Last Updated on October 26, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 9 maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 9 maths. In this article, you will find case study questions for CBSE Class 9 Maths Chapter 11 Surface Areas and Volumes. It is a part of Case Study Questions for CBSE Class 9 Maths Series.

Chapter | Surface Areas and Volumes |

Type of Questions | Case Study Questions |

Nature of Questions | Competency Based Questions |

Board | CBSE |

Class | 9 |

Subject | Maths |

Useful for | Class 9 Studying Students |

Answers provided | Yes |

Difficulty level | Mentioned |

Important Link | Class 9 Maths Chapterwise Case Study |

### Case Study Questions on Surface Areas and Volumes

**Case Study Questions**

**Question 1:**

For decoration purpose, Sneha bought 100 orbeez balls and put it in a cylindrical shaped box. After filling it with water, the orbeez ball swell up and completely filled the cylindrical shaped box. Behind the orbeez ball packet, the change in volume of each orbeez ball was mentioned and which was 32% increase. Suppose the volume of all orbeez ball is 9900 cm^{3}.

On the basis of the above information, solve the following questions:

Q 1. Find the volume of each orbeez ball.

Q 2. What is the volume of orbeez ball before swelling?

Q 3. How many orbeez balls before swelling was needed to completely fill the cylindrical shaped box?

Q 4. What is the cubic radius of a orbeez ball before swelling up?

Q 5. If the change in volume of orbeez ball is increased to 48%, then find the volume of orbeez ball after swell up.

**Solutions:**

1. Volume of one orbeez ball $\times 100=$ Volume of all orbeez ball

$\therefore$ Volume of one orbeez ball $=\frac{9900}{100}=99 \mathrm{~cm}^3$

2. Let original volume of orbeez ball before swelling be $x$.

Now, change in volume of orbeez ball

$$

=\frac{99-x}{x} \times 100

$$

Given, $\quad 32=\frac{(99-x) \times 100}{x}$

$$

\begin{array}{rlrl}

\Rightarrow & 32 x & =9900-100 x \\

\Rightarrow & 132 x & =9900 \\

\Rightarrow & x & =\frac{9900}{132} \\

& \therefore & x & =75 \mathrm{~cm}^3

\end{array}

$$

3. $n \times$ Volume of one orbeez ball before swell up

$$

\begin{array}{lr}

\Rightarrow & n \times 75=9900 \\

\Rightarrow & n=\frac{9900}{75} \\

\therefore & n=132 \text { balls }

\end{array}

$$

$$

=\text { Volume of container }

$$

4. Volume of a orbeez ball before swell up $=75 \mathrm{~cm}^3$

$\Rightarrow \quad \frac{4}{3} \pi r^3=75$

$\Rightarrow \quad \frac{4}{3} \times \frac{22}{7} \times r^3=75$

$\Rightarrow \quad r^3=\frac{75 \times 3 \times 7}{4 \times 22}$

$\therefore \quad r^3=17.89 \mathrm{~cm}^3$

5. Let the volume of orbeez ball after swell up be $x$.

Then, $\quad 48=\frac{x-75}{75} \times 100$

$$

\begin{aligned}

& \Rightarrow \quad 48=\frac{x-75}{3} \times 4 \\

& \Rightarrow \quad 48 \times 3=4 \mathrm{x}-300 \\

& \Rightarrow \quad 4 x=144+300 \Rightarrow 4 x=444 \mathrm{~cm}^3 \\

& \therefore \quad x=111 \mathrm{~cm}^3

\end{aligned}

$$

### Understanding Surface Areas and Volumes

**Solid Figures: **Three-dimensional figures having definite shape, size and occupying a fixed amount of space in three dimensions. e.g., cuboid, cylinder, cone, sphere etc.

**Surface Area: **The sum of areas of all the faces of a solid figure. Its unit is unit2 or square unit.

**Volume: **The space occupied by a solid. Its unit is unit3 or cubic unit.

**Right Circular Cone: **A solid generated by revolving a line segment that passes through a fixed point and makes a constant angle with a fixed

line. It has one curved face and one plane circular face. e.g., funnel, jokerâ€˜s cap, etc.

1. Volume $=\frac{1}{3} \pi r^2 h$

2. Curved Surface Area (CSA)

$$

\begin{aligned}

& =\frac{1}{2} \times l \times \text { Circumference of base } \\

& =\frac{1}{2} \times l \times 2 \pi r=\pi r l

\end{aligned}

$$

3. Total Surface Area (TSA) $=$ CSA + Area of the base

$$

=\pi r l+\pi r^2=\pi r(l+r)

$$

4. Slant height $(l)=\sqrt{h^2+r^2}$

**Sphere: **A solid figure made up of all points in space which lie at a constant distance called radius from a fixed point called the centre of the sphere. Let â€˜râ€™ be the radius of sphere.

1. Volume $=\frac{4}{3} \pi r^3$

2. Curved Surface Area (CSA) $=$ Total Surface Area $(\mathrm{TSA})=4 \pi r^2$

Hemisphere: A plane through the centre of a solid sphere cuts it into two equal parts, each part is called a hemisphere. Let â€˜râ€™ be the radius

of hemisphere.

1. Volume $=\frac{2}{3} \pi r^3$

2. Curved Surface Area $(C S A)=2 \pi r^2$

3. Total Surface Area $($ TSA $)=$ CSA + Area of the base $=2 \pi r^2+\pi r^2=3 \pi r^2$

Hemispherical Shell: Let â€˜Râ€™ be the external radius and â€˜râ€™ be the internal radius.

1. Area of base $=\pi\left(R^2-r^2\right)$

2. External CSA $=2 \pi R^2$

3. Internal CSA $=2 \pi r^2$

4. Volume of material $=\frac{2}{3} \pi\left(R^3-r^3\right)$

5. Thickness $=R-r$

**Boost your knowledge**

- The total surface area of any object is greater than its lateral surface area.
- When an object of certain volume is recast into a new shape, the volume of new shape formed will always be equal to the volume of the original object.
- The solids having the same curved surface do not necessarily occupy the same volume.
- When an object is dropped into a liquid, the volume of the displaced liquid is equal to the volume of the object that is dipped.
- The volume of the material in a hollow body is equal to the difference between the external volume and internal volume.
- To find the cost of polishing/covering/painting of a solid, we first find its surface area and then multiply by its per unit cost.
- To find the quantity of a substance contained in a solid, we find its volume.
- Volume of water accumulated on a roof after rain = Surface area of roof x Height of Rainfall.
- Volume of water released by a pipe in unit time = Rate of flow x Area of cross-section of pipe.

### Also check

- Statistics Class 9 Case Study Questions Maths Chapter 12
- Surface Areas and Volumes Class 9 Case Study Questions Maths Chapter 11
- Heron’s Formula Class 9 Case Study Questions Maths Chapter 10
- Circles Class 9 Case Study Questions Maths Chapter 9
- Quadrilaterals Class 9 Case Study Questions Maths Chapter 8
- Triangles Class 9 Case Study Questions Maths Chapter 7
- Lines and Angles Class 9 Case Study Questions Maths Chapter 6
- Introduction to Euclidâ€™s Geometry Class 9 Case Study Questions Maths Chapter 5
- Linear Equations in Two Variables Class 9 Case Study Questions Maths Chapter 4
- Coordinate Geometry Class 9 Case Study Questions Maths Chapter 3
- PolynomialsÂ Class 9 Case Study Questions Maths Chapter 2
- Number SystemsÂ Class 9 Case Study Questions Maths Chapter 1

### Topics from which case study questions may be asked

- Surface Area of Sphere
- Surface Area of Hemisphere
- Volume of Sphere and Hemisphere
- Surface Area of Cone
- Volume of Cone
- Surface Area and Volume of Hemispherical Shell

To find the quantity of a substance contained in a solid, we find its volume.

Case study questions from the above given topic may be asked.

### Helpful Links for CBSE Class 9 Preparation

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### Frequently Asked Questions (FAQs) on Surface Areas and Volumes Case Study

#### Q1: What are the main topics covered in the chapter on Surface Areas and Volumes?

A1: This chapter covers the surface areas and volumes of three-dimensional shapes like cones, spheres, and hemispheres. The chapter provides formulas and methods for calculating the surface area and volume of each of these solids.

#### Q2: What is the difference between total surface area and lateral surface area?

A2: The *total surface area* of a solid includes all the surfaces of the shape, while the *lateral surface area* includes only the side surfaces, excluding the top and bottom (bases). For example, in a cylinder, the lateral surface area includes only the curved surface, not the circular bases.

#### Q3: **What are the common units used in surface area and volume calculations?**

A3: Surface area is typically measured in square units (e.g., cmÂ², mÂ²), while volume is measured in cubic units (e.g., cmÂ³, mÂ³). Itâ€™s essential to use the correct units based on the shapeâ€™s measurements.

#### Q4: **How do you find the volume of a sphere?**

A4: The volume of a sphere with radius r is calculated using the formula:

Volume $=\frac{4}{3} \pi r^3$

#### Q5: What is the formula for the surface area of a hemisphere, and why is it different from a full sphere?

A5: The curved surface area of a hemisphere (half of a sphere) with radius r is:

CurvedÂ SurfaceÂ Area = 2Ï€r^{2}

The total surface area, which includes the curved surface and the flat circular base, is:

TotalÂ SurfaceÂ Area = 3Ï€r^{2}

It differs from a full sphere because a hemisphere has one flat circular base.

#### Q6: Why are formulas for surface area and volume important in real life?

A6: Understanding surface area and volume is essential in various fields, such as construction, packaging, and manufacturing. For example, calculating the surface area can help determine the amount of material needed to cover a shape, and knowing the volume can help assess the capacity of containers or spaces.

#### Q7: How can I remember the formulas for surface areas and volumes of different shapes?

A7: Practicing the formulas regularly, associating them with real-life examples, and understanding why each formula works (e.g., visualizing the shape and its components) can make it easier to remember. Using flashcards and solving problems also helps reinforce memory.

#### Q8: What are some common mistakes to avoid in surface area and volume calculations?

A8: Common mistakes include:

Using incorrect formulas for specific shapes.

Forgetting to square or cube units correctly.

Mixing up radius and diameter (especially with spheres and cylinders).

Not including all surfaces in total surface area calculations, such as forgetting the circular bases in cylinders.

#### Q9: **Are there any online resources or tools available for practicing surface areas and volumes case study questions?**

A9: We provide case study questions for CBSE Class 9 Maths on our website. Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit Physics Gurukul website. they are having a large collection of case study questions for all classes.