Chapter 12: Surface Areas and Volumes - Board Exam 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 of Solids ★★★★★

Surface Area

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

Example: A toy (Cone on Hemisphere).
TSA = CSA of Cone + CSA of Hemisphere
(Base of cone and top of hemisphere are hidden, so excluded).

Volume

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

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

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) = $\pi rl + 2\pi r^2$

$= \frac{22}{7} \times 3.5 (12.5 + 2 \times 3.5) = 11 \times 19.5 = \mathbf{214.5}$ cm².

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

$= \pi r^2 (h + \frac{4}{3}r) = \frac{22}{7} \times 1.4 \times 1.4 (2.2 + 1.87) \approx 25.05$ cm³.

Step 3: Total Syrup 1.5 Marks

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

Syrup = $30\%$ of Total Vol = $\frac{30}{100} \times 45 \times 25.05 \approx \mathbf{338}$ cm³.

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) = $\frac{1}{3}\pi (1)^2(1) + \frac{2}{3}\pi (1)^3 = \pi$ cm³.

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 (cm², m³) in the final answer. Missing units = -0.5 marks.

Self-Assessment Mock Test (10 Marks)

Q1 (1M): Volume of two spheres are in ratio 64:27. Find ratio of their surface areas.

Q2 (2M): Two cubes each of volume 64 cm³ are joined end to end. Find surface area of resulting cuboid.

Q3 (3M): From a solid cylinder (h=2.4 cm, d=1.4 cm), a conical cavity of same height and diameter is hollowed out. Find TSA of remaining solid.

Q4 (4M): 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.