Surface Areas and Volumes is one of the most scoring chapters in Class 10 Maths when students are clear with formulas, diagrams, and step-by-step calculation. At the same time, it is also one of the chapters where students lose marks because of formula confusion, wrong diagram reading, incomplete units, or mixing up total surface area, curved surface area, and volume.
This chapter is highly important for board exams because it regularly includes 2 mark, 3 mark, and 5 mark questions based on cylinders, cones, hemispheres, cubes, cuboids, and combinations of solids. Many questions are direct applications of formulas, while others test whether students can correctly interpret words such as surmounted, hollow, scooped out, or hemispherical ends.
At Deeksha Vedantu, we always encourage students to study this chapter visually. If you read the question carefully, draw the diagram, identify the visible surfaces, and choose the correct formula, even long questions become manageable.
Why Surface Areas and Volumes Is Important for Class 10 Boards
This chapter is important because it tests both formula knowledge and application skills.
Why Students Must Focus on This Chapter
- It is a regular part of the Class 10 board paper.
- It includes direct formula-based scoring questions.
- It features combination of solids and practical application problems.
- It helps students improve diagram-based understanding.
- It strengthens calculation accuracy and unit handling.
What Students Must Remember Before Solving Questions
Before attempting important questions from Surface Areas and Volumes, students must revise the formulas properly.
Basic Formula Reminder
Students should know these formulas clearly:
- Volume of cylinder = πr²h
- Curved surface area of cylinder = 2πrh
- Volume of cone = (1/3)πr²h
- Curved surface area of cone = πrl
- Volume of hemisphere = (2/3)πr³
- Curved surface area of hemisphere = 2πr²
- Total surface area of cube = 6a²
- Volume of cube = a³
Three Things That Matter Most in This Chapter
According to the transcript, students must be strong in three areas:
- formulas
- calculation accuracy
- proper reading of the question
If even one of these is weak, mistakes become common.
How to Approach Surface Areas and Volumes Questions
This chapter becomes easier when students follow a fixed method.
Step 1: Read the Language Carefully
Words like surmounted, hollow, scooped out, inner radius, outer radius, and total height completely change the solution.
Step 2: Draw the Diagram
Many questions become simple after a correct sketch.
Step 3: Identify What Is Asked
The question may ask for:
- total surface area
- curved surface area
- volume
- cost of painting
- ratio of volumes
Step 4: Apply the Correct Formula
Do not rush into calculation before confirming which surface or solid is involved.
Step 5: Write Units Properly
- cm² for area
- cm³ for volume
Important 2 Mark Question: Volume of a Right Circular Cylinder When Height Equals Radius
The volume of a right circular cylinder with height equal to the radius is 25 1/7 cm³. Find the height of the cylinder.
Given
- Volume = 25 1/7 cm³
- Height = radius
Formula Used
Volume of cylinder = πr²h
Since h = r, we get:
Volume = πh³
Solution
25 1/7 = 176/7
So,
176/7 = (22/7) × h³
h³ = 176/22
h³ = 8
h = 2 cm
Answer
The height of the cylinder is 2 cm.
Why This Is an Important 2 Mark Question
This is a classic board-style question where the condition h = r must be used immediately.
Important 3 Mark Question: Solid in the Form of a Cylinder with Hemispherical Ends
A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter is 7 cm. Find the total volume of the solid.
Step 1: Understand the Shape
The solid has:
- one cylinder in the middle
- two hemispheres at the ends
Two hemispheres together form one sphere.
Step 2: Identify Dimensions
- Diameter = 7 cm
- Radius = 3.5 cm
- Total height = 20 cm
Since the two hemispherical parts together contribute 7 cm, the height of the cylinder is:
20 – 7 = 13 cm
Step 3: Formula Used
Total volume = volume of cylinder + volume of two hemispheres
Total volume = πr²h + (4/3)πr³
Solution
Put r = 3.5 cm and h = 13 cm.
Using π = 22/7, the total volume comes to approximately 680.1 cm³.
Answer
The total volume of the solid is approximately 680.1 cm³.
Exam Insight
This is an important 3 mark type because students must first find the cylinder height before applying the formula.
Important 3 Mark Question: Wooden Article Made by Scooping Out a Hemisphere from Each End of a Solid Cylinder
A wooden article is made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10 cm and the radius is 3.5 cm, find the total surface area of the article.
Understanding the Shape
This is not a full cylinder anymore. Two hemispherical portions have been scooped out from its ends.
So the visible area includes:
- curved surface area of the cylinder
- curved surface area of one scooped hemisphere
- curved surface area of the other scooped hemisphere
Formula Used
Total surface area = curved surface area of cylinder + 2 × curved surface area of hemisphere
So,
Total surface area = 2πrh + 2 × 2πr²
Simplified Form
Total surface area = 2πr(h + 2r)
Solution
Put:
- r = 3.5 cm
- h = 10 cm
Using π = 22/7, the total surface area is:
374 cm²
Answer
The total surface area of the article is 374 cm².
Why Students Make Mistakes Here
Many students wrongly include circular faces. That is not correct because the ends are scooped out hemispheres.
Important 2 Mark Question: Ratio of Volumes of a Cylinder and a Cone
If the radii of the bases of a cylinder and a cone are in the ratio 3:4 and their heights are in the ratio 2:3, find the ratio of their volumes.
Step 1: Write the Ratios
- Radius ratio = 3:4
- Height ratio = 2:3
Step 2: Use Volume Formulas
Volume of cylinder = πr₁²h₁
Volume of cone = (1/3)πr₂²h₂
Step 3: Form the Ratio
Volume ratio = πr₁²h₁ ÷ [(1/3)πr₂²h₂]
This becomes:
3 × (r₁²/r₂²) × (h₁/h₂)
Step 4: Substitute the Given Ratios
3 × (3/4)² × (2/3)
= 3 × 9/16 × 2/3
= 9/8
Answer
The ratio of their volumes is 9:8.
Important 3 Mark Question: Cubical Block Surmounted by a Hemisphere
A cubical block of side 10 cm is surmounted by a hemisphere. Find the largest diameter the hemisphere can have. Also find the cost of painting the total surface area of the solid at the rate of 5 rupees per cm².
Part 1: Largest Diameter of the Hemisphere
The largest possible diameter of the hemisphere will be equal to the side of the cube.
So,
Diameter = 10 cm
Radius = 5 cm
Answer for Part 1
The largest diameter is 10 cm.
Part 2: Total Surface Area of the Solid
The painted area includes:
- total surface area of cube
- curved surface area of hemisphere
- but the circular base where the hemisphere touches the cube is hidden
Formula Used
Total surface area = TSA of cube + CSA of hemisphere – area of circular base
So,
Total surface area = 6a² + 2πr² – πr²
This becomes:
Total surface area = 6a² + πr²
Solution
Put:
- a = 10 cm
- r = 5 cm
Total surface area = 600 + 78.5
Total surface area = 678.5 cm²
Cost of Painting
Cost = area × rate
Cost = 678.5 × 5
Cost = 3392.5 rupees
Answer
- Largest diameter = 10 cm
- Cost of painting = 3392.5 rupees
Why This Is a Strong 3 Mark Question
It checks whether students can correctly subtract the circular base hidden under the hemisphere.
Important 5 Mark Question: Ice Cream Filled in a Cone with a Hemispherical Top
An empty cone has radius 3 cm and height 12 cm. Ice cream is filled in such a way that 1/6 of the cone remains unfilled. Find the total volume of the ice cream, including the hemispherical top.
Understanding the Question
The cone is not fully filled.
- Unfilled part = 1/6 of cone volume
- Filled part = 1 – 1/6 = 5/6 of cone volume
The total ice cream consists of:
- 5/6 of cone volume
- volume of hemispherical top
Formula Used
Volume of ice cream = volume of hemisphere + (5/6) × volume of cone
Step 1: Hemisphere Volume
Volume of hemisphere = (2/3)πr³
With r = 3 cm:
Volume of hemisphere = (2/3)π × 27 = 18π
Step 2: Cone Volume
Volume of cone = (1/3)πr²h
= (1/3)π × 9 × 12
= 36π
Filled cone portion = (5/6) × 36π = 30π
Step 3: Add Both Parts
Total volume of ice cream = 18π + 30π = 48π
Using π = 3.14:
Total volume = 150.72 cm³
Answer
The total volume of the ice cream is 150.72 cm³.
Why This Is an Important 5 Mark Question
This is a classic application problem involving partial filling and combination of solids.
Important 5 Mark Question: Hollow Cylinder Surmounted on a Hollow Hemisphere
The inner and outer radii of a hollow cylinder surmounted on a hollow hemisphere of the same radii are 3 cm and 4 cm respectively. If the height of the cylinder is 14 cm, find its total surface area, inner and outer.
Understanding the Shape
This is a hollow solid. So we must consider:
- inner curved surface of cylinder
- outer curved surface of cylinder
- inner curved surface of hemisphere
- outer curved surface of hemisphere
- circular ring portion at the top
Step 1: Curved Surface Area of Inner Cylinder
CSA = 2πrh
= 2π × 3 × 14
= 84π
Step 2: Curved Surface Area of Outer Cylinder
CSA = 2π × 4 × 14
= 112π
Step 3: Curved Surface Area of Inner Hemisphere
CSA = 2πr²
= 2π × 3²
= 18π
Step 4: Curved Surface Area of Outer Hemisphere
CSA = 2π × 4²
= 32π
Step 5: Circular Ring Area at the Top
Area = πR² – πr²
= π × 16 – π × 9
= 7π
Step 6: Total Surface Area
Total surface area = 84π + 112π + 18π + 32π + 7π
= 253π
Using π = 22/7 or 3.14, the total surface area is approximately 795 cm².
Answer
The total surface area is approximately 795 cm².
Why This Is a High-Value Board Question
Students often forget the top circular ring area. That missing step can reduce marks.
Important Board Question Patterns from Surface Areas and Volumes
The transcript highlights several repeated question types from previous year papers.
Pattern 1: Direct Formula with a Simple Condition
Example: cylinder where height equals radius.
Pattern 2: Combination of Solids
Example: cylinder with hemispherical ends, cone with hemispherical top.
Pattern 3: Scooped Out Solids
Example: a cylinder with hemispheres removed.
Pattern 4: Surmounted Solids
Example: hemisphere on cube, cylinder on hemisphere.
Pattern 5: Hollow Solids
These questions require inner and outer surface calculations separately.
Common Mistakes Students Make in This Chapter
Forgetting to Draw the Diagram
Many students try to solve combination of solids mentally and end up choosing the wrong formula.
Using Total Surface Area Instead of Curved Surface Area
This mistake is common in hemisphere and cylinder questions.
Ignoring Hidden Surfaces
When one solid is placed on another, the touching part is not visible and should not always be counted.
Missing the Difference Between Radius and Diameter
Students often use diameter directly in formulas without converting to radius.
Wrong Units
Area must be written in square units and volume in cubic units.
Smart Exam Tips for Surface Areas and Volumes
Memorise All Standard Formulas First
Without formulas, this chapter becomes difficult to handle in the exam hall.
Read the Words Carefully
Terms such as surmounted, hollow, scooped out, and total height are the key to solving the question correctly.
Keep Calculation Neat
This chapter involves long multiplication and fractions. Clean work reduces mistakes.
Mark What Is Visible
For surface area questions, always ask: which surfaces are actually exposed?
Practice Questions for Revision
Important 2 Mark Practice
- A cone has radius 4 cm and height 9 cm. Find its volume.
- The radius of a cylinder is equal to its height. If volume is given, find its height.
Important 3 Mark Practice
- A hemisphere is surmounted on a cube. Find the total surface area.
- A solid consists of a cylinder and two hemispherical ends. Find the total volume.
Important 5 Mark Practice
- A hollow cylinder with given inner and outer radius is attached to a hemisphere. Find total surface area.
- An ice cream cone with a hemispherical top is partially filled. Find total volume of ice cream.
FAQs
Q1. How can I score well in Surface Areas and Volumes in Class 10?
You can score well by memorising the formulas, drawing the diagram for every question, reading the wording carefully, and writing units correctly at the end.
Q2. What is the most common mistake students make in this chapter?
The most common mistake is using the wrong formula, especially confusing curved surface area with total surface area.
Q3. Why is drawing a diagram so important in Surface Areas and Volumes?
A diagram helps you identify the visible parts, hidden parts, radius, height, and shape combination clearly. Many questions become easier after a simple sketch.
Q4. What does surmounted mean in mensuration questions?
Surmounted means one solid is placed on top of another solid.
Q5. What does scooped out mean in these questions?
Scooped out means some part of the solid has been removed, usually in a curved shape such as a hemisphere.
Q6. When should I use curved surface area of a hemisphere instead of total surface area?
Use curved surface area when the flat circular face is not exposed or not part of the visible boundary.
Q7. How do I know whether to use radius or diameter in a formula?
Most mensuration formulas use radius. If diameter is given, divide it by 2 first.
Q8. How should I write units in Surface Areas and Volumes?
Write square units such as cm² for area and cubic units such as cm³ for volume.
Conclusion
Surface Areas and Volumes is one of the most practical and exam-relevant chapters in Class 10 Maths. It rewards students who combine formula knowledge with careful reading and neat calculation. Most 2 mark, 3 mark, and 5 mark questions in this chapter come from repeated board patterns such as cylinders, hemispheres, cones, hollow solids, and combined solids.
The best way to prepare is to revise formulas daily, solve previous year style questions, and train yourself to identify exposed surfaces correctly. At Deeksha Vedantu, we always remind students that in this chapter, visual clarity and formula accuracy are the real keys to high marks.
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