Class 10 Surface Areas & Volumes Notes

Exam Weightage & Blueprint

Total: 5-6 Marks

This chapter focuses on Combination of Solids (Surface Area & Volume) and Conversion of Solids. Calculation accuracy is key.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	High	Formulas, Ratios of Vol/Area
Short Answer	2 or 3	Medium	Combined Solids (Toy, Capsule)
Long Answer	4 or 5	Very High	Volume Conversion, Embankment, Flow of Water

⏰ Last 24-Hour Checklist

Formula Sheet: Memorize Cylinder, Cone, Sphere, Hemisphere formulas.
Slant Height: $l = \sqrt{h^2 + r^2}$ for Cone.
TSA of Combination: Add CSA of visible parts. Don't add base areas!

Volume of Combination: Simply add volumes of individual solids.
Conversion: Volume remains constant (Vol 1 = Vol 2).
Units: Ensure all units are same (cm vs m).

📐 Formulas Recap

Solid	CSA	TSA	Volume
Cylinder	$2\pi rh$	$2\pi r(r+h)$	$\pi r^2h$
Cone	$\pi rl$	$\pi r(l+r)$	$\frac{1}{3}\pi r^2h$
Hemisphere	$2\pi r^2$	$3\pi r^2$	$\frac{2}{3}\pi r^3$
Sphere	$4\pi r^2$	$4\pi r^2$	$\frac{4}{3}\pi r^3$

Concept: 📐 Combination & Conversion

Surface Area (Combination)

Logic: We calculate the area of the visible surface only.

Example: A toy (Cone on Hemisphere).
TSA = CSA(Cone) + CSA(Hemisphere). (Base areas hidden).

Volume (Combination)

Logic: Volume is the total space occupied. We simply add the volumes.

Example: A Gulab Jamun.
Total Vol = Vol(Cylinder) + 2 - Vol(Hemisphere).

Conversion of Solids

Logic: When a solid is melted and recast, Volume remains constant.

Formula: $V_1 = V_2 \text{ or } n \times v = V$ (if recast into $n$ smaller pieces).

Solved Examples (Board Marking Scheme)

Q1. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area. (3 Marks)

Step 1: Identify Dimensions 0.5 Mark

$r = 3.5$ cm. Total height = 15.5 cm.

Height of cone ($h$) = $15.5 - 3.5 = 12$ cm.

Step 2: Slant Height 1 Mark

$l = \sqrt{r^2 + h^2} = \sqrt{3.5^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5$ cm.

Step 3: Calculate Area 1.5 Marks

TSA = CSA(Cone) + CSA(Hemisphere) = $\pi rl + 2\pi r^2$

$= \frac{22}{7} \times 3.5 (12.5 + 2 \times 3.5) = 11 \times 19.5 = \mathbf{214.5}$ cm-.

Q2. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder. (2 Marks)

Step 1: Concept 0.5 Mark

Volume of Sphere = Volume of Cylinder (Since it is melted & recast).

Step 2: Equation 1 Mark

$\frac{4}{3}\pi R^3 = \pi r^2 h$

$\frac{4}{3} \times (4.2)^3 = (6)^2 \times h$

Step 3: Solve 0.5 Mark

$h = \frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6} = 1.4 \times 1.4 \times 1.4 \times 4 / 4$

$h = \frac{4 \times 74.088}{108} = 2.74$ cm.

Q3. A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximate syrup in 45 gulab jamuns, shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm. (4 Marks)

Step 1: Dimensions 1 Mark

Diameter = 2.8 cm $\Rightarrow r = 1.4$ cm. Length of cylindrical part $h = 5 - (1.4+1.4) = 2.2$ cm.

Step 2: Volume of 1 Piece 1.5 Marks

Vol = Vol(Cyl) + 2 $\times$ Vol(Hemi) = $\pi r^2 h + \frac{4}{3}\pi r^3$

$= \pi r^2 (h + \frac{4}{3}r) = \frac{22}{7} \times 1.4 \times 1.4 (2.2 + 1.87) \approx 25.05$ cm-.

Step 3: Total Syrup 1.5 Marks

Vol of 45 pieces = $45 \times 25.05$.

Syrup = $30\%$ of Total Vol = $\frac{30}{100} \times 45 \times 25.05 \approx \mathbf{338}$ cm-.

Previous Year Questions (PYQs)

2023 (3 Marks): A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.
Ans: Vol = Vol(Cone) + Vol(Hemi) = $\frac{1}{3}\pi (1)^2(1) + \frac{2}{3}\pi (1)^3 = \pi$ cm-.

2020 (5 Marks): A vessel is in the form of an inverted cone. Its height is 8 cm and radius of its top is 5 cm. It is filled with water. When lead shots (spheres of radius 0.5 cm) are dropped, one-fourth of water flows out. Find number of lead shots.
Ans: $n \times$ Vol(Sphere) = $\frac{1}{4} \times$ Vol(Cone). Answer: $n = 100$.

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Adding Bases: In TSA of combinations (e.g., cone on cylinder), do NOT add the area of the base where they join. Only add visible CSA.

Diameter vs Radius: Always check if $d$ or $r$ is given. Questions often give diameter to trap you.

Calculation: Don't multiply $\pi$ ($\frac{22}{7}$) early. Keep it common and cancel it out at the end if possible.

💡 Scoring Tips

Draw Diagrams: Always draw a rough sketch with dimensions labeled (radius, height). It gets marks.

Step Marks: Even if calculation is wrong, writing correct formula gets you marks.

Units: Write units (cm-, m-) in the final answer. Missing units = -0.5 marks.

📝 More Solved Board Questions

Q4. A well of diameter 3 m is dug 14 m deep. The earth taken out is spread evenly around it to form an embankment of width 4 m. Find the height of the embankment. 5 Marks

Sol. Vol of earth dug out = Vol of Cylinder = $\pi r^2 h = \pi (1.5)^2 \times 14 = 31.5\pi$ m³.

Outer radius of embankment ($R$) = $1.5 + 4 = 5.5$ m.

Inner radius ($r$) = 1.5 m.

Area of embankment = $\pi (R^2 - r^2) = \pi (5.5^2 - 1.5^2) = \pi (30.25 - 2.25) = 28\pi$ m².

Height of embankment ($H$) = $\frac{\text{Volume of earth}}{\text{Area of embankment}} = \frac{31.5\pi}{28\pi} = 1.125$ m.

Answer: 1.125 m

🎯 Board Pattern (2018–2025): Embankment and "Flow of water through a pipe" are the most common 5-mark questions. Remember: Area of embankment is a Ring (Outer Area - Inner Area).

📋 Board Revision Checklist

✅ Cube: $V = a^3$; $TSA = 6a^2$
✅ Cylinder: $V = \pi r^2h$; $CSA = 2\pi rh$
✅ Cone: $V = (1/3)\pi r^2h$; $CSA = \pi rl$ (where $l = \sqrt{h^2+r^2}$)
✅ Sphere: $V = (4/3)\pi r^3$; $SA = 4\pi r^2$
✅ Hemisphere: $V = (2/3)\pi r^3$; $TSA = 3\pi r^2$
✅ For Combined TSA: Only add the surfaces you can touch (visible CSA)
✅ Melting/Recasting: Volume $V_{old} = V_{new}$
✅ Deleted Topic: Frustum of a cone is NOT in syllabus

💡 Exam Tip:
Keep $\pi$ as it is in the middle steps. Many times $\pi$ from both sides of the equation will cancel out, saving you tedious calculations.

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. A solid piece of iron in the form of a cuboid of dimensions $49cm \times 33cm \times 24cm$, is moulded to form a solid sphere. The radius of the sphere is:

2. If two solid hemispheres of same base radius $r$ are joined together along their bases, then curved surface area of this new solid is:

3. The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is:

4. During conversion of a solid from one shape to another, the volume of the new shape will:

5. A metallic spherical shell of internal and external diameters 4 cm and 8 cm is melted and recast into the form of a cone of base diameter 8 cm. The height of the cone is:

Chapter 12: Surface Areas and Volumes

Overview