Surface Areas and Volumes Class 10 Previous Year Questions (Chapter 12)

1. 1

   Step 1: Draw the figure and find the radius

   1 mark (suggested)

   Let the radius of the hemisphere be r. Since the diameter of the hemisphere is equal to the side of the cube, r = 14/2 = 7 cm.

   Why? The radius of a hemisphere is half of its diameter; this fixes the only unknown before the surface-area work begins.

2. 2

   Step 2: Total surface area of the cube

   1 mark (suggested)

   TSA of cube = 6 × (14)² = 1176 cm².

   Why? A cube has six congruent square faces, so its total surface area is six times the area of one face.

3. 3

   Step 3: Area of the circular part removed

   1 mark (suggested)

   Area of circle removed = πr² = (22/7) × 7² = 154 cm².

   Why? The circular region of the top face disappears when the depression is scooped out, so its area is removed from the surface.

4. 4

   Step 4: Curved surface area of the hemisphere

   1 mark (suggested)

   CSA of hemisphere = 2πr² = 2 × (22/7) × 7² = 308 cm².

   Why? The hollow is a curved surface equal to half the surface of a sphere, so its area is half of 4πr².

5. 5

   Step 5: Total surface area of the remaining solid

   1 mark (suggested)

   Total surface area = 1176 − 154 + 308 = 1330 cm². Hence, the total surface area of the remaining solid block is 1330 cm².

   Why? The remaining surface is the cube's full surface, minus the circle covered by the depression, plus the curved surface exposed inside the depression.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Choose the option

   1 mark

   (D) Sphere, because the sphere has no flat faces, so its curved surface area is the total surface area.

   Why? The curved surface area counts only the curved parts of a solid, while the total surface area counts every part, so the two are equal exactly when the solid has no flat faces.

The full 1 mark is for the correct option.

1. 1

   Step 1: Choose the option

   1 mark

   (B) 1 : 3, because the volume of a cone is one-third of the volume of a cylinder with the same radius and height.

   Why? The cone's volume formula carries the factor 1/3, which the cylinder's formula does not, so for equal radius and height the cone is one-third of the cylinder.

The full 1 mark is for the correct option.

1. 1

   Step 1: Choose the option

   1 mark

   (B) 3 : 1, because equal volumes give (1/3)πr²H = πr²h, so H = 3h.

   Why? The volume of a cone is one-third of the volume of a cylinder with the same base and height, so to make the volumes equal the cone's height must be three times the cylinder's height.

The full 1 mark is for the correct option.

1. 1

   Step 1: Choose the option

   1 mark

   (A) 3πd², because the diameter is 2d, so the radius is d, and the total surface area of a solid hemisphere is 3πr².

   Why? The total surface area of a solid is the sum of the areas of all its faces. A solid hemisphere has two faces: the curved dome and the flat circular base, so both areas are included. Also, the radius of a circle is half its diameter.

The full 1 mark is for the correct option.

1. 1

   Step 1: Choose the option

   1 mark

   (C) 180, because (π × 6² × 8) ÷ ((1/3)π × 2² × 1.2) = 288 ÷ 1.6 = 180.

   Why? The number of conical bottles equals the quotient of the liquid volume in the cylinder by the volume of one conical bottle.

The full 1 mark is for the correct option.

(i) · 1 mark

1. 1

   Step 1: Draw the sphere and find its volume

   1 mark (suggested)

   Draw a sphere of diameter 7 cm. Radius r = 7/2 = 3.5 cm, so V = 4/3 × 22/7 × (3.5)³ = 539/3 cm³.

   Why? A ball is a sphere. Its volume is 4/3πr³, and the radius is half the diameter.

(ii) · 1 mark

2. 2

   Step 2: Find the painted surface area

   1 mark (suggested)

   Surface area of one ball = 4 × 22/7 × (3.5)² = 154 cm²; area for 10 balls = 154 × 10 = 1540 cm².

   Why? Neon paint covers the whole outer surface of each ball, so use the surface area of a sphere, 4πr², and multiply by the number of balls.

(iii)(a) · 2 marks

3. 3

   Step 3: Find the volume of the box

   1 mark (suggested)

   Volume of the box = 42 × 91 × 50 = 191100 cm³.

   Why? The box is a cuboid, so its volume is length × breadth × height.

4. 4

   Step 4: Find the empty space

   1 mark (suggested)

   Empty space = 191100 − 600 × (539/3) = 191100 − 107800 = 83300 cm³.

   Why? Empty space is the box volume minus the total volume occupied by all 600 balls.

Marks (iii)(a): 1 + 1 = 2

(iii)(b) · 2 marks

5. 5

   Step 5: Find the area covered by one ball

   1 mark (suggested)

   Each ball occupies a 7 cm × 7 cm square on the base, so area = 7 × 7 = 49 cm².

   Why? In an even layer that covers the base edge to edge, each ball's share of the base is a square of side equal to the ball's diameter.

6. 6

   Step 6: Find the number of balls in the lowermost layer

   1 mark (suggested)

   Balls along the length = 91/7 = 13; along the breadth = 42/7 = 6; total = 13 × 6 = 78 balls.

   Why? Along each edge, the number of balls is the edge length divided by one diameter; the layer count is the product of the two.

Marks (iii)(b): 1 + 1 = 2

Marks: 1 + 1 + 2 = 4 · Suggested split (CBSE’s usual pattern for 6-mark questions)

1. 1

   Step 1: Choose the option

   1 mark

   (B) 100, because the opening of each depression has diameter 2 × 0.35 = 0.7 cm, so 7 ÷ 0.7 = 10 fit along each edge, giving 10 × 10 = 100.

   Why? A hemispherical depression cut into a face has a circular opening whose diameter is twice its radius; dividing the side of the square face by this diameter gives the number along one edge, and the total is the square of that number.

The full 1 mark is for the correct option.

1. 1

   Step 1: Choose the option

   1 mark

   (A) 17 cm, because the cone's radius is √(13² − 12²) = 5 cm, so total height = cone height + hemisphere radius = 12 + 5 = 17 cm.

   Why? In a right circular cone the slant height, height and radius obey l² = r² + h², and the height of a hemisphere is its radius.

The full 1 mark is for the correct option.

1. 1

   Step 1: Choose the option

   1 mark

   (C) 2 : 1, because the greatest possible cone has the same base radius and height as the cylinder, so the remaining wood is 2/3 πr²h and the cone is 1/3 πr²h.

   Why? The largest cone that can be hollowed out of a cylinder has the cylinder's radius and height, and its volume is one-third of the cylinder's volume.

The full 1 mark is for the correct option.

1. 1

   Step 1: Choose the option

   1 mark

   (A) 25 cm, because CSA of cone = πrl = (22/7) × 7 × l = 550, so l = 550 ÷ 22 = 25.

   Why? By the formula for the curved surface area of a cone, CSA = πrl. Substituting the given values and solving for l gives the slant height.

The full 1 mark is for the correct option.

1. 1

   Step 1: Draw the figure and note the dimensions

   1 mark (suggested)

   Radius of the cylinder and cone: r = 4/2 = 2 m. Cylinder height h = 2.1 m. Cone slant height l = 2.8 m.

   Why? The radius of the cylinder equals the radius of the cone because the cone is mounted on the cylinder. The radius is half the diameter.

2. 2

   Step 2: Curved surface area of the cylinder

   1 mark (suggested)

   Curved surface area of cylinder = 2πrh = 2 × 22/7 × 2 × 2.1 = 26.4 m².

   Why? Curved surface area is the area of the side surface only. For a cylinder it is 2πrh; the top and bottom circular faces are not canvas because the top is covered by the cone and the bottom of the tent is not covered.

3. 3

   Step 3: Curved surface area of the cone

   1 mark (suggested)

   Curved surface area of cone = πrl = 22/7 × 2 × 2.8 = 17.6 m².

   Why? Curved surface area of a cone is πrl, where l is its slant height. The cone's circular base is hidden where it meets the cylinder, so it is not part of the canvas.

4. 4

   Step 4: Total canvas area

   1 mark (suggested)

   Area of canvas = 26.4 + 17.6 = 44 m².

   Why? The canvas covers only the two curved surfaces, so the total area is their sum.

5. 5

   Step 5: Find the cost of the canvas

   1 mark (suggested)

   Cost = 44 × 500 = ₹22000. Hence, the canvas used is 44 m² and its cost is ₹22000.

   Why? The total cost is the area of the canvas multiplied by the rate per square metre.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Draw the toy

   1 mark (suggested)

   Draw a cone of radius 7 cm mounted on a hemisphere of the same radius. Label the radius 7 cm and the total height 31 cm.

   Why? The drawing shows that only the curved surfaces of the cone and hemisphere are visible, so their curved surface areas have to be added.

2. 2

   Step 2: Find the height of the cone

   1 mark (suggested)

   Height of cone = total height − radius of hemisphere = 31 − 7 = 24 cm.

   Why? The radius of a hemisphere is also its height, so it is subtracted from the total height to get the cone's height.

3. 3

   Step 3: Find the slant height of the cone

   1 mark (suggested)

   l = √(r² + h²) = √(7² + 24²) = √625 = 25 cm.

   Why? The slant height of a cone is found from its radius and vertical height by the Pythagoras theorem.

4. 4

   Step 4: Find the curved surface area of the hemisphere

   1 mark (suggested)

   CSA of hemisphere = 2πr² = 2 × 22/7 × 7² = 308 cm².

   Why? A hemisphere has half the surface area of a sphere, so its curved surface area is 2πr².

5. 5

   Step 5: Find the total surface area

   1 mark (suggested)

   CSA of cone = πrl = 22/7 × 7 × 25 = 550 cm². Total surface area of the toy = 308 + 550 = 858 cm².

   Why? The surface area of the toy is the sum of the curved surface area of the hemisphere and the curved surface area of the cone, because the circular base is hidden inside the joint.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Draw the solid and find the cube's volume

   1 mark (suggested)

   Draw a cube of side 14 cm; from each of two opposite faces scoop out a hemisphere of diameter 14 cm. Volume of the cube = 14³ = 2744 cm³.

   Why? The cube is the whole solid before any wood is removed; its volume is edge × edge × edge.

2. 2

   Step 2: Radius of each hemisphere

   1 mark (suggested)

   The diameter of each hemisphere equals the side of the cube, so its radius is r = 14/2 = 7 cm.

   Why? A hemisphere's volume formula uses its radius, and the radius is half the diameter.

3. 3

   Step 3: Volume of one hemisphere

   1 mark (suggested)

   Volume of one hemisphere = 2/3 × 22/7 × 7³ = 2156/3 cm³.

   Why? A hemisphere is half a sphere, so its volume is half of the volume of a sphere of the same radius.

4. 4

   Step 4: Volume of two hemispheres

   1 mark (suggested)

   Volume of two hemispheres = 2 × 2156/3 = 4312/3 cm³.

   Why? The two depressions are identical, so their total removed volume is twice one hemisphere's volume.

5. 5

   Step 5: Volume of the remaining solid

   1 mark (suggested)

   Volume of remaining solid = 2744 − 4312/3 = 8232/3 − 4312/3 = 3920/3 cm³. Therefore the volume of the remaining wooden block is 3920/3 cm³.

   Why? The remaining wood is the cube's volume minus the total removed volume.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Draw the toy and set up the formula

   1 mark

   Let r be the common radius of the cylinder and each hemisphere, and h be the height of the cylinder. The total surface area of the toy = CSA of cylinder + CSA of one hemisphere + CSA of the other hemisphere.

   Why? When a hemisphere is scooped out, the flat circular end of the cylinder is removed. Only the curved surface of the cylinder and the curved surfaces of the two hemispheres remain visible, so the total surface area is their sum.

2. 2

   Step 2: Substitute the values in the formula

   1 mark

   Total surface area = 2πrh + 2πr² + 2πr² = 2 × 22/7 × 7 × 20 + 2 × 22/7 × 7² + 2 × 22/7 × 7² = 2 × 22/7 × 7 × (20 + 2 × 7) = 2 × 22/7 × 7 × 34.

   Why? The curved surface area of a cylinder is 2πrh, and of a hemisphere is 2πr². Substituting the radius 7 cm and the height 20 cm into the sum gives the expression with π = 22/7.

3. 3

   Step 3: Calculate and state the total surface area

   1 mark

   = 2 × 22 × 34 = 1496. Therefore, the total surface area of the toy is 1496 cm².

   Why? The 7 in the denominator cancels with the radius 7, leaving 2 × 22 × 34. The unit is cm² because all lengths are in cm.

Marks: 1 + 1 + 1 = 3 · Split as in the CBSE marking scheme, 2024 · Standard 30/1/2 Q29

1. 1

   Step 1: Choose the option

   1 mark

   (D) 25 cm, because the slant height l = √(r² + h²) = √(7² + 24²) = 25 cm.

   Why? In a right circular cone, the radius and the height are perpendicular, so the slant height is the hypotenuse of a right triangle and l² = r² + h².

The full 1 mark is for the correct option.

1. 1

   Step 1: Choose the option

   1 mark

   (C) 2400 cm², because the total surface area of a cube is 6 × (side)² = 6 × 20² = 2400 cm².

   Why? The total surface area of a cube is the sum of the areas of its six square faces, so TSA = 6a².

The full 1 mark is for the correct option.

1. 1

   Step 1: Choose the option

   1 mark

   (A) 2 cm, because CSA = 2πrh, so 2 × (22/7) × r × 14 = 88, giving r = 1 cm and diameter = 2 cm.

   Why? The curved surface area of a right circular cylinder is 2πrh, where r is the base radius and h is the height. Substituting the given values and solving for r gives the diameter as twice the radius.

The full 1 mark is for the correct option.

1. 1

   Step 1: Draw the figure and fix unknowns

   1 mark (suggested)

   Draw a right circular cylinder. Let the base radius be r cm and the height be h cm. Given: r + h = 37.

   Why? Naming the dimensions as variables lets the given sum and the surface area equation be written mathematically.

2. 2

   Step 2: Total surface area formula with numbers

   1 mark (suggested)

   Total surface area of a cylinder = 2πr(r + h). Using π = 22/7 and r + h = 37: 2 × (22/7) × r × 37 = 1628 cm².

   Why? The total surface area is the sum of the two circular ends and the curved surface; the factor r + h appears in this formula.

3. 3

   Step 3: Find the radius

   1 mark (suggested)

   r = 1628 ÷ (2 × (22/7) × 37) = 7.

   Why? Dividing both sides of the surface area equation by the coefficient of r isolates r.

4. 4

   Step 4: Find the height

   1 mark (suggested)

   h = 37 − r = 37 − 7 = 30.

   Why? The given condition relates the two variables, so subtracting the known radius from their sum gives the height.

5. 5

   Step 5: Volume of the cylinder

   1 mark (suggested)

   Volume = πr²h = (22/7) × 7² × 30 = 4620. The volume of the cylinder is 4620 cm³.

   Why? The volume of a right circular cylinder is the area of its circular base multiplied by its height.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Choose the option

   1 mark

   (C) 13.64 m², because total surface area = 2πr(r + h) = 2 × 22/7 × 0.7 × (0.7 + 2.4) = 13.64 m².

   Why? A closed hollow cylinder has a curved surface and two circular ends, so its total surface area is the curved surface area plus the areas of the two circular ends.

The full 1 mark is for the correct option.