# NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2

## NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes (Ex 13.2) Exercise 13.2

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### Some Formulas to Learn before Solving Exercise 13.2

Surface Area and Volume of Hemisphere:

A hemisphere is a three-dimensional geometric shape that is half of a sphere, with one side flat and the other round. It is made by cutting a sphere at the exact centre along its diameter, resulting in two equal hemispheres. The base or face of the hemisphere is the flat side of the hemisphere.

Surface Area and Volume of Hemisphere

• The volume of Hemisphere = (2πr3)/3

• Surface Area of Hemisphere = 3πr2

• Curved Surface Area of a Hemisphere = 2πr2

Surface Area and Volume of Right Circular Cone

A right circular cone is a cone with an axis perpendicular to the base plane. The base of a right circular cone is circular in shape.

The Formulas of Right Circular Cone are as follows:

• Curved surface area of a cone = πr√(r2 + h2)

• Volume of a Cone (V) = (1/3) × πr2 × h

• Total surface area of a cone = πr2 + πr√(r2 + h2)

Surface Area and Volume of Cylinder

A cylinder is a three-dimensional solid figure with two identical circular bases connected by a curved surface at a specified distance from the centre, which is the cylinder's height.

The Formulas for Cylinder are as Follows:

• Curved surface area of cylinder = 2πrh

• Total surface area of cylinder = 2πr(r+h)

• The volume of a cylinder, V = πr2h

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Exercise – 13.2

1. 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. From the given figure, we know that

Height (h) of conical part = Radius(r) of conical part ${\text{ = 1cm}}$

Radius(r) of hemispherical part = Radius of conical part (r) ${\text{ = 1cm}}$

Volume of solid = Volume of conical part + Volume of hemispherical part

${\text{ = }}\dfrac{1}{3}\pi {r^2}h + \dfrac{2}{3}\pi {r^3}$

${\text{ = }}\dfrac{1}{3}\pi {\left( 1 \right)^2}\left( 1 \right) + \dfrac{2}{3}\pi {\left( 1 \right)^3} = \dfrac{{2\pi }}{3} + \dfrac{1}{3} = \pi {\text{c}}{{\text{m}}^3}$

1. 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 aluminum sheet. The diameter of the model is ${\text{3cm}}$ and its length is${\text{12cm}}$. If each cone has a height of ${\text{2cm}}$, 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:

From the given figure, we know that

Height (h1) of each conical part ${\text{ = 2cm}}$

Height (h2) of cylindrical part ${\text{ = 12 - 2}} \times {\text{ Height of conical part}}$

${\text{ = 12 - 2}} \times {\text{2 = 8cm}}$

Radius (r) of cylindrical part = Radius of conical part ${\text{ = }}\dfrac{3}{2}{\text{cm}}$

Volume of air present in the model = Volume of cylinder $+ {\text{ }}2\, \times$Volume of cones

$= \pi {r^2}{h_2} + 2 \times \dfrac{1}{3}\pi {r^2}{h_1}$

$= \pi \times {\left( {\dfrac{3}{2}} \right)^2} \times 8 + 2 \times \dfrac{1}{3}\pi {\left( {\dfrac{3}{2}} \right)^2} \times 2$

$= \pi \times \dfrac{9}{4} \times 8 + \dfrac{2}{3}\pi \times \dfrac{9}{4} \times 2$

$= 18\pi + 3\pi = 21\pi = 66{\text{c}}{{\text{m}}^3}$

1. 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 given figure).

Ans:

We can observe that,

${\text{ = }}\dfrac{{{\text{2}}{\text{.8}}}}{{\text{2}}}{\text{ = 1}}{\text{.4cm}}$

Length of each hemispherical part = Radius of hemispherical part ${\text{ = 1}}{\text{.4cm}}$

Length (h) of cylindrical part ${\text{ = 5 - }}\,{\text{2}} \times {\text{Length of hemispherical part }}$

${\text{ = 5 - }}\,{\text{2}} \times {\text{1.4 = 2}}{\text{.2cm}}$

Volume of one gulab jamun = Vol. of cylindrical part + 2 × Vol. of hemispherical part

$= \pi {r^2}h + 2 \times \dfrac{2}{3}\pi {r^3} = \pi {r^2}h + \dfrac{4}{3}\pi {r^3}$

$= \pi {\left( {1.4} \right)^2}\left( {2.2} \right) + \dfrac{4}{3}\pi {\left( {1.4} \right)^3}$

$= \dfrac{{22}}{7} \times 1.4 \times 1.4 \times 2.2 + \dfrac{4}{3} \times \dfrac{{22}}{7} \times 1.4 \times 1.4 \times 1.4$

${\text{ = 13}}{\text{.552 + 11}}{\text{.498 = 25}}{\text{.05c}}{{\text{m}}^{\text{3}}}$

Volume of ${\text{45}}$ gulab jamuns ${\text{ = 45}} \times {\text{25}}{\text{.05 = 1127}}{\text{.25c}}{{\text{m}}^{\text{3}}}$

Volume of sugar syrup $= 30\%$  of volume

$= \dfrac{{30}}{{100}} \times 1127.25$

${\text{ = 338}}{\text{.17c}}{{\text{m}}^{\text{3}}}$

$\cong {\text{338c}}{{\text{m}}^{\text{3}}}$

1. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboids are ${\text{15cm}}$  by ${\text{10cm}}$  by ${\text{3}}{\text{.5cm}}$. The radius of each of the depressions is ${\text{0}}{\text{.5cm}}$ and the depth is ${\text{1}}{\text{.4cm}}$. Find the volume of wood in the entire stand (see the following figure).

Ans:

Depth (h) of each conical depression${\text{ = 1}}{\text{.4cm}}$

Radius (r) of each conical depression${\text{ = 0}}{\text{.5cm}}$

Volume of wood = Volume of cuboid $- {\text{ }}4 \times$ Volume of cones

${\text{ = l * b * h}}$$- 4 \times \dfrac{1}{3}\pi {r^2}h$

$= 15 \times 10 \times 3.5 - 4 \times \dfrac{1}{3} \times \dfrac{{22}}{7} \times {\left( {\dfrac{1}{2}} \right)^2} \times 1.4$

$= 525 - 1.47$

${\text{ = 523}}{\text{.53c}}{{\text{m}}^{\text{3}}}$

1. A vessel is in the form of an inverted cone. Its height is ${\text{8cm}}$ and the radius of its top, which is open, is ${\text{5cm}}$. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius ${\text{0}}{\text{.5cm}}$are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Ans:

Height (h) of conical vessel${\text{ = 8cm}}$

Radius (r1) of conical vessel${\text{ = 5cm}}$

Radius (r2) of lead shots${\text{ = 0}}{\text{.5cm}}$

Let the number of lead shots dropped in the vessels be n.

Volume of water spilled = Volume of dropped lead shots

$\dfrac{1}{4} \times$Volume of cone ${\text{ = n *}}\dfrac{{\text{4}}}{{\text{3}}}{{\text{r}}_{\text{2}}}^{\text{3}}$

$\dfrac{1}{4} \times \dfrac{1}{3}\pi {{\text{r}}_1}^2{\text{h = n * }}\dfrac{{\text{4}}}{{\text{3}}}{{\text{r}}_{\text{2}}}^{\text{3}}$

${{\text{r}}_1}^2{\text{h = n *16}}{{\text{r}}_{\text{2}}}^{\text{3}}$

${5^2} \times {\text{8 = n *16}} \times {\left( {0.5} \right)^{\text{3}}}$

$n = \dfrac{{25 \times 8}}{{16 \times {{\left( {\dfrac{1}{2}} \right)}^3}}} = 100$

Therefore, the number of lead shots dropped in the vessel is $100$

1. A solid iron pole consists of a cylinder of height ${\text{220cm}}$ and base diameter ${\text{24cm}}$, which is surmounted by another cylinder of height ${\text{60cm}}$ and radius ${\text{8cm}}$. Find the mass of the pole, given that ${\text{1c}}{{\text{m}}^3}$of iron has approximately ${\text{8g}}$ mass.

Ans:

From the above figure, we observe that

Height (h1) of larger cylinder ${\text{ = 220cm}}$

Radius (r1) of larger cylinder ${\text{ = }}\dfrac{{24}}{2}{\text{ = 12cm}}$

Height (h2) of smaller cylinder ${\text{ = 60cm}}$

Radius (r2) of smaller cylinder ${\text{ = 8cm}}$

Total volume of pole = Volume of larger cylinder + volume of smaller cylinder

${\text{ =}}\pi{{\text{r}}_{\text{1}}}^{\text{2}}{{\text{h}}_{\text{1}}}{\text{ +}} \pi {{\text{r}}_{\text{2}}}^{\text{2}}{{\text{h}}_{\text{2}}}$

${\text{ = }} \pi {\left( {12} \right)^{\text{2}}}* {\text{220 + }} \pi {\left( 8 \right)^{\text{2}}} * 60$

${\text{ = }}\pi\left[ {{\text{(144 * 220) + (64 * 60)}}} \right]$

${\text{ = 35520 * 3}}{\text{.14 = 111532}}{\text{.8c}}{{\text{m}}^{\text{3}}}$

Mass of ${\text{1c}}{{\text{m}}^3}$iron = ${\text{8g}}$

Mass of ${\text{111532}}{\text{.8c}}{{\text{m}}^{\text{3}}}$iron ${\text{ = 111532}}{\text{.8 * 8 = 892262}}{\text{.4g = 892}}{\text{.262kg}}$

1. A solid consisting of a right circular cone of height ${\text{120cm}}$ and radius ${\text{60cm}}$ standing on a hemisphere of radius ${\text{60cm}}$ 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 ${\text{60cm}}$ and its height is ${\text{180cm}}$.

Ans:

Radius (r) of hemispherical part = Radius (r) of conical part${\text{ = 60cm}}$

Height (h2) of conical part of solid${\text{ = 120cm}}$

Height (h1) of cylinder${\text{ = 180cm}}$

Radius (r) of cylinder${\text{ = 60cm}}$

Volume of water left = Volume of cylinder − Volume of solid

= Volume of cylinder – (Volume of cone + Volume of hemisphere)

${\text{ = }}\pi{{\text{r}}^{\text{2}}}{{\text{h}}_{\text{1}}}{\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{3}}}{\pi}{{\text{r}}^{\text{2}}}{{\text{h}}_{\text{2}}}{\text{ + }}\dfrac{{\text{2}}}{{\text{3}}} \pi {{\text{r}}^{\text{3}}}} \right)$

${\text{ = }}\pi {\left( {60} \right)^{\text{2}}} \times 180{\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{3}}}{\pi }{{\left( {60} \right)}^{\text{2}}} \times {\text{120 + }}\dfrac{{\text{2}}}{{\text{3}}}{\pi }{{\left( {60} \right)}^{\text{3}}}} \right)$

${\text{ = }}\pi{\left( {60} \right)^{\text{2}}}\left[ {\left( {180} \right) - \left( {40 + 40} \right)} \right]$

${\text{ = }}\pi\left( {{\text{3600}}} \right)\left( {{\text{100}}} \right){\text{ = 3,60,000 }}\pi{\text{c}}{{\text{m}}^{\text{3}}}{\text{ = 11311428}}{\text{.57c}}{{\text{m}}^{\text{3}}}{\text{ = 1}}{\text{.131}}{{\text{m}}^{\text{3}}}$

1. A spherical glass vessel has a cylindrical neck ${\text{8cm}}$ long, ${\text{2cm}}$ in diameter; the diameter of the spherical part is ${\text{8}}{\text{.5cm}}$. By measuring the amount of water it holds, a child finds its volume to be ${\text{345c}}{{\text{m}}^3}$. Check whether she is correct, taking the above as the inside measurements, and $\pi = 3.14$.

Ans:

Height (h) of cylindrical part ${\text{ = 8cm}}$

Radius (r2) of cylindrical part ${\text{ = }}\dfrac{2}{2}{\text{ = 1cm}}$

Radius (r1) spherical part ${\text{ = }}\dfrac{{8.5}}{2}{\text{ = 4}}{\text{.25cm}}$

Volume of vessel = Volume of sphere + Volume of cylinder ${\text{ = }}\dfrac{4}{3}\pi {\left( {\dfrac{{8.5}}{2}} \right)^3}{\text{ + }}\pi {\left( 1 \right)^2}\left( 8 \right)$

${\text{ = }}\dfrac{4}{3} \times 3.14 \times 76.765625{\text{ + 8}} \times {\text{3}}{\text{.14}}$

$= 321.392 + 25.12$

$= 346.512$

${\text{ = 346}}{\text{.51c}}{{\text{m}}^{\text{3}}}$

Therefore, she is wrong.

## NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2

Opting for the NCERT solutions for Ex 13.2 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.2 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Subject Surface Areas and Volumes textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 13 Exercise 13.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 10 Maths Chapter 13 Exercise 13.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 10 Maths Chapter 13 Exercise 13.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

### NCERT Solutions for Class 10 Maths Chapter 13 Exercises

 Chapter 13 - Surface Areas and Volumes All Exercises in PDF Format Exercise 13.1 9 Questions & Solutions (2 Short Answers, 7 Long Answers) Exercise 13.3 9 Questions & Solutions (9 Long Answers) Exercise 13.4 5 Questions & Solutions (5 Long Answers) Exercise 13.5 7 Questions & Solutions (7 Long Answers)

1. What will I study in this chapter?

Surface Areas and Volumes is one of the most fascinating chapters of Class 10 Maths. In CBSE Class 10 Maths Book Solutions, you will understand about the cuboid and its surface area, cube and its surface area, cylinder and it’s surface area, right circular cone, sphere and volume of cube, cuboid and cylinder. You will also come across the hemisphere and its surface area, combination of solids, shape conversion of solids and volume of a frustum. Our NCERT Solutions for Class 10 Maths Chapter 13 consists of 5 exercises with 38 questions in total and they are prepared by expert teachers based on the current Class 10 Maths syllabus.

2. How many questions are there in this exercise?

This exercise of Surface Areas and Volumes NCERT Solutions for Class 10 Maths Chapter 13 consist of 8 questions. Question 1, 2 4 and 7 are based on finding the volume of different shapes like the solid, the volume of air, the volume of wood, the volume of water. Question 3, 5 and 8 are scenario-based questions. Where incidents or situations are mentioned and you will have to find whether the given numbers are correct or not. Question 6 is based on mass, where you will have to find the mass of a pole.

3. Explain about Volume of Cube and Cylinder.

Volume of Cube

Here, in CBSE Class 10 Maths Chapter 13, you will learn that all the dimensions of the cube are identical, Volume = l3

Here, l will be the length of the cube of the edge

The volume of a cube = base area×height

Volume of Cuboid

Volume of a cuboid =(base area)×height=(lb)h=lbh

Volume of Cylinder

Volume of a cylinder = Base area × its height = (π×r2)×h=πr2h

In the later part of Class 10 Maths Surface Areas and Volumes, you will also come across the volume of a right circular cone, the volume of a sphere, the volume of a hemisphere, and hemisphere and it’s surface area.

4. How are the solutions for the exercise questions provided by Vedantu?

NCERT Solutions are explained in an easy to understand way. The solutions are to the extremity and suitable with high precision solutions for all the exercise questions. NCERT Solutions for Class 10 Maths are the most high-grade preparation materials which are equipped with the test point of representation. The answers are sketched as per the latest CBSE maths syllabus and CBSE NCERT Books.

All our NCERT answers will make learners build their conceptual knowledge of the concepts. They will also develop a strong foundation with the concepts as they are explained with examples and diagrams. NCERT Solutions by Vedantu are the most reliable study materials provided for smart education and effective answering of questions.

5. Which is the important section of Exercise 13.2 of Class 10 Maths?

Surface Area Combination Of Solids is the most important section for solving exercise 13.2 of NCERT class 10 Maths. But all other sections and exercises present in the chapter too are equally important from the exam point of view. To have a holistic understanding of the section, download NCERT Solutions Class 10 Maths Chapter 13 on Vedantu and get free of cost and full solutions to the 13.2 exercise. Practice these questions to improve your marks.

6. What to study in Exercise 13.2 of class 10 Mathematics?

In order to successfully solve exercise 13.2 of NCERT Class 10 Maths, you need to study the section called Surface Area Combination of Solids. This section will help you in solving the questions that have been asked in exercise 13.2 of the NCERT Maths textbook. For full step by step solutions of the exercise, download NCERT Solutions Class 10 Maths Chapter 13 Exercise 13.2 PDF available free of cost only on the Vedantu website (vedantu.com) and app.

7. Can you please brief me about the Class 10 Maths Chapter 13?

In Chapter 13 of Class 10 Maths, it's all about surface area and volume of 3-dimensional solids. On Vedantu, you can download the NCERT Solutions Class 10 Maths Chapter 13 PDF and get access to the full explanation and solutions of the Class 10 Maths Chapter 13 on Surface Area and Volume. All the exercises and examples have been solved step by step by experts.

8. What are the basics of Class 10 Maths Chapter 13?

The basics that you need to cover in Class 10 Maths Chapter 13 are Surface Area Of Combination Of Solids, Volume Of A Combination Of Solids, Conversion Of Shape From One Solid To Another, and Frustum Of A Cone. To get solutions to exercises related to these topics, you can download the NCERT Solutions Class 10 Maths Chapter 13 on Vedantu and practice daily to ensure good grades.

9. What are the important formulae to remember in Class 10 Maths Chapter 13?

Some of the most important formulae that you need to remember in completing your Class 10 Maths Chapter 13 are - Curved Surface Area Of Cone, Curved Surface Area Of Hemisphere, Height Of The Cone, Slant Height Of The Cone, Curved Surface Area Of Cylinder, Volume Of A Frustum Of Cone, Curved Surface Area Of A Frustum Of Cone, and Total Surface Area Of A Frustum Of Cone. SHARE TWEET SHARE SUBSCRIBE