What is the conservation principle when a solid is recast or melted into another shape?

When a solid is melted and recast without material loss, the **Volume remains constant**: \text{Volume of Original Solid} = \text{Volume of Recast Solid(s)}. If n identical small spheres are made from a large cylinder: \text{Vol}(\text{cylinder}) = n \times \text{Vol}(\text{sphere}).

What are the formulas for Curved Surface Area and Volume of a Right Circular Cone?

For radius r, vertical height h, and slant height l = \sqrt{r^2 + h^2}: - Slant height: l = \sqrt{r^2 + h^2} - \text{CSA} = \pi r l - \text{TSA} = \pi r (l + r) - \text{Volume} = \frac{1}{3}\pi r^2 h

What is the surface area and volume of a hollow hemispherical bowl with inner and outer radii?

For inner radius r and outer radius R: - Inner \text{CSA} = 2\pi r^2, Outer \text{CSA} = 2\pi R^2 - Area of top ring = \pi(R^2 - r^2) - \text{TSA} = 2\pi R^2 + 2\pi r^2 + \pi(R^2 - r^2) = \pi(3R^2 + r^2) - Material Volume = \frac{2}{3}\pi(R^3 - r^3)

Is the Frustum of a Cone included in the CBSE Class 10 2026–27 exam?

No, Frustum of a Cone has been deleted from the active CBSE syllabus. The syllabus covers combinations of cylinders, cones, spheres, hemispheres, and cuboids.

How do you calculate the speed of water flowing through a cylindrical pipe?

Volume of water flowing per unit time = \text{Cross-sectional area of pipe} \times \text{Flow speed} = (\pi r^2) \times v. To find the time taken to fill a tank: \text{Time} = \frac{\text{Volume of Tank}}{\text{Volume flowing per unit time}}. Ensure all dimensions are converted to identical units (e.g., meters) before calculation.

Why should you factor out common terms like $\pi$ and $r$ before substituting numerical values?

Factorising common terms (e.g., \text{TSA} = \pi r(2h + l) instead of evaluating \pi r^2 and 2\pi r h separately) reduces calculation steps by over 50% and prevents rounding errors.

All Revision Notes

MathematicsCh-12 15 min comprehensive revision

NCERT Class 10 Mathematics — Chapter 12

Surface Areas and Volumes

Surface areas and volumes of combinations of solids (cubes, cuboids, spheres, hemispheres, right circular cylinders, cones), and conversion of solids by melting.

Quick Key Takeaways:

Combination of Surface Areas: When two solids are joined base-to-base, the Total Surface Area (TSA) of the new solid is the sum of their Curved Surface Areas (CSA) — internal contact faces are NEVER added.

Volume of Combinations: Total Volume is the direct algebraic sum of the individual component volumes:

V_{\text{total}} = V_1 + V_2

Key Formulas: Sphere (

V = \frac{4}{3}\pi r^3, A = 4\pi r^2

), Hemisphere (

V = \frac{2}{3}\pi r^3, \text{CSA} = 2\pi r^2, \text{TSA} = 3\pi r^2

), Cylinder (

V = \pi r^2 h, \text{CSA} = 2\pi r h

), Cone (

V = \frac{1}{3}\pi r^2 h, \text{CSA} = \pi r l, l = \sqrt{r^2 + h^2}

Conversion by Melting: Volume remains strictly conserved:

n \times V_{\text{small}} = V_{\text{large}}

Mensuration Formula ReferenceReview surface areas and volumes of combined 3D solids

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Jump to Concept:1. 1. Mensuration Formulas Vault & Combination Geometry Principles 2. 2. Combinations of Solids & Volume Conservation Laws 3. 3. Important Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. CBSE Case-Study Modeling: Pharmaceutical Medicine Capsule Manufacture 5. 5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

1. Mensuration Formulas Vault & Combination Geometry Principles

Axioms & Foundational Theory

Comprehensive mathematical foundations, definitions, formulas, and visual conceptual model for Surface Areas and Volumes.

• Master Formulas Vault for 3D Solids

•

Cuboid:

V = lbh, \quad \text{LSA} = 2h(l+b), \quad \text{TSA} = 2(lb + bh + hl), \quad \text{Diagonal} = \sqrt{l^2 + b^2 + h^2}

•

Cube:

V = a^3, \quad \text{LSA} = 4a^2, \quad \text{TSA} = 6a^2, \quad \text{Diagonal} = a\sqrt{3}

•

Right Circular Cylinder:

V = \pi r^2 h, \quad \text{CSA} = 2\pi r h, \quad \text{TSA} = 2\pi r(r + h)

•

Right Circular Cone: Slant height

l = \sqrt{r^2 + h^2}, \quad V = \frac{1}{3}\pi r^2 h, \quad \text{CSA} = \pi r l, \quad \text{TSA} = \pi r(r + l)

•

Sphere:

V = \frac{4}{3}\pi r^3, \quad \text{Surface Area} = 4\pi r^2

•

Hemisphere:

V = \frac{2}{3}\pi r^3, \quad \text{CSA} = 2\pi r^2, \quad \text{TSA} = 3\pi r^2

📊 Surface Areas & Volumes: Solid Combinations & MeltingVisual Model

Visual schematic mapping combined solids (cone + hemisphere, cylinder + hemispheres) and the volumetric conservation law during melting.

2. Combinations of Solids & Volume Conservation Laws

Theorem Proofs & Derivations

Step-by-step rigorous geometric and algebraic theorem proofs and formula derivations for Surface Areas and Volumes.

• The Golden Rule for Joined Solids Surface Area

•

When a cone is mounted on a hemisphere of the same radius

r

\text{TSA of Toy} = \text{CSA of Cone} + \text{CSA of Hemisphere} = \pi r l + 2\pi r^2 = \pi r(l + 2r)

(Note: The flat circular bases touch each other internally and disappear from the exposed outer boundary).

•

When cylindrical capsule has two hemispherical ends:

\text{TSA} = \text{CSA of Cylinder} + 2(\text{CSA of Hemisphere}) = 2\pi r h + 2(2\pi r^2) = 2\pi r(h + 2r)

3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

Topper Step Solutions

Standard CBSE board exam numericals and riders with complete step-by-step mathematical working and justifications.

• 3-Mark Standard Board Question: 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

•

Step 1 (Edge of single cube):

V = a^3 = 64\text{ cm}^3 \implies a = \sqrt[3]{64} = 4\text{ cm}

•

Step 2 (Dimensions of resulting cuboid):
When two cubes are joined end to end:
Length

l = 4 + 4 = 8

cm; Breadth

b = 4

cm; Height

h = 4

cm.

•

Step 3 (Surface Area of Cuboid):

\text{TSA} = 2(lb + bh + hl) = 2(8 \times 4 + 4 \times 4 + 4 \times 8)

\text{TSA} = 2(32 + 16 + 32) = 2(80) = \mathbf{160\text{ cm}^2}

•

Final Boxed Answer:

\mathbf{\text{Surface Area of Cuboid} = 160\text{ cm}^2}

• 5-Mark Heavyweight Board Problem / Rider: A solid toy is in the form of a hemisphere surmounted by a right circular cone of the same radius. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. (Take $\pi = 3.14$ )

•

Step 1 (Dimensions):
Radius

r = \frac{4}{2} = 2

cm. Height of cone

h = 2

cm.

•

Step 2 (Volume Formulation):

\text{Volume of Toy} = \text{Volume of Cone} + \text{Volume of Hemisphere}

V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2(h + 2r)

•

Step 3 (Calculation):

V = \frac{1}{3} \times 3.14 \times (2)^2 \times [2 + 2(2)] = \frac{1}{3} \times 3.14 \times 4 \times [2 + 4]

V = \frac{1}{3} \times 3.14 \times 4 \times 6 = 3.14 \times 4 \times 2 = 3.14 \times 8 = \mathbf{25.12\text{ cm}^3}

•

Final Boxed Answer:

\mathbf{\text{Volume of Toy} = 25.12\text{ cm}^3}

4. CBSE Case-Study Modeling: Pharmaceutical Medicine Capsule Manufacture

Case Study Mastery (4 Marks)

Real-world mathematical modeling scenario with structured multi-part questions and full step solutions.

• Practical Application Context: Pharmaceutical Medicine Capsule Manufacture

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm.

•

Q1: Find the radius and the length of the cylindrical portion.

\rightarrow

Radius

r = \frac{5}{2} = 2.5

mm. Length of cylindrical portion

h = 14 - (2.5 + 2.5) = 14 - 5 = \mathbf{9\text{ mm}}

•

Q2: Calculate the total surface area of the capsule.

\rightarrow

\text{TSA} = \text{CSA of cylinder} + 2(\text{CSA of hemisphere}) = 2\pi r h + 4\pi r^2 = 2\pi r(h + 2r) = 2 \times \frac{22}{7} \times \frac{5}{2} \times [9 + 2(2.5)] = \frac{110}{7} \times 14 = 110 \times 2 = \mathbf{220\text{ mm}^2}

•

Q3: Calculate the internal volume of medicine the capsule can hold.

\rightarrow

V = \pi r^2 h + \frac{4}{3}\pi r^3 = \pi r^2(h + \frac{4}{3}r) = \frac{22}{7} \times 6.25 \times [9 + \frac{10}{3}] = \frac{22}{7} \times 6.25 \times \frac{37}{3} \approx \mathbf{242.26\text{ mm}^3}

5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Important Solved Board Questions

Examiner step-marking allocations, common algebraic/arithmetic deduction traps, and presentation guidelines.

• Step-by-Step Marking Rubric & Presentation Guidelines

•

1 Mark: Stating the correct component formulas (cone, cylinder, hemisphere).

•

1 Mark: Extracting common terms (e.g.

\pi r^2

) before plugging in numbers.

•

2 Marks: Systematic calculation without arithmetic slips.

•

1 Mark: Boxed answer with correct 3D volume units (cm³, mm³, m³).

• Common Calculation Traps & Verification Checklist

•

Trap 1: Adding the flat circular base areas when computing the surface area of combined solids.

•

Trap 2: Forgetting to subtract two radii from total capsule length to find cylindrical height (

h = L - 2r

•

Trap 3: Mixing up diameter and radius.

Authentic Board Question (3 Marks)Topic: Surface Areas and Volumes Mathematical Proof & Computations

State the governing mathematical theorem/formula, show complete geometric or computational steps, and find the exact result for Surface Areas and Volumes.

Official CBSE Step-by-Step Marking Breakdown:

Step 1: Given / To Prove / Formula Setup: State the governing theorem (BPT, Tangents, Pythagoras), mensuration formula, or trigonometric ratios with given values.

1 Mark

Step 2: Step-by-Step Derivation / Arithmetic Working: Show complete deductive steps or numerical calculations with

\pi = 22/7

or trigonometric standard values.

1 Mark

Step 3: Final Boxed Answer with Units / Q.E.D.: State the final numerical result with proper square/cubic units (

\text{cm}^2, \text{m}^3

) or formal proof conclusion.

1 Mark

Model Student Answer (Target: Full 3/3 Marks):

To score full 3 marks on Surface Areas and Volumes in CBSE Mathematics:

1. Theorem / Formula: Write the standard governing formula or state the geometric theorem clearly.
2. Calculation / Proof: Substitute dimensions systematically or provide deductive step justifications.
3. Final Result: Box the final answer with required units (

\text{cm}, \text{cm}^2, \text{m}^3

) or conclude with "Hence Proved".

Examiner Mark Deduction Traps:

•Always write intermediate calculation lines to earn partial credit under CBSE step-marking.

•State reasons in parentheses (e.g., "[Tangents from an external point are equal]") during geometric proofs.

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.

The Total Surface Area (TSA) of the combined solid is NOT the sum of their individual TSAs. When two solids are joined, the overlapping contact surfaces are hidden.

\text{TSA}(\text{combined}) = \text{CSA}_1 + \text{CSA}_2

(sum of exposed curved surface areas).

Surface Areas and Volumes

1. Mensuration Formulas Vault & Combination Geometry Principles

2. Combinations of Solids & Volume Conservation Laws

3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

4. CBSE Case-Study Modeling: Pharmaceutical Medicine Capsule Manufacture

5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Official CBSE Step-by-Step Marking Breakdown:

High-Frequency Conceptual Doubts & FAQs

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