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Surface Areas and Volumes: Class 10 Maths Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 10 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

Riya, a craft teacher at a government school in Jaipur, prepared a decorative model for the annual science exhibition by joining a solid cone mounted on top of a solid hemisphere, both having the same base radius of 7 cm. The height of the cone is 24 cm and the radius of the hemisphere is also 7 cm. The model is to be painted and displayed as a showpiece representing a traditional lamp design.

Riya, a craft teacher at a government school in Jaipur, prepared a decorative model for the annual science exhibition by joining a solid cone mounted on top of a solid hemisphere, both having the same base radius of 7 cm. The height of the cone is 24 cm and the radius of the hemisphere is also 7 cm. The model is to be painted and displayed as a showpiece representing a traditional lamp design.

Diagram for question 1: Surface Areas and Volumes
Show answer
(i)

Slant height of cone: l = √(r² + h²)

⟹ l = √(7² + 24²) = √(49 + 576) = √625

∴ Slant height of the cone = 25 cm

---

(ii)

Curved surface area of hemisphere = 2πr²

⟹ 2 × (22/7) × 7² = 2 × (22/7) × 49 = 2 × 154

∴ Curved surface area of hemisphere = 308 cm²

---

(iii)

Total surface area of the model = Curved surface area of cone + Curved surface area of hemisphere

(The base of the cone coincides with the flat face of the hemisphere; neither the base of the cone nor the flat base of the hemisphere is exposed.)

Curved surface area of cone = πrl = (22/7) × 7 × 25 = 22 × 25 = 550 cm²

Curved surface area of hemisphere = 2πr² = 308 cm² (calculated above)

⟹ Total surface area = 550 + 308

∴ Total surface area of the model = 858 cm²
Q2Case-based4 marks

A school is organising a Science Fair. Students in the craft club decide to make decorative items using solid clay. Priya takes a solid clay cylinder of radius 6 cm and height 12 cm. She reshapes the entire clay into small solid spheres, each of radius 1 cm, without any wastage.

A school is organising a Science Fair. Students in the craft club decide to make decorative items using solid clay. Priya takes a solid clay cylinder of radius 6 cm and height 12 cm. She reshapes the entire clay into small solid spheres, each of radius 1 cm, without any wastage.

(i) What is the volume of the original solid clay cylinder? [1 mark]
(ii) What is the volume of each small solid sphere? [1 mark]
(iii) How many complete small solid spheres can Priya make from the clay? [2 marks]

Show answer
(i) Volume of cylinder = πr²h

⟹ V = π × (6)² × 12

⟹ V = π × 36 × 12

∴ Volume of cylinder = 432π cm³

---

(ii) Volume of each small sphere = (4/3)πr³

⟹ V = (4/3) × π × (1)³

∴ Volume of each sphere = (4/3)π cm³

---

(iii) Since the clay is completely reshaped without wastage, the volume is conserved.

∴ Number of spheres = Volume of cylinder ÷ Volume of each sphere

⟹ Number of spheres = 432π ÷ (4/3)π

⟹ Number of spheres = 432π × (3/4π)

⟹ Number of spheres = (432 × 3) / 4

⟹ Number of spheres = 1296 / 4

∴ Number of small solid spheres = 324
Q3Case-based4 marks

A school is organising a Science Fair and students are making decorative pieces. Riya decides to make a pencil-shaped showpiece by joining a cylinder and a cone. The cylinder has radius 1.4 cm and height 10 cm. The cone is placed on top of the cylinder and has the same base radius as the cylinder and a height of 3 cm. (Take π = 22/7)

A school is organising a Science Fair and students are making decorative pieces. Riya decides to make a pencil-shaped showpiece by joining a cylinder and a cone. The cylinder has radius 1.4 cm and height 10 cm. The cone is placed on top of the cylinder and has the same base radius as the cylinder and a height of 3 cm.

(i) Find the volume of the cylindrical part of the showpiece.
(ii) Find the volume of the conical part of the showpiece.
(iii) Find the total volume of the showpiece. OR Find how much cardboard (curved surface area) is needed to cover only the lateral surface of the cylinder and the slant surface of the cone. (Take π = 22/7)

Diagram for question 3: Surface Areas and Volumes
Show answer
Given: Radius of cylinder = Radius of cone = r = 1.4 cm; Height of cylinder H = 10 cm; Height of cone h = 3 cm; π = 22/7.

(i) Volume of cylindrical part:

Formula: V<sub>cylinder</sub> = πr²H

⟹ V<sub>cylinder</sub> = (22/7) × (1.4)² × 10

⟹ V<sub>cylinder</sub> = (22/7) × 1.96 × 10

⟹ V<sub>cylinder</sub> = (22 × 1.96 × 10) / 7

⟹ V<sub>cylinder</sub> = 431.2 / 7

∴ Volume of cylindrical part = 61.6 cm³ [1 mark]

(ii) Volume of conical part:

Formula: V<sub>cone</sub> = (1/3)πr²h

⟹ V<sub>cone</sub> = (1/3) × (22/7) × (1.4)² × 3

⟹ V<sub>cone</sub> = (1/3) × (22/7) × 1.96 × 3

⟹ V<sub>cone</sub> = (22/7) × 1.96 [the 3s cancel]

⟹ V<sub>cone</sub> = (22 × 1.96) / 7

⟹ V<sub>cone</sub> = 43.12 / 7

∴ Volume of conical part = 6.16 cm³ [1 mark]

(iii) Total volume of the showpiece:

Total Volume = V<sub>cylinder</sub> + V<sub>cone</sub>

⟹ Total Volume = 61.6 + 6.16

∴ Total volume of the showpiece = 67.76 cm³ [2 marks]

OR

Slant height of cone:

Formula: l = √(r² + h²)

⟹ l = √[(1.4)² + (3)²]

⟹ l = √[1.96 + 9]

⟹ l = √10.96

⟹ l ≈ 3.31 cm

Curved Surface Area of cylinder = 2πrH

⟹ CSA<sub>cylinder</sub> = 2 × (22/7) × 1.4 × 10

⟹ CSA<sub>cylinder</sub> = 2 × 22 × 0.2 × 10

⟹ CSA<sub>cylinder</sub> = 88 cm²

Curved Surface Area of cone = πrl

⟹ CSA<sub>cone</sub> = (22/7) × 1.4 × 3.31

⟹ CSA<sub>cone</sub> = 22 × 0.2 × 3.31

⟹ CSA<sub>cone</sub> ≈ 14.56 cm²

Total lateral cardboard area = CSA<sub>cylinder</sub> + CSA<sub>cone</sub>

⟹ Total area = 88 + 14.56

∴ Total cardboard needed ≈ 102.56 cm² [2 marks]
Q4Case-based4 marks

A decorative pen stand is made by joining a hollow cylinder (open at both ends) with a hollow cone fixed at its bottom. The cylindrical part has height 10 cm and radius 3.5 cm. The conical part has height 4 cm and the same base radius of 3.5 cm. The outer surface, including the base circle of the cone, is to be painted.

A decorative pen stand is made by joining a hollow cylinder open at both ends with a hollow cone fixed at its bottom (closed), as shown in the figure. The pen stand is to be gifted after painting its entire outer surface (including the base of the cone).

The dimensions are:
• Cylindrical part: height = 10 cm, radius = 3.5 cm
• Conical part: height = 4 cm, radius = 3.5 cm (same as cylinder)

Based on this, answer the following:
(i) What is the slant height of the conical part? [1 mark]
(ii) What is the curved surface area of the cylindrical part? [1 mark]
(iii) Find the total surface area to be painted (curved surface of cylinder + curved surface of cone + base of cone). [2 marks]

(Use π = 22/7)

Diagram for question 4: Surface Areas and Volumes
Show answer
(i) Slant height of the conical part:

Formula: l = √(r² + h²)

⟹ l = √((3.5)² + (4)²)
⟹ l = √(12.25 + 16)
⟹ l = √28.25
⟹ l = √(113/4)
⟹ l = √113 / 2

Using approximate calculation: √28.25 ≈ 5.315 cm

However, note: let us check if a cleaner value exists. With r = 3.5 = 7/2 and h = 4:
l = √((7/2)² + 4²) = √(49/4 + 64/4) = √(113/4) = √113/2 ≈ 5.32 cm

∴ Slant height of the conical part, l = √113/2 ≈ 5.32 cm

(ii) Curved surface area of the cylindrical part:

Formula: CSA of cylinder = 2πrh

⟹ CSA = 2 × (22/7) × 3.5 × 10
⟹ CSA = 2 × (22/7) × (7/2) × 10
⟹ CSA = 2 × 22 × (1/2) × 10
⟹ CSA = 2 × 11 × 10
⟹ CSA = 220 cm²

∴ Curved surface area of the cylindrical part = 220 cm²

(iii) Total surface area to be painted:

Total area = CSA of cylinder + CSA of cone + Base area of cone

CSA of cylinder = 220 cm² [from part (ii)]

CSA of cone = πrl
= (22/7) × (7/2) × (√113/2)
= 11 × (√113/2)
= (11√113)/2
≈ (11 × 10.630)/2
≈ 116.93/2
≈ 58.46 cm²

However, for a clean CBSE computation, we retain the exact value or use l ≈ 5.315:

CSA of cone = (22/7) × 3.5 × 5.315
= (22/7) × (7/2) × 5.315
= 11 × 5.315
= 58.46 cm²

Base area of cone = πr²
= (22/7) × (3.5)²
= (22/7) × (7/2)²
= (22/7) × (49/4)
= (22 × 7)/4
= 154/4
= 38.5 cm²

⟹ Total surface area = 220 + 58.46 + 38.5
⟹ Total surface area ≈ 316.96 cm²
⟹ Total surface area ≈ 317 cm²

∴ The total surface area to be painted ≈ 317 cm²

[Marking breakup:
(i) Correct formula and l = √113/2 or ≈ 5.32 cm — 1 mark
(ii) Correct formula and CSA = 220 cm² — 1 mark
(iii) Correct CSA of cone (1 mark) + correct base area and final addition (1 mark) — 2 marks]
Q5Case-based4 marks

A school is organising a Science Fair. Priya makes a Rocket Model by placing a solid cone on top of a solid cylinder. The cylinder has height 21 cm and radius 7 cm. The cone has the same base radius 7 cm and height 9 cm. The model is to be painted on all outer surfaces except the bottom circular base. [Use π = 22/7]

A school is organising a Science Fair and students are asked to make models using everyday objects. Priya decides to make a 'Rocket Model' by placing a cone on top of a cylinder. She uses a solid wooden cylinder of height 21 cm and radius 7 cm. On top of the cylinder, she fixes a solid cone of the same base radius (7 cm) and height 9 cm. The entire outer surface of the model (except the bottom circular base) is to be painted.

(i) What is the slant height of the cone fixed on top? (1 mark)
(ii) What is the curved surface area of the cylindrical part that is to be painted? (1 mark)
(iii) Find the total surface area of the rocket model that is to be painted (excluding the bottom base). (2 marks)

[Use π = 22/7]

Diagram for question 5: Surface Areas and Volumes
Show answer
(i) Finding the slant height of the cone:

The slant height l of a cone is given by:
l = √(r² + h²)

Here r = 7 cm, h = 9 cm.

⟹ l = √(7² + 9²)
⟹ l = √(49 + 81)
⟹ l = √130

∴ Slant height of the cone = √130 cm ≈ 11.4 cm

(ii) Finding the Curved Surface Area (CSA) of the cylinder:

CSA of cylinder = 2πrh

Here r = 7 cm, h = 21 cm.

⟹ CSA of cylinder = 2 × (22/7) × 7 × 21
⟹ CSA of cylinder = 2 × 22 × 21
⟹ CSA of cylinder = 924 cm²

∴ Curved surface area of the cylindrical part = 924 cm²

(iii) Finding the total surface area to be painted:

The surfaces to be painted are:
(a) Curved surface area of the cylinder
(b) Curved surface area of the cone
(Note: The top circular face of the cylinder is covered by the base of the cone, so it is NOT painted. The bottom circular base of the cylinder is also NOT painted as stated.)

CSA of cone = πrl
⟹ CSA of cone = (22/7) × 7 × √130
⟹ CSA of cone = 22 × √130
⟹ CSA of cone = 22 × 11.4 (approx.)
⟹ CSA of cone ≈ 250.8 cm²

Using exact value √130:
Total painted surface area = CSA of cylinder + CSA of cone
⟹ Total area = 924 + 22√130
⟹ Total area = 924 + 22 × 11.402
⟹ Total area = 924 + 250.84
⟹ Total area ≈ 1174.84 cm²

∴ Total surface area of the rocket model to be painted = (924 + 22√130) cm² ≈ 1174.84 cm²
Q6Case-based4 marks

A school's science club makes paperweights by fixing a solid hemisphere on top of a solid cylinder. Both shapes share the same radius of 3.5 cm. The height of the cylinder is 5 cm. Use π = 22/7.

A school's science club is making decorative paperweights to sell at the annual fair. Each paperweight is made by placing a solid hemisphere exactly on top of a solid cylinder, both having the same radius of 3.5 cm. The height of the cylinder is 5 cm.

(i) What is the curved surface area of the hemisphere? [1 mark]
(ii) What is the total surface area of the cylindrical part (excluding the top face, since the hemisphere sits on it)? [1 mark]
(iii) Find the total surface area of the combined solid (paperweight). [2 marks]

Diagram for question 6: Surface Areas and Volumes
Show answer
(i) Curved Surface Area of the hemisphere:

Formula: CSA of hemisphere = 2πr²

⟹ CSA = 2 × (22/7) × (3.5)²

⟹ CSA = 2 × (22/7) × 12.25

⟹ CSA = 2 × 38.5

∴ CSA of hemisphere = 77 cm²

---

(ii) Total surface area of the cylindrical part (excluding the top face):

The cylinder contributes: curved surface area (lateral) + base circle (bottom face).

The top face of the cylinder is covered by the hemisphere and is NOT included.

Formula: Required area = 2πrh + πr²

⟹ = 2 × (22/7) × 3.5 × 5 + (22/7) × (3.5)²

⟹ = 2 × (22/7) × 3.5 × 5 + (22/7) × 12.25

⟹ = 110 + 38.5

∴ Required area of cylindrical part = 148.5 cm²

---

(iii) Total surface area of the combined solid (paperweight):

The combined solid exposes the following surfaces:
• Curved surface area of the hemisphere (top curved dome)
• Curved surface area of the cylinder (lateral surface)
• Base circle of the cylinder (bottom face)

Note: The flat circular face of the hemisphere and the top face of the cylinder are joined together — neither is exposed.

Total Surface Area = CSA of hemisphere + CSA of cylinder + Base of cylinder

⟹ Total SA = 2πr² + 2πrh + πr²

⟹ Total SA = 3πr² + 2πrh

⟹ Total SA = πr(3r + 2h)

⟹ Total SA = (22/7) × 3.5 × (3 × 3.5 + 2 × 5)

⟹ Total SA = (22/7) × 3.5 × (10.5 + 10)

⟹ Total SA = (22/7) × 3.5 × 20.5

⟹ Total SA = 11 × 20.5

∴ Total Surface Area of the paperweight = 225.5 cm²
Q7Case-based4 marks

A decorative table lamp shade is made by combining a hollow cone and a hollow cylinder, both open at their ends. The conical top has a slant height of 10 cm and base radius of 6 cm. The cylindrical base has the same radius (6 cm) and height of 8 cm. A craftsman wants to cover the entire outer curved surface of the shade with metallic foil costing ₹15 per 100 cm². He also melts a solid sphere of radius 6 cm to cast solid cylindrical rods of radius 1 cm and height 4 cm.

A decorative table lamp shade is made by combining a hollow cone and a hollow cylinder, both open at their ends, as shown in the description below. The conical top has a slant height of 10 cm and base radius of 6 cm. The cylindrical base has the same radius as the cone (6 cm) and a height of 8 cm. A craftsman wants to cover the entire outer surface (curved surfaces only) of the shade with a special metallic foil that costs ₹15 per 100 cm². Additionally, the craftsman melts a solid sphere of radius 6 cm to cast solid cylindrical rods of radius 1 cm and height 4 cm to be used as lamp stand supports.

Based on the above information, answer the following questions:
(i) Find the curved surface area of the conical part of the shade. [1 mark]
(ii) Find the total curved surface area of the lamp shade (cone + cylinder) that needs to be covered with foil. [1 mark]
(iii) Find the total cost of covering the shade with metallic foil. [2 marks]
OR
(iii) Find the number of solid cylindrical rods that can be cast from the melted solid sphere of radius 6 cm. [2 marks]

Diagram for question 7: Surface Areas and Volumes
Show answer
(i) Curved Surface Area of the conical part:

The formula for Curved Surface Area of a cone = πrl, where r = base radius and l = slant height.

Here, r = 6 cm and l = 10 cm.

⟹ CSA of cone = π × 6 × 10

⟹ CSA of cone = 60π

∴ CSA of cone = 60 × (22/7) = 1320/7 ≈ 188.57 cm²

(ii) Total Curved Surface Area of the lamp shade:

The Curved Surface Area of a cylinder = 2πrh, where r = 6 cm and h = 8 cm.

⟹ CSA of cylinder = 2 × (22/7) × 6 × 8

⟹ CSA of cylinder = 2 × (22/7) × 48

⟹ CSA of cylinder = 2112/7 ≈ 301.71 cm²

Total CSA of shade = CSA of cone + CSA of cylinder

⟹ Total CSA = 1320/7 + 2112/7

⟹ Total CSA = 3432/7

∴ Total CSA = 490.29 cm² (approximately) = 3432/7 cm²

(iii) [Main Option] Total cost of covering the shade with metallic foil:

Total Curved Surface Area = 3432/7 cm²

Cost of foil = ₹15 per 100 cm²

⟹ Total cost = (Total CSA / 100) × 15

⟹ Total cost = (3432/7 × 1/100) × 15

⟹ Total cost = (3432 × 15) / (7 × 100)

⟹ Total cost = 51480 / 700

⟹ Total cost = 73.54 (approximately)

∴ Total cost of covering the lamp shade with metallic foil = ₹73.54 (approximately).

OR

(iii) [Alternate Option] Number of cylindrical rods cast from the melted sphere:

Volume is conserved when a solid is melted and recast.

Volume of sphere = (4/3)πr³, where r = 6 cm.

⟹ Volume of sphere = (4/3) × (22/7) × 6³

⟹ Volume of sphere = (4/3) × (22/7) × 216

⟹ Volume of sphere = (4 × 22 × 216) / (3 × 7)

⟹ Volume of sphere = 19008 / 21 = 6336/7 cm³

Volume of each cylindrical rod = πr²h, where r = 1 cm and h = 4 cm.

⟹ Volume of one cylinder = (22/7) × 1² × 4

⟹ Volume of one cylinder = 88/7 cm³

Let the number of rods = n.

According to the question,

n × (88/7) = 6336/7

⟹ n × 88 = 6336

⟹ n = 6336 / 88

⟹ n = 72

∴ The number of solid cylindrical rods that can be cast = 72.
Q8Case-based4 marks

A school craft club is making decorative paperweights to sell at the annual fair. Each paperweight is made by placing a solid metallic cone on top of a solid metallic cylinder, both sharing the same base radius of 7 cm. The height of the cylinder is 15 cm and the slant height of the cone is 25 cm.

A school craft club is making decorative paperweights to sell at the annual fair. Each paperweight is made by placing a solid metallic cone on top of a solid metallic cylinder, both sharing the same base radius of 7 cm. The height of the cylinder is 15 cm and the slant height of the cone is 25 cm.

(i) Find the height of the cone.
(ii) Find the curved surface area of the cone.
(iii) Find the total surface area of the paperweight (the circular base of the cylinder is the bottom of the paperweight, and the cone sits exactly on top of the cylinder, so the top circular face of the cylinder is not exposed).

OR

(iii) If 20 such paperweights are to be painted, and the cost of painting is ₹5 per cm², find the total cost of painting all 20 paperweights. (Use π = 22/7)

Diagram for question 8: Surface Areas and Volumes
Show answer
(i) Finding the height of the cone:

For the cone, radius r = 7 cm, slant height l = 25 cm.

By Pythagoras theorem:
l² = r² + h²
⟹ 25² = 7² + h²
⟹ 625 = 49 + h²
⟹ h² = 576
⟹ h = 24 cm

∴ Height of the cone = 24 cm. [1 mark]

─────────────────────────────────────────

(ii) Curved surface area of the cone:

CSA of cone = πrl
⟹ CSA = (22/7) × 7 × 25
⟹ CSA = 22 × 25
⟹ CSA = 550 cm²

∴ Curved surface area of the cone = 550 cm². [1 mark]

─────────────────────────────────────────

(iii) Total surface area of the paperweight:

The paperweight consists of:
• Curved surface area of the cone (top)
• Curved surface area of the cylinder (side)
• Circular base of the cylinder (bottom, 1 circle)

Note: The top face of the cylinder is covered by the base of the cone, so it is NOT included. The bottom face of the cylinder (1 circle) is the base of the paperweight.

CSA of cylinder = 2πrh
⟹ = 2 × (22/7) × 7 × 15
⟹ = 2 × 22 × 15
⟹ = 660 cm²

Area of circular base = πr²
⟹ = (22/7) × 7 × 7
⟹ = 22 × 7
⟹ = 154 cm²

Total surface area = CSA of cone + CSA of cylinder + Area of base
⟹ = 550 + 660 + 154
⟹ = 1364 cm²

∴ Total surface area of the paperweight = 1364 cm². [2 marks]

─────────────────────────────────────────

OR

(iii) Total cost of painting 20 paperweights:

Total surface area of one paperweight = 1364 cm² (from above working)

Total surface area of 20 paperweights = 20 × 1364
⟹ = 27280 cm²

Cost of painting = Total area × rate
⟹ = 27280 × 5
⟹ = ₹1,36,400

∴ Total cost of painting all 20 paperweights = ₹1,36,400. [2 marks]
Q9MCQ1 mark

A solid cylinder of radius r and height h is melted and recast into a solid hemisphere of the same radius r. The ratio of the curved surface area of the cylinder (excluding the two circular ends) to the curved surface area of the hemisphere is:

Show answer
Option (A) is correct.

Explanation: Curved surface area of cylinder (excluding circular ends) = 2πrh; curved surface area of hemisphere = 2πr². ⟹ Ratio = 2πrh : 2πr² = h : r.
Q10MCQ1 mark

The curved surface area of a cone is 550 cm² and its slant height is 25 cm. What is the radius of the base of the cone? (Take π = 22/7)

Show answer
Option (A) is correct.

Explanation: CSA of a cone = πrl ⟹ r = CSA/(πl) = 550 ÷ ((22/7) × 25) = (550 × 7)/(22 × 25) = 3850/550 = 7 cm.
Q11MCQ1 mark

A decorative cap is in the shape of a cone with slant height 25 cm and base radius 10 cm. What is the curved surface area of the cap? (Take π = 22/7)

Show answer
Option (B) is correct.

Explanation: Curved surface area of a cone = πrl ⟹ (22/7) × 10 × 25 = (22 × 250)/7 = 5500/7 cm².
Q12MCQ1 mark

A solid sphere of radius 6 cm is cut into two equal hemispheres. The ratio of the curved surface area of the sphere to the sum of the curved surface areas of the two hemispheres taken together is:

Show answer
Option (A) is correct.

Explanation: CSA of sphere = 4πr² = 4π(6)² = 144π cm². Each hemisphere has CSA (curved part only) = 2πr² = 2π(6)² = 72π cm², so sum of CSAs of two hemispheres = 2 × 72π = 144π cm². ∴ Ratio = 144π : 144π = 1 : 1.
Q13MCQ1 mark

The total surface area of a solid hemisphere is 675π cm². What is the radius of the hemisphere?

Diagram for question 13: Surface Areas and Volumes
Show answer
Option (A) is correct.

Explanation: Total surface area of a solid hemisphere = 3πr². Given 3πr² = 675π ⟹ r² = 675/3 = 225 ⟹ r = 15 cm.
Q14MCQ1 mark

A decorative cap is in the shape of a cone with slant height 25 cm and base radius 7 cm. What is the curved surface area of the cap? (Take π = 22/7)

Show answer
Option (A) is correct.

Explanation: CSA of a cone = πrl ⟹ (22/7) × 7 × 25 = 22 × 25 = 550 cm².
Q15MCQ1 mark

The curved surface area of a cylinder of height 14 cm is 528 cm². What is the radius of the cylinder? (Take π = 22/7)

Show answer
Option (A) is correct.

Explanation: CSA of cylinder = 2πrh ⟹ 2 × (22/7) × r × 14 = 528 ⟹ 88r = 528 ⟹ r = 6 cm.
Q16MCQ1 mark

The ratio of the total surface area of a solid cone (excluding the base) to the product of its slant height and base radius is:

Show answer
Option (A) is correct.

Explanation: The total surface area of a solid cone excluding the base equals its curved surface area = πrl, where r is the base radius and l is the slant height. The product of slant height and base radius = l × r. ∴ Required ratio = πrl ÷ (l × r) = π.
Q17MCQ1 mark

The curved surface area of a cone having height 40 cm and base radius 9 cm is (Take π = 22/7):

Show answer
Option (B) is correct.

Explanation: Slant height l = √(h² + r²) = √(40² + 9²) = √(1600 + 81) = √1681 = 41 cm. CSA of cone = πrl = (22/7) × 9 × 41 = 8118/7 ≈ 1159.71 cm².
Q18MCQ1 mark

A cone has a base diameter of 28 cm and a perpendicular height of 48 cm. What is the curved surface area of the cone? (Take π = 22/7)

Show answer
Option (C) is correct.

Explanation: Slant height l = √(r² + h²); here r = 28/2 = 14 cm and h = 48 cm, so l = √(14² + 48²) = √(196 + 2304) = √2500 = 50 cm. CSA = πrl = (22/7) × 14 × 50 = 22 × 2 × 50 = 2200 cm².
Q19MCQ1 mark

A heap of grain is in the form of a cone of diameter 18 m and height 12 m. What is the curved surface area of the cone that represents the heap?

Show answer
Option (A) is correct.

Explanation: Slant height l = √(r² + h²); here r = 18/2 = 9 m, h = 12 m ⟹ l = √(81 + 144) = √225 = 15 m. CSA = πrl = π × 9 × 15 = 135π m².
Q20MCQ1 mark

A solid sphere of radius r is cut into two equal hemispheres. The ratio of the curved surface area of the sphere to the total surface area of one hemisphere is:

Show answer
Option (C) is correct.

Explanation: Curved surface area of the sphere = 4πr²; total surface area of one hemisphere = 2πr² + πr² = 3πr². ∴ Ratio = 4πr² : 3πr² = 4 : 3.
Q21MCQ1 mark

The curved surface area of a cone of slant height 13 cm is 204.1 cm². What is the radius of the base of the cone? (Take π = 3.14)

Show answer
Option (B) is correct.

Explanation: CSA of a cone = πrl. Substituting CSA = 204.1 cm², π = 3.14, l = 13 cm ⟹ r = 204.1 ÷ (3.14 × 13) = 204.1 ÷ 40.82 = 5 cm. ∴ the radius of the base is 5 cm.
Q22MCQ1 mark

A conical tent has a vertical height of 8 m and a base radius of 6 m. What is the curved surface area of the tent? (Take π = 3.14)

Show answer
Option (A) is correct.

Explanation: Slant height l = √(r² + h²) ⟹ l = √(6² + 8²) = √(36 + 64) = √100 = 10 m. CSA of cone = πrl = 3.14 × 6 × 10 = 188.4 m².
Q23MCQ1 mark

The ratio of the curved surface area of a solid sphere to the square of its radius is:

Show answer
Option (B) is correct.

Explanation: The curved surface area of a solid sphere = 4πr². Dividing by r², the ratio = 4πr²/r² = 4π, which is independent of r.
Q24MCQ1 mark

The curved surface area of a cylinder of radius 4 cm is 176 cm². What is the height of the cylinder? (Take π = 22/7)

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Option (A) is correct.

Explanation: Curved surface area of a cylinder = 2πrh. Substituting the given values: 2 × (22/7) × 4 × h = 176 ⟹ (176h/7) = 176 ⟹ h = 7 cm.
Q25MCQ1 mark

A solid cylinder is melted and recast into a solid cone of the same base radius. If the height of the cylinder is 9 cm, what is the height of the cone formed?

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Option (A) is correct. Explanation: Volume is conserved when a solid is recast. Volume of cylinder = πr²× 9; Volume of cone = (1/3)πr²h. Setting equal: πr²× 9 = (1/3)πr²h ⟹ h = 3 × 9 = 27 cm.
Q26MCQ1 mark

The curved surface area of a cylinder of radius 3 cm is 132 cm². What is the height of the cylinder? (Take π = 22/7)

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Option (A) is correct.

Explanation: CSA of cylinder = 2πrh. Substituting CSA = 132 cm² and r = 3 cm:

2 × (22/7) × 3 × h = 132 ⟹ (132/7) × h = 132 ⟹ h = 132 × 7/132 ∴ h = 7 cm.
Q27MCQ1 mark

The curved surface area of a cylinder of radius 7 cm is 308 cm². What is the height of the cylinder? (Take π = 22/7)

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Option (A) is correct.

Explanation: CSA of a cylinder = 2πrh ⟹ 308 = 2 × (22/7) × 7 × h ⟹ 308 = 44h ⟹ h = 308 ÷ 44 = 7 cm.
Q28MCQ1 mark

The ratio of the total surface area of a solid hemisphere to the curved surface area of a sphere of the same radius is:

Show answer
Option (B) is correct.

Explanation: Total surface area of a solid hemisphere = 2πr² + πr² = 3πr²; Curved surface area of a sphere of the same radius = 4πr². ∴ Required ratio = 3πr² : 4πr² = 3 : 4.
Q29MCQ1 mark

A solid toy is in the shape of a hemisphere of radius 4 cm placed on top of a cylinder of the same base radius and height 6 cm. The total surface area of the combined solid is:

Diagram for question 29: Surface Areas and Volumes
Show answer
Option (A) is correct.

Explanation: Total Surface Area = CSA of hemisphere + CSA of cylinder + base area of cylinder. The circular face where the hemisphere sits on the cylinder is internal and not exposed, so it is excluded. CSA of hemisphere = 2πr² = 2π(4)² = 32π; CSA of cylinder = 2πrh = 2π(4)(6) = 48π; base of cylinder = πr² = π(4)² = 16π. ∴ Total Surface Area = 32π + 48π + 16π = 96π cm².
Q30Short Answer3 marks

A solid metallic hemisphere of base radius 9 cm is melted and recast into small solid cylinders, each of base radius 1 cm and height cm. Find the number of cylinders so formed. (Use )

Show answer
Volume of hemisphere (R = 9 cm):

V<sub>hemisphere</sub> = (2/3)πR³ = (2/3) × (22/7) × 9³

⟹ = (2 × 22 × 729) / (3 × 7) = 32076/21 = 10692/7 cm³

Volume of each cylinder (r = 1 cm, h = 3/2 cm):

V<sub>cylinder</sub> = πr²h = (22/7) × 1² × (3/2)

⟹ = 66/14 = 33/7 cm³

Since the hemisphere is melted and recast, total volume is conserved.

Number of cylinders formed:

n = V<sub>hemisphere</sub> / V<sub>cylinder</sub> = (10692/7) ÷ (33/7)

⟹ = 10692/33

∴ n = 324

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Surface Areas and Volumes Class 10 Maths Questions