Focus Keyword: NCERT Solutions Class 10 Math Chapter 12 Surface Areas and Volumes
Secondary Keywords & LSI: CBSE Class 10 Maths Chapter 12 solutions, Class 10 Maths Exercise 12.1 solutions, Class 10 Maths Exercise 12.2 solutions, combination of solids surface area volume, CBSE Class 10 Maths board exam preparation
SEO Meta Description: Comprehensive NCERT Solutions for Class 10 Math Chapter 12 Surface Areas and Volumes. Step-by-step solutions for Exercises 12.1 & 12.2, formulas, diagrams, & FAQs.
H1 Title: NCERT Solutions for Class 10 Math Chapter 12: Surface Areas and Volumes (Complete Step-by-Step Guide)
Navigating through the CBSE Class 10 Mathematics curriculum requires an in-depth understanding of three-dimensional spatial geometry, composite solid boundary metrics, internal hollowed cavities, and additive volumetric conservation. Chapter 12 of Class 10 Mathematics, “Surface Areas and Volumes”, forms the foundation of structural architecture, industrial packaging design, mechanical casting, and fluid storage analysis. It investigates the total surface area exposed when standard geometric solids are joined or excavated; explores the difference between visible boundary areas and interior contact interfaces; and details the quantitative computation of combined capacities when cones, cylinders, hemispheres, spheres, and cuboids merge. To help students master every aspect of this high-weightage chapter, this comprehensive guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE evaluation standards.
Every question presented in the official NCERT textbook—ranging from foundational textbook examples to the complete question sets of Exercise 12.1 and Exercise 12.2 and an expanded set of 15 board-level FAQs—has been solved with exhaustive detail. Key scoring terms, systematic four-step mathematical workflows (Given Data $\rightarrow$ Formula Stated $\rightarrow$ Step-by-Step LaTeX Substitution $\rightarrow$ Final Answer with Units), and clear inline geometric diagrams have been highlighted to ensure students secure maximum marks in their CBSE Board Examinations.
Chapter 12: Surface Areas and Volumes
Master Chapter Summary & Formula Blueprint
In the rationalised NCERT syllabus for Class 10 Mathematics, Chapter 12 focuses strictly on two core domains: finding the surface area of a combination of solids and determining the volume of a combination of solids. Topics regarding the conversion of solids and frustums of cones have been removed to align with the rationalised curriculum.
| Solid Shape | Curved / Lateral Surface Area (CSA / LSA) | Total Surface Area (TSA) | Volume ($V$) | Essential Geometric Parameters |
|---|---|---|---|---|
| Cuboid | $2h(l + b)$ | $2(lb + bh + hl)$ | $l \times b \times h$ | Length $l$, Breadth $b$, Height $h$ |
| Cube | $4a^2$ | $6a^2$ | $a^3$ | Edge length $a$ |
| Right Circular Cylinder | $2\pi rh$ | $2\pi r(r + h)$ | $\pi r^2 h$ | Base radius $r$, Height $h$ |
| Right Circular Cone | $\pi rl$ | $\pi r(r + l)$ | $\frac{1}{3}\pi r^2 h$ | Radius $r$, Vertical height $h$, Slant height $l = \sqrt{r^2 + h^2}$ |
| Sphere | $4\pi r^2$ | $4\pi r^2$ | $\frac{4}{3}\pi r^3$ | Radius $r$ |
| Hemisphere | $2\pi r^2$ | $3\pi r^2$ | $\frac{2}{3}\pi r^3$ | Radius $r$ |
🧠 Examiner’s Secret: A universal rule for surface areas of composite solids: Never simply add the total surface areas of the individual constituent solids. When two solids are joined together, their contact faces become internal and are no longer exposed. Total Surface Area of composite solid = $\sum (\text{Exposed Curved Surface Areas}) + \text{Any Exposed Base Areas}$. Conversely, volumes are purely scalar and additive: $\text{Total Volume} = \sum (\text{Volumes of Individual Components})$, or subtracted in the case of hollowed cavities.
Foundational Geometric Concepts and Principles
Surface Area of Combined Solids
The surface area of a combination of solids represents the total measure of the exterior exposed two-dimensional surfaces visible and tactile to an observer. Hemisphere CSA: 2πr² Cone CSA: πrl Total Exposed SA = 2πr² + πrl
When two solids are joined together along a common boundary (such as a cone mounted on a hemisphere), the flat circular bases coincide and form an internal interface. Because this circular junction is enclosed inside the solid, it does not contribute to the exposed surface area.
$$\text{TSA of Toy} = \text{CSA of Hemispherical Base} + \text{CSA of Conical Top} = 2\pi r^2 + \pi rl$$
Volume of Combined Solids
The volume of a combination of solids is the total three-dimensional space enclosed within the exterior boundaries of the merged physical body.
Unlike surface area, volume is an intrinsic scalar property. The volume of a solid composed of two or more distinct geometric shapes equals the direct sum of the volumes of each component shape:
$$\text{Total Volume} = V_1 + V_2 + \dots + V_n$$
When a solid contains an internal hollowed depression or carved cavity (such as a cylinder with a cone drilled out), the remaining volume is obtained by subtraction:
$$\text{Remaining Volume} = V_{\text{original solid}} – V_{\text{carved cavity}}$$
💡 Did You Know?: The ancient Greek mathematician Archimedes requested that a sphere inscribed within a right cylinder of equal diameter and height be carved onto his tombstone. He proved that the volume of the sphere is exactly $\frac{2}{3}$ the volume of the enclosing cylinder, and its surface area is also exactly $\frac{2}{3}$ of the cylinder’s total surface area.
[👉 Also Read: Class 10 Math Chapter 11 Areas Related to Circles NCERT Solutions]
Step-by-Step Solutions: NCERT Class 10 Mathematics Chapter 12 Solved Examples
Example 1 (Page 164) [CBSE 2017, 2020 Standard]
Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is $5\text{ cm}$ in height and the diameter of the top is $3.5\text{ cm}$. Find the area he has to colour. (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Identify Given Data
- Total height of playing top, $H = 5\text{ cm}$.
- Diameter of hemispherical and conical base, $d = 3.5\text{ cm}$.
- Common radius, $r = \frac{d}{2} = \frac{3.5}{2} = 1.75\text{ cm} = \frac{7}{4}\text{ cm}$.
- Height of hemispherical part = radius $r = 1.75\text{ cm}$.
- Height of conical part, $h = H – r = 5 – 1.75 = 3.25\text{ cm} = \frac{13}{4}\text{ cm}$.
Step 2: Calculate Slant Height ($l$) of the Conical Part
$$l = \sqrt{r^2 + h^2} = \sqrt{\left(\frac{7}{4}\right)^2 + \left(\frac{13}{4}\right)^2} = \sqrt{\frac{49 + 169}{16}} = \sqrt{\frac{218}{16}} = \frac{\sqrt{218}}{4} \approx \frac{14.76}{4} \approx 3.7\text{ cm}$$
Step 3: State the Formulation for Surface Area to be Coloured
The flat circular base is common to both the hemisphere and the cone, so it is internal:
$$\text{Total Surface Area to Colour} = \text{CSA of Hemisphere} + \text{CSA of Cone} = 2\pi r^2 + \pi rl = \pi r(2r + l)$$
Step 4: Step-by-Step LaTeX Substitution
$$\text{TSA} = \frac{22}{7} \times \frac{7}{4} \times \left(2 \times \frac{7}{4} + 3.7\right)$$
$$\text{TSA} = \frac{11}{2} \times (3.5 + 3.7) = 5.5 \times 7.2 = 39.6\text{ cm}^2$$
Final Answer:
The total surface area Rasheed has to colour is approximately $39.6\text{ cm}^2$.
Example 2 (Page 165) [CBSE 2016, 2019, 2023 Set-1]
The decorative block shown in Fig. 12.7 is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge $5\text{ cm}$, and the hemisphere fixed on the top has a diameter of $4.2\text{ cm}$. Find the total surface area of the block. (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Identify Given Data
- Edge of the cube, $a = 5\text{ cm}$.
- Diameter of the hemisphere, $d = 4.2\text{ cm}$.
- Radius of the hemisphere, $r = \frac{4.2}{2} = 2.1\text{ cm} = \frac{21}{10}\text{ cm}$.
Step 2: Formulation for Total Surface Area
The hemisphere is mounted on one face of the cube. It covers a circular area of the top face, while adding its curved hemispherical surface:
$$\text{TSA of Block} = \text{TSA of Cube} – \text{Base Area of Hemisphere} + \text{CSA of Hemisphere}$$
$$\text{TSA of Block} = 6a^2 – \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2$$
Step 3: Step-by-Step Substitution
- Total surface area of the cube:
$$6a^2 = 6 \times (5)^2 = 6 \times 25 = 150\text{ cm}^2$$ - Net exposed area from the hemisphere ($\pi r^2$):
$$\pi r^2 = \frac{22}{7} \times (2.1)^2 = \frac{22}{7} \times 4.41 = 22 \times 0.63 = 13.86\text{ cm}^2$$
Combine the two parts:
$$\text{TSA of Block} = 150 + 13.86 = 163.86\text{ cm}^2$$
Final Answer:
The total surface area of the decorative block is $163.86\text{ cm}^2$.
Example 3 (Page 166) [CBSE 2015, 2018, 2022 Term-2]
A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in Fig. 12.8. The height of the entire rocket is $26\text{ cm}$, while the height of the conical part is $6\text{ cm}$. The base of the conical portion has a diameter of $5\text{ cm}$, while the base diameter of the cylindrical portion is $3\text{ cm}$. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take $\pi = 3.14$)
Answer:
Step 1: Partition Geometric Dimensions
- Total height of rocket = $26\text{ cm}$.
- Height of conical part, $h = 6\text{ cm}$.
- Height of cylindrical part, $H = 26 – 6 = 20\text{ cm}$.
- Conical base diameter $D = 5\text{ cm} \implies$ Conical radius $R = 2.5\text{ cm}$.
- Cylindrical base diameter $d = 3\text{ cm} \implies$ Cylindrical radius $r = 1.5\text{ cm}$.
Step 2: Slant Height of the Conical Portion ($l$)
$$l = \sqrt{R^2 + h^2} = \sqrt{(2.5)^2 + 6^2} = \sqrt{6.25 + 36} = \sqrt{42.25} = 6.5\text{ cm}$$
Step 3: Calculate Area to be Painted Orange (Conical Portion)
The conical base sits on the smaller cylindrical base. The exposed conical area includes its curved surface plus the base ring not covered by the cylinder:
$$\text{Area to be painted orange} = \text{CSA of Cone} + (\text{Area of Conical Base} – \text{Area of Cylindrical Base})$$
$$\text{Area} = \pi R l + (\pi R^2 – \pi r^2) = \pi [R l + (R^2 – r^2)]$$
Substitute $R = 2.5\text{ cm}$, $r = 1.5\text{ cm}$, $l = 6.5\text{ cm}$:
$$\text{Area} = 3.14 \times [2.5 \times 6.5 + (2.5^2 – 1.5^2)]$$
$$\text{Area} = 3.14 \times [16.25 + (6.25 – 2.25)] = 3.14 \times [16.25 + 4] = 3.14 \times 20.25 = 63.585\text{ cm}^2$$
Step 4: Calculate Area to be Painted Yellow (Cylindrical Portion)
The cylindrical part is exposed along its curved surface and its bottom circular base:
$$\text{Area to be painted yellow} = \text{CSA of Cylinder} + \text{Area of Bottom Base} = 2\pi r H + \pi r^2 = \pi r(2H + r)$$
Substitute $r = 1.5\text{ cm}$, $H = 20\text{ cm}$:
$$\text{Area} = 3.14 \times 1.5 \times [2(20) + 1.5] = 4.71 \times (40 + 1.5) = 4.71 \times 41.5 = 195.465\text{ cm}^2$$
Final Answer:
The area painted orange is $63.585\text{ cm}^2$, and the area painted yellow is $195.465\text{ cm}^2$.
Example 4 (Page 167) [CBSE 2014, 2020 Standard]
Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end (see Fig. 12.9). The height of the cylinder is $1.45\text{ m}$ and its radius is $30\text{ cm}$. Find the total surface area of the bird-bath. (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Align Units and Identify Given Data
- Radius of cylinder and hemispherical depression, $r = 30\text{ cm} = 0.3\text{ m}$.
- Height of cylinder, $h = 1.45\text{ m} = 145\text{ cm}$.
Step 2: Formulation for Total Surface Area
The bird-bath stands on the ground; its exterior includes the curved surface of the cylinder and the curved interior of the scooped-out hemisphere:
$$\text{TSA of Bird-bath} = \text{CSA of Cylinder} + \text{CSA of Hemisphere} = 2\pi rh + 2\pi r^2 = 2\pi r(h + r)$$
Step 3: Step-by-Step Substitution
Using centimetres ($r = 30\text{ cm}, h = 145\text{ cm}$):
$$\text{TSA} = 2 \times \frac{22}{7} \times 30 \times (145 + 30) = \frac{44}{7} \times 30 \times 175$$
Divide $175$ by $7$ ($175 / 7 = 25$):
$$\text{TSA} = 44 \times 30 \times 25 = 33000\text{ cm}^2$$
Convert to square metres ($1\text{ m}^2 = 10000\text{ cm}^2$):
$$\text{TSA} = \frac{33000}{10000} = 3.3\text{ m}^2$$
Final Answer:
The total surface area of the bird-bath is $3.3\text{ m}^2$ (or $33000\text{ cm}^2$).
Example 5 (Page 168) [CBSE 2016, 2019 Set-2]
A shed is in the shape of a cuboid surmounted by a half-cylinder (see Fig. 12.12). If the base of the shed is of dimension $7\text{ m} \times 15\text{ m}$, and the height of the cuboidal portion is $8\text{ m}$, find the volume of air that the shed can hold. Further, suppose the interior of the shed contains machinery that occupies a total space of $300\text{ m}^3$, and there are $20$ workers, each of whom occupies about $0.08\text{ m}^3$ space on an average. Then, how much air is in the shed? (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Identify Given Data
- Cuboid base: Length $l = 15\text{ m}$, Breadth $b = 7\text{ m}$, Height $h = 8\text{ m}$.
- The half-cylinder sits on top along breadth $b = 7\text{ m}$.
- Diameter of half-cylinder = $7\text{ m} \implies$ Radius $r = \frac{7}{2}\text{ m} = 3.5\text{ m}$.
- Length of half-cylinder, $H = l = 15\text{ m}$.
Step 2: Calculate Total Volume Capacity of the Shed
$$\text{Total Volume} = \text{Volume of Cuboid} + \text{Volume of Half-Cylinder}$$
- Volume of Cuboidal portion:
$$V_{\text{cuboid}} = l \times b \times h = 15 \times 7 \times 8 = 840\text{ m}^3$$ - Volume of Half-Cylindrical portion:
$$V_{\text{half-cylinder}} = \frac{1}{2}\pi r^2 H = \frac{1}{2} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times 15 = \frac{1}{2} \times \frac{22}{7} \times \frac{49}{4} \times 15$$
$$V_{\text{half-cylinder}} = \frac{11 \times 7 \times 15}{4} = \frac{1155}{4} = 288.75\text{ m}^3$$
Total internal volume:
$$\text{Total Volume} = 840 + 288.75 = 1128.75\text{ m}^3$$
Step 3: Calculate Net Air Volume with Obstructions
- Space occupied by machinery = $300\text{ m}^3$.
- Space occupied by $20$ workers = $20 \times 0.08 = 1.6\text{ m}^3$.
- Total occupied space = $300 + 1.6 = 301.6\text{ m}^3$.
Net air available:
$$\text{Volume of Air} = 1128.75 – 301.6 = 827.15\text{ m}^3$$
Final Answer:
The shed can hold $1128.75\text{ m}^3$ of air when empty, and $827.15\text{ m}^3$ of air when machinery and workers are present.
Example 6 (Page 170) [CBSE 2013, 2018]
A juice seller was serving his customers using glasses as shown in Fig. 12.13. The inner diameter of the cylindrical glass was $5\text{ cm}$, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was $10\text{ cm}$, find the apparent capacity of the glass and its actual capacity. (Use $\pi = 3.14$)
Answer:
Step 1: Identify Given Data
- Inner diameter of glass, $d = 5\text{ cm} \implies$ Radius $r = \frac{5}{2} = 2.5\text{ cm}$.
- Height of glass, $h = 10\text{ cm}$.
- Radius of raised hemispherical bottom, $r = 2.5\text{ cm}$.
- Value of $\pi = 3.14$.
Step 2: Compute Apparent Capacity (Full Cylinder)
$$\text{Apparent Capacity} = \pi r^2 h = 3.14 \times (2.5)^2 \times 10 = 3.14 \times 6.25 \times 10 = 3.14 \times 62.5 = 196.25\text{ cm}^3$$
Step 3: Compute Volume of Hemispherical Raised Portion
$$V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 = \frac{2}{3} \times 3.14 \times (2.5)^3 = \frac{2}{3} \times 3.14 \times 15.625 = \frac{98.125}{3} \approx 32.71\text{ cm}^3$$
Step 4: Compute Actual Capacity
$$\text{Actual Capacity} = \text{Apparent Capacity} – V_{\text{hemisphere}} = 196.25 – 32.71 = 163.54\text{ cm}^3$$
Final Answer:
The apparent capacity of the glass is $196.25\text{ cm}^3$, and its actual capacity is $163.54\text{ cm}^3$.
Example 7 (Page 171) [CBSE 2015, 2017, 2021 Term-2]
A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is $2\text{ cm}$ and the diameter of the base is $4\text{ cm}$. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take $\pi = 3.14$)
Answer:
Step 1: Identify Given Data
- Diameter of cone and hemisphere, $d = 4\text{ cm} \implies$ Radius $r = 2\text{ cm}$.
- Height of the cone, $h = 2\text{ cm}$.
- Height of the hemisphere = radius $r = 2\text{ cm}$.
- Total height of the toy = $h + r = 2 + 2 = 4\text{ cm}$.
- For the circumscribing cylinder: Radius $R = r = 2\text{ cm}$, Height $H = 4\text{ cm}$.
Step 2: Compute Volume of the Toy
$$\text{Volume of Toy} = \text{Volume of Cone} + \text{Volume of Hemisphere} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2(h + 2r)$$
Substitute $r = 2\text{ cm}$, $h = 2\text{ cm}$:
$$\text{Volume of Toy} = \frac{1}{3} \times 3.14 \times 2^2 \times [2 + 2(2)] = \frac{1}{3} \times 3.14 \times 4 \times 6 = 3.14 \times 8 = 25.12\text{ cm}^3$$
Step 3: Compute Volume of the Circumscribing Cylinder
$$V_{\text{cylinder}} = \pi R^2 H = 3.14 \times 2^2 \times 4 = 3.14 \times 16 = 50.24\text{ cm}^3$$
Step 4: Determine the Difference in Volumes
$$\text{Difference} = V_{\text{cylinder}} – \text{Volume of Toy} = 50.24 – 25.12 = 25.12\text{ cm}^3$$
Final Answer:
The volume of the toy is $25.12\text{ cm}^3$, and the difference between the volumes of the cylinder and the toy is $25.12\text{ cm}^3$.
[👉 Also Read: Class 10 Math Chapter 13 Statistics NCERT Solutions]
Step-by-Step Solutions: NCERT Class 10 Mathematics Exercise 12.1
Question 1 (Page 172) [CBSE 2013, 2017, 2020 Standard]
2 cubes each of volume $64\text{ cm}^3$ are joined end to end. Find the surface area of the resulting cuboid.
Answer:
Step 1: Determine Edge Length of Each Cube
Let the edge of each cube be $a$.
$$\text{Volume of Cube} = a^3 = 64\text{ cm}^3 \implies a = \sqrt[3]{64} = 4\text{ cm}$$
Step 2: Dimensions of the Resulting Cuboid
When two identical cubes of edge $4\text{ cm}$ are joined end to end:
- Length of cuboid, $l = 4 + 4 = 8\text{ cm}$.
- Breadth of cuboid, $b = 4\text{ cm}$.
- Height of cuboid, $h = 4\text{ cm}$.
Step 3: Calculate Total Surface Area of Cuboid
$$\text{Surface Area} = 2(lb + bh + hl)$$
$$\text{Surface Area} = 2(8 \times 4 + 4 \times 4 + 4 \times 8) = 2(32 + 16 + 32) = 2(80) = 160\text{ cm}^2$$
Final Answer:
The surface area of the resulting cuboid is $160\text{ cm}^2$.
Question 2 (Page 172) [CBSE 2014, 2019, 2023]
A vessel is in the form of a hollow hemisphere surmounted by a hollow cylinder. The diameter of the hemisphere is $14\text{ cm}$ and the total height of the vessel is $13\text{ cm}$. Find the inner surface area of the vessel. (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Identify Given Data
- Diameter of hemisphere, $d = 14\text{ cm} \implies$ Common radius $r = 7\text{ cm}$.
- Total height of vessel, $H = 13\text{ cm}$.
- Height of hemispherical part = radius $r = 7\text{ cm}$.
- Height of cylindrical part, $h = H – r = 13 – 7 = 6\text{ cm}$.
Step 2: Formula for Inner Surface Area
Because both shapes are hollow and open at the junction:
$$\text{Inner Surface Area} = \text{CSA of Cylinder} + \text{CSA of Hemisphere} = 2\pi rh + 2\pi r^2 = 2\pi r(h + r)$$
Step 3: Step-by-Step Substitution
$$\text{Area} = 2 \times \frac{22}{7} \times 7 \times (6 + 7) = 44 \times 13 = 572\text{ cm}^2$$
Final Answer:
The inner surface area of the vessel is $572\text{ cm}^2$.
Question 3 (Page 172) [CBSE 2012, 2016, 2020 Standard, 2024]
A toy is in the form of a cone of radius $3.5\text{ cm}$ mounted on a hemisphere of same radius. The total height of the toy is $15.5\text{ cm}$. Find the total surface area of the toy. (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Identify Given Data
- Radius of cone and hemisphere, $r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$.
- Total height of toy, $H = 15.5\text{ cm}$.
- Height of hemispherical portion = radius $r = 3.5\text{ cm}$.
- Height of conical portion, $h = H – r = 15.5 – 3.5 = 12\text{ cm}$.
Step 2: Calculate Slant Height ($l$) of the Cone
$$l = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\text{ cm}$$
Step 3: Compute Total Surface Area of Toy
$$\text{TSA of Toy} = \text{CSA of Cone} + \text{CSA of Hemisphere} = \pi rl + 2\pi r^2 = \pi r(l + 2r)$$
$$\text{TSA} = \frac{22}{7} \times \frac{7}{2} \times [12.5 + 2(3.5)] = 11 \times [12.5 + 7] = 11 \times 19.5 = 214.5\text{ cm}^2$$
Final Answer:
The total surface area of the toy is $214.5\text{ cm}^2$.
Question 4 (Page 172) [CBSE 2015, 2018, 2021]
A cubical block of side $7\text{ cm}$ is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Determine Greatest Diameter
The hemisphere sits on the top square face of side $7\text{ cm}$. The circular boundary cannot extend past the edges of the face:
$$\text{Greatest Diameter } d = \text{Edge of Cube } a = 7\text{ cm}$$
$$\text{Radius } r = \frac{7}{2}\text{ cm} = 3.5\text{ cm}$$
Step 2: Formulation for Total Surface Area
$$\text{Surface Area} = \text{TSA of Cube} – \text{Base Area of Hemisphere} + \text{CSA of Hemisphere}$$
$$\text{Surface Area} = 6a^2 – \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2$$
Step 3: Step-by-Step Substitution
$$\text{Surface Area} = 6(7)^2 + \frac{22}{7} \times \left(\frac{7}{2}\right)^2$$
$$\text{Surface Area} = 6(49) + \frac{22}{7} \times \frac{49}{4} = 294 + \frac{77}{2} = 294 + 38.5 = 332.5\text{ cm}^2$$
Final Answer:
The greatest diameter the hemisphere can have is $7\text{ cm}$, and the surface area of the solid is $332.5\text{ cm}^2$.
Question 5 (Page 172) [CBSE 2014, 2017, 2020 Standard]
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter $l$ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Answer:
Step 1: Identify Given Data
- Edge of the cube = $l$.
- Diameter of the carved hemisphere, $d = l \implies$ Radius $r = \frac{l}{2}$.
Step 2: Formulate Surface Area of the Carved Block
When a depression is scooped out, the top square face loses a circle of area $\pi r^2$, but gains the curved interior surface of the hemisphere ($2\pi r^2$):
$$\text{Remaining Surface Area} = \text{TSA of Cube} – \text{Base Area of Hemisphere} + \text{Internal CSA of Hemisphere}$$
$$\text{Surface Area} = 6l^2 – \pi r^2 + 2\pi r^2 = 6l^2 + \pi r^2$$
Step 3: Algebraic Substitution
Substitute $r = \frac{l}{2}$:
$$\text{Surface Area} = 6l^2 + \pi \left(\frac{l}{2}\right)^2 = 6l^2 + \frac{\pi l^2}{4} = \frac{l^2}{4}(24 + \pi)$$
Final Answer:
The surface area of the remaining solid is $\frac{l^2}{4}(24 + \pi)\text{ sq. units}$ (or $\frac{1}{4}l^2(\pi + 24)$).
Question 6 (Page 172) [CBSE 2012, 2016, 2019, 2023 Set-2]
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10). The length of the entire capsule is $14\text{ mm}$ and the diameter of the capsule is $5\text{ mm}$. Find its surface area. (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Identify Given Data
- Total length of capsule = $14\text{ mm}$.
- Diameter of capsule, $d = 5\text{ mm} \implies$ Radius $r = \frac{5}{2} = 2.5\text{ mm}$.
- The capsule ends consist of two hemispheres of radius $2.5\text{ mm}$ each.
- Length of cylindrical portion, $h = 14 – (2.5 + 2.5) = 14 – 5 = 9\text{ mm}$.
Step 2: Formulation for Total Surface Area
$$\text{Total Surface Area} = \text{CSA of Cylinder} + 2 \times (\text{CSA of Hemisphere})$$
$$\text{Total Surface Area} = 2\pi rh + 2(2\pi r^2) = 2\pi rh + 4\pi r^2 = 2\pi r(h + 2r)$$
Step 3: Step-by-Step Substitution
$$\text{Surface Area} = 2 \times \frac{22}{7} \times \frac{5}{2} \times [9 + 2(2.5)] = \frac{110}{7} \times (9 + 5) = \frac{110}{7} \times 14 = 110 \times 2 = 220\text{ mm}^2$$
Final Answer:
The surface area of the medicine capsule is $220\text{ mm}^2$.
Question 7 (Page 173) [CBSE 2015, 2018, 2022 Term-2]
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1\text{ m}$ and $4\text{ m}$ respectively, and the slant height of the top is $2.8\text{ m}$, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹$500\text{ per m}^2$. (Note that the base of the tent will not be covered with canvas). (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Identify Given Data
- Cylindrical height, $h = 2.1\text{ m}$.
- Diameter, $d = 4\text{ m} \implies$ Radius $r = 2\text{ m}$.
- Slant height of conical top, $l = 2.8\text{ m}$.
- Rate of canvas = ₹$500\text{ per m}^2$.
Step 2: Area of Canvas Required
$$\text{Area of Canvas} = \text{CSA of Cylinder} + \text{CSA of Cone} = 2\pi rh + \pi rl = \pi r(2h + l)$$
Step 3: Step-by-Step Substitution
$$\text{Area} = \frac{22}{7} \times 2 \times [2(2.1) + 2.8] = \frac{44}{7} \times [4.2 + 2.8] = \frac{44}{7} \times 7 = 44\text{ m}^2$$
Step 4: Compute Total Cost
$$\text{Cost} = \text{Area} \times \text{Rate} = 44 \times 500 = ₹22000$$
Final Answer:
The area of the canvas is $44\text{ m}^2$, and the total cost is ₹$22000$.
Question 8 (Page 173) [CBSE 2014, 2019 Set-3, 2024]
From a solid cylinder whose height is $2.4\text{ cm}$ and diameter $1.4\text{ cm}$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $\text{cm}^2$. (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Identify Given Data
- Height of cylinder and conical cavity, $h = 2.4\text{ cm}$.
- Diameter, $d = 1.4\text{ cm} \implies$ Radius $r = 0.7\text{ cm} = \frac{7}{10}\text{ cm}$.
Step 2: Calculate Slant Height ($l$) of the Conical Cavity
$$l = \sqrt{r^2 + h^2} = \sqrt{(0.7)^2 + (2.4)^2} = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5\text{ cm}$$
Step 3: Formulate Remaining Total Surface Area
The remaining solid has three exposed boundary components: the outer curved cylinder, the intact solid circular base, and the inner conical cavity:
$$\text{Total Surface Area} = \text{CSA of Cylinder} + \text{Area of Circular Base} + \text{CSA of Conical Cavity}$$
$$\text{TSA} = 2\pi rh + \pi r^2 + \pi rl = \pi r(2h + r + l)$$
Step 4: Step-by-Step Substitution
$$\text{TSA} = \frac{22}{7} \times 0.7 \times [2(2.4) + 0.7 + 2.5] = 2.2 \times [4.8 + 0.7 + 2.5] = 2.2 \times 8.0 = 17.6\text{ cm}^2$$
Rounding to the nearest $\text{cm}^2$:
$$17.6\text{ cm}^2 \approx 18\text{ cm}^2$$
Final Answer:
The total surface area of the remaining solid is $17.6\text{ cm}^2$ (or $18\text{ cm}^2$ to the nearest $\text{cm}^2$).
Question 9 (Page 173) [CBSE 2013, 2018, 2023 Set-3]
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. If the height of the cylinder is $10\text{ cm}$, and its base is of radius $3.5\text{ cm}$, find the total surface area of the article. (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Identify Given Data
- Height of cylinder, $h = 10\text{ cm}$.
- Radius of cylinder and hemispherical scoops, $r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$.
Step 2: Formulate Total Surface Area
Scooping out two hemispheres removes the two flat base disks and replaces them with two curved hemispherical cavities:
$$\text{Total Surface Area} = \text{CSA of Cylinder} + 2 \times (\text{CSA of Hemisphere})$$
$$\text{TSA} = 2\pi rh + 2(2\pi r^2) = 2\pi rh + 4\pi r^2 = 2\pi r(h + 2r)$$
Step 3: Step-by-Step Substitution
$$\text{TSA} = 2 \times \frac{22}{7} \times \frac{7}{2} \times [10 + 2(3.5)] = 22 \times [10 + 7] = 22 \times 17 = 374\text{ cm}^2$$
Final Answer:
The total surface area of the article is $374\text{ cm}^2$.
Step-by-Step Solutions: NCERT Class 10 Mathematics Exercise 12.2
Question 1 (Page 176) [CBSE 2012, 2017, 2021]
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1\text{ cm}$ and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.
Answer:
Step 1: Identify Given Data
- Radius of hemisphere and cone, $r = 1\text{ cm}$.
- Height of cone, $h = r = 1\text{ cm}$.
Step 2: Formula for Total Volume
$$\text{Total Volume} = \text{Volume of Cone} + \text{Volume of Hemisphere} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2(h + 2r)$$
Step 3: Substitution in Terms of $\pi$
$$\text{Total Volume} = \frac{1}{3}\pi (1)^2 [1 + 2(1)] = \frac{1}{3}\pi (1)[3] = \pi\text{ cm}^3$$
Final Answer:
The volume of the solid is $\pi\text{ cm}^3$.
Question 2 (Page 176) [CBSE 2015, 2019, 2023]
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is $3\text{ cm}$ and its length is $12\text{ cm}$. If each cone has a height of $2\text{ cm}$, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same). (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Identify Given Data
- Total length of model = $12\text{ cm}$.
- Height of each conical end, $h_c = 2\text{ cm}$.
- Length of cylindrical portion, $h = 12 – (2 + 2) = 12 – 4 = 8\text{ cm}$.
- Diameter of model, $d = 3\text{ cm} \implies$ Common radius $r = \frac{3}{2} = 1.5\text{ cm}$.
Step 2: Formula for Total Air Volume
$$\text{Total Volume} = \text{Volume of Cylinder} + 2 \times (\text{Volume of Cone})$$
$$\text{Total Volume} = \pi r^2 h + 2\left(\frac{1}{3}\pi r^2 h_c\right) = \pi r^2 \left(h + \frac{2}{3}h_c\right)$$
Step 3: Step-by-Step Substitution
$$\text{Total Volume} = \frac{22}{7} \times \left(\frac{3}{2}\right)^2 \times \left(8 + \frac{2}{3} \times 2\right) = \frac{22}{7} \times \frac{9}{4} \times \left(8 + \frac{4}{3}\right)$$
$$\text{Total Volume} = \frac{22}{7} \times \frac{9}{4} \times \frac{28}{3}$$
Divide $28$ by $7$ ($28/7 = 4$), and $9$ by $3$ ($9/3 = 3$):
$$\text{Total Volume} = \frac{22 \times 3 \times 4}{4} = 22 \times 3 = 66\text{ cm}^3$$
Final Answer:
The volume of air contained in the model is $66\text{ cm}^3$.
Question 3 (Page 176) [CBSE 2014, 2018, 2020 Standard, 2024]
A gulab jamun, contains sugar syrup up to about $30%$ of its volume. Find approximately how much syrup would be found in $45\text{ gulab jamuns}$, each shaped like a cylinder with two hemispherical ends with length $5\text{ cm}$ and diameter $2.8\text{ cm}$ (see Fig. 12.15). (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Identify Given Data
- Total length of one gulab jamun = $5\text{ cm}$.
- Diameter, $d = 2.8\text{ cm} \implies$ Radius $r = 1.4\text{ cm} = \frac{7}{5}\text{ cm}$.
- The ends are hemispheres of radius $1.4\text{ cm}$ each.
- Length of cylindrical portion, $h = 5 – (1.4 + 1.4) = 5 – 2.8 = 2.2\text{ cm}$.
Step 2: Volume of One Gulab Jamun
$$\text{Volume} = \text{Volume of Cylinder} + 2 \times (\text{Volume of Hemisphere}) = \pi r^2 h + \frac{4}{3}\pi r^3 = \pi r^2 \left(h + \frac{4}{3}r\right)$$
Substitute values:
$$\text{Volume} = \frac{22}{7} \times (1.4)^2 \times \left(2.2 + \frac{4}{3} \times 1.4\right) = \frac{22}{7} \times 1.96 \times \left(2.2 + \frac{5.6}{3}\right)$$
$$\text{Volume} = 22 \times 0.28 \times \left(\frac{6.6 + 5.6}{3}\right) = 6.16 \times \frac{12.2}{3} = \frac{75.152}{3}\text{ cm}^3 \approx 25.05\text{ cm}^3$$
Step 3: Total Volume of 45 Gulab Jamuns
$$\text{Total Volume} = 45 \times \frac{75.152}{3} = 15 \times 75.152 = 1127.28\text{ cm}^3$$
Step 4: Compute Quantity of Sugar Syrup ($30%$)
$$\text{Volume of Syrup} = 30% \times 1127.28 = \frac{30}{100} \times 1127.28 = 0.3 \times 1127.28 = 338.184\text{ cm}^3$$
Rounding to the nearest whole number:
$$338.184\text{ cm}^3 \approx 338\text{ cm}^3$$
Final Answer:
The volume of sugar syrup found in 45 gulab jamuns is approximately $338\text{ cm}^3$.
Question 4 (Page 177) [CBSE 2016, 2020 Standard]
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are $15\text{ cm}$ by $10\text{ cm}$ by $3.5\text{ cm}$. The radius of each of the depressions is $0.5\text{ cm}$ and the depth is $1.4\text{ cm}$. Find the volume of wood in the entire stand (see Fig. 12.16). (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Identify Given Data
- Cuboid dimensions: $l = 15\text{ cm}$, $b = 10\text{ cm}$, $h = 3.5\text{ cm}$.
- Four conical depressions: radius $r = 0.5\text{ cm} = \frac{1}{2}\text{ cm}$, depth/height $h_c = 1.4\text{ cm} = \frac{7}{5}\text{ cm}$.
Step 2: Calculate Volume of the Cuboid
$$V_{\text{cuboid}} = l \times b \times h = 15 \times 10 \times 3.5 = 525\text{ cm}^3$$
Step 3: Calculate Volume of Four Conical Depressions
$$V_{\text{depressions}} = 4 \times \left(\frac{1}{3}\pi r^2 h_c\right) = 4 \times \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4$$
$$V_{\text{depressions}} = \frac{88}{21} \times 0.25 \times 1.4 = \frac{88}{21} \times 0.35 = \frac{88 \times 0.05}{3} = \frac{4.4}{3} \approx 1.47\text{ cm}^3$$
Step 4: Compute Volume of Wood Remaining
$$\text{Volume of Wood} = V_{\text{cuboid}} – V_{\text{depressions}} = 525 – 1.47 = 523.53\text{ cm}^3$$
Final Answer:
The volume of wood in the entire stand is $523.53\text{ cm}^3$.
Question 5 (Page 177) [CBSE 2013, 2017, 2020 Standard, 2023]
A vessel is in the form of an inverted cone. Its height is $8\text{ cm}$ and the radius of its top, which is open, is $5\text{ cm}$. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius $0.5\text{ cm}$ are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped into the vessel.
Answer:
Step 1: Identify Given Data
- Cone height, $h = 8\text{ cm}$, radius $R = 5\text{ cm}$.
- Spherical lead shot radius, $r = 0.5\text{ cm} = \frac{1}{2}\text{ cm}$.
- Volume of water displaced = $\frac{1}{4} \times \text{Volume of the Cone}$.
Step 2: Calculate Total Volume of Water in Cone
$$V_{\text{cone}} = \frac{1}{3}\pi R^2 h = \frac{1}{3}\pi (5)^2 (8) = \frac{200\pi}{3}\text{ cm}^3$$
Step 3: Calculate Volume of Water Displaced
$$V_{\text{displaced}} = \frac{1}{4} \times V_{\text{cone}} = \frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3}\text{ cm}^3$$
Step 4: Calculate Volume of One Spherical Lead Shot
$$V_{\text{shot}} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left(\frac{1}{2}\right)^3 = \frac{4}{3}\pi \times \frac{1}{8} = \frac{\pi}{6}\text{ cm}^3$$
Step 5: Determine the Number of Lead Shots ($n$)
By Archimedes’ Principle, the volume of $n$ lead shots equals the volume of water displaced:
$$n \times V_{\text{shot}} = V_{\text{displaced}}$$
$$n \times \frac{\pi}{6} = \frac{50\pi}{3}$$
Divide both sides by $\pi$:
$$n \times \frac{1}{6} = \frac{50}{3} \implies n = \frac{50 \times 6}{3} = 50 \times 2 = 100$$
Final Answer:
The number of lead shots dropped into the vessel is $100$.
Question 6 (Page 177) [CBSE 2012, 2018, 2024]
A solid iron pole consists of a cylinder of height $220\text{ cm}$ and base diameter $24\text{ cm}$, which is surmounted by another cylinder of height $60\text{ cm}$ and radius $8\text{ cm}$. Find the mass of the pole, given that $1\text{ cm}^3$ of iron has approximately $8\text{ g}$ mass. (Use $\pi = 3.14$)
Answer:
Step 1: Identify Dimensions of Both Cylinders
- Lower cylinder: Height $H = 220\text{ cm}$, Diameter $D = 24\text{ cm} \implies$ Radius $R = 12\text{ cm}$.
- Upper cylinder: Height $h = 60\text{ cm}$, Radius $r = 8\text{ cm}$.
- Value of $\pi = 3.14$.
Step 2: Calculate Total Volume of the Pole
$$\text{Total Volume} = V_{\text{lower}} + V_{\text{upper}} = \pi R^2 H + \pi r^2 h = \pi (R^2 H + r^2 h)$$
$$\text{Total Volume} = 3.14 \times [(12)^2 \times 220 + (8)^2 \times 60] = 3.14 \times [144 \times 220 + 64 \times 60]$$
$$\text{Total Volume} = 3.14 \times [31680 + 3840] = 3.14 \times 35520 = 111532.8\text{ cm}^3$$
Step 3: Calculate Mass of the Pole
$$\text{Mass} = \text{Volume} \times \text{Density} = 111532.8 \times 8 = 892262.4\text{ g}$$
Convert grams to kilograms ($1\text{ kg} = 1000\text{ g}$):
$$\text{Mass} = \frac{892262.4}{1000} \approx 892.26\text{ kg}$$
Final Answer:
The mass of the iron pole is approximately $892.26\text{ kg}$ (or $892262.4\text{ g}$).
Question 7 (Page 177) [CBSE 2015, 2019 Set-1, 2023]
A solid consisting of a right circular cone of height $120\text{ cm}$ and radius $60\text{ cm}$ standing on a hemisphere of radius $60\text{ cm}$ is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is $60\text{ cm}$ and its height is $180\text{ cm}$. (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Identify Given Data
- Common radius for all shapes, $r = 60\text{ cm}$.
- Cone: Height $h = 120\text{ cm}$.
- Hemisphere: Radius $r = 60\text{ cm}$.
- Cylinder: Height $H = 180\text{ cm}$, Radius $r = 60\text{ cm}$.
- Notice that $H = h + r = 120 + 60 = 180\text{ cm}$, so the solid fits the cylinder’s height completely.
Step 2: Calculate Volume of the Immersed Solid
$$V_{\text{solid}} = \text{Volume of Cone} + \text{Volume of Hemisphere} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2(h + 2r)$$
$$V_{\text{solid}} = \frac{1}{3}\pi (60)^2 [120 + 2(60)] = \frac{1}{3}\pi (3600)[240] = 288000\pi\text{ cm}^3$$
Step 3: Calculate Volume of the Cylinder
$$V_{\text{cylinder}} = \pi r^2 H = \pi (60)^2 (180) = \pi (3600)(180) = 648000\pi\text{ cm}^3$$
Step 4: Compute Volume of Water Left
$$\text{Volume Left} = V_{\text{cylinder}} – V_{\text{solid}} = 648000\pi – 288000\pi = 360000\pi\text{ cm}^3$$
Substitute $\pi = \frac{22}{7}$:
$$\text{Volume Left} = 360000 \times \frac{22}{7} = \frac{7920000}{7} \approx 1131428.57\text{ cm}^3$$
Convert to cubic metres ($1\text{ m}^3 = 1000000\text{ cm}^3 = 10^6\text{ cm}^3$):
$$\text{Volume Left} = \frac{1131428.57}{10^6} \approx 1.131\text{ m}^3$$
Final Answer:
The volume of water left in the cylinder is approximately $1.131\text{ m}^3$ (or $\frac{7920000}{7}\text{ cm}^3$).
Question 8 (Page 178) [CBSE 2014, 2018, 2020 Standard]
A spherical glass vessel has a cylindrical neck $8\text{ cm}$ long, $2\text{ cm}$ in diameter; the diameter of the spherical part is $8.5\text{ cm}$. By measuring the amount of water it holds, a child finds its volume to be $345\text{ cm}^3$. Check whether she is correct, taking the above as the inside measurements, and $\pi = 3.14$.
Answer:
Step 1: Identify Given Data
- Cylindrical neck: Height $h = 8\text{ cm}$, Diameter $d = 2\text{ cm} \implies$ Radius $r_1 = 1\text{ cm}$.
- Spherical body: Diameter $D = 8.5\text{ cm} \implies$ Radius $r_2 = \frac{8.5}{2} = 4.25\text{ cm} = \frac{17}{4}\text{ cm}$.
- Value of $\pi = 3.14$.
- Child’s measured volume = $345\text{ cm}^3$.
Step 2: Calculate Volume of the Cylindrical Neck
$$V_{\text{cylinder}} = \pi r_1^2 h = 3.14 \times (1)^2 \times 8 = 25.12\text{ cm}^3$$
Step 3: Calculate Volume of the Spherical Body
$$V_{\text{sphere}} = \frac{4}{3}\pi r_2^3 = \frac{4}{3} \times 3.14 \times (4.25)^3 = \frac{4}{3} \times 3.14 \times 76.765625$$
$$V_{\text{sphere}} = \frac{964.17625}{3} \approx 321.392\text{ cm}^3$$
Step 4: Compute Total Volume of the Vessel
$$\text{Total Volume} = V_{\text{cylinder}} + V_{\text{sphere}} = 25.12 + 321.392 = 346.512\text{ cm}^3 \approx 346.51\text{ cm}^3$$
Step 5: Verify the Child’s Finding
The child found the volume to be $345\text{ cm}^3$. However, the actual calculated internal capacity is $346.51\text{ cm}^3$.
$$346.51\text{ cm}^3 \neq 345\text{ cm}^3$$
Final Answer:
No, the child is not correct. The true volume of the vessel is $346.51\text{ cm}^3$.
[👉 Also Read: Class 10 Math Chapter 14 Probability NCERT Solutions]
Master High-Yield Board FAQs (Rank Math Schema Ready)
Why is the total surface area of a combined solid not equal to the sum of the total surface areas of its parts?
When two or more solids are joined together to create a composite solid, the surfaces along which they make contact are merged internally. These joining faces are no longer exposed on the outside of the shape. Therefore, the total surface area includes only the outer exposed curved and flat surfaces, and adding individual total surface areas double-counts the hidden contact interfaces.
What is the difference between total surface area and curved surface area?
Curved Surface Area (CSA) or Lateral Surface Area (LSA) measures only the area of the curved or side faces of a solid, excluding its top and bottom flat bases. Total Surface Area (TSA) measures the complete boundary of the solid, combining the curved or lateral area with all flat circular or polygonal base areas.
How do you find the slant height of a cone when radius and vertical height are given?
The vertical height $h$, base radius $r$, and slant height $l$ of a right circular cone form a right-angled triangle. By applying the Pythagoras Theorem, the slant height is calculated as $l = \sqrt{r^2 + h^2}$.
What happens to the volume and surface area when a cavity is hollowed out of a solid?
When a cavity (such as a cone or hemisphere) is scooped out of a solid, the total volume always decreases by the volume of the carved-out portion ($\text{Volume}{\text{remaining}} = \text{Volume}{\text{original}} – \text{Volume}_{\text{cavity}}$). However, the total surface area typically increases because the newly formed internal walls create extra exposed surface area.
When should students substitute 3.14 instead of 22/7 for pi?
Students must use $\pi = 3.14$ only when the examination question paper explicitly states “Use $\pi = 3.14$”. Under CBSE marking schemes, if no value is mentioned, students are required to use $\pi = \frac{22}{7}$.
How do you calculate the mass of a solid from its volume?
The mass of an object is obtained by multiplying its calculated total volume by its material density ($\text{Mass} = \text{Volume} \times \text{Density}$). Ensure units match before multiplying; for example, volume in $\text{cm}^3$ multiplied by density in $\text{g/cm}^3$ gives mass in grams, which can then be divided by $1000$ to express the final answer in kilograms.
How many cubic centimetres are there in one cubic metre?
One cubic metre equals one million cubic centimetres ($1\text{ m}^3 = 100\text{ cm} \times 100\text{ cm} \times 100\text{ cm} = 1000000\text{ cm}^3 = 10^6\text{ cm}^3$). To convert cubic centimetres to cubic metres, divide the volume by $10^6$.
How many litres are in one cubic metre and one cubic centimetre?
One cubic metre contains $1000\text{ litres}$ ($1\text{ m}^3 = 1000\text{ L}$). One litre contains $1000\text{ cm}^3$, which means that $1\text{ cm}^3 = \frac{1}{1000}\text{ L} = 0.001\text{ L} = 1\text{ mL}$.
Why is Archimedes Principle used in solid geometry problems?
Archimedes’ Principle states that an object submerged in a fluid displaces an amount of fluid equal to its own submerged volume. In math problems like Exercise 12.2 Question 5, dropping spherical shots into a full container causes water to overflow equal to the total volume of all dropped spheres ($n \times V_{\text{sphere}} = V_{\text{overflow}}$).
What is the greatest diameter a hemisphere can have when mounted on a cube of edge a?
The circular base of the hemisphere cannot extend beyond the flat edges of the square face of the cube. Therefore, the greatest diameter the hemisphere can possess is equal to the edge length of the cube ($d = a$), meaning its maximum radius is $r = \frac{a}{2}$.
What common mistake do students make when solving the medicine capsule problem?
A common mistake in the capsule problem (Exercise 12.1 Question 6) is using the total length of the capsule as the height of the cylindrical section. The total length of the capsule includes the two hemispherical ends. To find the cylinder’s height, subtract twice the radius from the total length ($h = \text{Total Length} – 2r$).
How do you calculate the volume of wood remaining in a pen stand with conical depressions?
The volume of wood remaining in a pen stand is found by calculating the total volume of the solid rectangular cuboid ($V = l \times b \times h$) and subtracting the combined volume of all the conical drilled holes ($4 \times \frac{1}{3}\pi r^2 h_c$).
What units must be used for surface areas and volumes in board answers?
Surface areas represent two-dimensional spaces and must always be written in square units, such as $\text{cm}^2$, $\text{m}^2$, or $\text{mm}^2$. Volumes measure three-dimensional space and must be written in cubic units, such as $\text{cm}^3$, $\text{m}^3$, $\text{mm}^3$, or liquid volume units like $\text{L}$ and $\text{mL}$. Omitting units results in a half-mark deduction in CBSE examinations.
How does the volume of a sphere relate to the volume of a cylinder with the same radius and height?
A cylinder circumscribing a sphere has radius $r$ and height $h = 2r$. The volume of the cylinder is $V_{\text{cylinder}} = \pi r^2(2r) = 2\pi r^3$. The volume of the sphere is $V_{\text{sphere}} = \frac{4}{3}\pi r^3$. Dividing the two yields $\frac{V_{\text{sphere}}}{V_{\text{cylinder}}} = \frac{4/3}{2} = \frac{2}{3}$, showing that the sphere occupies exactly two-thirds of the cylinder’s volume.
What are the key presentation steps to score 100% on combination of solids questions?
To earn full marks on 5-mark combination of solids questions: (1) state the given parameters with uniform physical units, (2) write down the individual geometric formulas algebraically before plugging in numbers, (3) factor out common terms like $\pi$ or $r^2$ to simplify arithmetic and avoid repeated rounding, and (4) conclude with a final statement highlighting the numerical value and correct unit.