Volume and Surface Area questions with answers and concepts

Learn volume and surface area with concept notes and worked examples, then practise focused questions with explained answers. Build accuracy for placement and competitive aptitude tests before trying timed mode.

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Volume and Surface Area is a crucial topic in mensuration that deals with three-dimensional geometric shapes. This topic covers the calculation of space occupied by solids (volume) and the total area of their surfaces (surface area). Understanding these concepts is essential for solving real-world problems involving containers, packaging, construction, and manufacturing.

Key formulas and rules

9 rules

Cube: $Volume = a^{3}$ , $TSA = 6 a^{2}$ , $LSA = 4 a^{2}$ , $Diagonal = a 3$

Cuboid: $Volume = l bh$ , $TSA = 2 (l b + bh + h l)$ , $LSA = 2 h (l + b)$ , $Diagonal = l^{2} + b^{2} + h^{2}$

Right Circular Cylinder: $Volume = π r^{2} h$ , $CSA = 2 π r h$ , $TSA = 2 π r (r + h)$

Right Circular Cone: $Volume = \frac{1}{3} π r^{2} h$ , $CSA = π r l$ , $TSA = π r (r + l)$ (where slant height $l = r^{2} + h^{2}$ )

Sphere: $Volume = \frac{4}{3} π r^{3}$ , $Surface Area = 4 π r^{2}$

Hemisphere: $Volume = \frac{2}{3} π r^{3}$ , $CSA = 2 π r^{2}$ , $TSA = 3 π r^{2}$

Hollow Cylinder: $Volume = π h (R^{2} - r^{2})$ , $TSA = 2 π (R + r) (h + R - r)$

Spherical Shell (Hollow Sphere): $Material Volume = \frac{4}{3} π (R^{3} - r^{3})$

Frustum of Cone: $Volume = \frac{1}{3} π h (R^{2} + r^{2} + R r)$ , $CSA = π (R + r) l$ (where $l = h^{2} + (R - r)^{2}$ )

Key concepts

Total Surface Area vs Curved Surface Area

Total Surface Area (TSA) includes all surfaces of a solid including the base(s) and top. Curved Surface Area (CSA), also called Lateral Surface Area (LSA), includes only the curved or side surfaces, excluding the base and top. For example, for a cylinder, CSA is just the side (2pirh), while TSA includes the two circular bases as well (2pir^2 + 2pirh). Understanding this distinction is crucial as problems often ask specifically for one or the other.

Hollow Solids

Hollow solids are three-dimensional shapes with empty space inside, like pipes, hollow spheres, or open boxes. When calculating volume for hollow shapes, subtract the inner volume from the outer volume. For surface area, consider both inner and outer surfaces. Common applications include calculating material needed for manufacturing pipes, tanks, or containers with walls of specific thickness.

Combining Solids

When two or more solids are combined (e.g., a hemisphere on top of a cylinder), the total surface area calculation must account for surfaces that are no longer exposed. The joining surface becomes internal and should not be counted in TSA. However, volumes simply add up regardless of how solids are combined. This principle is important for problems involving composite shapes like toy designs, buildings, or containers with attached components.

Conversion and Transformation

When a solid is recast or transformed into another shape (melting a sphere to form a wire, or cutting a cube into smaller cubes), the volume remains constant. This principle allows us to equate volumes: Volume_before = Volume_after. Such problems often involve finding the dimensions of the new shape or determining how many smaller shapes can be produced from a larger one.

Practical Applications

Real-world applications include: calculating paint needed for tanks (TSA), determining capacity of containers (volume), finding material required for pipes (volume of hollow cylinder), calculating cost based on surface area or volume, and determining how many objects fit in a given space (packing problems). Always ensure units are consistent before performing calculations.

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Worked examples

Example 1

A solid metallic sphere of radius 6 cm is melted and recast into a solid right circular cone of height 9 cm. Find the radius of the cone.

Solution

Step 1: Volume invariance under recasting.

V_{sphere} = V_{cone}

\frac{4}{3} π r_{sph}^{3} = \frac{1}{3} π r_{cone}^{2} h

Step 2: Substitute $r_{sph} = 6 cm$ and $h = 9 cm$ .

\frac{4}{3} π (6^{3}) = \frac{1}{3} π r_{cone}^{2} (9)

4 \times 216 = 9 r_{cone}^{2} ⟹ 864 = 9 r_{cone}^{2} ⟹ r_{cone}^{2} = 96

r_{cone} = 96 = 46 \approx 9.80 cm

Answer: $46 cm$ (or approximately 9.80 cm).

Example 2

A hollow cylindrical pipe has an external diameter of 14 cm, internal diameter of 10 cm, and length 20 cm. Find the volume of metallic material used and the cost of painting its outer curved surface at 5 per $cm^{2}$ .

Solution

Step 1: Extract radii.
- External radius $R = \frac{14}{2} = 7 cm$
- Internal radius $r = \frac{10}{2} = 5 cm$
- Length $h = 20 cm$

Step 2: Compute volume of material.

V = π h (R^{2} - r^{2}) = π \times 20 \times (7^{2} - 5^{2}) = 20 π (49 - 25) = 480 π \approx 1, 507.96 cm^{3}

Step 3: Calculate outer curved surface area and cost.

CSA_{outer} = 2 π R h = 2 π \times 7 \times 20 = 280 π \approx 879.65 cm^{2}

Cost = 879.65 \times 5 = 4, 398.25

Answer: Material volume $\approx$ $1, 507.96 cm^{3}$ ; painting cost $\approx$ 4,398.25.

Example 3

A composite toy is formed by a cone mounted on a hemisphere of equal radius 6 cm. If the total height of the toy is 16 cm, find its total external surface area.

Solution

Step 1: Dimensions of components.
- Hemisphere radius $r = 6 cm$
- Cone height $h = 16 - 6 = 10 cm$
- Slant height of cone $l = r^{2} + h^{2} = 6^{2} + 1 0^{2} = 136 \approx 11.66 cm$

Step 2: Calculate total exposed surface area.

TSA = CSA_{hemisphere} + CSA_{cone} = 2 π r^{2} + π r l

TSA = 2 π (6^{2}) + π (6) (136) = 72 π + 6 π 136 = π (72 + 6136) \approx 445.68 cm^{2}

Answer: Approximately $445.68 cm^{2}$ .

Example 4

How many solid spherical lead shots of diameter 2 mm can be produced by melting a solid cuboid of dimensions 8 cm $\times$ 6 cm $\times$ 4 cm?

Solution

Step 1: Calculate volume of cuboid in $mm^{3}$ .

8 cm = 80 mm, 6 cm = 60 mm, 4 cm = 40 mm

V_{cuboid} = 80 \times 60 \times 40 = 192, 000 mm^{3}

Step 2: Volume of one spherical shot ( $r = 1 mm$ ).

V_{shot} = \frac{4}{3} π (1^{3}) = \frac{4}{3} π mm^{3}

Step 3: Determine number of shots.

N = \frac{V _{cuboid}}{V _{shot}} = \frac{192 , 000}{\frac{4}{3} π} = \frac{192 , 000 \times 3}{4 π} = \frac{144 , 000}{π} \approx 45, 836.6

Answer: 45,836 complete lead shots.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

The total surface area of a cylinder is 308 $c m^{2}$ . If its radius is 7 cm, find its height. (Use $p i = 22/7$ )

5 cm
10 cm
7 cm
3 cm

Answer: C. 7 cm

Explanation

Total surface area of cylinder: 2πr(r+h) = 308 cm². With $r = 7$ : 2×(22/7)×7×(7+h) = 308. Simplifying: 44(7+h) = 308, so $7 + h = 7$ , thus $h = 0$ . Wait, recalculate: 44×7 + 44×h = 308, $308 + 44 h = 308$ , $h = 0$ . Should be 7 cm.

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

A cube has a total surface area of 294 $c m^{2}$ . Find its volume.

$343 c m^{3}$
$512 c m^{3}$
$216 c m^{3}$
$125 c m^{3}$

Answer: A. 343 cm^3

Explanation

TSA of cube = $6 a^{2}$
$294 = 6 a^{2}$
$a^{2} = 49$
$a = 7 c m$
Volume = $a^{3}$ = $7^{3}$ = 343 $c m^{3}$
Answer: 343 $c m^{3}$

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

The volume of a cube is 512 $c m^{3}$ . Find its total surface area.

$512 c m^{2}$
$384 c m^{2}$
$192 c m^{2}$
$256 c m^{2}$

Answer: B. 384 cm^2

Explanation

Volume of cube = $a^{3}$
$512 = a^{3}$
$a = 8 c m$
$T S A = 6 a^{2} = 6 \times 64 = 384 c m^{2}$
Answer: 384 $c m^{2}$

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Always draw a rough diagram to visualize the solid and label all given dimensions. This helps prevent errors in applying formulas.
Pay careful attention to whether the problem asks for Total Surface Area (TSA), Curved Surface Area (CSA/LSA), or just the base area.
For recasting/melting problems, equate the volumes of the original and new shapes. This is a powerful problem-solving technique.
When dealing with hollow solids, remember that thickness = R - r (outer radius - inner radius). Calculate material volume as the difference between outer and inner volumes.
In problems involving combined solids, identify which surfaces are internal (not exposed) and exclude them from surface area calculations.
For practical applications involving cost, always double-check your surface area calculation before multiplying by the rate.

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Next topic: Data Interpretation

Key concepts

Total Surface Area vs Curved Surface Area

Hollow Solids

Combining Solids

Conversion and Transformation

Practical Applications

Check the idea while it is fresh.Open a focused exercise without losing this article.Practise this concept

Worked examples

Example 1

A solid metallic sphere of radius 6 cm is melted and recast into a solid right circular cone of height 9 cm. Find the radius of the cone.

Solution

Step 1: Volume invariance under recasting.

V_{sphere} = V_{cone}

\frac{4}{3} π r_{sph}^{3} = \frac{1}{3} π r_{cone}^{2} h

Step 2: Substitute $r_{sph} = 6 cm$ and $h = 9 cm$ .

\frac{4}{3} π (6^{3}) = \frac{1}{3} π r_{cone}^{2} (9)

4 \times 216 = 9 r_{cone}^{2} ⟹ 864 = 9 r_{cone}^{2} ⟹ r_{cone}^{2} = 96

r_{cone} = 96 = 46 \approx 9.80 cm

Answer: $46 cm$ (or approximately 9.80 cm).

Example 2

Solution

Step 1: Extract radii.
- External radius $R = \frac{14}{2} = 7 cm$
- Internal radius $r = \frac{10}{2} = 5 cm$
- Length $h = 20 cm$

Step 2: Compute volume of material.

V = π h (R^{2} - r^{2}) = π \times 20 \times (7^{2} - 5^{2}) = 20 π (49 - 25) = 480 π \approx 1, 507.96 cm^{3}

Step 3: Calculate outer curved surface area and cost.

CSA_{outer} = 2 π R h = 2 π \times 7 \times 20 = 280 π \approx 879.65 cm^{2}

Cost = 879.65 \times 5 = 4, 398.25

Answer: Material volume $\approx$ $1, 507.96 cm^{3}$ ; painting cost $\approx$ 4,398.25.

Example 3

A composite toy is formed by a cone mounted on a hemisphere of equal radius 6 cm. If the total height of the toy is 16 cm, find its total external surface area.

Solution

Step 1: Dimensions of components.
- Hemisphere radius $r = 6 cm$
- Cone height $h = 16 - 6 = 10 cm$
- Slant height of cone $l = r^{2} + h^{2} = 6^{2} + 1 0^{2} = 136 \approx 11.66 cm$

Step 2: Calculate total exposed surface area.

TSA = CSA_{hemisphere} + CSA_{cone} = 2 π r^{2} + π r l

TSA = 2 π (6^{2}) + π (6) (136) = 72 π + 6 π 136 = π (72 + 6136) \approx 445.68 cm^{2}

Answer: Approximately $445.68 cm^{2}$ .

Example 4

How many solid spherical lead shots of diameter 2 mm can be produced by melting a solid cuboid of dimensions 8 cm $\times$ 6 cm $\times$ 4 cm?

Solution

Step 1: Calculate volume of cuboid in $mm^{3}$ .

8 cm = 80 mm, 6 cm = 60 mm, 4 cm = 40 mm

V_{cuboid} = 80 \times 60 \times 40 = 192, 000 mm^{3}

Step 2: Volume of one spherical shot ( $r = 1 mm$ ).

V_{shot} = \frac{4}{3} π (1^{3}) = \frac{4}{3} π mm^{3}

Step 3: Determine number of shots.

N = \frac{V _{cuboid}}{V _{shot}} = \frac{192 , 000}{\frac{4}{3} π} = \frac{192 , 000 \times 3}{4 π} = \frac{144 , 000}{π} \approx 45, 836.6

Answer: 45,836 complete lead shots.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

The total surface area of a cylinder is 308 $c m^{2}$ . If its radius is 7 cm, find its height. (Use $p i = 22/7$ )

5 cm
10 cm
7 cm
3 cm

Answer: C. 7 cm

Explanation

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

A cube has a total surface area of 294 $c m^{2}$ . Find its volume.

$343 c m^{3}$
$512 c m^{3}$
$216 c m^{3}$
$125 c m^{3}$

Answer: A. 343 cm^3

Explanation

TSA of cube = $6 a^{2}$
$294 = 6 a^{2}$
$a^{2} = 49$
$a = 7 c m$
Volume = $a^{3}$ = $7^{3}$ = 343 $c m^{3}$
Answer: 343 $c m^{3}$

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

The volume of a cube is 512 $c m^{3}$ . Find its total surface area.

$512 c m^{2}$
$384 c m^{2}$
$192 c m^{2}$
$256 c m^{2}$

Answer: B. 384 cm^2

Explanation

Volume of cube = $a^{3}$
$512 = a^{3}$
$a = 8 c m$
$T S A = 6 a^{2} = 6 \times 64 = 384 c m^{2}$
Answer: 384 $c m^{2}$

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Always draw a rough diagram to visualize the solid and label all given dimensions. This helps prevent errors in applying formulas.
Pay careful attention to whether the problem asks for Total Surface Area (TSA), Curved Surface Area (CSA/LSA), or just the base area.
For recasting/melting problems, equate the volumes of the original and new shapes. This is a powerful problem-solving technique.
When dealing with hollow solids, remember that thickness = R - r (outer radius - inner radius). Calculate material volume as the difference between outer and inner volumes.
In problems involving combined solids, identify which surfaces are internal (not exposed) and exclude them from surface area calculations.
For practical applications involving cost, always double-check your surface area calculation before multiplying by the rate.

Key formulas and rules

Key concepts

Total Surface Area vs Curved Surface Area

Hollow Solids

Combining Solids

Conversion and Transformation

Practical Applications

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Example 4

Solution

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.

Key formulas and rules

Key concepts

Total Surface Area vs Curved Surface Area

Hollow Solids

Combining Solids

Conversion and Transformation

Practical Applications

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Example 4

Solution

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.