# NCERT Solutions Class 10 Maths Chapter 13

## NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes

Class 10 is a  turning point in the life of every student. Whether a student wants to pursue subjects of his choice or attend the college of his dreams, a lot of it depends on their performance in the Class 10 board exam. Chapter 13 is one of those chapters that often confuses students, and they end up losing marks in the examination. Practising questions related to Chapter 13 regularly can help students have a better understanding of the chapter. Referring to NCERT Solutions for Class 10 Mathematics Chapter 13 by Extramarks will enable students to solve practice questions given in their NCERT books with precision, and master the topic. Solutions are there not just for Mathematics, but for all other subjects as well so that students don’t have to look elsewhere for any assistance.

NCERT Solutions Class 10 Mathematics Chapter 13 are curated by subject matter experts. These have answers to every question given at the end of NCERT Class 10 Chapter 13 textbook.

When it comes to solving questions from NCERT Class 10 Mathematics Chapter 13, students are expected to understand theoretical concepts like what surface area & volume is, along with other practical elements like calculating them. Extramarks’ NCERT Solutions for Class 10 Mathematics Chapter 13 are  as per the latest syllabus of Class 10 CBSE pattern. . Thus, making it an ideal solution  for their CBSE Class 10 Mathematics board exam preparations.

## NCERT Solutions for Class 10 Maths

Mathematics is one of those subjects in Class 10 that helps to raise the overall percentage of students. To increase the chances of scoring full marks in Class 10 Mathematics,  students need not only practise but do so with precision. A good place to begin for Class 10 Mathematics would be with NCERT Solutions  by Extramarks. Every chapter-wise solution comes with solved answers to textbook questions and in-depth explanations of the same. All the chapters are listed below: :

•       Chapter 1 – Real Numbers
•       Chapter 2 – Polynomials
•       Chapter 3 – Pair of Linear Equations in Two Variables
•       Chapter 4 – Quadratic Equations
•       Chapter 5 – Arithmetic Progressions
•       Chapter 6 – Triangles
•       Chapter 7 – Coordinate Geometry
•       Chapter 8 – Introduction to Trigonometry
•       Chapter 9 – Some Applications of Trigonometry
•       Chapter 10 – Circles
•       Chapter 11 – Constructions
•       Chapter 12 – Areas Related to Circles
•       Chapter 13 – Surface Areas and Volumes
•       Chapter 14 – Statistics
•       Chapter 15 – Probability

### Chapter 13 – Surface Area and Volumes

Chapter 13 – Surface Area and Volumes of Class 10 Mathematics is a lesson loaded with formulas. It teaches the students how to calculate areas and volumes of different shapes. The chapter comprises 5 exercises 13.1 – 13.5. NCERT Solutions for Class 10 Mathematics Chapter 13 has solved answers to every question in exercise 13.1 to 13.5.

#### 13.1 Introduction

Just like any other introduction from any other chapter, in Chapter 13 of Class 10 Mathematics, you would be required to recall  what you studied in Class 9 about cubes, cylinders, circles, etc. This is why at Extramarks we never recommend memorising anything. We always suggest understanding a particular concept.  The sole reason is that, by memorising things,you are not able to retain them in your mind for a long time. But if you understand a concept, chances are you will remember it forever. That’s the cardinal rule in Mathematics.

#### 13.2 Surface Area of a Combination of Solids

Some items are a combination of two or more solid shapes. At the end of this chapter, students will also learn how to calculate the surface area and volumes of a combination of solids using formulas.

#### 13.3 Volume of a Combination of Solids

For calculating the volume of shapes that are a combination of two or more solid shapes, students must find out the volume area of individual shapes first.

#### 13.4 Conversion of a Solid From One Form to Another

 Shape Formula Total surface area of sphere = curved surface area of sphere 4 π r2 Total surface area of cone πr(r+l) Curved surface area of cone πrl Total surface area of cuboid 2(lb+bh+hl) Total surface area of cylinder 2 πr(h+r)

Volumes of 3D objects:

 Shape Formula Volume of sphere 4/3 πr³ Volume of hemisphere 2πr3/3 Volume of cone (⅓)πr2h Volume of cube s3 Volume of cuboid lbh Volume of cylinder πr2h

Even if a solid is converted from one form to another, then the volume remains the same. Let us understand this with an example:

If there is water in the cuboid shape (which has dimensions of 20 * 22) which is transferred into a cylinder that has a height of 3.5 m and 2 m then what is the height of the water level in the cuboid shape if the water (once transferred into the cylinder) fills the cylinder to the brim?

Solution – From the theorem on volumes, we know that the volume of the water in the cylinder and the volume of the water in the cuboid would be the same. If the water fills the cylinder to the brim, then:

Volume of water in cylinder = π * r² * h = 22/7 * 1 * 3.5 = 11.

Volume of water in cuboid = l * b * h = 20 * 22 * h.

Since 20 * 22 * h = 3.5 * π, we get h = 11/(20 * 22) = 2.5 cm.

## Related Questions:

Q1. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Solution: It is known that the coins are cylindrical in shape.

So, height (h1) of the cylinder = 2 mm = 0.2 cm

Radius (r) of circular end of coins = 1.75/2 = 0.875 cm

Now, the number of coins to be melted to form the required cuboids be “n”

So, Volume of n coins = Volume of cuboids

n × π × r2 × h1 = l × b × h

n×π×(0.875)2×0.2 = 5.5×10×3.5

Or, n = 400

Q2. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Solution: Radius (r1) of the upper base = 4/2 = 2 cm

Radius (r2) of lower the base = 2/2 = 1 cm

Height = 14 cm

Now, Capacity of glass = Volume of frustum of cone

So, Capacity of glass = (⅓)×π×h(r12+r22+r1r2)

= (⅓)×π×(14)(22+12+ (2)(1))

∴ The capacity of the glass = 102×(⅔) cm3

### Key Features of NCERT Solutions for Class 9 Maths Chapter 13

• NCERT Solutions for Class 10 Mathematics Chapter 13 is a study material to help Class 10 students solve the exercise questions given in NCERT Chapter 13 with accuracy.
• NCERT Solutions Class 10 Mathematics Chapter 13 is prepared by experienced faculty as per the latest CBSE Class 10 syllabus.
• A student doesn’t require any aid from teachers or parents to understand the answers given NCERT Solutions Class 10 Mathematics Chapter 13. The language used in it is uncomplicated and designed in a way that everything becomes self-explanatory.
• All numerical problems are solved step-wise with appropriate diagrams and explanations as and when required.

Q.1 Two cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Ans. $\begin{array}{l}\text{Given that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}volume of each cube}={\text{64 cm}}^{3}\\ \text{or}{\left(\text{edge}\right)}^{3}=\text{64}\\ \text{or edge\hspace{0.17em}of each cube}=4\text{cm}\\ \text{The two cubes are joined end to end to form a cuboid of}\\ \text{dimensions 4 cm, 4 cm and 8 cm.}\\ \therefore \text{surface area of the cuboid}=2\left(\mathrm{lb}+\mathrm{bh}+\mathrm{lh}\right)\\ \text{}=2\left(4×4+4×8+4×8\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=160{\text{cm}}^{2}\end{array}$

Q.2 A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Ans. $\begin{array}{l}\text{Height of cylindrical part}=\mathrm{h}=13-7=6\text{cm}\\ \text{Inner surface area of the vessel}\\ \text{}=\mathrm{Curved}\text{}\mathrm{surface}\text{}\mathrm{area}\text{}\mathrm{of}\text{cylindrical part}\\ \text{}-\mathrm{Curved}\text{surface area of hemispherical part}\\ \text{}=2\mathrm{\pi rh}+2{\mathrm{\pi r}}^{2}=2×\frac{22}{7}×7\left(6+7\right)=572{\text{cm}}^{2}\end{array}$

Q.3 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 of the toy.

Ans. $\begin{array}{l}\text{Radius of cone}=\text{Radius of hemisphere}=\mathrm{r}=3.5\text{cm}\\ \text{Height of hemisphere}=\text{Radius of hemisphere}=3.5\text{cm}\\ \text{Height of cone}=\mathrm{h}=15.5-3.5=12\\ \text{Slant height of cone}=\mathrm{l}=\sqrt{{\mathrm{h}}^{2}+{\mathrm{r}}^{2}}=\sqrt{{12}^{2}+{\left(3.5\right)}^{2}}=\frac{25}{2}\text{cm}\\ \text{Total surface area of toy}=\text{CSA of conical part}\\ \text{}+\text{CSA of hemispherical part}\\ \text{}=\mathrm{\pi rl}+2{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{22}{7}×3.5×\frac{25}{2}+2×\frac{22}{7}×\text{\hspace{0.17em}}{\left(3.5\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=214.5{\text{\hspace{0.17em}cm}}^{2}\end{array}$

Q.4 A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Ans. $\begin{array}{l}\text{From figure it is obvious that greatest diameter of the}\\ \text{hemisphere is 7 cm.}\\ \therefore \text{Radius of hemisphere}=\mathrm{r}=\frac{7}{2}\text{cm}\\ \text{Total surface area}=\text{Surface area of cubical part}\\ \text{}+\text{CSA of hemispherical part}\\ \text{}-\text{area of base of hemispherical part}\\ \text{}=6×{\left(7\right)}^{2}+2{\mathrm{\pi r}}^{2}-{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6×{\left(7\right)}^{2}\text{\hspace{0.17em}\hspace{0.17em}}+\frac{22}{7}×{\left(\frac{7}{2}\right)}^{2}\\ \text{}=332.5{\text{cm}}^{2}\end{array}$

Q.5 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.

Ans. $\begin{array}{l}\text{Diameter of hemisphere}=\text{edge of cube}=\mathrm{l}\\ \therefore \text{Radius of hemisphere}=\frac{\mathrm{l}}{2}\text{}\\ \text{Total surface area of solid}\\ \text{}=\text{Surface area of cubical part}\\ \text{}+\text{CSA of hemispherical part}\\ \text{}-\text{area of base of hemispherical part}\\ \text{}=6×{\mathrm{l}}^{2}+2{\mathrm{\pi r}}^{2}-{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6{\mathrm{l}}^{2}+\mathrm{\pi }{\frac{\mathrm{l}}{4}}^{2}\text{}\\ \text{}=\frac{1}{4}{\mathrm{l}}^{2}\left(24+\mathrm{\pi }\right){\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}unit}}^{2}\end{array}$

Q.6 A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see the following figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. Ans. $\begin{array}{l}\text{Surface area of capsule}\\ \text{}=2×\text{CSA of hemispherical part}\\ \text{}+\text{CSA of cylindrical part}\\ \text{}=4\mathrm{\pi }{\left(\frac{5}{2}\right)}^{2}+2\mathrm{\pi }\frac{5}{2}×9\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=220{\text{mm}}^{2}\end{array}$

Q.7 From a solid cylinder whose height is 2.4 cm and diameter 1.4 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 cm2.

Ans. $\begin{array}{l}\text{Slant height of conical part}=\mathrm{l}=\sqrt{{\mathrm{r}}^{2}+{\mathrm{h}}^{2}}=\sqrt{{\left(0.7\right)}^{2}+{\left(2.4\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2.5\text{cm}\\ \text{Required surface area}\\ \text{}=\text{curved surface area of cylindrical portion}\\ \text{}+\text{curved surface area of conical portion}\\ \text{}+\text{\hspace{0.17em}\hspace{0.17em}area of cylindrical base}\\ \text{}=\text{2}\mathrm{\pi }\text{rh}+\mathrm{\pi rl}+\mathrm{\pi }{\text{r}}^{2}\\ \text{}=2×\frac{22}{7}×0.7×2.4+\frac{22}{7}×0.7×2.5+\frac{22}{7}×{\left(0.7\right)}^{2}\\ \text{}=17.60{\text{cm}}^{2}\approx 18{\text{cm}}^{2}\end{array}$

Q.8 A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the following figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article. Ans.

$\begin{array}{l}\text{Surface area of the given article}\\ \text{}=\text{curved surface area of cylindrical portion}\\ \text{}+2×\text{curved surface area of hemispherical portion}\\ \text{}=\text{2}\mathrm{\pi }\text{rh}+2×2{\mathrm{\pi r}}^{2}\\ \text{}=2×\frac{22}{7}×3.5×10+2×2×\frac{22}{7}×{\left(3.5\right)}^{2}\\ \text{}=374{\text{cm}}^{2}\end{array}$

Q.9 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 π.

Ans.

$\begin{array}{l}\text{Volume of the given solid}=\text{Volume of cone}\\ \text{}+\text{Volume of hemisphere}\\ \text{}=\frac{1}{3}×{\mathrm{\pi r}}^{2}\mathrm{h}+\frac{2}{3}{\mathrm{\pi r}}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}×\mathrm{\pi }{\left(1\right)}^{2}\left(1\right)+\frac{2}{3}\mathrm{\pi }{\left(1\right)}^{3}=\mathrm{\pi }{\text{cm}}^{3}\text{}\end{array}$

Q.10 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 cm and its length is 12 cm. If each cone has a height of 2 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.)

Ans. $\begin{array}{l}\text{Volume of air present in the model}=\text{Volume of cylinder}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+2×\text{Volume of cone}\\ \text{}={\mathrm{\pi r}}^{2}\mathrm{h}+2×\frac{1}{3}×{\mathrm{\pi r}}^{2}\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{\pi }{\left(\frac{3}{2}\right)}^{2}\left(8\right)+\frac{2}{3}\mathrm{\pi }{\left(\frac{3}{2}\right)}^{2}\left(2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=66{\text{cm}}^{2}\text{}\end{array}$

Q.11 A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see the following figure) Ans. $\begin{array}{l}\text{Volume of one gulab jamun}=\text{Volume of cylinderical part}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+2×\text{Volume of hemispherical part}\\ \text{}={\mathrm{\pi r}}^{2}\mathrm{h}+2×\frac{2}{3}×{\mathrm{\pi r}}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{\pi }{\left(1.4\right)}^{2}\left(2.2\right)+\frac{4}{3}\mathrm{\pi }{\left(1.4\right)}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25.05{\text{cm}}^{3}\\ \text{Volume of 45 gulab jamuns}=45×25.05{\text{cm}}^{3}=1,127.25{\text{cm}}^{3}\\ \text{Volume of sugar syrup}=30\mathrm{%}\text{of the volume of 45 gulab jamuns}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{30}{100}×1,127.25\approx 338{\text{cm}}^{3}\text{}\end{array}$

Q.12 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 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see the following figure). Ans.

$\begin{array}{l}\text{Volume of wood}=\text{Volume of cuboid}\\ \text{}-4×\text{Volume cones}\\ \text{}=\mathrm{lbh}-4×\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=15×10×3.5-\frac{4}{3}\mathrm{\pi }{\left(0.5\right)}^{2}×1.4\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=523.53{\text{cm}}^{3}\text{}\end{array}$

Q.13 A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Ans.

$\begin{array}{l}\text{Let n lead shots were dropped in the vessel.}\\ \text{Volume of water spilled}=\text{Volume of dropped lead shots}\\ \text{or}\frac{1}{4}×\text{Volume cone}=\mathrm{n}×\frac{4}{3}{{\mathrm{\pi r}}_{2}}^{3}\\ \text{or}\frac{1}{4}×\frac{1}{3}{{\mathrm{\pi r}}_{1}}^{2}\mathrm{h}=\mathrm{n}×\frac{4}{3}{{\mathrm{\pi r}}_{2}}^{3}\\ \text{\hspace{0.17em}or}{5}^{2}×8=\mathrm{n}×16{\left(0.5\right)}^{3}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}or}\mathrm{n}=100\text{\hspace{0.17em}}\end{array}$

Q.14 A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use π = 3.14)

Ans.

$\begin{array}{l}\text{Volume of the iron pole}\\ =\text{Volume of a cylinder of height 220 cm and radius 12 cm}\\ \text{}+\text{Volume of a cylinder of height 60 cm and radius 8 cm}\\ =\mathrm{\pi }{\left(12\right)}^{2}×220+\mathrm{\pi }{\left(8\right)}^{2}×60\\ =1,11,532.8{\text{cm}}^{3}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Mass of 1 cm}}^{3}\text{iron}=8\text{gm}\\ \therefore \text{Mass of}1,11,532.8{\text{cm}}^{3}\text{iron}=8×1,11,532.8\text{gm}=892.262\text{kg}\end{array}$

Q.15 A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 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 cm and its height is 180 cm.

Ans. $\begin{array}{l}\text{Volume of water left}\\ =\text{Volume of a cylinder}-\text{Volume of a soloid}\\ =\text{Volume of a cylinder}-\text{Volume of cone}-\text{Volume of hemisphere}\\ ={\mathrm{\pi r}}^{2}{\mathrm{h}}_{1}-\left(\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}+\frac{2}{3}{\mathrm{\pi r}}^{3}\right)\\ =\mathrm{\pi }{\left(60\right)}^{2}×180-\left[\frac{1}{3}\mathrm{\pi }{\left(60\right)}^{2}×120+\frac{2}{3}\mathrm{\pi }{\left(60\right)}^{3}\right]\\ =\mathrm{\pi }{\left(60\right)}^{2}×180-\frac{1}{3}\mathrm{\pi }{\left(60\right)}^{2}×120-\frac{2}{3}\mathrm{\pi }{\left(60\right)}^{3}\\ =\frac{1}{3}\mathrm{\pi }{\left(60\right)}^{2}\left(540-120-120\right)\\ =\frac{1}{3}×\frac{22}{7}×3600×300\\ =\frac{22}{7}×360000{\text{cm}}^{3}\\ =\frac{22}{7}×0.36{\text{m}}^{3}\\ =1.131{\text{m}}^{3}\text{(approx.)}\end{array}$

Q.16 A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Ans. $\begin{array}{l}\text{Volume of vessel}\\ =\text{Volume of sphere}+\text{Volume of cylinder}\\ =\frac{4}{3}{{\mathrm{\pi r}}_{1}}^{3}+{{\mathrm{\pi r}}_{2}}^{2}\mathrm{h}\\ =\frac{4}{3}\mathrm{\pi }{\left(4.25\right)}^{3}+\mathrm{\pi }{\left(1\right)}^{2}×8\\ =346.51{\text{cm}}^{3}\text{(approx.)}\\ \ne 345\\ \text{Hence the child is not correct.\hspace{0.17em}\hspace{0.17em}}\end{array}$

Q.17 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.

Ans.

$\begin{array}{l}\text{Radius of sphere}={\mathrm{r}}_{1}=4.2\text{cm}\\ \text{Radius of cylinder}={\mathrm{r}}_{2}=6\text{cm}\\ \text{Height of cylinder}=\mathrm{h}\\ \text{Recasted cylinder and sphere will be of same volume.}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of cylinder}=\text{Volume of sphere}\\ \text{or}{{\mathrm{\pi r}}_{2}}^{2}\mathrm{h}=\frac{4}{3}{{\mathrm{\pi r}}_{1}}^{3}\\ \text{or}{6}^{2}\mathrm{h}=\frac{4}{3}{\left(4.2\right)}^{3}\\ \text{or}\mathrm{h}=2.74\text{cm\hspace{0.17em}}\end{array}$

Q.18 Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Ans.

$\begin{array}{l}\text{Radius of 1st sphere}={\mathrm{r}}_{1}=6\text{cm}\\ \text{Radius of 2nd sphere}={\mathrm{r}}_{2}=8\text{cm}\\ \text{Radius of 3rd sphere}={\mathrm{r}}_{3}=10\text{cm}\\ \text{Let radius of the resulting sphere be}\mathrm{r}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of the resulting sphere}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{Sum of volumes of the given 3 spheres}\\ \text{or}\frac{4}{3}{\mathrm{\pi r}}^{3}=\frac{4}{3}{{\mathrm{\pi r}}_{1}}^{3}+\frac{4}{3}{{\mathrm{\pi r}}_{2}}^{3}+\frac{4}{3}{{\mathrm{\pi r}}_{3}}^{3}\\ \text{or}{\mathrm{r}}^{3}={{\mathrm{r}}_{1}}^{3}+{{\mathrm{r}}_{2}}^{3}+{{\mathrm{r}}_{3}}^{3}={6}^{3}+{8}^{3}+{\left(10\right)}^{3}\\ \text{or}\mathrm{r}=12\text{cm\hspace{0.17em}}\end{array}$

Q.19 A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Ans.

$\begin{array}{l}\text{Depth of the well}=\mathrm{d}=20\text{m}\\ \text{Radius of the well}=\mathrm{r}=\frac{7}{2}\text{m}\\ \text{Length of the platform}=\mathrm{l}=22\text{m}\\ \text{Width of the platform}=\mathrm{b}=14\text{m}\\ \text{Let height of the platform be}\mathrm{h}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of the platform}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{Volume of the well}\\ \text{or}\mathrm{lbh}={\mathrm{\pi r}}^{2}\mathrm{d}\\ \text{or}22×14\mathrm{h}=\mathrm{\pi }{\left(\frac{7}{2}\right)}^{2}×20\\ \text{or}\mathrm{h}=2.5\text{m\hspace{0.17em}}\end{array}$

Q.20 A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Ans.

$\begin{array}{l}\text{Depth of the well}=\mathrm{d}=14\text{m}\\ \text{Radius of the well}=\mathrm{r}=\frac{3}{2}\text{m}=1.5\text{\hspace{0.17em}\hspace{0.17em}m}\\ \text{Width of the circular ring}=\mathrm{b}=4\text{m}\\ \text{Radius of the inner circle of the circular ring}\\ \text{}=\text{Radius of the well}=\mathrm{r}=\frac{3}{2}\text{m}=1.5\text{\hspace{0.17em}\hspace{0.17em}m}\\ \text{Radius of the outer circle of the circular ring}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{r}}_{1}=\text{Radius of the well\hspace{0.17em}}+\text{Width of the circular ring}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{2}+4=5.5\text{m}\\ \text{Shape of the embankment is cylindrical.}\\ \text{Let height of the embankment be}\mathrm{h}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of the embankment}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{Volume of the well}\\ \text{or}{{\mathrm{\pi r}}_{1}}^{2}\mathrm{h}-{\mathrm{\pi r}}^{2}\mathrm{h}={\mathrm{\pi r}}^{2}\mathrm{d}\\ \text{or}{\left(5.5\right)}^{2}-{\left(1.5\right)}^{2}\right]\mathrm{h}={\left(1.5\right)}^{2}×14\\ \text{or}\mathrm{h}=1.125\text{m\hspace{0.17em}}\end{array}$

Q.21 A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Ans.

$\begin{array}{l}\text{Height of cylindrical container}=\mathrm{h}=15\text{cm}\\ \text{Radius of cylindrical container}=\mathrm{r}=\frac{12}{2}\text{cm}=6\text{cm}\\ \text{Height of cone}={\mathrm{h}}_{1}=12\text{cm}\\ \text{Radius of cone}={\mathrm{r}}_{1}=\frac{6}{2}\text{cm}=3\text{cm}\\ \text{Radius of hemisphere}=\text{Radius of cone}={\mathrm{r}}_{1}=3\text{cm}\\ \text{Volume of one ice-cream cone}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{Volume of conical part}+\text{Volume of hemispherical part}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}{{\mathrm{\pi r}}_{1}}^{2}{\mathrm{h}}_{1}+2{{\mathrm{\pi r}}_{1}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\mathrm{\pi }×{3}^{2}×12+2\mathrm{\pi }×{3}^{2}=54\mathrm{\pi }\\ \text{Let the number of cones be}\mathrm{n}\text{.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of cylindrical container}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{n}×\text{Volume of one ice-cream cone}\\ \text{or}{\mathrm{\pi r}}^{2}\mathrm{h}=\mathrm{n}×54\mathrm{\pi }\\ \text{or}{6}^{2}×15=54\mathrm{n}\\ \text{or}\mathrm{n}=10\text{\hspace{0.17em}}\end{array}$

Q.22 How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Ans.

$\begin{array}{l}\text{Thickness of coin}=\mathrm{h}=2\text{mm}=0.2\text{cm}\\ \text{Radius of coin}=\mathrm{r}=\frac{1.75}{2}\text{cm}\\ \text{Cuboid is of the dimensions 5.5 cm × 10 cm × 3.5 cm.}\\ \text{Let number of coines be}\mathrm{n}\text{.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of cuboid}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{n}×\text{Volume of one coin}\\ \text{or 5.5 cm × 10 cm × 3.5 cm}=\mathrm{n}×{\mathrm{\pi r}}^{2}\mathrm{h}=\mathrm{n\pi }{\left(\frac{1.75}{2}\right)}^{2}\left(0.2\right)\\ \text{or}\mathrm{n}=400\end{array}$

Q.23 A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Ans. $\begin{array}{l}\text{Height of cylindrical bucket}={\mathrm{h}}_{1}=32\text{\hspace{0.17em}\hspace{0.17em}cm}\\ \text{Radius of circular end of bucket}={\mathrm{r}}_{1}=\frac{18}{2}\text{cm}=9\text{\hspace{0.17em}\hspace{0.17em}cm}\\ \text{Height of conical heap}={\mathrm{h}}_{2}=24\text{\hspace{0.17em}\hspace{0.17em}cm}\\ \text{Let radius of circular end of conical heap\hspace{0.17em}\hspace{0.17em}be}{\mathrm{r}}_{2}\text{.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of sand in the cylindrical bucket}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{Volume of sand in conical heap}\\ \text{or}{{\mathrm{\pi r}}_{1}}^{2}{\mathrm{h}}_{1}=\frac{1}{3}{{\mathrm{\pi r}}_{2}}^{2}{\mathrm{h}}_{2}\\ \text{or}{\left(\text{18}\right)}^{2}×32=\frac{1}{3}{{\mathrm{r}}_{2}}^{2}×24\\ \text{or}{\mathrm{r}}_{2}=36\text{cm}\\ \text{Slant height}=\sqrt{{24}^{2}+{36}^{2}}=12\sqrt{13}\text{cm}\\ \text{So, radius of circular end of conical heap}=36\text{cm}\\ \text{and slant height of conical heap}=12\sqrt{13}\text{cm}\end{array}$

Q.24 Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Ans.

$\begin{array}{l}\text{Canal is 6 m wide and 1.5 m deep.}\\ \therefore \text{Cross sectional area of the canal}=\text{6}×1.5=9{\text{m}}^{2}\\ \text{Speed of the water}=\text{10 km/h}=\frac{10,000}{60}\text{m/minute}\\ \text{}=\frac{500}{3}\text{\hspace{0.17em}\hspace{0.17em}m/minute}\\ \text{Volume of water flown out of canal in 1 minute}=\frac{500}{3}×9=1500{\text{m}}^{3}\\ \text{Volume of water flown out of canal in 30 minutes}=1500×30\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=45000{\text{m}}^{3}\\ \text{Let the canal irrigate area A with 8 cm of standing water in}\\ \text{30 minutes.}\\ \text{So,}\\ \text{Volume of water flown out of canal in 30 minutes}=\text{A}×{\text{0.08 m}}^{3}\\ \text{or,}45000{\text{m}}^{3}=\text{A}×{\text{0.08 m}}^{3}\\ \text{or, A}=\frac{45000}{\text{0.08}}=562500{\text{m}}^{2}\\ \text{Therefore, area irrigated in 30 minutes is}562500{\text{m}}^{2}.\end{array}$

Q.25 A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Ans.

$\begin{array}{l}\text{Internal diameter of pipe}=\text{20 cm}\\ \therefore \text{Internal radius of pipe}=\frac{\text{20}}{2}\text{cm}=10\text{cm}=0.1\text{m}\\ \therefore \text{Cross sectional area of pipe}=\mathrm{\pi }×{\left(0.1\right)}^{2}=0.01\mathrm{\pi }{\text{m}}^{2}\\ \text{Speed of the water}=\text{3 km/h}=\frac{3000}{60}\text{m/minute}\\ \text{}=50\text{\hspace{0.17em}\hspace{0.17em}m/minute}\\ \text{Volume of water flown out of pipe in 1 minute}=50×0.01\mathrm{\pi }=\frac{\mathrm{\pi }}{2}{\text{m}}^{3}\\ \text{Cylindrical tank is 2 m deep and 10 m in diameter.}\\ \text{Volume of cylindrical tank}={\mathrm{\pi r}}^{2}\mathrm{h}=50\mathrm{\pi }{\text{\hspace{0.17em}\hspace{0.17em}m}}^{3}\\ \text{Let the pipe fills the tank in}\mathrm{t}\text{minutes.}\\ \text{So,}\\ \text{Volume of water flown out of pipe in}\mathrm{t}\text{minutes}\\ \text{}=\text{Volume of cylindrical tank}\\ \text{or,}\frac{\mathrm{\pi }}{2}\mathrm{t}{\text{m}}^{3}=50\mathrm{\pi }{\text{\hspace{0.17em}\hspace{0.17em}m}}^{3}\\ \text{or,}\mathrm{t}=100\text{minutes}\\ \text{Therefore, pipe fills the tank in 100 minutes.}\end{array}$

Q.26 A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Ans. $\begin{array}{l}\text{Radius of upper base of glass}={\mathrm{r}}_{1}=\frac{4}{2}=2\text{\hspace{0.17em}\hspace{0.17em}cm}\\ \text{Radius of lower base of glass}={\mathrm{r}}_{2}=\frac{2}{2}=1\text{\hspace{0.17em}\hspace{0.17em}cm}\\ \mathrm{Capacity}\text{of glass}=\mathrm{Volume}\text{of frustum of cone}\\ \text{}=\frac{1}{3}\mathrm{\pi h}\left({{\mathrm{r}}_{1}}^{2}+{{\mathrm{r}}_{2}}^{2}+{\mathrm{r}}_{1}{\mathrm{r}}_{2}\right)\\ \text{}=\frac{14}{3}\mathrm{\pi }\left({2}^{2}+{1}^{2}+2×1\right)=102\frac{2}{3}{\text{cm}}^{3}\end{array}$

Q.27 The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Ans.

$\begin{array}{l}\text{S}=\mathrm{l}=\text{}\\ \text{P i.e.,}\\ \text{2}\mathrm{\pi }{\text{r}}_{1}=18\text{cm and 2}\mathrm{\pi }{\text{r}}_{2}=6\text{cm}\\ {\text{i.e., r}}_{1}=\frac{9}{\mathrm{\pi }}{\text{cm and r}}_{2}=\frac{3}{\mathrm{\pi }}\text{cm}\\ \text{C}=\mathrm{\pi }\text{}\left({\mathrm{r}}_{1}+{\mathrm{r}}_{2}\right)\mathrm{l}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{\pi }\text{}\left(\frac{9}{\mathrm{\pi }}+\frac{3}{\mathrm{\pi }}\right)×\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=48{\text{cm}}^{2}\end{array}$

Q.28 A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the following figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it. Ans.

$\begin{array}{l}\text{R}\mathbf{adius}\text{}\mathbf{on}\text{}\mathbf{the}\text{}\mathbf{open}\text{}\mathbf{side}={\mathrm{r}}_{1}=\mathbf{10}\text{}\mathbf{cm}\\ \text{R}\mathbf{adius}\text{}\mathbf{at}\text{}\mathbf{the}\text{}\mathbf{upper}\text{}\mathbf{base}={\mathrm{r}}_{2}=\mathbf{4}\text{}\mathbf{cm}\\ \text{S}\mathbf{lant}\text{}\mathbf{height}\text{}\mathbf{of}\text{}\mathbf{frustum}=\mathrm{l}=15\text{}\mathbf{cm}\\ \text{A}\mathbf{rea}\text{}\mathbf{of}\text{}\mathbf{material}\text{}\mathbf{used}\text{}\mathbf{for}\text{}\mathbf{making}\text{fez}\\ \text{}=\text{CSA}\mathbf{of}\text{}\mathbf{frustum}+\text{area of the}\mathbf{upper}\text{}\mathbf{base}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{\pi }\left({\mathrm{r}}_{1}+{\mathrm{r}}_{2}\right)\mathrm{l}+{{\mathrm{\pi r}}_{2}}^{2}=\mathrm{\pi }\left(10+4\right)×15+\mathrm{\pi }{4}^{2}=710\frac{2}{7}{\text{cm}}^{2}\end{array}$

Q.29

$\begin{array}{l}\text{A container, opened from the top and made up of a metal sheet,}\\ \text{is in the form of a frustum of a cone of height 16 cm with radii of its}\\ \text{lower and upper ends as 8 cm and 20 cm, respectively.}\\ \text{Find the cost of the milk which can completely fill the container, at}\\ \text{the rate of â‚¹ 20 per litre. Also find the cost of metal sheet used to}\\ {\text{make the container, if it costs â‚¹\hspace{0.17em}8 per 100 cm}}^{\text{2}}\text{. (Take}\mathrm{\pi }\text{= 3.14)}\end{array}$

Ans.

$\begin{array}{l}\text{Height of frustum=h=16 cm}\\ {\text{Radius of lower end\hspace{0.17em}=r}}_{\text{1}}\text{=8 cm}\\ {\text{Radius of upper end=r}}_{\text{2}}\text{=20 cm}\\ \text{Slant height of frustum=l=}\sqrt{{\left({\text{r}}_{\text{1}}{\text{-r}}_{\text{2}}\right)}^{\text{2}}{\text{+h}}^{\text{2}}}\\ \text{=}\sqrt{{\left(\text{8-20}\right)}^{\text{2}}{\text{+16}}^{\text{2}}}\text{\hspace{0.17em}=20 cm}\\ \text{Capacity of container=Volume of frustum}\\ \text{=}\frac{\text{1}}{\text{3}}{{\text{πh (r}}^{\text{2}}}_{\text{1}}{{\text{+r}}_{\text{2}}}^{\text{2}}{\text{+r}}_{\text{1}}{\text{r}}_{\text{2}}\text{)}\\ \text{=}\frac{\text{16}}{\text{3}}{\text{π (8}}^{\text{2}}{\text{+20}}^{\text{2}}\text{+8×20)}\\ \text{=10.45 litres}\\ \text{Cost of 10.45 litre milk=20×10.45= â‚¹ 209}\\ \text{Area of metal sheet used to make the container}\\ {\text{=π(r}}_{\text{1}}{\text{+r}}_{\text{2}}{{\text{)l+πr}}_{\text{1}}}^{\text{2}}\\ {\text{=π(8+20)×20+π×8}}^{\text{2}}\\ {\text{=624π cm}}^{\text{2}}\\ {\text{Cost of 624π cm}}^{\text{2}}\text{metal sheet=624×3.14×}\frac{\text{8}}{\text{100}}\text{=â‚¹156.75}\end{array}$

Q.30 A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.

Ans. $\begin{array}{l}\text{In}\mathrm{\Delta }\text{AEO,}\\ \text{}\frac{\mathrm{EO}}{\mathrm{AO}}=\mathrm{tan}30\mathrm{°}\\ \text{or}\mathrm{EO}=\frac{1}{\sqrt{3}}\mathrm{AO}=\frac{10}{\sqrt{3}}=\frac{10\sqrt{3}}{3}\text{cm}\\ \text{In}\mathrm{\Delta }\text{ABD,}\\ \text{}\frac{\mathrm{BD}}{\mathrm{AD}}=\mathrm{tan}30\mathrm{°}\\ \text{or}\mathrm{BD}=\frac{1}{\sqrt{3}}\mathrm{AD}=\frac{20}{\sqrt{3}}=\frac{20\sqrt{3}}{3}\text{cm}\\ \text{R}\mathrm{adius}\text{of upper end of frustum}={\mathrm{r}}_{1}=\frac{10\sqrt{3}}{3}\text{}\mathrm{cm}\\ \text{R}\mathrm{adius}\text{of lower end of frustum}={\mathrm{r}}_{2}=\frac{20\sqrt{3}}{3}\text{}\mathrm{cm}\\ \text{Height of frustum}=\mathrm{h}=10\text{cm}\end{array}$

$\begin{array}{l}\text{Volume of}\mathrm{frustum}=\frac{1}{3}\mathrm{\pi h}\left({{\mathrm{r}}^{2}}_{1}+{{\mathrm{r}}_{2}}^{2}+{\mathrm{r}}_{1}{\mathrm{r}}_{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{10}{3}\mathrm{\pi }\left({\left(\frac{10\sqrt{3}}{3}\right)}^{2}+{\left(\frac{20\sqrt{3}}{3}\right)}^{2}+\frac{10\sqrt{3}}{3}×\frac{20\sqrt{3}}{3}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{22000}{9}{\text{cm}}^{3}\\ \text{R}\mathrm{adius}\text{of wire}=\mathrm{r}=\frac{1}{16}×\frac{1}{2}=\frac{1}{32}\text{cm}\\ \text{Let length of wire be}\mathrm{l}.\\ \text{Volume of wire}=\text{Area of cross section}×\text{length}=\mathrm{\pi }{\left(\frac{1}{32}\right)}^{2}\mathrm{l}\\ \text{\hspace{0.17em} Volume of}\mathrm{frustum}=\text{Volume of wire}\\ \text{or}\frac{22000}{9}=\frac{22}{7}×{\left(\frac{1}{32}\right)}^{2}\mathrm{l}\\ \text{or}\mathrm{l}=7964.44\text{m}\end{array}$

Q.31 A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.

Ans.

$\begin{array}{l}\mathrm{One}\text{round of wire covers 3 mm height of cylinder.}\\ \text{Number of rounds}=\frac{\mathrm{Height}\text{of cylinder}}{\mathrm{Diameter}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{wire}}=\frac{12\text{cm}}{3\text{mm}}=\frac{12\text{cm}}{0.3\text{cm}}=40\\ \mathrm{Length}\text{of wire required in one round}=\mathrm{Circumference}\text{of cylinder}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{\pi r}=2\mathrm{\pi }×\frac{10}{2}=10\mathrm{\pi }\text{cm}\\ \text{Length of wire in 40 rounds}=40×10\mathrm{\pi }=40×10×3.14=1256\text{\hspace{0.17em}â€‹\hspace{0.17em}}\mathrm{c}\text{m}\\ \text{Radius of wire}=\frac{0.3}{2}\text{cm}=0.15\text{cm}\\ \mathrm{Volume}\text{of wire}=\text{Area of cross section of wire}×\mathrm{Length}\text{of wire}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3.14×{\left(0.15\right)}^{2}×1256\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}{\text{m}}^{3}=88.7364{\text{cm}}^{3}\\ \mathrm{Mass}=\text{Volume}×\text{Density}=88.7364×8.88=787.979232\text{gm = 788 gm (approx.)}\end{array}$

Q.32 A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate.)

Ans. $\begin{array}{l}\text{The double cone formed by revolving right-angled}\mathrm{\Delta }\text{ABC}\\ \text{about hypotenus AC is shown above.}\\ \text{AC}=\sqrt{{3}^{2}+{4}^{2}}=5\text{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Area of}\mathrm{\Delta }\text{ABC}=\frac{1}{2}×\mathrm{AB}×\mathrm{BC}=\frac{1}{2}×3×4=6{\text{cm}}^{2}\\ \text{or,}\frac{1}{2}×\mathrm{AC}×\mathrm{OB}=6\\ \text{or,}\frac{1}{2}×5×\mathrm{OB}=6\\ \text{or,}\mathrm{OB}=\frac{12}{5}=2.4\text{cm}\\ \text{Volume of double cone}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{Volume of cone\hspace{0.17em}1}+\text{Volume of cone 2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}{\mathrm{\pi r}}^{2}{\mathrm{h}}_{1}+\frac{1}{3}{\mathrm{\pi r}}^{2}{\mathrm{h}}_{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\mathrm{\pi }{\left(2.4\right)}^{2}\left({\mathrm{h}}_{1}+{\mathrm{h}}_{2}\right)\text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\mathrm{\pi }{\left(2.4\right)}^{2}×5=30.14{\text{cm}}^{3}\\ \text{Surface area of double cone}=\text{Surface area of cone 1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{Surface area of cone 2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{\pi rl}}_{1}+{\mathrm{\pi rl}}_{2}\\ \text{}=3.14×2.4\left(3+4\right)=52.75{\text{cm}}^{2}\end{array}$

Q.33 A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?

Ans.

$\begin{array}{l}\text{Volume of cistern}=150×120×110=1980000{\text{cm}}^{3}\\ \text{Volume to be filled in cistern}=1980000-129600=1850400{\text{cm}}^{3}\\ \text{Let}\mathrm{n}\text{number of porous bricks were placed in cistern.}\\ \text{So, volume of}\mathrm{n}\text{bricks}=\mathrm{n}×22.5×7.5×6.5=1096.875\mathrm{n}\\ \text{Each brick absorbs one-seventeenth of its volume.}\\ \therefore \text{Volume absorbed by}\mathrm{n}\text{bricks}=\frac{\mathrm{n}}{17}\left(1096.875\right)\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume to be filled in cistern}+\text{Volume absorbed by}\mathrm{n}\text{bricks}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{Volume of}\mathrm{n}\text{bricks}\\ \text{or,}1850400+\frac{\mathrm{n}}{17}\left(1096.875\right)=1096.875\mathrm{n}\\ \text{or,}\mathrm{n}=1792.41\\ \text{So, 1792 bricks can be put in the given cistern.}\end{array}$

Q.34 In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Ans.

$\begin{array}{l}\text{Rainfall=10 cm=0.1m}\\ {\text{Areaofthevalley=7280km}}^{\text{2}}\\ {\text{Vaolumeofrainfallinthevalley=7280×1000×1000×0.1m}}^{\text{3}}\\ {\text{=728000000 m}}^{\text{3}}\\ {\text{Volumeofwaterin3revers=3×1072000×75×3=723600000 m}}^{\text{3}}\\ \text{Thus,thetotalrainfallisapproximatelysameasthe}\mathrm{total}\\ \text{volumeof thethreerivers.}\end{array}$

Q.35 An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see the following figure). Ans. $\begin{array}{l}\text{Height of frustum}={\mathrm{h}}_{1}=22-10=12\text{cm}\\ \text{Height of cylindrical part}={\mathrm{h}}_{2}=10\text{cm}\\ \text{R of lower end}={\mathrm{r}}_{1}=\frac{8}{2}\text{}=4\text{cm}\\ \text{R of end}={\mathrm{r}}_{2}=\frac{18}{2}\text{}=9\text{cm}\\ \text{S}=\mathrm{l}=\sqrt{{\left({\mathrm{r}}_{1}-{\mathrm{r}}_{2}\right)}^{2}+{{\mathrm{h}}_{1}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(4-9\right)}^{2}+{12}^{2}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}13\text{}\\ \mathrm{Areaoftinsheetrequired}\\ \text{\hspace{0.17em}}=\text{CSA of frustum part}+\text{CSA of cylindrical part}\\ \text{\hspace{0.17em}}=\mathrm{\pi }\left({\mathrm{r}}_{1}+{\mathrm{r}}_{2}\right)\mathrm{l}+2{\mathrm{\pi r}}_{1}{\mathrm{h}}_{2}\\ =\frac{22}{7}\left(4+9\right)×13+2×\frac{22}{7}×4×10\\ =782\frac{4}{7}{\text{cm}}^{2}\end{array}$

Q.36 Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Ans. $\begin{array}{l}\text{Let ABC be a cone. A frustum DECB is cut by a plane parallel}\\ \text{to its base.}\\ \text{In}\mathrm{\Delta }\text{ADF and}\mathrm{\Delta }\text{ABG,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{DF}}{\mathrm{BG}}=\frac{\mathrm{AF}}{\mathrm{AG}}=\frac{\mathrm{AD}}{\mathrm{AB}}\\ \text{or}\frac{{\mathrm{r}}_{2}}{{\mathrm{r}}_{1}}=\frac{{\mathrm{h}}_{1}-\mathrm{h}}{{\mathrm{h}}_{1}}=\frac{{\mathrm{l}}_{1}-\mathrm{l}}{{\mathrm{l}}_{1}}\\ \text{or}\frac{{\mathrm{r}}_{2}}{{\mathrm{r}}_{1}}=1-\frac{\mathrm{h}}{{\mathrm{h}}_{1}}=1-\frac{\mathrm{l}}{{\mathrm{l}}_{1}}\\ \text{or}1-\frac{\mathrm{l}}{{\mathrm{l}}_{1}}=\frac{{\mathrm{r}}_{2}}{{\mathrm{r}}_{1}}\\ \text{or}\frac{\mathrm{l}}{{\mathrm{l}}_{1}}=1-\frac{{\mathrm{r}}_{2}}{{\mathrm{r}}_{1}}=\frac{{\mathrm{r}}_{1}-{\mathrm{r}}_{2}}{{\mathrm{r}}_{1}}\\ \text{or}\frac{{\mathrm{l}}_{1}}{\mathrm{l}}=\frac{{\mathrm{r}}_{1}}{{\mathrm{r}}_{1}-{\mathrm{r}}_{2}}\end{array}$

$\begin{array}{l}\text{CSA of frustum DECB}=\text{CSA of cone ABC}-\text{CSA of cone ADE}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{\pi r}}_{1}{\mathrm{l}}_{1}-{\mathrm{\pi r}}_{2}\left({\mathrm{l}}_{1}-\mathrm{l}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}={\mathrm{\pi r}}_{1}\left(\frac{{\mathrm{lr}}_{1}}{{\mathrm{r}}_{1}-{\mathrm{r}}_{2}}\right)-{\mathrm{\pi r}}_{2}\left(\frac{{\mathrm{lr}}_{1}}{{\mathrm{r}}_{1}-{\mathrm{r}}_{2}}-\mathrm{l}\right)\\ \text{}=\frac{{{\mathrm{\pi r}}_{1}}^{2}\mathrm{l}}{{\mathrm{r}}_{1}-{\mathrm{r}}_{2}}-{\mathrm{\pi r}}_{2}\left(\frac{{\mathrm{lr}}_{2}}{{\mathrm{r}}_{1}-{\mathrm{r}}_{2}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{{\mathrm{\pi r}}_{1}}^{2}\mathrm{l}}{{\mathrm{r}}_{1}-{\mathrm{r}}_{2}}-\frac{{{\mathrm{\pi r}}_{2}}^{2}\mathrm{l}}{{\mathrm{r}}_{1}-{\mathrm{r}}_{2}}=\mathrm{\pi }\left({\mathrm{r}}_{1}+{\mathrm{r}}_{2}\right)\mathrm{l}\\ \text{CSA of frustum}=\mathrm{\pi }\left({\mathrm{r}}_{1}+{\mathrm{r}}_{2}\right)\mathrm{l}\\ \text{Total surface area of frustum}=\text{CSA of frustum}+\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{CSA of upper circular end}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{CSA of lower circular end}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{\pi }\left({\mathrm{r}}_{1}+{\mathrm{r}}_{2}\right)\mathrm{l}+{{\mathrm{\pi r}}_{1}}^{2}+{{\mathrm{\pi r}}_{2}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{\pi }\left[\left({\mathrm{r}}_{1}+{\mathrm{r}}_{2}\right)\mathrm{l}+{{\mathrm{r}}_{1}}^{2}+{{\mathrm{r}}_{2}}^{2}\right]\end{array}$

Q.37 Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Ans. $\begin{array}{l}\text{Let ABC be a cone. A frustum DECB is cut by a plane parallel}\\ \text{to its base.}\\ \text{In}\mathrm{\Delta }\text{ADF and}\mathrm{\Delta }\text{ABG,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{DF}}{\mathrm{BG}}=\frac{\mathrm{AF}}{\mathrm{AG}}=\frac{\mathrm{AD}}{\mathrm{AB}}\\ \text{or}\frac{{\mathrm{r}}_{2}}{{\mathrm{r}}_{1}}=\frac{{\mathrm{h}}_{1}-\mathrm{h}}{{\mathrm{h}}_{1}}=\frac{{\mathrm{l}}_{1}-\mathrm{l}}{{\mathrm{l}}_{1}}\\ \text{or}\frac{{\mathrm{r}}_{2}}{{\mathrm{r}}_{1}}=1-\frac{\mathrm{h}}{{\mathrm{h}}_{1}}=1-\frac{\mathrm{l}}{{\mathrm{l}}_{1}}\\ \text{or \hspace{0.17em}}\frac{\mathrm{h}}{{\mathrm{h}}_{1}}=\frac{{\mathrm{r}}_{1}-{\mathrm{r}}_{2}}{{\mathrm{r}}_{1}}\text{}\end{array}$

$\begin{array}{l}\text{Volume of frustum DECB}=\text{Volume of cone ABC}-\text{\hspace{0.17em}Volume of cone ADE}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}{{\mathrm{\pi r}}_{1}}^{2}{\mathrm{h}}_{1}-\frac{1}{3}{{\mathrm{\pi r}}_{2}}^{2}\left({\mathrm{h}}_{1}-\mathrm{h}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{\pi }}{3}\left[{{\mathrm{r}}^{2}}_{1}{\mathrm{h}}_{1}-{{\mathrm{r}}_{2}}^{2}\left({\mathrm{h}}_{1}-\mathrm{h}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{\pi }}{3}\left[{{\mathrm{r}}^{2}}_{1}\frac{{\mathrm{r}}_{1}\mathrm{h}}{{\mathrm{r}}_{1}-{\mathrm{r}}_{2}}-{{\mathrm{r}}_{2}}^{2}\left(\frac{{\mathrm{r}}_{1}\mathrm{h}}{{\mathrm{r}}_{1}-{\mathrm{r}}_{2}}-\mathrm{h}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{\pi }}{3}\left[{{\mathrm{r}}^{2}}_{1}\frac{{\mathrm{r}}_{1}\mathrm{h}}{{\mathrm{r}}_{1}-{\mathrm{r}}_{2}}-{{\mathrm{r}}_{2}}^{2}\left(\frac{{\mathrm{r}}_{2}\mathrm{h}}{{\mathrm{r}}_{1}-{\mathrm{r}}_{2}}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{\pi h}}{3}\left[\frac{{{\mathrm{r}}_{1}}^{3}-{{\mathrm{r}}_{2}}^{3}}{{\mathrm{r}}_{1}-{\mathrm{r}}_{2}}\right]=\frac{\mathrm{\pi h}}{3}\left({{\mathrm{r}}^{2}}_{1}+{{\mathrm{r}}^{2}}_{2}+{\mathrm{r}}_{1}{\mathrm{r}}_{2}\right)\end{array}$

Q.38 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 m and 4 m respectively, and the slant height of the top is 2.8 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 â‚¹ 500 per m2. (Note that the base of the tent will not be covered with canvas.)

Ans.

$\begin{array}{l}\text{Total canvas used}=\text{Total curved surface area of tent}\\ \text{Total curved surface area of tent}\\ \text{}=\text{curved surface area of cylindrical portion}\\ \text{}+\text{curved surface area of conical portion}\\ \text{}=\text{2}\mathrm{\pi }\text{rh}+\mathrm{\pi rl}\\ \text{}=2×\frac{22}{7}×2×2.1+\frac{22}{7}×2×2.8\\ \text{}=44{\text{m}}^{2}\\ \mathrm{Cost}{\text{of 44 m}}^{2}\text{convas}=44×500=\text{\hspace{0.17em}}â‚¹22000\end{array}$

## 1. What are the topics covered in NCERT Solutions for Class 10 Mathematics Chapter 13 by Extramarks?

The topics covered in NCERT Solutions for Class 10 Mathematics Chapter 13 by Extramarks include:

–         Introduction

–         Surface area of a combination of solids

–         Volume of a combination of solids

–        Conversion of solid from one shape to another

–        Frustum of a cone

–        Summary

## 2. From where can I access NCERT Solutions Class 10 Mathematics Chapter 13?

You can access NCERT Solutions Class 10 Mathematics Chapter 13 from the Extramarks website or app.