# NCERT Solutions For Class 10 Maths Chapter 13 Surface Areas And Volumes (Ex 13.3)

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Surface Areas and Volumes is an important Chapter of Mathematics for Class 10. The visible area of a surface occupied by the edges of a particular object’s surface is known as its surface area. This is usually calculated using formulas that can be used for different objects of different shapes. Surfaces can be computed for both 2D and 3D objects. There are two types of surface area, total surface area and curved surface area, which are commonly calculated for solid objects. All the solutions can be downloaded from NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3 from the website of Extramarks. The Ex 13.3 Class 10 Maths NCERT Solutions can be easily downloaded in PDF format for offline access.

**NCERT Solutions For Class 10 Maths Chapter 13 Surface Areas And Volumes (Ex 13.3) Exercise 13.3 **

Area and Volume – Combinations of Solids. In daily life, people come across many solids of different shapes and sizes, among which they can calculate both the Area and the Volume. It is necessary to compute the area and volume of the surrounding objects. But what if these basic shapes come together and form a different shape than the original one. Now the question is, how will someone calculate the Volume, Surface and Area of the new objects. When calculating the Area and Volume of these new shapes, one must observe the new shape. A thorough discussion is given below, which will give a clearer picture of these objects and how they are calculated. Solids consisting of conventional geometrical solids are called composite solids. These include Pyramids, Prisms, Cylinders, Pairs and Cones to know the Volume and Surface of a Solid. Students need to know the Volumes and Areas Of Prisms, Cones, Spheres, Pyramids and Cylinders. The total area of the individual solids that make up the combined solid excluding the overlapping parts of each figure is the total area of a combined solid while the sum of the volumes of the individual solids that make up the combined solid is the mass of a combined solid. All the explanations are provided on NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3

Combining Solids – When calculating the Surface Area of combined Solids, only the visible area needs to be calculated. For example, students should get a (C.S.A.) Curved Surface Area. They should calculate the hemisphere and cone separately and add them together if the hemisphere is above the cone. The surface area of an object can be said to be the total area of the surface occupied by the object or the total area of all surfaces of a three-dimensional figure. The surface area of any shape other than a Cube or Cuboid can be calculated as the area of the sides of the shape plus the base. If the Prism and Cylinder are the same, it can be considered twice the area of the base. The area of any given figure can be calculated as a three-dimensional figure using the example of a gift, and its area can be the wrapping paper, so the amount of wrapping paper covering the gift is determined by the given cubic figure. Surfaces can be further divided into two types: B.

- Total Surface Area – The area including the base and the curved part is called the Total Surface Area.
- Curved Surface Area – The area of the curved portion excluding the base is called the Curved Surface Area.

The Volume of a particular object can be specified as the amount of liquid that can be contained within the object. Basically, the amount enclosed by a given three-dimensional object is called the Volume of that three-dimensional object. The Volume of one-dimensional objects (such as lines) and two-dimensional objects (such as squares) is considered zero since Volume is considered a quantity. The basic properties for finding the volume of a given object are: A given object has a volume of length * width * height (V = lwh). The total Volume of a given object is the sum of all non-overlapping areas. Even if the overlapping shapes have the same Volume, each polyhedral region has a unique volume according to the unit cube.

Combining Solids: Students have encountered a variety of shapes that combine various Solids with one or more basic shapes. The most common examples of combining Solids are ice cream Cones, Capsules, Tents, and Capsule-Shaped Trucks carrying gasoline or LPG. Basically, a combination of Solids can be described as a solid that can be decomposed into two or more distinct Solids with sides. A combination of Solids is also called a compound shape because it is formed by merging two or more different shapes into a new shape. To calculate the Surface Area or Volume of these types of Solids, students first need to ascertain the number of Solid shapes that form these shapes. A one-dimensional shape of six squares forms. The Surface Area of a given composite shape is the sum of the areas of all its faces of that Solid. To understand the Combination of Solids, students can take the example of an ice cream-filled cone, which is a fusion of a Cone and Hemispherical ice cream. Therefore, the Total Surface Area of a Cone filled with ice cream is equal to the Sum of the Curved Area of the Hemisphere and the Curved Area of the Cone. All these formulas are available on NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3

For computing the volume of the combined shape, students first need to figure out the different shapes needed to form the composite shape. The Volume of a combined shape can be calculated by calculating the Volumes of the specific shapes that make up the new combined shape and summing them to form the total volume of the composite shape. Similarly, for Volume, the Volume of a Cone filled with ice cream can be found by calculating the Volume of the hemisphere and the Volume of the Cone separately and adding them to the Volume of the Cone filled with ice cream. Students are free to download all the materials like the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3 from the Extramarks website.

**NCERT Solutions For Class 10 Maths Chapter 13 Surface Areas And Volumes Exercise 13.3**

The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3 contains a wide range of precisely designed exercises that are solved for excellent student comprehension. The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3 have been created taking into account the latest CBSE syllabus published by the board. The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3 are developed to help students solve their questions spontaneously and effectively. The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3 are prepared to meet students’ objectives by faculty and subject-matter experts with several years of teaching experience.

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When converting one Solid shape to another, its Volume remains the same regardless of the newly formed shape. Students should equalise the Volume of both Shapes to find a solution. For example, if someone melts 1 large cylindrical candle into 10 small cylindrical candles, the sum of the Volumes of the small cylindrical candles equals the volume of the melted large cylindrical candles.

The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3 are considered to be the best option for CBSE students when it comes to exam preparation. This chapter consists of many exercises. From these, Extramarks provides the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3 in PDF format. Students can download this solution according to their convenience or study online directly from the Extramarks website

Extramarks subject-matter experts solve the exercise problems/questions meticulously and in compliance with the CBSE guidelines. When students are familiar with all the concepts of the Subject Surfaces and Volumes and all of the exercises they contain, Class 10 students can easily achieve the highest possible scores in their final exams. With the NCERT Solutions For Class 10 Mathematics Chapter 13 Exercise 13.3, students can easily understand patterns of questions that may appear on exams in this chapter, and also learn how to weigh chapters of any particular subject, so that they can prepare accordingly for the final exam.

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**Q.1 **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**

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

**Q.2 **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.3 **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**

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

**Q.4** 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**

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

**Q.5 **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}\times {3}^{2}\times 12+2\mathrm{\pi}\times {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}\times \text{Volume of one ice-cream cone}\\ \text{or}{\mathrm{\pi r}}^{2}\mathrm{h}=\mathrm{n}\times 54\mathrm{\pi}\\ \text{or}{6}^{2}\times 15=54\mathrm{n}\\ \text{or}\mathrm{n}=10\text{\hspace{0.17em}}\end{array}$

**Q.6** 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**

$\begin{array}{l}\text{Thickness of coin}=\mathrm{h}=2\text{mm}=0.2\text{cm}\\ \text{Radius of coin}=\mathrm{r}=\frac{1.75}{2}\text{cm}\\ \text{Cuboid is of the dimensions 5.5 cm \xd7 10 cm \xd7 3.5 cm.}\\ \text{Let number of coines be}\mathrm{n}\text{.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of cuboid}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{n}\times \text{Volume of one coin}\\ \text{or 5.5 cm \xd7 10 cm \xd7 3.5 cm}=\mathrm{n}\times {\mathrm{\pi r}}^{2}\mathrm{h}=\mathrm{n\pi}{\left(\frac{1.75}{2}\right)}^{2}(0.2)\\ \text{or}\mathrm{n}=400\end{array}$

**Q.7** 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**

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

**Q.8 **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}\times 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}\times 9=1500{\text{m}}^{3}\\ \text{Volume of water flown out of canal in 30 minutes}=1500\times 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}\times {\text{0.08 m}}^{3}\\ \text{or,}45000{\text{m}}^{3}=\text{A}\times {\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.9** 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**

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

## FAQs (Frequently Asked Questions)

### 1. What is Surface Area, as described in NCERT Solutions For Class 10 Mathematics Chapter 13 Exercise 13.3?

The visible area of a surface occupied by the edges of a particular object’s surface is known as its Surface Area. This is usually calculated using formulas that can be used for different objects of different shapes. Surfaces can be computed for both 2D and 3D objects. There are two types of surface area, Total Surface Area and Curved Surface Area, which are commonly calculated for solid objects.

### 2. How can one find the Surface Area and Volume according to NCERT Solutions For Class 10 Mathematics Chapter 13 Exercise 13.3?

The area occupied by the perimeter of a particular object’s surface is called Surface Area and can be calculated using various geometry formulas. Each shape has its own formula for calculating the surface area of that particular shape. Volume is the total amount of space enclosed in the space of a three-dimensional object and can be calculated using various formulas for a given shape and size.

### 3. Why are the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3 reliable?

The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3 will focus on Fixed Transformations of Surfaces and Volumes. For example, one can melt a cylindrical candle and pour it into a cubic container. Candles have a new look. The essence of this is that the Volume of candles is kept constant. As a result of these NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3 students will be able to determine the volume of that object, even if the object is transformed from solids to another. Transforming solids from one form to another is an interesting idea with numerous applications in everyday life. Students can learn more about this subject by engaging in interesting activities such as modelling clay and moulding into different solids. These topics are covered in detail in the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.3.