrl = (22/7) × 10.5 × 17.5 = 577.5 m².
4. Length of canvas = 577.5 ÷ 1.1 = 525 m.

Teacher's Note:
1. The floor area gives the radius.
2. Length of canvas = area ÷ width.

Answer: (c) larger than r

Explanation:
1. The centres of the three bases form an equilateral triangle of side 2r.
2. The vertices lie directly above the centres, so the circle through them has the same radius as the circle through the three centres.
3. That circumradius is 2r/√3, which is about 1.15r. So the radius is larger than r.

Teacher's Note:
1. The height of the cones does not matter for the circle through the vertices.
2. Circumradius of an equilateral triangle of side s is s/√3.

Answer: (c) more than 50%

Explanation:
1. The centre of the ball lies in the plane of the rim of the cone, so half of the ball is above the rim and outside.
2. The lower half is a hemisphere of the same radius as the cone's base, while the cone narrows towards the vertex, so part of this hemisphere also lies outside the slanting walls.
3. Hence more than 50% of the ball is outside the cone.

Teacher's Note:
1. Compare a hemisphere with the narrowing cone below it.
2. A cone is slimmer than a hemisphere of the same base.

Answer: (b) 3.5

Explanation:
1. Volume of slab = 8 × 11 × 2 = 176 cubic inches.
2. Rod radius = 4 inches, so π × 4² × l = 176.
3. l = 176 × 7 / (22 × 16) = 3.5 inches.

Teacher's Note:
1. The volume stays the same after melting.
2. Use V = πr²l for the rod.

Answer: (a) 2 : 1

Explanation:
1. πrl = 2πrh.
2. Cancelling πr gives l = 2h, so l : h = 2 : 1.

Teacher's Note:
1. Write the curved surface areas of both solids and equate them.
2. Cancel the common factor πr.

Answer: (b) 150 cm²

Explanation:
1. Perimeter of a square face = 4a = 20, so a = 5 cm.
2. Surface area = 6a² = 6 × 25 = 150 cm².

Teacher's Note:
1. Find the edge from the perimeter of one face.
2. A cube has six equal faces.

Answer: (b) 8

Explanation:
1. The big cube is cut into 4 × 4 × 4 = 64 small cubes.
2. Cubes without any painted face lie in the inner 2 × 2 × 2 block, which is 8 cubes.

Teacher's Note:
1. Each edge holds 4 small cubes.
2. Remove the outer layer on every side, leaving (4 − 2) cubes per edge.

Answer: (c) 4

Explanation:
1. Surface area = 4πr² depends on r². Doubling the diameter doubles r.
2. New area = 4π(2r)² = 4 × (4πr²), so x = 4.

Teacher's Note:
1. Surface area is proportional to the square of the radius.
2. Doubling the diameter doubles the radius.

Answer: (c) 700%

Explanation:
1. Increasing by 100% doubles the radius: r becomes 2r.
2. Volume becomes (4/3)π(2r)³ = 8 × (4/3)πr³.
3. Increase = 8 − 1 = 7 times the original, that is 700%.

Teacher's Note:
1. Volume is proportional to the cube of the radius.
2. The new volume is 800% of the old one, so the increase is 700%.

Answer: (b) \( \left(\frac{8}{3}\right)^{1/3} \)

Explanation:
1. Half the liquid forms 11 cubes of side a and the other half forms 7 spheres of radius r, so 11a³ = 7 × (4/3)πr³.
2. a³/r³ = (28π)/(33) = 28 × 22/(7 × 33) = 8/3.
3. So a/r = (8/3)^{1/3}.

Teacher's Note:
1. Both halves have the same volume.
2. Substitute π = 22/7 and simplify.

Answer: (a) remains the same

Explanation:
1. Melting and recasting does not add or remove material.
2. So the volume remains the same.

Teacher's Note:
1. Think of melting a candle and moulding it again.
2. Surface area may change, volume does not.

Answer: (b) \( 117\pi\ \text{cm}^3 \)

Explanation:
1. Radius = 3 cm and height = 15 cm.
2. Volume of cylinder = π × 3² × 15 = 135π cm³.
3. Volume of hemispherical raised portion = (2/3)π × 3³ = 18π cm³.
4. Juice = 135π − 18π = 117π cm³.

Teacher's Note:
1. The raised portion takes up space inside the glass.
2. Subtract its volume from the cylinder.

Answer: (c) R : r

Explanation:
1. Let the common height be h and radii R and r.
2. Ratio = 2πRh : 2πrh = R : r.

Teacher's Note:
1. Equal heights cancel.
2. The ratio then depends only on the radii.

Answer: (c) R² : r²

Explanation:
1. Volumes are πR²h and πr²h with the same h.
2. Ratio = R² : r².

Teacher's Note:
1. Cancel π and the common height.
2. Volume depends on the square of the radius.

Answer: (a) \( h_1 : h_2 \)

Explanation:
1. V₁ = (1/3)πr²h₁ and V₂ = (1/3)πr²h₂ with the same r.
2. Ratio = h₁ : h₂.

Teacher's Note:
1. Cancel the common factors (1/3)πr².
2. Only the heights are left.

Answer: (d) \( \frac{4}{3}\pi(R^3-r^3) \)

Explanation:
1. Volume of shell = volume of outer sphere − volume of inner sphere.
2. = (4/3)πR³ − (4/3)πr³ = (4/3)π(R³ − r³).

Teacher's Note:
1. Subtract the hollow inner sphere from the solid outer sphere.
2. Volume of a sphere is (4/3)πr³.

Answer: (c) \( \frac{4\pi}{3}\ \text{m}^3 \)

Explanation:
1. The diameter of the sphere cannot exceed the diameter of the base, 2 m. So the largest sphere has radius 1 m.
2. Volume = (4/3)π(1)³ = 4π/3 m³.

Teacher's Note:
1. Compare the base diameter (2 m) with the height (4 m); the smaller one limits the sphere.
2. The sphere has radius equal to the base radius.

Answer: (b) 8 cm

Explanation:
1. 4 × 2πr² = 2 × 2πrh.
2. 8r² = 4rh, so r = h/2 = 4 cm.
3. Diameter = 8 cm.

Teacher's Note:
1. The two circular faces have total area 2πr².
2. Cancel πr on both sides.

Answer: (a) 3.5

Explanation:
1. Height = 40 cm, so the 22 cm side becomes the circumference.
2. 2πr = 22 gives r = 22 × 7/(2 × 22) = 3.5 cm.

Teacher's Note:
1. The side that is not the height wraps around the base.
2. Circumference = 2πr.

Answer: (c) 127.5

Explanation:
1. Let the cylinder have radius r and height 2r. By similar triangles in the axial section: 2r/12 = (5 − r)/5.
2. 10r = 60 − 12r, so r = 30/11 cm and h = 60/11 cm.
3. Volume = (22/7) × (30/11)² × (60/11) ≈ 127.5 cm³.

Teacher's Note:
1. Use similar triangles in the axial cross-section.
2. The top rim of the cylinder touches the slant side of the cone.

Answer: (c) 9 : 16

Explanation:
1. a₁³ : a₂³ = 27 : 64 gives a₁ : a₂ = 3 : 4.
2. Surface areas are in the ratio of the squares of the edges: 9 : 16, that is 9 : 16.

Teacher's Note:
1. Take cube roots to get the ratio of edges.
2. Square the edge ratio to get the area ratio.

Answer: (b) 25 : 16

Explanation:
1. r₁³ : r₂³ = 125 : 64 gives r₁ : r₂ = 5 : 4.
2. Surface areas are in the ratio r₁² : r₂² = 25 : 16.

Teacher's Note:
1. Take cube roots for the radii.
2. Surface area depends on r².

Answer: (d) \( 3\frac{1}{3}\ \text{kg} \)

Explanation:
1. Let the side of the square base be x m. Then 9x² = 1, so x = 1/3 m.
2. The biggest cube cut from the bar has side 1/3 m and volume 1/27 m³.
3. Weight = 90 × 1/27 = 10/3 = 3⅓ kg.

Teacher's Note:
1. The biggest cube has the side of the square base.
2. Weight is proportional to volume.

Answer: (d) 384

Explanation:
1. Let the sides be x, 2x and 3x (whole numbers of unit cubes). Volume = 6x³ ≤ 400, so x³ ≤ 66.6.
2. The largest whole x is 4, because 4³ = 64 and 5³ = 125.
3. Cubes used = 6 × 64 = 384.

Teacher's Note:
1. The sides must be a whole number of unit cubes.
2. Find the largest whole x with 6x³ not more than 400.

Answer: (c) 27 cm

Explanation:
1. Let the heights of cylinders A and B be h and H. The volume of water is the same in both positions.
2. Upright: B is full and A holds water up to 15 cm in all: π(3)²H + π(1.5)²(15 − H). Upside down: A is full and B holds water up to 24 cm in all: π(1.5)²h + π(3)²(24 − h).
3. Equating and simplifying: 4H + 15 − H = h + 4(24 − h), which gives 3H + 3h = 81, so H + h = 27 cm.

Teacher's Note:
1. The water volume does not change when the bottle is turned over.
2. The radii are in the ratio 1 : 2, so areas are in the ratio 1 : 4.

Answer: (c) 50

Explanation:
1. Let the cones have radii r₁ and r₂. Volume of cylinder = π × 7² × 10 = (π/3) × 10 (r₁² + r₂²), so r₁² + r₂² = 147.
2. Flat surface areas: cylinder 2 × π × 49 = 98π; cones π(r₁² + r₂²) = 147π.
3. Increase = (147π − 98π)/98π × 100 = 50%.

Teacher's Note:
1. Flat surfaces of the cylinder are two circles; each cone has one circular base.
2. Use the volume equation to get r₁² + r₂².

Answer: (b) Total surface area of a cube is 4l²

Explanation:
1. A cube has six square faces, so its total surface area is 6l², not 4l².
2. The other three statements are correct. So the incorrect statement is (b).

Teacher's Note:
1. Count the faces of a cube.
2. Four walls form a rectangle of height h and length equal to the perimeter 2(l + b).

Answer: (a) 38.8 cm³

Explanation:
1. The water that flows out equals the volume of the marble.
2. V = (4/3)π(2.1)³ = (4/3) × (22/7) × 9.261 = 38.8 cm³.

Teacher's Note:
1. A submerged object pushes out its own volume of water.
2. The marble is fully inside the cup.

Answer: (c) 363

Explanation:
1. Volume of brick = 22³ = 10648 cm³.
2. Volume of one cone = (1/3)π r² h = (1/3)(22/7)(4)(7) = 88/3 cm³.
3. Number of cones = 10648 ÷ (88/3) = 363.

Teacher's Note:
1. Divide the total volume by the volume of one cone.
2. Use π = 22/7.

Answer: (d) 19.4 cm³

Explanation:
1. The largest cone has base diameter 4.2 cm (radius 2.1 cm) and height 4.2 cm.
2. V = (1/3)(22/7)(2.1)²(4.2) = 19.4 cm³.

Teacher's Note:
1. The base circle fits inside a face of the cube.
2. The height equals the edge of the cube.

Answer: (a) \( 4\pi r^2 \)

Explanation:
1. Two hemispheres joined along their bases make a sphere of radius r.
2. Curved surface area = 2πr² + 2πr² = 4πr².

Teacher's Note:
1. The joined flat faces are hidden inside.
2. Each hemisphere gives 2πr².

Answer: (b) 2r cm

Explanation:
1. A sphere that just fits in a cylinder touches its curved surface, so its radius equals r.
2. Its diameter is 2r cm.

Teacher's Note:
1. Since h > 2r, the width of the cylinder decides the size.
2. The sphere touches the curved surface all round.