Madhyamik Mathematics Suggestion – Right Circular Cone লম্ব বৃত্তাকার শঙ্কু Chapter 16 Question and Answer

Madhyamik Mathematics Chapter 16 focuses on the Right Circular Cone and its important geometrical and mensuration-based applications. This chapter is especially useful for practising formula-based problems involving radius, height, slant height, curved surface area, total surface area and volume.

The following resource presents the chapter-wise suggestion in a clear, exam-friendly format, covering MCQ, True or False, Fill in the Blanks, Short Answer and Long Answer questions.

16 Chapter

1 Mark Objective Questions

2 & 5 Marks Written Problems

Right Circular Cone – Important Formulae

Before solving the questions, keep the basic formulae of a right circular cone ready. These formulae are repeatedly used in the numerical problems from this chapter.

Quantity	Formula
Volume	V = 1/3 πr²h
Curved Surface Area	CSA = πrl
Total Surface Area	TSA = πr(l + r)
Slant Height	l² = h² + r²
Height	h = √(l² − r²)

Multiple Choice Questions – MCQ

1 If the height of a right circular cone remains unchanged and its radius is increased by 20 percent, the percentage increase in its volume will be

(a) 44% (b) 33% (c) 22% (d) 11%

Answer: (a) 44%

2 The radius of the base of a cone is 1.5 m and its slant height is 2 m. Its curved surface area is

(a) 2π m² (b) 3π m² (c) 4π m² (d) 5π m²

Answer: (b) 3π m²

3 The ratio of the volumes of two cones is 1 : 4 and the ratio of their diameters is 4 : 5. The ratio of their heights is

(a) 5 : 4 (b) 25 : 8 (c) 5 : 64 (d) 25 : 64

Answer: (d) 25 : 64

True or False

1 The height, radius and slant height of a right circular cone always represent the three sides of a right-angled triangle.

Answer: True

2 The height, radius and slant height of a right circular cone represent the three sides of a right-angled triangle.

Answer: True

3 The base of a right circular cone is elliptical.

Answer: False

4 If the two sides adjacent to the right angle of a right-angled triangle are used separately as axes of revolution, two cones can be generated.

Answer: True

Fill in the Blanks

1 A right circular cylinder and a right circular cone have equal base radii and equal heights. Their volumes are in the ratio ______.

Answer: 3 : 1

2 If the volume of a right circular cone is V cubic units and the area of its base is A square units, its height is ______.

V = 1/3 Ah
Therefore, h = 3V/A

Answer: 3V/A

3 A semicircular sheet has centre O and diameter AB. When OA and OB are joined appropriately to form the required solid, a ______ is obtained.

Answer: Cone

4 If the numerical value of the volume of a cone is equal to the numerical value of the area of its base, the height of the cone is ______ units.

V = 1/3 Ah
Given V = A
Therefore, A = 1/3 Ah
Hence, h = 3 units

Source note: The source text displays the answer as “13”. Using the stated formula and condition, the mathematical result is 3 units. The source appears to contain a typographical inconsistency here.

5 The total number of surfaces of a right circular cone is ______.

Answer: Two

6 If the volume of a right circular cone is V cubic units and the area of its base is A square units, its height is ______.

Answer: 3V/A

7 Volume of a cone = 1/3 × ______ × height.

Answer: Area of the base

8 In right-angled triangle ABC, AC is the hypotenuse. When AB is used as the axis of one complete revolution, the radius of the generated right circular cone is ______.

Answer: BC

Short Answer Questions – 2 Marks

Question 1 – Height of a Cone-Shaped Mountain

A cone-shaped mountain has a slant height of 2.5 km and a base area of 1.54 square km. Find the height of the mountain.

```

Given:
Slant height, l = 2.5 km

Base area = 1.54 km²
πr² = 1.54

Taking π = 22/7:
r² = 1.54 × 7/22 = 0.49
Therefore, r = 0.7 km

For a right circular cone:
l² = h² + r²
h² = l² − r²
h² = 2.5² − 0.7²
h² = 6.25 − 0.49 = 5.76

Therefore,
h = 2.4 km

Answer: 2.4 km ```

Question 2 – Ratio of Height and Radius

The curved surface area of a right circular cone is √5 times its base area. Find the ratio of the height to the radius of the cone.

```

Curved surface area:
πrl

Base area:
πr²

According to the question:
πrl = √5 × πr²
Therefore, l = √5r

Using the relation:
l² = h² + r²
5r² = h² + r²
h² = 4r²
h = 2r

Answer: h : r = 2 : 1 ```

Question 3 – Expression Involving Volume, Base Area and Height

A right circular cone has volume x, base area y and height z. Find the value of (yz + x).

```

For a cone:
x = 1/3 yz

Therefore:
yz = 3x

Hence:
yz + x = 3x + x = 4x

Answer: 4x ```

Long Answer Questions – 5 Marks

Question 1 – Solid Formed by Revolving a Right-Angled Triangle

The two sides adjacent to the right angle of a right-angled triangle are 4 cm and 3 cm. If the triangle is revolved once about the longer of these two sides, calculate the curved surface area, total surface area and volume of the solid formed.

```

The longer side is used as the axis.
Therefore, height of the cone, h = 4 cm
Radius, r = 3 cm

Slant height:
l = √(h² + r²)
= √(4² + 3²)
= √25 = 5 cm

Curved Surface Area:
CSA = πrl
= π × 3 × 5
= 15π cm²

Total Surface Area:
TSA = πr(l + r)
= π × 3 × (5 + 3)
= 24π cm²

Volume:
V = 1/3 πr²h
= 1/3 × π × 3² × 4
= 12π cm³

Answer: CSA = 15π cm², TSA = 24π cm², Volume = 12π cm³ ```

Question 2 – Cone-Shaped Traditional Headgear

A cone-shaped traditional headgear made of shola has an outer base diameter of 21 cm. Covering its upper surface with metallic foil costs Rs.57.75 at the rate of 10 paise per square centimetre. Find its height and slant height.

```

Diameter = 21 cm
Therefore, r = 21/2 = 10.5 cm

Cost = Rs.57.75
Rate = 10 paise per cm² = Rs.0.10 per cm²
Required curved surface area = 57.75 ÷ 0.10
= 577.5 cm²

For the cone:
πrl = 577.5
Taking π = 22/7:
22/7 × 10.5 × l = 577.5
33l = 577.5
l = 17.5 cm

Now:
l² = h² + r²
h² = 17.5² − 10.5²
= 306.25 − 110.25
= 196
Therefore, h = 14 cm

Answer: Height = 14 cm, Slant height = 17.5 cm ```

Question 3 – Base Area of a Cone-Shaped Tent

A cone-shaped tent requires 77 square metres of tarpaulin. If the slant height of the tent is 7 metres, find the area of its base.

```

The tarpaulin covers the curved surface of the cone.
Therefore:
πrl = 77

Given l = 7 m:
7πr = 77
Taking π = 22/7:
22r = 77
r = 3.5 m

Base area:
πr²
= 22/7 × 3.5²
= 38.5 m²

Answer: 38.5 m² ```

Question 4 – Number of Cones Made by Melting a Solid Cylinder

A solid iron right circular cylinder has a cross-sectional diameter of 16 cm and a length of 1 metre. It is melted to make solid right circular cones of height 8 cm and base radius 5 cm. How many such cones can be made.

```

Cylinder radius = 16/2 = 8 cm
Cylinder height = 1 m = 100 cm

Volume of the cylinder:
V = πr²h
= π × 8² × 100
= 6400π cm³

Volume of one cone:
V = 1/3 πr²h
= 1/3 × π × 5² × 8
= 200π/3 cm³

Number of cones:
= 6400π ÷ (200π/3)
= 96

Answer: 96 cones ```

Question 5 – Proving the Height Relationship Between a Cylinder and Cone

A right circular cylinder and a cone have equal bases and their volumes are in the ratio 3 : 2. Prove that the height of the cone is twice the height of the cylinder.

```

Let the common base area be A.
Let the height of the cylinder be h₁ and the height of the cone be h₂.

Volume of cylinder:
V₁ = Ah₁

Volume of cone:
V₂ = 1/3 Ah₂

According to the question:
V₁ : V₂ = 3 : 2
Therefore:
Ah₁ : 1/3 Ah₂ = 3 : 2

Cancelling A:
h₁ : h₂/3 = 3 : 2
Therefore:
2h₁ = h₂

Hence proved: h₂ = 2h₁ ```

Question 6 – Hollow Cylinder Melted into a Solid Cone

A hollow iron cylinder is 20 cm high. Its outer and inner radii are respectively 5 cm and 4 cm. The cylinder is melted to form a solid right circular cone whose height is one-third of the cylinder's height. Find the diameter of the base of the cone.

```

Height of hollow cylinder = 20 cm
Outer radius = 5 cm
Inner radius = 4 cm

Volume of hollow cylinder:
V = π(R² − r²)h
= π(5² − 4²) × 20
= π(25 − 16) × 20
= 180π cm³

Height of cone:
H = 20/3 cm

Let the radius of the cone be x cm.
Volume of cone:
V = 1/3 πx² × 20/3
= 20πx²/9

Since the same iron is used:
20πx²/9 = 180π
x² = 81
x = 9 cm

Therefore, diameter = 2 × 9 = 18 cm

Answer: 18 cm ```

Chapter 16 Quick Revision Table

Topic	Key Point
Volume	V = 1/3 πr²h
Curved Surface Area	πrl
Total Surface Area	πr(l + r)
Slant Height	l² = h² + r²
Volume Ratio with Same Base	Depends directly on height, with the cone carrying the factor 1/3
Melting Problems	Volume of original solid = Total volume of newly formed solids

Revision Tip: For this chapter, practise the relationship between radius, height and slant height first. Then revise curved surface area, total surface area and volume. Problems involving melting and revolution should be solved carefully by identifying the solid and writing its volume formula before substituting values.

Madhyamik Mathematics Suggestion Resources

For additional subject-wise preparation, students can explore the Cademy Madhyamik Suggestion 2027 – All Subjects resource, which includes a dedicated Mathematics suggestion section along with the other WBBSE Class 10 subjects.

Study Note: This Chapter 16 resource is designed for focused revision of the Right Circular Cone topic. Students should work through the prescribed textbook and practise numerical problems along with these suggested questions.

The following resource presents the chapter-wise suggestion in a clear, exam-friendly format, covering MCQ, True or False, Fill in the Blanks, Short Answer and Long Answer questions.

16 Chapter

1 Mark Objective Questions

2 & 5 Marks Written Problems

Right Circular Cone – Important Formulae

Before solving the questions, keep the basic formulae of a right circular cone ready. These formulae are repeatedly used in the numerical problems from this chapter.

Quantity	Formula
Volume	V = 1/3 πr²h
Curved Surface Area	CSA = πrl
Total Surface Area	TSA = πr(l + r)
Slant Height	l² = h² + r²
Height	h = √(l² − r²)

Multiple Choice Questions – MCQ

1 If the height of a right circular cone remains unchanged and its radius is increased by 20 percent, the percentage increase in its volume will be

(a) 44% (b) 33% (c) 22% (d) 11%

Answer: (a) 44%

2 The radius of the base of a cone is 1.5 m and its slant height is 2 m. Its curved surface area is

(a) 2π m² (b) 3π m² (c) 4π m² (d) 5π m²

Answer: (b) 3π m²

3 The ratio of the volumes of two cones is 1 : 4 and the ratio of their diameters is 4 : 5. The ratio of their heights is

(a) 5 : 4 (b) 25 : 8 (c) 5 : 64 (d) 25 : 64

Answer: (d) 25 : 64

True or False

1 The height, radius and slant height of a right circular cone always represent the three sides of a right-angled triangle.

Answer: True

2 The height, radius and slant height of a right circular cone represent the three sides of a right-angled triangle.

Answer: True

3 The base of a right circular cone is elliptical.

Answer: False

4 If the two sides adjacent to the right angle of a right-angled triangle are used separately as axes of revolution, two cones can be generated.

Answer: True

Fill in the Blanks

1 A right circular cylinder and a right circular cone have equal base radii and equal heights. Their volumes are in the ratio ______.

Answer: 3 : 1

2 If the volume of a right circular cone is V cubic units and the area of its base is A square units, its height is ______.

V = 1/3 Ah
Therefore, h = 3V/A

Answer: 3V/A

3 A semicircular sheet has centre O and diameter AB. When OA and OB are joined appropriately to form the required solid, a ______ is obtained.

Answer: Cone

4 If the numerical value of the volume of a cone is equal to the numerical value of the area of its base, the height of the cone is ______ units.

V = 1/3 Ah
Given V = A
Therefore, A = 1/3 Ah
Hence, h = 3 units

5 The total number of surfaces of a right circular cone is ______.

Answer: Two

6 If the volume of a right circular cone is V cubic units and the area of its base is A square units, its height is ______.

Answer: 3V/A

7 Volume of a cone = 1/3 × ______ × height.

Answer: Area of the base

8 In right-angled triangle ABC, AC is the hypotenuse. When AB is used as the axis of one complete revolution, the radius of the generated right circular cone is ______.

Answer: BC

Short Answer Questions – 2 Marks

Question 1 – Height of a Cone-Shaped Mountain

A cone-shaped mountain has a slant height of 2.5 km and a base area of 1.54 square km. Find the height of the mountain.

```

Given:
Slant height, l = 2.5 km

Base area = 1.54 km²
πr² = 1.54

Taking π = 22/7:
r² = 1.54 × 7/22 = 0.49
Therefore, r = 0.7 km

For a right circular cone:
l² = h² + r²
h² = l² − r²
h² = 2.5² − 0.7²
h² = 6.25 − 0.49 = 5.76

Therefore,
h = 2.4 km

Answer: 2.4 km ```

Question 2 – Ratio of Height and Radius

The curved surface area of a right circular cone is √5 times its base area. Find the ratio of the height to the radius of the cone.

```

Curved surface area:
πrl

Base area:
πr²

According to the question:
πrl = √5 × πr²
Therefore, l = √5r

Using the relation:
l² = h² + r²
5r² = h² + r²
h² = 4r²
h = 2r

Answer: h : r = 2 : 1 ```

Question 3 – Expression Involving Volume, Base Area and Height

A right circular cone has volume x, base area y and height z. Find the value of (yz + x).

```

For a cone:
x = 1/3 yz

Therefore:
yz = 3x

Hence:
yz + x = 3x + x = 4x

Answer: 4x ```

Long Answer Questions – 5 Marks

Question 1 – Solid Formed by Revolving a Right-Angled Triangle

```

The longer side is used as the axis.
Therefore, height of the cone, h = 4 cm
Radius, r = 3 cm

Slant height:
l = √(h² + r²)
= √(4² + 3²)
= √25 = 5 cm

Curved Surface Area:
CSA = πrl
= π × 3 × 5
= 15π cm²

Total Surface Area:
TSA = πr(l + r)
= π × 3 × (5 + 3)
= 24π cm²

Volume:
V = 1/3 πr²h
= 1/3 × π × 3² × 4
= 12π cm³

Answer: CSA = 15π cm², TSA = 24π cm², Volume = 12π cm³ ```

Question 2 – Cone-Shaped Traditional Headgear

```

Diameter = 21 cm
Therefore, r = 21/2 = 10.5 cm

Cost = Rs.57.75
Rate = 10 paise per cm² = Rs.0.10 per cm²
Required curved surface area = 57.75 ÷ 0.10
= 577.5 cm²

For the cone:
πrl = 577.5
Taking π = 22/7:
22/7 × 10.5 × l = 577.5
33l = 577.5
l = 17.5 cm

Now:
l² = h² + r²
h² = 17.5² − 10.5²
= 306.25 − 110.25
= 196
Therefore, h = 14 cm

Answer: Height = 14 cm, Slant height = 17.5 cm ```

Question 3 – Base Area of a Cone-Shaped Tent

A cone-shaped tent requires 77 square metres of tarpaulin. If the slant height of the tent is 7 metres, find the area of its base.

```

The tarpaulin covers the curved surface of the cone.
Therefore:
πrl = 77

Given l = 7 m:
7πr = 77
Taking π = 22/7:
22r = 77
r = 3.5 m

Base area:
πr²
= 22/7 × 3.5²
= 38.5 m²

Answer: 38.5 m² ```

Question 4 – Number of Cones Made by Melting a Solid Cylinder

```

Cylinder radius = 16/2 = 8 cm
Cylinder height = 1 m = 100 cm

Volume of the cylinder:
V = πr²h
= π × 8² × 100
= 6400π cm³

Volume of one cone:
V = 1/3 πr²h
= 1/3 × π × 5² × 8
= 200π/3 cm³

Number of cones:
= 6400π ÷ (200π/3)
= 96

Answer: 96 cones ```

Question 5 – Proving the Height Relationship Between a Cylinder and Cone

A right circular cylinder and a cone have equal bases and their volumes are in the ratio 3 : 2. Prove that the height of the cone is twice the height of the cylinder.

```

Let the common base area be A.
Let the height of the cylinder be h₁ and the height of the cone be h₂.

Volume of cylinder:
V₁ = Ah₁

Volume of cone:
V₂ = 1/3 Ah₂

According to the question:
V₁ : V₂ = 3 : 2
Therefore:
Ah₁ : 1/3 Ah₂ = 3 : 2

Cancelling A:
h₁ : h₂/3 = 3 : 2
Therefore:
2h₁ = h₂

Hence proved: h₂ = 2h₁ ```

Question 6 – Hollow Cylinder Melted into a Solid Cone

```

Height of hollow cylinder = 20 cm
Outer radius = 5 cm
Inner radius = 4 cm

Volume of hollow cylinder:
V = π(R² − r²)h
= π(5² − 4²) × 20
= π(25 − 16) × 20
= 180π cm³

Height of cone:
H = 20/3 cm

Let the radius of the cone be x cm.
Volume of cone:
V = 1/3 πx² × 20/3
= 20πx²/9

Since the same iron is used:
20πx²/9 = 180π
x² = 81
x = 9 cm

Therefore, diameter = 2 × 9 = 18 cm

Answer: 18 cm ```

Chapter 16 Quick Revision Table

Topic	Key Point
Volume	V = 1/3 πr²h
Curved Surface Area	πrl
Total Surface Area	πr(l + r)
Slant Height	l² = h² + r²
Volume Ratio with Same Base	Depends directly on height, with the cone carrying the factor 1/3
Melting Problems	Volume of original solid = Total volume of newly formed solids

Madhyamik Mathematics Suggestion – Right Circular Cone লম্ব বৃত্তাকার শঙ্কু Chapter 16 Question and Answer

Right Circular Cone – Important Formulae

Multiple Choice Questions – MCQ

True or False

Fill in the Blanks

Short Answer Questions – 2 Marks

Question 1 – Height of a Cone-Shaped Mountain

Question 2 – Ratio of Height and Radius

Question 3 – Expression Involving Volume, Base Area and Height

Long Answer Questions – 5 Marks

Question 1 – Solid Formed by Revolving a Right-Angled Triangle

Question 2 – Cone-Shaped Traditional Headgear

Question 3 – Base Area of a Cone-Shaped Tent

Question 4 – Number of Cones Made by Melting a Solid Cylinder

Question 5 – Proving the Height Relationship Between a Cylinder and Cone

Question 6 – Hollow Cylinder Melted into a Solid Cone

Chapter 16 Quick Revision Table

Madhyamik Mathematics Suggestion Resources

Frequently Asked Questions

Madhyamik Mathematics Suggestion – Right Circular Cone লম্ব বৃত্তাকার শঙ্কু Chapter 16 Question and Answer

Right Circular Cone – Important Formulae

Multiple Choice Questions – MCQ

True or False

Fill in the Blanks

Short Answer Questions – 2 Marks

Question 1 – Height of a Cone-Shaped Mountain

Question 2 – Ratio of Height and Radius

Question 3 – Expression Involving Volume, Base Area and Height

Long Answer Questions – 5 Marks

Question 1 – Solid Formed by Revolving a Right-Angled Triangle

Question 2 – Cone-Shaped Traditional Headgear

Question 3 – Base Area of a Cone-Shaped Tent

Question 4 – Number of Cones Made by Melting a Solid Cylinder

Question 5 – Proving the Height Relationship Between a Cylinder and Cone

Question 6 – Hollow Cylinder Melted into a Solid Cone

Chapter 16 Quick Revision Table

Madhyamik Mathematics Suggestion Resources

Frequently Asked Questions