Surface Areas and Volumes | Grade 10th - Mathematics Chapter Summary & NCERT CBSE Study Resources

Study Materials

10th

10th - Mathematics

Surface Areas and Volumes

Overview

This chapter introduces students to the concepts of surface areas and volumes of various three-dimensional shapes such as cubes, cuboids, cylinders, cones, and spheres. It covers the derivation of formulas and their applications in real-life scenarios.

Surface Area and Volume of a Cuboid

A cuboid is a three-dimensional shape with six rectangular faces. The surface area is the total area of all its faces, while volume is the space occupied by the cuboid.

Surface Area (SA): 2(lb + bh + hl), where l = length, b = breadth, h = height.

Volume (V): l × b × h

Surface Area and Volume of a Cube

A cube is a special case of a cuboid where all edges are equal.

Surface Area (SA): 6a², where a = edge length.

Volume (V): a³

Surface Area and Volume of a Cylinder

A cylinder has two circular bases and a curved surface. It can be a right circular cylinder if the axis is perpendicular to the base.

Total Surface Area (TSA): 2πr(h + r), where r = radius, h = height.

Curved Surface Area (CSA): 2πrh

Volume (V): πr²h

Surface Area and Volume of a Cone

A cone has a circular base and a curved surface tapering to a point called the apex.

Total Surface Area (TSA): πr(l + r), where r = radius, l = slant height.

Curved Surface Area (CSA): πrl

Volume (V): (1/3)πr²h

Surface Area and Volume of a Sphere

A sphere is a perfectly round three-dimensional object where every point on the surface is equidistant from the center.

Surface Area (SA): 4πr², where r = radius.

Volume (V): (4/3)πr³

Combination of Solids

This section discusses the calculation of surface areas and volumes when two or more solids are combined. Examples include a cylinder surmounted by a cone or a hemisphere attached to a cylinder.

Conversion of Units

Students learn to convert between different units of measurement (e.g., cm³ to m³) for volume and surface area calculations.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:

Find the total surface area of a cube with side 4 cm.

Answer:

96 cm²

Question 2:

What is the volume of a sphere with radius 3 cm?

Answer:

113.1 cm³

Question 3:

Calculate the curved surface area of a cylinder with radius 7 cm and height 10 cm.

Answer:

440 cm²

Question 4:

A cone has a slant height of 5 cm and radius 3 cm. Find its total surface area.

Answer:

75.4 cm²

Question 5:

What is the volume of a cuboid with dimensions 5 cm × 4 cm × 3 cm?

Answer:

60 cm³

Question 6:

Find the lateral surface area of a cube with edge 6 cm.

Answer:

144 cm²

Question 7:

A hemispherical bowl has a radius of 7 cm. Calculate its curved surface area.

Answer:

308 cm²

Question 8:

What is the total surface area of a cylinder with radius 2 cm and height 5 cm?

Answer:

88 cm²

Question 9:

Find the volume of a cone with radius 6 cm and height 8 cm.

Answer:

301.6 cm³

Question 10:

A sphere has a surface area of 154 cm². Find its radius.

Answer:

3.5 cm

Question 11:

Calculate the lateral surface area of a cone with slant height 10 cm and radius 6 cm.

Answer:

188.5 cm²

Question 12:

A cuboidal tank is 10 m long, 8 m wide, and 6 m high. Find its capacity in litres.

Answer:

480,000 litres

Question 13:

Find the total surface area of a cube with side length 5 cm.

Answer:

The total surface area of a cube is calculated using the formula: 6 × (side)².
Given side = 5 cm,
Total surface area = 6 × (5)²
= 6 × 25
= 150 cm².

Question 14:

Calculate the volume of a cylinder with radius 7 cm and height 10 cm. (Use π = 22/7)

Answer:

The volume of a cylinder is given by: πr²h.
Given r = 7 cm, h = 10 cm,
Volume = (22/7) × (7)² × 10
= (22/7) × 49 × 10
= 22 × 7 × 10
= 1540 cm³.

Question 15:

What is the lateral surface area of a cone with slant height 13 cm and radius 5 cm? (Use π = 3.14)

Answer:

The lateral surface area of a cone is: πrl.
Given r = 5 cm, l = 13 cm,
Lateral surface area = 3.14 × 5 × 13
= 3.14 × 65
= 204.1 cm².

Question 16:

A sphere has a diameter of 14 cm. Find its volume. (Use π = 22/7)

Answer:

The volume of a sphere is: (4/3)πr³.
Given diameter = 14 cm, so radius r = 7 cm.
Volume = (4/3) × (22/7) × (7)³
= (4/3) × (22/7) × 343
= (4/3) × 22 × 49
= 1437.33 cm³ (approx).

Question 17:

Find the total surface area of a hemisphere with radius 3 cm. (Use π = 3.14)

Answer:

The total surface area of a hemisphere is: 3πr².
Given r = 3 cm,
Total surface area = 3 × 3.14 × (3)²
= 3 × 3.14 × 9
= 84.78 cm².

Question 18:

A cuboidal tank is 6 m long, 5 m wide, and 4.5 m deep. How much water can it hold?

Answer:

The volume of a cuboid is: length × width × height.
Given dimensions: 6 m × 5 m × 4.5 m,
Volume = 6 × 5 × 4.5
= 30 × 4.5
= 135 m³.
Thus, the tank can hold 135,000 liters (since 1 m³ = 1000 liters).

Question 19:

The curved surface area of a cylinder is 440 cm² and its height is 10 cm. Find its radius. (Use π = 22/7)

Answer:

The curved surface area of a cylinder is: 2πrh.
Given CSA = 440 cm², h = 10 cm,
440 = 2 × (22/7) × r × 10
440 = (440/7) × r
r = (440 × 7) / 440
r = 7 cm.

Question 20:

A cone has a base radius of 6 cm and height 8 cm. Find its slant height.

Answer:

The slant height (l) of a cone is: √(r² + h²).
Given r = 6 cm, h = 8 cm,
l = √(6² + 8²)
= √(36 + 64)
= √100
= 10 cm.

Question 21:

The volume of a cube is 216 cm³. Find its edge length.

Answer:

The volume of a cube is: (side)³.
Given volume = 216 cm³,
side = ∛216
= 6 cm.

Question 22:

A metallic sphere of radius 4.2 cm is melted and recast into a cylinder of radius 6 cm. Find the height of the cylinder. (Use π = 22/7)

Answer:

Since the volume remains the same,
Volume of sphere = Volume of cylinder.
(4/3)πr³ = πR²h
(4/3) × (22/7) × (4.2)³ = (22/7) × (6)² × h
h = [(4/3) × (4.2)³] / 36
h = (4 × 74.088) / 108
h = 2.744 cm (approx).

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:

Find the total surface area of a cube with edge length 5 cm.

Answer:

The total surface area of a cube is given by 6 × (edge)².
Edge length = 5 cm.
Total surface area = 6 × (5 cm)²
= 6 × 25 cm²
= 150 cm².

Question 2:

A cone has a base radius of 3 cm and slant height of 5 cm. Find its curved surface area.

Answer:

The curved surface area of a cone is given by πrl.
Radius (r) = 3 cm, Slant height (l) = 5 cm.
Curved surface area = (22/7) × 3 cm × 5 cm
= (22/7) × 15 cm²
= 47.14 cm² (approx).

Question 3:

Determine the lateral surface area of a cuboid with dimensions 4 cm × 5 cm × 6 cm.

Answer:

The lateral surface area of a cuboid is given by 2h(l + b).
Dimensions: l = 6 cm, b = 5 cm, h = 4 cm.
Lateral surface area = 2 × 4 cm × (6 cm + 5 cm)
= 8 cm × 11 cm
= 88 cm².

Question 4:

Find the volume of a sphere with diameter 14 cm. (Use π = 22/7)

Answer:

The volume of a sphere is given by (4/3)πr³.
Diameter = 14 cm ⇒ Radius (r) = 7 cm.
Volume = (4/3) × (22/7) × (7 cm)³
= (4/3) × (22/7) × 343 cm³
= 1437.33 cm³ (approx).

Question 5:

A hemisphere has a radius of 10 cm. Calculate its total surface area.

Answer:

The total surface area of a hemisphere is 3πr².
Radius (r) = 10 cm.
Total surface area = 3 × (22/7) × (10 cm)²
= 3 × (22/7) × 100 cm²
= 942.86 cm² (approx).

Question 6:

The height of a cylinder is 15 cm and its curved surface area is 660 cm². Find its radius.

Answer:

The curved surface area of a cylinder is 2πrh.
Given: CSA = 660 cm², h = 15 cm.
660 cm² = 2 × (22/7) × r × 15 cm
r = 660 / (2 × (22/7) × 15)
r = 7 cm.

Question 7:

A rectangular sheet of paper 22 cm × 12 cm is rolled along its length to form a cylinder. Find its volume.

Answer:

When rolled along length, height (h) = 12 cm, circumference = 22 cm.
Circumference = 2πr ⇒ 22 cm = 2 × (22/7) × r
r = 3.5 cm.
Volume = πr²h = (22/7) × (3.5 cm)² × 12 cm
= 462 cm³.

Question 8:

The radius and height of a cone are in the ratio 3:4. If its volume is 301.44 cm³, find the radius. (Use π = 3.14)

Answer:

Let radius (r) = 3x, height (h) = 4x.
Volume of cone = (1/3)πr²h ⇒ 301.44 = (1/3) × 3.14 × (3x)² × 4x
301.44 = 3.14 × 12x³
x³ = 8 ⇒ x = 2.
Radius = 3 × 2 = 6 cm.

Question 9:

A metallic sphere of radius 8 cm is melted and recast into small cubes of edge 2 cm. How many cubes can be made?

Answer:

Volume of sphere = (4/3)πr³ = (4/3) × (22/7) × 512 cm³ ≈ 2145.52 cm³.
Volume of one cube = (edge)³ = 8 cm³.
Number of cubes = Volume of sphere / Volume of cube
= 2145.52 / 8 ≈ 268 cubes.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:

A cylindrical tank has a radius of 7 m and a height of 3 m. Find its total surface area.

Answer:

The total surface area of a cylinder is given by the formula:
2πr(h + r), where r is the radius and h is the height.

Given: r = 7 m, h = 3 m.

Step 1: Substitute the values into the formula.
Total Surface Area = 2 × (22/7) × 7 × (3 + 7)

Step 2: Simplify the expression.
= 2 × 22 × 10

Step 3: Calculate the final value.
= 440 m².

Note: The total surface area includes both the curved surface and the two circular bases.

Question 2:

A cone has a slant height of 10 cm and a base radius of 6 cm. Calculate its curved surface area.

Answer:

The curved surface area of a cone is given by the formula:
πrl, where r is the radius and l is the slant height.

Given: r = 6 cm, l = 10 cm.

Step 1: Substitute the values into the formula.
Curved Surface Area = (22/7) × 6 × 10

Step 2: Simplify the expression.
= (22/7) × 60

Step 3: Calculate the final value.
≈ 188.57 cm².

Note: The curved surface area refers only to the lateral part of the cone, excluding the base.

Question 3:

A sphere has a diameter of 14 cm. Determine its volume.

Answer:

The volume of a sphere is given by the formula:
(4/3)πr³, where r is the radius.

Given: Diameter = 14 cm, so r = 7 cm.

Step 1: Substitute the values into the formula.
Volume = (4/3) × (22/7) × 7³

Step 2: Simplify the expression.
= (4/3) × (22/7) × 343

Step 3: Calculate the final value.
≈ 1437.33 cm³.

Note: The volume represents the space occupied by the sphere in three-dimensional space.

Question 4:

A cuboidal water tank is 5 m long, 4 m wide, and 3 m high. How much water can it hold in liters?

Answer:

The volume of a cuboid is given by the formula:
l × b × h, where l is length, b is breadth, and h is height.

Given: l = 5 m, b = 4 m, h = 3 m.

Step 1: Calculate the volume in cubic meters.
Volume = 5 × 4 × 3 = 60 m³.

Step 2: Convert cubic meters to liters.
1 m³ = 1000 liters.

Step 3: Multiply to find the capacity.
Capacity = 60 × 1000 = 60,000 liters.

Note: This calculation is essential for real-world applications like water storage.

Question 5:

A hemispherical bowl has a radius of 3.5 cm. Find its total surface area.

Answer:

The total surface area of a hemisphere is given by the formula:
3πr², where r is the radius.

Given: r = 3.5 cm.

Step 1: Substitute the values into the formula.
Total Surface Area = 3 × (22/7) × (3.5)²

Step 2: Simplify the expression.
= 3 × (22/7) × 12.25

Step 3: Calculate the final value.
≈ 115.5 cm².

Note: The total surface area includes the curved part and the circular base of the hemisphere.

Question 6:

A cylindrical tank has a diameter of 14 m and height of 5 m. Calculate its lateral surface area.

Answer:

The lateral surface area of a cylinder is given by the formula: 2πrh.
Given, diameter = 14 m, so radius (r) = 14/2 = 7 m.
Height (h) = 5 m.

Step 1: Substitute values into the formula.
Lateral Surface Area = 2 × π × 7 × 5

Step 2: Simplify the calculation.
= 2 × 22/7 × 7 × 5
= 2 × 22 × 5
= 220 m².

Thus, the lateral surface area is 220 m².

Question 7:

A cone has a slant height of 10 cm and a base radius of 6 cm. Find its total surface area.

Answer:

The total surface area of a cone is given by: πr(l + r).
Given, radius (r) = 6 cm, slant height (l) = 10 cm.

Step 1: Substitute values into the formula.
Total Surface Area = π × 6 × (10 + 6)

Step 2: Simplify the calculation.
= 22/7 × 6 × 16
= 22/7 × 96
= 301.71 cm² (approx).

Thus, the total surface area is 301.71 cm².

Question 8:

A sphere has a volume of 4851 cm³. Calculate its radius (use π = 22/7).

Answer:

The volume of a sphere is given by: (4/3)πr³.
Given, Volume = 4851 cm³.

Step 1: Substitute values into the formula.
4851 = (4/3) × (22/7) × r³

Step 2: Simplify the equation.
4851 × 3/4 = 22/7 × r³
3638.25 = 22/7 × r³

Step 3: Solve for r³.
r³ = 3638.25 × 7/22
r³ = 1157.625

Step 4: Find the cube root.
r = ∛1157.625 ≈ 10.5 cm.

Thus, the radius is 10.5 cm.

Question 9:

A cuboidal water tank is 6 m long, 5 m wide, and 4.5 m deep. How many litres of water can it hold?

Answer:

The volume of a cuboid is given by: l × b × h.
Given, length (l) = 6 m, breadth (b) = 5 m, height (h) = 4.5 m.

Step 1: Calculate volume in cubic metres.
Volume = 6 × 5 × 4.5
= 135 m³.

Step 2: Convert to litres (1 m³ = 1000 litres).
Volume in litres = 135 × 1000
= 135,000 litres.

Thus, the tank can hold 135,000 litres of water.

Question 10:

A hemispherical bowl has a radius of 3.5 cm. Find its curved surface area.

Answer:

The curved surface area of a hemisphere is given by: 2πr².
Given, radius (r) = 3.5 cm.

Step 1: Substitute values into the formula.
Curved Surface Area = 2 × π × (3.5)²

Step 2: Simplify the calculation.
= 2 × 22/7 × 12.25
= 2 × 22 × 1.75
= 77 cm².

Thus, the curved surface area is 77 cm².

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:

A cylindrical tank has a radius of 7 m and height 10 m. Calculate its total surface area and volume. Explain the steps with proper units.

Answer:

Introduction

We studied that a cylinder has two circular bases and a curved surface. The given radius (r) is 7 m and height (h) is 10 m.

Argument 1

Total Surface Area (TSA): TSA = 2πr(h + r) = 2 × (22/7) × 7 × (10 + 7) = 748 m².
Volume: V = πr²h = (22/7) × 7² × 10 = 1540 m³.

Conclusion

Thus, the TSA is 748 m² and volume is 1540 m³, following NCERT formulas.

Question 2:

A hemispherical dome has a radius of 14 cm. Find its curved surface area and total surface area. Compare it with NCERT Example 2.

Answer:

Introduction

Our textbook shows that a hemisphere has half the surface area of a sphere. Here, radius (r) = 14 cm.

Argument 1

Curved Surface Area (CSA): CSA = 2πr² = 2 × (22/7) × 14² = 1232 cm².
Total Surface Area (TSA): TSA = 3πr² = 3 × (22/7) × 14² = 1848 cm².

Conclusion

Like NCERT Example 2, CSA is 1232 cm² and TSA is 1848 cm², verified with π = 22/7.

Question 3:

A cylindrical vessel has an internal radius of 7 cm and height 10 cm. Calculate its total surface area (including the base and top). Explain the steps with proper units.

Answer:

Introduction

We studied that the total surface area of a cylinder includes its curved surface and two circular bases.

Argument 1

Given: radius (r) = 7 cm, height (h) = 10 cm
Curved Surface Area (CSA) = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm²

Argument 2

Area of two bases = 2πr² = 2 × (22/7) × 7 × 7 = 308 cm²
Total Surface Area = CSA + Base Areas = 440 + 308 = 748 cm²

Conclusion

Thus, the total surface area is 748 cm², as derived from NCERT formulas.

Question 4:

A hemispherical bowl has a radius of 3.5 cm. Find its outer curved surface area and total surface area. Compare the results with a similar NCERT problem.

Answer:

Introduction

Our textbook shows that a hemisphere has a curved surface and a flat circular base.

Argument 1

Given: radius (r) = 3.5 cm
Outer Curved Surface Area = 2πr² = 2 × (22/7) × 3.5 × 3.5 = 77 cm²

Argument 2

Total Surface Area = 3πr² (including base) = 3 × (22/7) × 3.5 × 3.5 = 115.5 cm²
Similar to NCERT Example 2, where r = 7 cm, but our answer scales proportionally.

Conclusion

The outer curved surface area is 77 cm², and the total surface area is 115.5 cm².

Question 5:

A cone has a base radius of 5 cm and slant height 13 cm. Calculate its total surface area and explain how it differs from a cylinder's surface area.

Answer:

Introduction

We learned that a cone has a circular base and a curved surface, unlike a cylinder.

Argument 1

Given: radius (r) = 5 cm, slant height (l) = 13 cm
Curved Surface Area = πrl = (22/7) × 5 × 13 = 204.28 cm²

Argument 2

Base Area = πr² = (22/7) × 5 × 5 = 78.57 cm²
Total Surface Area = CSA + Base = 204.28 + 78.57 = 282.85 cm²
A cylinder has two circular bases, but a cone has only one.

Conclusion

The total surface area of the cone is 282.85 cm², differing from a cylinder due to its single base.

Question 6:

A cylindrical tank has a radius of 7 m and height 10 m. Calculate its total surface area and volume. Compare these values with a cubical tank of the same volume (Use π = 22/7).

Answer:

Introduction

We studied that surface area and volume calculations are essential for real-life applications like storage tanks. Here, we compare a cylinder and cube.

Argument 1

Cylinder: Radius = 7 m, Height = 10 m
Total Surface Area = 2πr(h + r) = 2 × (22/7) × 7 × (10 + 7) = 748 m²
Volume = πr²h = (22/7) × 7² × 10 = 1540 m³

Argument 2

Cube Volume = 1540 m³ → Side = ∛1540 ≈ 11.52 m
Cube Surface Area = 6 × (11.52)² ≈ 796.26 m²

Conclusion

The cylindrical tank has less surface area than the cubical tank for the same volume, making it more efficient for material usage.

Question 7:

Derive the formula for the curved surface area of a cone using its slant height (l) and base radius (r). Verify it with an example where r = 5 cm and l = 13 cm.

Answer:

Introduction

Our textbook shows that a cone's curved surface can be unfolded into a sector. We derive its area formula.

Argument 1

Unfolded cone forms a sector with radius = slant height (l) and arc length = 2πr.
Sector area = (θ/360°) × πl², where θ = (arc length/circumference) × 360° = (2πr/2πl) × 360° = (r/l) × 360°.
Substituting θ: Area = (r/l × 360°/360°) × πl² = πrl.

Argument 2

Example: r = 5 cm, l = 13 cm.
Curved Surface Area = πrl = (22/7) × 5 × 13 ≈ 204.29 cm².

Conclusion

The derived formula πrl matches the example, confirming its validity for real-world calculations.

Question 8:

A cylindrical tank has a radius of 7 m and height 10 m. Calculate its total surface area and volume. Compare it with a cubical tank of the same volume.

Answer:

Introduction

We studied that a cylinder's total surface area (TSA) is 2πr(r + h) and volume is πr²h. A cube's volume is a³.

Argument 1

Cylinder TSA = 2 × (22/7) × 7 × (7 + 10) = 748 m²
Volume = (22/7) × 7² × 10 = 1540 m³

Argument 2

For the cube with volume 1540 m³, side = ∛1540 ≈ 11.52 m. Its TSA = 6 × (11.52)² ≈ 796.6 m².

Conclusion

The cube has a larger surface area than the cylinder for the same volume, showing shapes impact material usage.

Question 9:

Derive the formula for the curved surface area of a cone using its slant height (l) and base radius (r). Verify with NCERT Example 4.

Answer:

Introduction

Our textbook shows that a cone's curved surface area (CSA) is πrl. Let's derive it stepwise.

Argument 1

Unfolding a cone gives a sector with radius l and arc length 2πr.
Sector area = (θ/360°) × πl², where θ = (2πr/2πl) × 360° = (r/l) × 360°.

Argument 2

Substituting θ, CSA = (r/l × 360°/360°) × πl² = πrl. NCERT Example 4 confirms this for r=7 cm, l=25 cm (CSA=550 cm²).

Conclusion

The derivation matches the formula, proving its validity for real applications like making conical tents.

Question 10:

Derive the formula for the curved surface area of a cone using its net. Assume slant height l and base radius r.

Answer:

Introduction

Our textbook shows that a cone's net consists of a sector and a circle. We derive its curved surface area.

Argument 1

Net: A sector (radius = slant height l) and a base circle (radius = r).
Sector arc length = Circumference of base ⇒ 2πr.

Argument 2

Full circle circumference = 2πl.
Sector area = (Arc length/Circumference) × πl² = (2πr/2πl) × πl² = πrl.

Conclusion

Thus, the curved surface area of a cone is πrl, derived from its net.

Question 11:

A cylindrical vessel of height 14 cm and radius 7 cm is full of water. If the water is poured into a hemispherical bowl of radius 10.5 cm, find if the bowl will overflow. (Use π = 22/7)

Answer:

Introduction

We studied that volume calculations help compare capacities. Here, we compare volumes of a cylinder and hemisphere.

Argument 1

Cylinder volume = πr²h = (22/7) × 7² × 14 = 2156 cm³
Hemisphere volume = (2/3)πr³ = (2/3) × (22/7) × 10.5³ ≈ 2425.5 cm³

Argument 2

Since 2156 cm³ (water) < 2425.5 cm³ (bowl), the bowl won’t overflow. Our textbook shows similar problems.

Conclusion

Thus, the hemispherical bowl can accommodate all water without overflowing.

Question 12:

Derive the formula for the total surface area of a cone with radius 'r', slant height 'l', and height 'h'. Verify it using an example where r = 5 cm and l = 13 cm.

Answer:

Introduction

We studied that a cone’s surface area includes its base and curved part. Let’s derive the formula.

Argument 1

Base area = πr²
Curved surface area = πrl
Total surface area = πr² + πrl = πr(r + l)

Argument 2

For r = 5 cm, l = 13 cm: TSA = (22/7) × 5 × (5 + 13) ≈ 282.86 cm². NCERT confirms this.

Conclusion

The derived formula πr(r + l) is verified with the example.

Question 13:

A cylindrical tank has a radius of 7 m and height 10 m. Calculate its total surface area and volume. Explain the steps with units.

Answer:

Introduction

We studied that a cylinder has two circular bases and a curved surface. The given dimensions are radius (r) = 7 m and height (h) = 10 m.

Argument 1

Total Surface Area (TSA): TSA = 2πr(h + r) = 2 × (22/7) × 7 × (10 + 7) = 748 m².

Argument 2

Volume: Volume = πr²h = (22/7) × 7² × 10 = 1540 m³.

Conclusion

Thus, the TSA is 748 m² and volume is 1540 m³, as per NCERT examples.

Question 14:

A hemispherical dome has an inner radius of 14 m. Find its curved surface area and volume. Relate it to real-life applications.

Answer:

Introduction

Our textbook shows that a hemisphere is half of a sphere. Here, radius (r) = 14 m.

Argument 1

Curved Surface Area (CSA): CSA = 2πr² = 2 × (22/7) × 14² = 1232 m².

Argument 2

Volume: Volume = (2/3)πr³ = (2/3) × (22/7) × 14³ ≈ 5749.33 m³.

Conclusion

Such calculations are used in constructing domes like planetariums, combining basic applications and real-life utility.

Question 15:

Derive the formula for the curved surface area of a cone using its slant height (l) and base radius (r). Verify it for a cone with r = 5 cm and l = 13 cm.

Answer:

Introduction

Our textbook shows that a cone's curved surface can be unfolded into a sector. We derive its area formula.

Argument 1

Unfolded sector radius = slant height (l).
Sector arc length = cone base circumference (2πr).
Sector area = (arc length/circle circumference)×πl² = (2πr/2πl)×πl² = πrl.

Argument 2

For r = 5 cm, l = 13 cm: CSA = πrl = (22/7)×5×13 ≈ 204.29 cm².
Manual calculation matches the formula.

Conclusion

The derivation proves CSA = πrl, verified for the given cone (204.29 cm²).

Question 16:

A cylindrical vessel has an internal radius of 7 cm and height 10 cm. Calculate its total surface area and volume. (Use π = 22/7)

Answer:

Introduction

We studied that a cylinder has two circular bases and a curved surface. Its total surface area includes both.

Argument 1

Volume = πr²h = (22/7) × 7² × 10 = 1540 cm³
Curved Surface Area = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm²

Argument 2

Base Area = πr² = (22/7) × 7² = 154 cm²
Total Surface Area = CSA + 2×Base Area = 440 + 308 = 748 cm²

Conclusion

Thus, the volume is 1540 cm³ and total surface area is 748 cm², matching NCERT examples.

Question 17:

A hemispherical bowl has radius 5 cm. Find its curved surface area and total surface area. (Use π = 3.14)

Answer:

Introduction

Our textbook shows that a hemisphere is half of a sphere, with one flat circular face.

Argument 1

Curved Surface Area = 2πr² = 2 × 3.14 × 5² = 157 cm²
Flat Circular Area = πr² = 3.14 × 5² = 78.5 cm²

Argument 2

Total Surface Area = CSA + Flat Area = 157 + 78.5 = 235.5 cm²

Conclusion

Hence, the curved surface area is 157 cm² and total surface area is 235.5 cm², as per basic applications.

Question 18:

A cylindrical tank has a diameter of 14 m and height of 3.5 m. Calculate its capacity in litres. (Use π = 22/7)

Answer:

To find the capacity of the cylindrical tank, we first calculate its volume using the formula for the volume of a cylinder: V = πr²h, where r is the radius and h is the height.

Given:
Diameter = 14 m → Radius (r) = 14/2 = 7 m
Height (h) = 3.5 m
π = 22/7

Step 1: Calculate the volume in cubic meters (m³):
V = πr²h
= (22/7) × (7)² × 3.5
= (22/7) × 49 × 3.5
= 22 × 7 × 3.5
= 22 × 24.5
= 539 m³

Step 2: Convert cubic meters to litres (since 1 m³ = 1000 litres):
Capacity = 539 × 1000
= 5,39,000 litres

Thus, the tank can hold 5,39,000 litres of liquid.

Question 19:

A cone and a hemisphere share the same base radius of 7 cm. The height of the cone is equal to the radius of the hemisphere. Find the total surface area of the combined solid. (Use π = 22/7)

Answer:

The combined solid consists of a cone and a hemisphere with the same base radius. To find the total surface area, we add the curved surface area (CSA) of the cone and the hemisphere, excluding the common base area.

Given:
Radius (r) = 7 cm
Height of cone (h) = Radius of hemisphere = 7 cm
π = 22/7

Step 1: Calculate the slant height (l) of the cone:
l = √(r² + h²)
= √(7² + 7²)
= √(49 + 49)
= √98
= 7√2 cm

Step 2: Find the CSA of the cone:
CSA_cone = πrl
= (22/7) × 7 × 7√2
= 22 × 7√2
= 154√2 cm²

Step 3: Find the CSA of the hemisphere:
CSA_hemisphere = 2πr²
= 2 × (22/7) × 7²
= 2 × (22/7) × 49
= 2 × 22 × 7
= 308 cm²

Step 4: Total surface area of the combined solid:
Total SA = CSA_cone + CSA_hemisphere
= 154√2 + 308
= 308 + 154√2 cm²

Thus, the total surface area is 308 + 154√2 cm².

Question 20:

A metallic sphere of radius 10.5 cm is melted and recast into smaller cones, each of radius 3.5 cm and height 3 cm. Find the number of cones formed.

Answer:

To determine the number of cones formed, we compare the volume of the original sphere with the volume of one cone and divide the two.

Given:
Sphere radius (R) = 10.5 cm
Cone radius (r) = 3.5 cm
Cone height (h) = 3 cm
π = 22/7

Step 1: Calculate the volume of the sphere:
V_sphere = (4/3)πR³
= (4/3) × (22/7) × (10.5)³
= (4/3) × (22/7) × 1157.625
= (4/3) × 22 × 165.375
= (4/3) × 3638.25
= 4851 cm³

Step 2: Calculate the volume of one cone:
V_cone = (1/3)πr²h
= (1/3) × (22/7) × (3.5)² × 3
= (1/3) × (22/7) × 12.25 × 3
= (1/3) × 22 × 5.25
= (1/3) × 115.5
= 38.5 cm³

Step 3: Determine the number of cones:
Number of cones = V_sphere / V_cone
= 4851 / 38.5
= 126 cones

Thus, 126 cones can be formed from the melted sphere.

Question 21:

A cylindrical tank has a diameter of 14 m and a height of 3.5 m. Calculate the cost of painting its curved surface area at the rate of ₹20 per m². Also, explain why the base and top are excluded in this calculation.

Answer:

To solve the problem, follow these steps:

Step 1: Find the radius of the cylindrical tank.
Given diameter = 14 m
Radius (r) = Diameter / 2 = 14 / 2 = 7 m

Step 2: Calculate the curved surface area (CSA) of the cylinder.
Formula: CSA = 2πrh
Height (h) = 3.5 m
CSA = 2 × (22/7) × 7 × 3.5
CSA = 2 × 22 × 3.5 = 154 m²

Step 3: Compute the cost of painting.
Rate = ₹20 per m²
Total cost = CSA × Rate = 154 × 20 = ₹3080

Explanation for excluding base and top: The question specifically asks for the curved surface area to be painted, which does not include the circular base and top. If the entire surface area (including base and top) were to be painted, we would use the formula: Total Surface Area = 2πr(r + h).

Question 22:

A hemispherical dome of a building needs to be painted. If the circumference of the base is 17.6 m, find the cost of painting it at ₹5 per 100 cm². Explain the significance of using the curved surface area formula for a hemisphere in this context.

Answer:

Here’s the step-by-step solution:

Step 1: Find the radius of the hemisphere.
Given circumference of base = 17.6 m
Circumference = 2πr
17.6 = 2 × (22/7) × r
r = (17.6 × 7) / (2 × 22) = 2.8 m

Step 2: Calculate the curved surface area (CSA) of the hemisphere.
Formula: CSA = 2πr²
CSA = 2 × (22/7) × (2.8)²
CSA = 2 × (22/7) × 7.84 = 49.28 m²

Step 3: Convert CSA to cm² and compute the cost.
1 m² = 10,000 cm²
CSA = 49.28 × 10,000 = 492,800 cm²
Cost per 100 cm² = ₹5
Total cost = (492,800 / 100) × 5 = ₹24,640

Significance of using CSA formula: A hemisphere is half of a sphere, and only the outer curved part is painted. The flat circular base is not part of the surface being painted, so we use CSA = 2πr² instead of the total surface area (which would include the base).

Question 23:

A cylindrical tank has a diameter of 14 meters and a height of 3 meters. It is full of water. The water is emptied into a cuboidal tank of dimensions 11 m × 10 m × 7 m. Calculate the height of the water left in the cuboidal tank after pouring. Justify each step of your calculation.

Answer:

To solve this problem, we need to find the volume of water transferred from the cylindrical tank to the cuboidal tank and then determine the remaining height of water in the cuboidal tank.

Step 1: Calculate the volume of the cylindrical tank
Given:
Diameter (d) = 14 m ⇒ Radius (r) = d/2 = 7 m
Height (h) = 3 m

Volume of cylinder = πr²h
= (22/7) × 7 × 7 × 3
= 22 × 7 × 3
= 462 m³

Step 2: Calculate the volume of the cuboidal tank
Given dimensions: 11 m × 10 m × 7 m
Volume of cuboid = l × b × h
= 11 × 10 × 7
= 770 m³

Step 3: Determine the height of water filled in the cuboidal tank
Volume of water poured = 462 m³
Let the height of water in the cuboidal tank be H.
Volume occupied = l × b × H
⇒ 11 × 10 × H = 462
⇒ H = 462 / 110
⇒ H = 4.2 m

Step 4: Find the remaining height of the cuboidal tank
Total height of cuboidal tank = 7 m
Height of water filled = 4.2 m
Remaining height = 7 - 4.2 = 2.8 m

Thus, the height of the water left in the cuboidal tank is 2.8 meters. This means the cuboidal tank is not completely filled, and there is still space left.

Question 24:

A cylindrical tank has a diameter of 14 meters and a height of 3 meters. It is filled with water up to a height of 2 meters. Calculate the volume of water in the tank. Also, find the total surface area of the wet portion of the tank. (Use π = 22/7)

Answer:

To solve this problem, we need to calculate two things: the volume of water in the cylindrical tank and the total surface area of the wet portion of the tank.

Given:
Diameter of the tank = 14 meters
Radius (r) = Diameter / 2 = 14 / 2 = 7 meters
Height of the tank (H) = 3 meters
Height of water (h) = 2 meters

Step 1: Calculate the volume of water in the tank
The formula for the volume of a cylinder is V = πr²h.
Substituting the given values:
V = (22/7) × (7)² × 2
V = (22/7) × 49 × 2
V = 22 × 7 × 2
V = 308 cubic meters

Step 2: Calculate the total surface area of the wet portion
The wet portion includes the base of the cylinder and the curved surface area up to the water level.

Base area (A₁) = πr² = (22/7) × 7 × 7 = 154 square meters
Curved surface area (A₂) = 2πrh = 2 × (22/7) × 7 × 2 = 88 square meters

Total wet surface area = A₁ + A₂ = 154 + 88 = 242 square meters

Final Answers:
Volume of water = 308 m³
Total wet surface area = 242 m²

Question 25:

A cylindrical tank has a diameter of 14 meters and a height of 3 meters. It is full of water. Calculate the total surface area of the tank and determine how much water (in liters) it can hold when completely filled. (Use π = 22/7)

Answer:

To solve this problem, we need to find two things: the total surface area of the cylindrical tank and the volume of water it can hold.

Given:
Diameter of the tank = 14 meters
Height of the tank = 3 meters
π = 22/7

Step 1: Find the radius of the tank
Radius (r) = Diameter / 2 = 14 / 2 = 7 meters

Step 2: Calculate the total surface area of the tank
The total surface area of a closed cylinder includes the curved surface area and the areas of the two circular bases.
Total Surface Area = 2πr(h + r)
= 2 × (22/7) × 7 × (3 + 7)
= 2 × 22 × 10
= 440 square meters

Step 3: Calculate the volume of water the tank can hold
Volume of a cylinder = πr²h
= (22/7) × 7 × 7 × 3
= 22 × 7 × 3
= 462 cubic meters

Step 4: Convert cubic meters to liters
1 cubic meter = 1000 liters
462 cubic meters = 462 × 1000 = 462,000 liters

Final Answers:
Total Surface Area = 440 square meters
Volume of water = 462,000 liters

Additional Insight: The total surface area helps in estimating the material required to construct the tank, while the volume indicates its storage capacity. Such calculations are crucial in real-life applications like water storage and industrial tank design.

Question 26:

A cylindrical tank has a diameter of 14 m and a height of 3.5 m. Calculate the cost of painting its curved surface area at the rate of ₹20 per m². Also, explain why painting the surface of such tanks is necessary.

Answer:

To solve the problem, we first need to find the curved surface area of the cylindrical tank and then calculate the cost of painting it.

Step 1: Find the radius of the cylinder
Given diameter = 14 m
Radius (r) = Diameter / 2 = 14 / 2 = 7 m

Step 2: Calculate the curved surface area (CSA)
Formula for CSA of a cylinder = 2πrh
Given height (h) = 3.5 m
CSA = 2 × (22/7) × 7 × 3.5
CSA = 2 × 22 × 3.5 = 154 m²

Step 3: Calculate the cost of painting
Cost per m² = ₹20
Total cost = CSA × Rate = 154 × 20 = ₹3080

Why painting is necessary: Painting the surface of cylindrical tanks, especially those used for water storage, prevents rusting and corrosion. It also provides a protective layer against environmental factors, ensuring the tank's longevity and maintaining hygiene.

Question 27:

A cylindrical container has a diameter of 14 cm and a height of 20 cm. It is filled with water up to 16 cm. Calculate the volume of water in the container. If this water is poured into another cylindrical container with a diameter of 10 cm, find the height of the water in the second container. (Use π = 22/7)

Answer:

To solve this problem, we will follow these steps:

Step 1: Calculate the radius of both containers
Radius (r) = Diameter / 2
For the first container: r₁ = 14 cm / 2 = 7 cm
For the second container: r₂ = 10 cm / 2 = 5 cm

Step 2: Calculate the volume of water in the first container
Volume of water (V) = π × r₁² × height of water
V = (22/7) × (7 cm)² × 16 cm
V = (22/7) × 49 cm² × 16 cm
V = 22 × 7 cm² × 16 cm
V = 2464 cm³

Step 3: Find the height of water in the second container
Volume remains the same when water is poured into the second container.
V = π × r₂² × new height (h)
2464 cm³ = (22/7) × (5 cm)² × h
2464 cm³ = (22/7) × 25 cm² × h
h = (2464 cm³ × 7) / (22 × 25 cm²)
h = 17248 cm³ / 550 cm²
h = 31.36 cm

Thus, the height of the water in the second container is 31.36 cm.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:

A cylindrical water tank has a radius of 3.5 m and height of 7 m. Calculate its lateral surface area and total surface area. (Use π = 22/7)

Answer:

Problem Interpretation

We need to find the lateral and total surface area of a cylindrical tank.

Mathematical Modeling

Lateral Surface Area (LSA) = 2πrh
Total Surface Area (TSA) = 2πr(r + h)

Solution

Given: r = 3.5 m, h = 7 m. LSA = 2 × (22/7) × 3.5 × 7 = 154 m². TSA = 2 × (22/7) × 3.5 × (3.5 + 7) = 231 m².

Question 2:

A conical tent has a base radius of 6 m and slant height of 10 m. Find its curved surface area and the canvas required to make it.

Answer:

Problem Interpretation

We studied conical shapes in NCERT. Here, we calculate the curved surface area and canvas needed.

Mathematical Modeling

Curved Surface Area (CSA) = πrl

Solution

Given: r = 6 m, l = 10 m. CSA = (22/7) × 6 × 10 ≈ 188.57 m². Since canvas covers the entire tent, required canvas = CSA = 188.57 m².

Question 3:

A cone has a base radius of 6 cm and slant height of 10 cm. Find its curved surface area and total surface area.

Answer:

Problem Interpretation

We studied cones in NCERT. Here, we find curved and total surface areas.

Mathematical Modeling

Curved Surface Area (CSA) = πrl
Total Surface Area (TSA) = πr(r + l)

Solution

Given r = 6 cm, l = 10 cm. CSA = (22/7) × 6 × 10 ≈ 188.57 cm². TSA = (22/7) × 6 (6 + 10) ≈ 301.71 cm².

Question 4:

A conical tent has a slant height of 10 m and base radius of 6 m. Find its curved surface area and volume. (Use π = 3.14)

Answer:

Problem Interpretation

We studied conical shapes in NCERT. Here, we calculate CSA and volume using given dimensions.

Mathematical Modeling

CSA = πrl
Volume = (1/3)πr²h

Solution

Given: r = 6 m, l = 10 m. CSA = 3.14 × 6 × 10 = 188.4 m². Height h = √(10² - 6²) = 8 m. Volume = (1/3) × 3.14 × 6² × 8 = 301.44 m³.

Question 5:

A cylindrical water tank has a diameter of 2.8 m and height 3.5 m. Calculate its lateral surface area and total surface area. (Use π = 22/7)

Answer:

Problem Interpretation

We need to find the lateral and total surface area of a cylindrical tank.

Mathematical Modeling

Radius (r) = Diameter/2 = 1.4 m
Height (h) = 3.5 m

Solution

Lateral Surface Area = 2πrh = 2 × (22/7) × 1.4 × 3.5 = 30.8 m²
Total Surface Area = 2πr(r + h) = 2 × (22/7) × 1.4(1.4 + 3.5) = 43.12 m²

Question 6:

A conical tent has a base radius of 7 m and slant height 25 m. Find the canvas area required and volume of air inside. (Use π = 22/7)

Answer:

Problem Interpretation

We studied conical shapes in NCERT. Here, we calculate canvas area (CSA) and volume.

Mathematical Modeling

Radius (r) = 7 m
Slant height (l) = 25 m

Solution

Canvas Area = πrl = (22/7) × 7 × 25 = 550 m²
Height (h) = √(l² - r²) = √(625 - 49) = 24 m
Volume = (1/3)πr²h = (1/3) × (22/7) × 49 × 24 = 1232 m³

Question 7:

A cylindrical water tank has a diameter of 1.4 m and height 2.1 m. How much metal sheet is required to make it, including the top and bottom? (Use π = 22/7)

Answer:

Problem Interpretation

We need to find the total surface area of a closed cylinder, which includes the curved surface and two circular bases.

Mathematical Modeling

Radius (r) = Diameter/2 = 1.4/2 = 0.7 m
Height (h) = 2.1 m

Solution

Total surface area = 2πr(h + r) = 2 × (22/7) × 0.7 × (2.1 + 0.7) = 12.32 m². Thus, 12.32 m² metal sheet is needed.

Question 8:

An ice-cream cone has a hemispherical top. The cone's height is 9 cm and base radius is 3.5 cm. Find its total volume. (Use π = 22/7)

Answer:

Problem Interpretation

We must calculate the combined volume of a cone and a hemisphere sharing the same radius.

Mathematical Modeling

Cone: r = 3.5 cm, h = 9 cm
Hemisphere: r = 3.5 cm

Solution

Volume = (1/3)πr²h + (2/3)πr³ = (1/3 × 22/7 × 3.5² × 9) + (2/3 × 22/7 × 3.5³) = 115.5 + 89.83 = 205.33 cm³.

Question 9:

A cylindrical water tank has a radius of 3.5 m and height of 7 m. Find its lateral surface area and total surface area. (Use π = 22/7)

Answer:

Problem Interpretation

We need to find the lateral and total surface area of a cylindrical water tank.

Mathematical Modeling

Radius (r) = 3.5 m
Height (h) = 7 m

Solution

Lateral Surface Area = 2πrh = 2 × (22/7) × 3.5 × 7 = 154 m². Total Surface Area = 2πr(r + h) = 2 × (22/7) × 3.5 × (3.5 + 7) = 231 m².

Question 10:

A conical tent has a base radius of 6 m and slant height of 10 m. Calculate the canvas area required and the volume of air inside it.

Answer:

Problem Interpretation

We studied conical tents in our textbook. Here, we find the canvas area (lateral surface) and volume.

Mathematical Modeling

Radius (r) = 6 m
Slant height (l) = 10 m

Solution

Canvas Area = πrl = (22/7) × 6 × 10 ≈ 188.57 m². Height (h) = √(l² - r²) = √(100 - 36) = 8 m. Volume = (1/3)πr²h = (1/3) × (22/7) × 36 × 8 ≈ 301.71 m³.

Question 11:

A cylindrical water tank has a diameter of 1.4 m and height 2.1 m. Calculate its lateral surface area and total surface area. (Use π = 22/7)

Answer:

Problem Interpretation

We need to find the lateral and total surface area of a cylinder. Our textbook shows the formulas: LSA = 2πrh and TSA = 2πr(r + h).

Mathematical Modeling

Radius (r) = Diameter/2 = 1.4/2 = 0.7 m
Height (h) = 2.1 m

Solution

LSA = 2 × (22/7) × 0.7 × 2.1 = 9.24 m². TSA = 2 × (22/7) × 0.7 (0.7 + 2.1) = 12.32 m². Both answers include units.

Question 12:

A cone has a slant height of 10 cm and base radius 6 cm. Find its curved surface area and total surface area. (Use π = 3.14)

Answer:

Problem Interpretation

We studied that CSA of a cone = πrl and TSA = πr(r + l). Here, r = 6 cm, l = 10 cm.

Mathematical Modeling

CSA = π × 6 × 10
TSA = π × 6 (6 + 10)

Solution

CSA = 3.14 × 6 × 10 = 188.4 cm². TSA = 3.14 × 6 × 16 = 301.44 cm². Both values are rounded to two decimal places.

Question 13:

[CBSE-aligned, NCERT-referenced] A cylindrical tank has a radius of 7 m and a height of 3 m. It is filled with water up to 2 m. Calculate the volume of water in the tank. (Use π = 22/7)

Answer:

Given:
Radius (r) = 7 m
Height of tank (h) = 3 m
Water level = 2 m

Volume of water = πr²h
= (22/7) × (7)² × 2
= (22/7) × 49 × 2
= 22 × 7 × 2
= 308 m³

Thus, the volume of water in the tank is 308 cubic meters.

Question 14:

[CBSE-aligned, NCERT-referenced] A cone and a hemisphere share the same base radius of 5 cm. If the total height of the combined solid (cone mounted on hemisphere) is 13 cm, find its total surface area. (Use π = 3.14)

Answer:

Given:
Radius (r) = 5 cm
Total height = 13 cm
Height of hemisphere = radius = 5 cm

Height of cone (h) = Total height − Radius
= 13 cm − 5 cm = 8 cm

Slant height of cone (l) = √(r² + h²)
= √(5² + 8²)
= √(25 + 64)
= √89 ≈ 9.43 cm

Total Surface Area = CSA of cone + CSA of hemisphere
= πrl + 2πr²
= (3.14 × 5 × 9.43) + (2 × 3.14 × 25)
= 148.03 + 157
= 305.03 cm²

Question 15:

[CBSE-aligned, NCERT-referenced] A metallic sphere of radius 4.2 cm is melted and recast into small cones of radius 1.4 cm and height 2.1 cm each. How many such cones can be made?

Answer:

Given:
Sphere radius (R) = 4.2 cm
Cone radius (r) = 1.4 cm
Cone height (h) = 2.1 cm

Volume of sphere = (4/3)πR³
= (4/3) × (22/7) × (4.2)³
= (4/3) × (22/7) × 74.088 ≈ 310.464 cm³

Volume of one cone = (1/3)πr²h
= (1/3) × (22/7) × (1.4)² × 2.1
= (1/3) × 22 × 0.28 × 2.1 ≈ 4.312 cm³

Number of cones = Volume of sphere ÷ Volume of one cone
= 310.464 ÷ 4.312 ≈ 72 cones

Question 16:

A cylindrical tank has a diameter of 14 m and a height of 3.5 m. It is filled with water up to a height of 2.1 m. Calculate:

The volume of water in the tank.
The internal curved surface area of the tank that is in contact with water.

Answer:

Given: Diameter of cylindrical tank = 14 m ⇒ Radius (r) = 7 m
Height of tank (h) = 3.5 m
Height of water (h_w) = 2.1 m

Volume of water:
Volume = πr²h_w
= (22/7) × 7 × 7 × 2.1
= 22 × 7 × 2.1 = 323.4 m³

Curved surface area in contact with water:
CSA = 2πrh_w
= 2 × (22/7) × 7 × 2.1
= 44 × 2.1 = 92.4 m²

Note: The curved surface area only considers the side contact, not the base/top.

Question 17:

A conical tent has a base radius of 6 m and a slant height of 10 m. Calculate:

The canvas required to make the tent (assuming no wastage).
The volume of air inside the tent.

Answer:

Given: Radius of cone (r) = 6 m
Slant height (l) = 10 m

Canvas required (lateral surface area):
LSA = πrl
= (22/7) × 6 × 10
= (22 × 60)/7 ≈ 188.57 m²

Volume of air:
First, find height (h) using Pythagoras theorem:
h = √(l² - r²) = √(100 - 36) = 8 m
Volume = (1/3)πr²h
= (1/3) × (22/7) × 36 × 8
= (22 × 96)/7 ≈ 301.71 m³

Application: Such calculations help in material estimation for real-life structures.

Question 18:

A cylindrical tank has a diameter of 14 m and height of 3.5 m. It is full of water. The water is emptied into a cuboidal tank of dimensions 11 m × 8 m × 7 m.

(i) Calculate the volume of the cylindrical tank.
(ii) Will the cuboidal tank be able to hold all the water from the cylindrical tank? Justify your answer.

Answer:

(i) Volume of the cylindrical tank:

Given, diameter = 14 m ⇒ radius (r) = 7 m, height (h) = 3.5 m
Volume = πr²h
= (22/7) × 7 × 7 × 3.5
= 22 × 7 × 3.5
= 539 m³

(ii) Checking capacity of cuboidal tank:

Volume of cuboidal tank = l × b × h
= 11 × 8 × 7
= 616 m³
Since 539 m³ (cylindrical tank) < 616 m³ (cuboidal tank), the cuboidal tank can hold all the water.

Additional note: Always verify units and compare volumes precisely in such problems.

Question 19:

An ice cream cone consists of a hemispherical scoop on top of a conical base. The height of the conical part is 12 cm and its base radius is 3 cm.

(i) Find the volume of the ice cream (including hemispherical part).
(ii) If the ice cream melts completely, will it fill a cylindrical container of radius 3 cm and height 7 cm? Show calculations.

Answer:

(i) Total volume of ice cream:

Conical part: radius (r) = 3 cm, height (h) = 12 cm
Volume = (1/3)πr²h
= (1/3) × (22/7) × 3 × 3 × 12
= 113.14 cm³
Hemispherical part: radius (r) = 3 cm (same as cone's top)
Volume = (2/3)πr³
= (2/3) × (22/7) × 3 × 3 × 3
= 56.57 cm³
Total volume = 113.14 + 56.57 = 169.71 cm³

(ii) Checking cylindrical container capacity:

Volume of cylinder = πr²h
= (22/7) × 3 × 3 × 7
= 198 cm³
Since 169.71 cm³ (ice cream) < 198 cm³ (cylinder), the melted ice cream will fit.

Note: The hemispherical scoop shares the same radius as the cone's top, a common problem scenario.

Question 20:

A cylindrical tank has a diameter of 14 m and height of 3.5 m. It is full of water. The water is emptied into a cuboidal tank of dimensions 10 m × 7 m × 5 m.

(i) Calculate the volume of the cylindrical tank.
(ii) Will the cuboidal tank be able to hold all the water from the cylindrical tank? Justify your answer.

Answer:

(i) Volume of the cylindrical tank:
Given, diameter = 14 m ⇒ radius (r) = 7 m, height (h) = 3.5 m.
Volume of cylinder = πr²h
= (22/7) × 7 × 7 × 3.5
= 22 × 7 × 3.5
= 539 m³.

(ii) Checking capacity of cuboidal tank:
Volume of cuboid = l × b × h = 10 × 7 × 5 = 350 m³.
Since 539 m³ (cylinder) > 350 m³ (cuboid), the cuboidal tank cannot hold all the water. Water will overflow by 189 m³ (539 - 350).

Key concept: Comparison of volumes requires consistent units. Always verify dimensions before transferring contents.

Question 21:

An ice cream cone consists of a hemisphere of diameter 7 cm mounted on a cone of height 12 cm.

(i) Find the total surface area of the ice cream cone.
(ii) If the cone is filled completely with ice cream, what is the volume of ice cream served? (Use π = 22/7)

Answer:

(i) Total surface area:
Hemisphere radius (r) = 3.5 cm.
Hemisphere CSA = 2πr² = 2 × (22/7) × 3.5 × 3.5 = 77 cm².
Cone slant height (l) = √(r² + h²) = √(3.5² + 12²) = 12.5 cm.
Cone CSA = πrl = (22/7) × 3.5 × 12.5 = 137.5 cm².
Total surface area = Hemisphere CSA + Cone CSA = 77 + 137.5 = 214.5 cm².

(ii) Volume of ice cream:
Hemisphere volume = (2/3)πr³ = (2/3) × (22/7) × 3.5³ ≈ 89.83 cm³.
Cone volume = (1/3)πr²h = (1/3) × (22/7) × 3.5² × 12 ≈ 154 cm³.
Total volume = 89.83 + 154 = 243.83 cm³ (approx).

Note: The base area of the cone is not added in surface area as the hemisphere covers it.

Question 22:

A cylindrical tank has a diameter of 14 m and height of 3.5 m. It is filled with water up to 2 m height. Calculate the volume of water in the tank. Also, find the total surface area of the wet portion of the tank.

Answer:

Given:
Diameter of the cylindrical tank = 14 m
Radius (r) = Diameter / 2 = 7 m
Height of the tank (H) = 3.5 m
Height of water (h) = 2 m

Volume of water in the tank is calculated using the formula for volume of a cylinder:
V = πr²h
V = (22/7) × 7 × 7 × 2
V = 22 × 7 × 2
V = 308 m³

Total surface area of the wet portion includes the base area and the curved surface area up to the water level:
Base area = πr² = (22/7) × 7 × 7 = 154 m²
Curved surface area = 2πrh = 2 × (22/7) × 7 × 2 = 88 m²
Total wet surface area = Base area + Curved surface area = 154 + 88 = 242 m²

Question 23:

A conical tent has a base radius of 6 m and a slant height of 10 m. Find the cost of canvas required to make the tent at the rate of ₹50 per m². Also, calculate the volume of air inside the tent.

Answer:

Given:
Base radius (r) of the conical tent = 6 m
Slant height (l) = 10 m
Cost of canvas = ₹50 per m²

Curved surface area of the conical tent is calculated using the formula:
CSA = πrl
CSA = (22/7) × 6 × 10
CSA = (22/7) × 60 ≈ 188.57 m²
Cost of canvas = CSA × Rate = 188.57 × 50 ≈ ₹9428.57

To find the volume of air inside the tent, we first need the height (h) of the cone:
Using Pythagoras' theorem: l² = r² + h²
10² = 6² + h²
100 = 36 + h²
h² = 64
h = 8 m
Volume = (1/3)πr²h = (1/3) × (22/7) × 6 × 6 × 8 ≈ 301.71 m³

Question 24:

A cylindrical container has a radius of 7 cm and a height of 15 cm. It is filled with water up to 10 cm. A solid metallic sphere of radius 3.5 cm is completely immersed into the container. Calculate the rise in the water level. (Use π = 22/7)

Answer:

To find the rise in water level after immersing the sphere, we first calculate the volume of the sphere and then determine how much the water level increases in the cylinder.

Step 1: Calculate the volume of the sphere
Volume of sphere = (4/3)πr³
= (4/3) × (22/7) × (3.5)³
= (4/3) × (22/7) × 42.875
= (4/3) × 22 × 6.125
= (4/3) × 134.75
= 179.666... cm³ ≈ 179.67 cm³

Step 2: Find the rise in water level
The volume of the sphere equals the volume of water displaced in the cylinder.
Let the rise in water level be h cm.
Volume of displaced water = πR²h (where R = radius of cylinder)
179.67 = (22/7) × 7 × 7 × h
179.67 = 154 × h
h = 179.67 / 154 ≈ 1.166 cm

The water level rises by approximately 1.17 cm.

Question 25:

A conical tent has a base radius of 6 m and a slant height of 10 m. Calculate the cost of canvas required to make the tent at the rate of ₹50 per m². (Assume no wastage and use π = 3.14)

Answer:

To find the cost of the canvas, we first calculate the curved surface area (CSA) of the conical tent and then multiply it by the given rate.

Step 1: Calculate the curved surface area of the cone
CSA of cone = πrl
where r = radius = 6 m, l = slant height = 10 m
= 3.14 × 6 × 10
= 188.4 m²

Step 2: Compute the cost of canvas
Cost = CSA × Rate per m²
= 188.4 × 50
= ₹9,420

The total cost of the canvas required is ₹9,420.

Surface Areas and Volumes – CBSE NCERT Study Resources

Overview

Surface Area and Volume of a Cuboid

Surface Area and Volume of a Cube

Surface Area and Volume of a Cylinder

Surface Area and Volume of a Cone

Surface Area and Volume of a Sphere

Combination of Solids

Conversion of Units

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