# NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2

The National Council of Educational Research, also known as NCERT, is self-governing. It was formed in 1961 by the Indian government. Its principal purpose is to assist the central and state governments in achieving the objective of raising the standard of education in the nation. The NCERT is tasked with conducting, organising, and promoting research in areas linked to academic study. They are responsible for producing all written material needed for educational purposes, including textbooks, bulletins, journals, educational kits, multimedia digital content, etc.

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

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**Access NCERT Solutions for Class-10 Maths Chapter 13 – Surface Areas and Volumes **

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

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**Q.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**

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

**Q.2 **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\times \text{Volume of cone}\\ \text{}={\mathrm{\pi r}}^{2}\mathrm{h}+2\times \frac{1}{3}\times {\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.3 **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\times \text{Volume of hemispherical part}\\ \text{}={\mathrm{\pi r}}^{2}\mathrm{h}+2\times \frac{2}{3}\times {\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}{(1.4)}^{2}(2.2)+\frac{4}{3}\mathrm{\pi}{(1.4)}^{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\times 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}\times 1,127.25\approx 338{\text{cm}}^{3}\text{}\end{array}$

**Q.4 **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**

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

**Q.5** 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**

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

**Q.6 **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 cm^{3} of iron has approximately 8g mass. (Use π = 3.14)

**Ans**

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

**Q.7 **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**

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

**Q.8 **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 cm^{3}. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

**Ans**

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

## FAQs (Frequently Asked Questions)

### 1. What is Chapter 13 of Class 10 Mathematics about?

Chapter 13 of Class 10 Mathematics is divided into six sections from 13.1 to 13.6 and these all cover the sub-topics taught under the chapter. Chapter 13 teaches the topic ‘Surface Areas and Volumes’. The area or region that an object’s surface occupies is known as its Surface Area. Volume, on the other hand, refers to how much room an object has. There are numerous shapes and dimensions in Geometry, including Spheres, Cubes, Cuboids, Cones, Cylinders, etc. Each form has its Volume and Surface Area. Chapter 13 talked about all these and more.

The NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 are the solutions to the questions based on the topic covered in Exercise 13.2 of Chapter 13. On the Extramarks website, the students can find the solutions for all six exercises in Chapter 13.

### 2. Can Extramarks’ NCERT Solutions be relied upon?

These solutions are compiled keeping in mind the most recent CBSE guidelines. Theyu are compiled by experts with years of experience. The professionals at Extramarks have developed a variety of useful materials that students may quickly access on the Extramarks website to aid in their understanding and mastery of Mathematics. Students can also access a wide variety of other study material and learning tools via the Extramarks’ website.