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# NCERT Solutions for Class 10 Maths Chapter 12 Surface Area and Volume

Last updated date: 11th Sep 2024
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## Complete Resource of NCERT Solutions for Class 10 Maths Chapter 12 - Free PDF Download

NCERT Solutions for Maths Chapter 12 Class 10 Surface Area and Volume will clear your concepts at the root level. Once you go through the lucid explanations of Class 10 Surface Area and Volume Solutions provided by our subject matter experts, you don't need to mug up formulas. There are many tips and tricks taught here for quickly recalling important points of Chapter 12 Class 10 Maths. Surface Area and Volume Class 10 solution is designed considering the latest CBSE curriculum. Also, you can revise and solve the updated NCERT Class 10 Maths solutions provided by Vedantu.

Table of Content
1. Complete Resource of NCERT Solutions for Class 10 Maths Chapter 12 - Free PDF Download
2. Glance of NCERT Solutions for Maths Chapter 12 Class 10 Surface Area and Volume Class 10
3. Exercises Under NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes
4. Access NCERT Solutions for Class 10 Maths Chapter 12 - Surface Area and Volume
4.1Exercise 12.1
4.2Exercise 12.2
5. Overview of Deleted Syllabus for CBSE Class 10 Maths Surface Area and Volume
6. Class 10 Maths Chapter 12 : Exercises Breakdown
7. Other Study Material for CBSE Class 10 Maths Chapter 12
8. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

## Glance of NCERT Solutions for Maths Chapter 12 Class 10 Surface Area and Volume Class 10

• Calculating the surface area and volume for various 3D shapes.

• Shapes covered include spheres, hemispheres, cones, cylinders, cubes, and cuboids.

• Important formulas are surface area and volume of a sphere, hemisphere, cone, cylinder, and cube, total surface area and curved surface area calculations.

• This article contains chapter notes and important questions for Chapter 12 - Surface Areas and Volumes, which you can download as PDFs.

• There are exercise links provided. It has solutions for each question from the Surface area and volume of a sphere, hemispheres, cones, cylinders, cubes, and cuboids.

• There are two exercises (17 fully solved questions) in Class 10th Maths Chapter 12 Surface.

 S.No. Current Syllabus Exercises of Class 10 Maths Chapter 3 1 NCERT Solutions of Class 10 Maths Surface Area and Volume Exercise 12.1 2 NCERT Solutions of Class 10 Maths Surface Area and Volume Exercise 12.2
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## Exercises Under NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes

### 12.1 Introduction

In the introduction part of Class 10 Chapter 12, you would recall what you studied in class IX about solids like a cube, cylinder, cuboid, etc. You had learned the formulas for finding the surface area and volumes of these solids earlier.

We come across multiple objects in our daily lives that are a combination of many of these shapes. For example, a truck carrying oil which is in the shape of a cylinder that has 2 hemispheres at its end. In the Surface Area Volume Class 10 you would learn how to calculate the surface area and volumes of such solids which are a combination of two or more solid shapes.

### 12.2 Surface Area of a Combination of Solids

The outer part of any 3-D figure is the surface area of that figure. To find out the surface area of a solid which is a combination of solid shapes, we would need to find out the surface area of individual solid shapes separately to find the surface area of the entire 3-D solid shape.

TSA or total surface area of the solid = Curved surface area of (Cone + Cylinder + Hemisphere) = $\pi r \times \sqrt{(l^{2} + r^{2})} (cone) + 2\pi rl (cylinder) + 2\pi r^{2} (hemisphere)$.

### 12.3 Volume of a Combination of Solids

The volume of solids by joining two or more basic solids is the sum of the volumes of individual solids. The total volume of the solid is obtained by adding up the volume of two constituent solids.

Volume of cuboid = l * b * h.

Volume of pyramid = ⅓ * area of the base * height.

Volume of the solid = Volume of cuboid + Volume of pyramid.

Exercise 12.1: This exercise involves questions that are based on finding the surface areas and volumes of cuboids and cubes. The exercise contains a total of six questions that test the student's understanding of the formulas and concepts related to finding the surface area and volume of cuboids and cubes. The solutions to each question provide step-by-step instructions on how to find the surface area and volume of the given figures.

Exercise 12.2: This exercise involves questions based on finding the surface areas and volumes of the right circular cylinders and cones. The exercise contains a total of six questions that test the student's understanding of the concepts related to the surface area and volume of cylinders and cones. The solutions to each question provide step-by-step instructions and diagrams to help students understand the concept better.

## Access NCERT Solutions for Class 10 Maths Chapter 12 - Surface Area and Volume

### Exercise 12.1

1. $2$ cubes of each of volume $64\text{ c}{{\text{m}}^{3}}$ are joined end to end. Find the surface area of the resulting cuboids.

Ans:

Given that,

$2$ cubes are joined end to end as given in the following diagram.

To find the surface area of the resulting cuboid.

Volume of each cube $=64\text{ c}{{\text{m}}^{3}}$

We know that,

Volume of a cube $={{a}^{3}}$

${{a}^{3}}=64$

$a=4\text{ cm}$

Thus the dimension of the resulting cuboid is of $4\text{ cm, }4\text{ cm}$ and $8\text{ cm}$when they are joined end to end. That is, $l=4\text{ cm, b=}4\text{ cm}$and $h=8\text{ cm}$

Then,

Surface area of cuboid 

$=2\left( \left( 4\times 4 \right)+\left( 4\times 8 \right)+\left( 4\times 8 \right) \right)$

$=2\left( 16+32+32 \right)$

$=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 3 \right)}^{2}}}$

$=160\text{ c}{{\text{m}}^{2}}$

The surface area of the resultant cuboid is $160\text{ c}{{\text{m}}^{2}}$

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $14\text{ cm}$ and the total height of the vessel is $13\text{ cm}$. Find the inner surface area of the vessel. (Use $\pi =\frac{22}{7}$)

Ans:

Given that,

The diameter of the hemisphere $=14\text{ cm}$

The total height of the vessel 

To find,

The inner surface area of the vessel.

Thus the radius of the hollow hemisphere $=\frac{d}{2}$

$=\frac{14}{2}$

$=7\text{ cm}$

From the diagram, it can be observed that the radius of the cylindrical part and that of the hemispherical part is the same.

Thus, height of hemispherical part $=$ Radius $=7\text{ cm}$

Height of the cylindrical part$=13-7$

$=6\text{ cm}$

Inner surface area of the vessel$=$ CSA of cylindrical part + CSA of hemispherical part

CSA of cylindrical part $=2\pi rh$

$=2\times \frac{22}{7}\times 7\times 6$

$=\left( 2\times 22\times 6 \right)$

$=44\times 6$

CSA of hemispherical part$=2\pi {{r}^{2}}$

$=\left( 2\times \frac{22}{7}\times \left( {{7}^{2}} \right) \right)$

$AB,BC,CD=\left( 2\times 22\times 7 \right)$

$=44\times 7$

Inner surface area of the vessel$=44\left( 6+7 \right)$

$=44\left( 13 \right)$

$=572\text{ c}{{\text{m}}^{2}}$

$\therefore$The inner surface area of the vessel is $572\text{ c}{{\text{m}}^{2}}$

3. A toy is in the form of a cone of radius $3.5\text{ cm}$ mounted on a hemisphere of the same radius. The total height of the toy is $15.5\text{ cm}$. Find the total surface area of the toy. (Use $\pi =\frac{22}{7}$)

Ans:

Given that,

Radius of the cone $=3.5\text{ cm}$

Radius of the hemisphere$=3.5\text{ cm}$

The total height of the toy$=15.5\text{ cm}$

To find,

The total surface area of the toy

From the diagram, it can be observed that the radius of both conical and hemispherical part is the same.

Height of the hemispherical part

$=\frac{7}{2}$

Height of the conical part$=15.5-3.5$

$=12\text{ cm}$

Slant height of the conical part$\left( l \right)=\sqrt{{{r}^{2}}+{{h}^{2}}}$

$=\sqrt{{{\left( \frac{7}{2} \right)}^{2}}+{{\left( 12 \right)}^{2}}}$

$=\sqrt{\left( \frac{49}{4} \right)+144}$

$=\sqrt{\frac{625}{4}}$

$=\frac{25}{2}D\left( -3,0 \right)$

Total surface area of the toy$=$ CSA of conical part + CSA of hemispherical part

CSA of conical part   $=\pi rl$

$=\left( \frac{22}{7}\times \frac{7}{2}\times \frac{25}{2} \right)$

=137.5

CSA of hemispherical part  $=2\pi {{r}^{2}}$

$=\left( 2\times \frac{22}{7}\times {{\left( \frac{7}{2} \right)}^{2}} \right)$

$=77$

Total surface area of the toy$=137.5+77$

$=214.5\text{ c}{{\text{m}}^{2}}$

$\therefore$The total surface area of the toy is $214.5\text{ c}{{\text{m}}^{2}}$.

4. A cubical block of side $7\text{ cm}$ is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. (Use $\pi =\frac{22}{7}$)

Ans:

Given that,

Side of a cube $=7\text{ cm}$

To find,

• The greatest diameter of the hemisphere.

• The surface area of the solid.

It can be observed from the diagram that the greatest possible diameter of the hemisphere is equal to the cube’s edge.

Thus the greatest diameter of the hemispherical part$=7\text{ cm}$

So the radius of the hemispherical part $=\frac{7}{2}$

$=3.5\text{ cm}$

Total surface area of the solid = Surface area of cubical part + CSA of hemispherical part – Area of hemispherical part

$=6{{a}^{2}}+2\pi {{r}^{2}}-\pi {{r}^{2}}$

$=6{{a}^{2}}+\pi {{r}^{2}}$

$=6{{\left( 7 \right)}^{2}}+\left( \frac{22}{7}\times {{\left( \frac{7}{2} \right)}^{2}} \right)$

$=294+38.5$

$=332.5\text{ c}{{\text{m}}^{2}}$

$\therefore$The greatest diameter of the hemisphere is $7\text{ cm}$ and the surface area of the solid is $332.5\text{ c}{{\text{m}}^{2}}$.

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:

Given that,

Diameter of the hemisphere = Edge of the cube

To determine,

The surface area of the remaining solid.

Diameter of the hemisphere = Edge of the cube 

Radius of the hemisphere $=\frac{l}{2}$

Total surface area of the solid = Surface area of cubical part + CSA of hemispherical part – Area of base of hemispherical part

$=6{{a}^{2}}+2\pi {{r}^{2}}-\pi {{r}^{2}}$

$=6{{a}^{2}}+\pi {{r}^{2}}$

$=6{{l}^{2}}+\pi {{\left( \frac{l}{2} \right)}^{2}}$

$=6{{l}^{2}}+\frac{\pi {{l}^{2}}}{4}$

$=\frac{24{{l}^{2}}+\pi {{l}^{2}}}{4}$

$=\frac{1}{4}\left( 24+\pi \right){{l}^{2}}\text{ uni}{{\text{t}}^{2}}$

$\therefore$The area of the remaining solid is $\frac{1}{4}\left( 24+\pi \right){{l}^{2}}\text{ uni}{{\text{t}}^{2}}B\left( 3,1 \right)$.

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see the given figure). The length of the entire capsule is $14\text{ mm}$ and the diameter of the capsule is $5\text{ mm}$. Find its surface area. (Use $\pi =\frac{22}{7}$)

Ans:

Given that,

Length of the capsule $=15\text{ mm}$

Diameter of the capsule $=5\text{ mm}$

To find,

The surface area of the capsule

From the diagram,

$=\frac{\text{Diameter of the capsule}}{2}$

$=\frac{5}{2}$

Length of the cylindrical part (h) = Length of the entire capsule - $2r$

$=14-2\left( \frac{5}{2} \right)$

$=9\text{ mm}$

Surface area of capsule = 2CSA of hemispherical part + CSA of cylindrical part

2CSA of hemispherical part  $=2\left( 2\pi {{r}^{2}} \right)$

$=4\pi {{\left( \frac{5}{2} \right)}^{2}}$

$=25\pi$

CSA of cylindrical part  $=2\pi rh$

$=2\pi \left( \frac{5}{2} \right)\times 9$

Surface area of the capsule$=25\pi +45\pi$

$=70\pi$

$=70\times \frac{22}{7}$

$=10\times 22$

$=220\text{ m}{{\text{m}}^{2}}$

∴The surface area of the capsule is $220\text{ m}{{\text{m}}^{2}}$.

7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1\text{ m}$ and $4\text{ m}$ respectively, and the slant height of the top is $2.8\text{ m}$, 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 $Rs.500$ per ${{m}^{2}}$. (Note that the base of the tent will not be covered with canvas.)

Ans:

Given that,

Height of the cylindrical part $=2.1\text{ m}$

Diameter of the cylindrical part $=4\text{ m}$

Radius of the cylindrical part$=2\text{ m}$

Slant height of conical part $=2.8\text{ m}$

Cost of $1\text{ }{{\text{m}}^{2}}$ canvas $=Rs.500$

Area of the canvas used = CSA of conical part + CSA os cylindrical part

$=\pi rl+2\pi rh$

$=\left( \pi \times 2\times 2.8 \right)+\left( 2\pi \times 2\times 2.1 \right)$

$=2\pi \left( 2.8+4.2 \right)$

$=2\times \frac{22}{7}\times 7$

$=44\text{ }{{\text{m}}^{2}}$

Cost of $1\text{ }{{\text{m}}^{2}}$ canvas $=Rs.500$

Cost of  canvas$=44\times 500$

$=Rs.22,000$

$\therefore$The cost the canvas that is used to cover the tent is $Rs.22,000$.

8. From a solid cylinder whose height is $2.4\text{ cm}$ and diameter $1.4\text{ cm}$, 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 $c{{m}^{2}}$. (Use $\pi =\frac{22}{7}$)

Ans:

Height of the cylindrical part = Height of the conical part$=2.4\text{ cm}$

Diameter of the cylindrical part = Diameter of the conical part $=1.4\text{ cm}$

To find,

The total surface area of the remaining solid.

Diameter of the cylindrical part $=1.4\text{ cm}$

Radius of the cylindrical part $\frac{1.4}{2}$

$=0.7\text{ cm}$

Slant height of conical part $=\sqrt{{{r}^{2}}+{{h}^{2}}}$

$=\sqrt{0.49+5.76}$

$=3\sqrt{2}=\sqrt{6.25}$

$=2.5$

The total surface area of the solid = CSA of cylindrical part + CSA of conical part + Area of cylindrical base

$=2\pi rh+\pi rl+\pi {{r}^{2}}C\left( 4,3 \right)$

$=\left( 2\times \frac{22}{7}\times 0.7\times 2.4 \right)+\left( \frac{22}{7}\times 0.7\times 2.5 \right)+\left( \frac{22}{7}\times {{\left( 0.7 \right)}^{2}} \right)$

$=\left( 4.4\times 2.4 \right)+\left( 2.2\times 2.5 \right)+\left( 2.2\times 0.7 \right)$

$=10.56+5.5+1.54$

$=17.6\text{ c}{{\text{m}}^{2}}$

$\therefore$The total surface area of the remaining solid to the nearest $c{{m}^{2}}$ is $18\text{ c}{{\text{m}}^{2}}$.

9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the given figure. If the height of the cylinder is $10\text{ cm}$ and its base is of radius $3.5\text{ cm}$, find the total surface area of the article. (Use $\pi =\frac{22}{7}$)

Ans:

Given,

Height of the cylindrical part$=10\text{ cm}$

Radius of the cylindrical part = Radius of the hemispherical part $=3.5$

The total surface area of the article = CSA of cylindrical part + 2CSA of hemispherical part

$=2\pi rh+2\times 2\pi {{r}^{2}}$

$=\left( 2\pi \times 3.5\times 10 \right)+4\pi {{\left( 3.5 \right)}^{2}}$

$=70\pi +49\pi$

$=119\pi$

$=119\left( \frac{22}{7} \right)$

$=374\text{ c}{{\text{m}}^{2}}$

$\therefore$The total surface area of the article is $374\text{ c}{{\text{m}}^{2}}$.

### Exercise 12.2

1.A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1\text{ cm}$ and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.

Ans:

Given that,

Radius of cone and the hemisphere $=1\text{ cm}$

Height of the cone = Radius of the cone $=1\text{ cm}$

To find,

The volume of the solid in terms of $\pi$.

Volume of the given solid = Volume of the conical solid + Volume of the Hemispherical solid

$=\frac{1}{3}\pi {{r}^{2}}h+\frac{2}{3}\pi {{r}^{3}}{{\left( x-2 \right)}^{2}}+25={{\left( x+2 \right)}^{2}}+81$

$=\frac{1}{3}\pi {{\left( 1 \right)}^{2}}\left( 1 \right)+\frac{2}{3}\pi {{\left( 1 \right)}^{3}}$

$=\frac{1}{3}\pi +\frac{2}{3}\pi$

$=\frac{3}{3}\pi$

$=\pi$

$\therefore$ The volume of the solid in terms of  is $\pi \text{ c}{{\text{m}}^{3}}$.

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 aluminum sheet. The diameter of the model is $3\text{ cm}$ and its length is $x=\pm 412\text{ cm}$. If each cone has a height of $2\text{ cm}$, 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). (Use $\pi =\frac{22}{7}$)

Ans:

Given that,

Diameter of the cylindrical part $=3\text{ cm}$

Length of the cylindrical part $=12\text{ cm}$

Height of the conical part = 2cm

To find,

The volume of the air contained in the model

It can be observed from the figure that the,

Height of each conical part $\left( {{h}_{1}} \right)=2\text{ cm}$

Height of cylindrical part $\left( {{h}_{2}} \right)=12-2\times$Height of the conical part

$=12-\left( 2\times 2 \right)$

$=12-4$

$=8\text{ cm}$

Diameter of the cylindrical part $=3\text{ cm}$

$=\frac{3}{2}$

Volume of the air in the model = Volume of cylindrical part + 2(Volume of cones)

Volume of the cylinder 

$=\pi {{\left( \frac{3}{2} \right)}^{2}}\left( 8 \right)$

$=\pi \times \frac{9}{4}\times 8$

$=18\pi$

Volume of $2$ cones$=2\times \frac{1}{3}\pi {{r}^{2}}h$

$=2\times \frac{1}{3}\pi \times {{\left( \frac{3}{2} \right)}^{2}}\times 2$

$=\frac{2}{3}\pi \times \frac{9}{4}\left( 2 \right)$

$=3\pi$

Volume of the air in cuboid $=18\pi +3\pi$

$=21\pi$

$=21\left( \frac{22}{7} \right)$

=66 cm$^{3}$

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 $5\text{ cm}$ and diameter $2.8\text{ cm}$(see the given figure). (Use $\pi =\frac{22}{7}$)

Ans:

Given that,

The length of  the gulab jamun $=5\text{ cm}$

Diameter of the gulab jamun $=2.8\text{ cm}$

The volume of syrup in gulab jamun is $30%$ to its volume.

To find,

The volume of syrup in $45$ gulab jamun.

The diagram of gulab jamun shaped like a cylinder with two hemispherical ends is shown in the following diagram.

From the diagram,

Radius of the cylindrical part$\left( {{r}_{1}} \right)$ = Radius of hemispherical part $\left( {{r}_{2}} \right)$

$=\frac{2.8}{2}$

$=1.4\text{ cm}$

The length of the hemispherical part is the same as that of the radius of the hemispherical part.

Length of each hemispherical part $=1.4\text{ cm}$

Height of the cylindrical part$=5-2\times$Length of hemispherical part

$=5-2\left( 1.4 \right)$

$=5-2.8$

$=2.2\text{ cm}$

Volume of one gulab jamun = Volume of cylindrical part$+2$(Volume of hemispherical part

Volume of cylindrical part$=\pi {{r}^{2}}h$

$=\pi {{\left( 1.4 \right)}^{2}}\left( 2.2 \right)$

$=\frac{22}{7}{{\left( 1.4 \right)}^{2}}\times 2.2$

$=13.552$

$2$(Volume of hemispherical part) $=2\times \frac{2}{3}\pi {{r}^{3}}$

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

$=\frac{4}{3}\pi {{\left( 1.4 \right)}^{3}}$

$=\frac{4}{3}\times \frac{22}{7}\times {{\left( 1.4 \right)}^{3}}$

$=11.498$

Volume of one gulab jamun$=13.552+11.498$

$=25.05\text{ c}{{\text{m}}^{3}}$

Volume of $45$ gulab jamuns$=45\times 25.05$

$=1,127.25\text{ c}{{\text{m}}^{3}}$

olume of sugar syrup $=30%$ of volume

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

$=338.17\text{ c}{{\text{m}}^{3}}$

$\therefore$The volume of sugar syrup found in $45$ gulab jamuns is approximately $338\text{ c}{{\text{m}}^{3}}$.

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 cuboids are $15\text{ cm}$ by $10\text{ cm}$ by $3.5\text{ cm}$. The radius of each of the depressions is $0.5\text{ cm}$ and the depth is $1.4\text{ cm}$. Find the volume of wood in the entire stand (see the following figure. (Use $\pi =\frac{22}{7}$)

Ans:

Length of the cuboid$=15\text{ cm}$

Breadth of the cuboid$=10\text{ cm}$

Height of the cuboid$=3.5\text{ cm}$

Radius of conical depression $=0.5\text{ cm}$

Height of conical depression $=1.4\text{ cm}$

To find,

The volume of wood in entire stand

Volume of the wood = Volume of cuboid - $4\times$Volume of cones

Volume of cuboid $=lbh$

$=15\times 10\times 3.5$

$=525\text{ c}{{\text{m}}^{3}}$

$4\times$Volume of cones$=4\times \frac{1}{3}\pi {{r}^{2}}h$

$=\frac{4}{3}\times \frac{22}{7}\times {{\left( \frac{1}{2} \right)}^{2}}\times 1.4$

$=1.47\text{ c}{{\text{m}}^{3}}$

Volume of the wood$=525-1.47$

$=523.53\text{ c}{{\text{m}}^{3}}$

$\therefore$The volume of the wood in the entire stand is $523.53\text{ c}{{\text{m}}^{3}}$.

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

Ans:

Given that,

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

Radius of conical vessel$\left( {{r}_{1}} \right)=5\text{ cm}$

Radius of lead shots $\left( {{r}_{2}} \right)=0.5\text{ cm}$

One-fourth of water flows out from the vessel.

To find,

The number of lead shots dropped in the vessel.

Let the number of lead shots that has been dropped in the vessel be $n$.

Volume of water flows out = Volume of lead shots dropped in the vessel

$\frac{1}{4}\times$Volume of the cone $=n\times \frac{4}{3}\pi {{r}_{2}}^{3}$

$\frac{1}{4}\times \frac{1}{3}\pi {{r}_{1}}^{2}h=n\times \frac{4}{3}\pi {{r}_{2}}^{3}$

${{r}_{1}}^{2}h=16n{{r}_{2}}^{3}$

Substituting the values we known, we obtain,

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

$n=\frac{200}{16\times {{\left( 0.5 \right)}^{3}}}$

$n=100$

$\therefore$The number of lead shots dropped in the vessel is $100$.

6. A solid iron pole consists of a cylinder of height $220\text{ cm}$ and base diameter $24\text{ cm}$, which is surmounted by another cylinder of height $60\text{ cm}$and radius $8\text{ cm}$. Find the mass of the pole, given that $1\text{ c}{{\text{m}}^{3}}$ of iron has approximately $8\text{ g}$mass. (Use $\pi =3.14$)

Ans:

Let there be two cylinders, one is of larger and the other is smaller

Given that,

Height of the larger cylinder ${{h}_{1}}=220\text{ cm}$

Diameter of the larger cylinder ${{d}_{1}}=24\text{ cm}$

Height of the smaller cylinder${{h}_{2}}=60\text{ cm}$

Radius of the smaller cylinder ${{r}_{2}}=8\text{ cm}$

To find,

The mass of iron pole.

Radius of larger cylinder $\left( {{r}_{1}} \right)=\frac{24}{2}$

$=12\text{ cm}$

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

Volume of the larger cylinder$=\pi {{r}_{1}}^{2}{{h}_{1}}$

$=\pi {{\left( 12 \right)}^{2}}\times 220$

$=31,680\pi$

Volume of the smaller cylinder$=\pi {{r}_{2}}^{2}{{h}_{2}}$

$=3840\pi$

Volume of the iron pole $=31,680\pi +3,840\pi$

$=35,520\pi$

$=35,520\left( 3.14 \right)$

$=111,532.8\text{ c}{{\text{m}}^{3}}$

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

Mass of $111,532.8\text{ c}{{\text{m}}^{3}}$ iron $=111,532.8\times 8$

$=892262.4\text{ g}$

We know that $1000\text{ g}=1\text{ kg}$

$892262.4\text{ g}=892262.4\times 1000\text{ kg}$

$=892.262\text{ kg}$

$\therefore$The mass of the iron is $892.262\text{ kg}$.

7. A solid consisting of a right circular cone of height $120\text{ cm}$ and radius $60\text{ cm}$ standing on a hemisphere of radius $60\text{ cm}$ is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of the water left in the cylinder, if the radius of the cylinder is $60\text{ cm}$ and its height is $180\text{ cm}$. (Use $\pi =\frac{22}{7}$).

Ans:

Given that,

A solid with a right circular cone and a hemisphere.

Height of the conical part of the cylinder $=120\text{ cm}$

Radius of the conical part of the cylinder$=60\text{ cm}$

Radius of hemispherical part of the cylinder $=60\text{ cm}$

Radius of the outer cylinder $=60\text{ cm}$

Height of the outer cylinder $=180\text{ cm}$

To find,

The volume of the water left in the cylinder.

Volume of the water left = Volume of cylinder – Volume of the solid

Volume of the cylinder$=\pi {{r}^{2}}h$

$=\pi {{\left( 60 \right)}^{2}}\left( 180 \right)$

$=\pi \times 3600\times 180$

$=648,000\pi \text{ c}{{\text{m}}^{3}}$

Volume of the solid = Volume of cone + Volume of the hemisphere

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

$=\frac{1}{3}\pi {{\left( 60 \right)}^{2}}120+\frac{2}{3}\pi {{\left( 60 \right)}^{3}}$

$=\frac{1}{3}\pi \left( 432,000 \right)+\frac{2}{3}\pi \left( 216,000 \right)$

$=288,000\pi$

Volume of the water left$=648,000\pi -288,000\pi$

$=360,000\pi$

$=360,000\times \frac{22}{7}$

$=11311428.57\text{ c}{{\text{m}}^{3}}$

$=1.131\text{ }{{\text{m}}^{3}}$

$\therefore$The volume of the water left in the cylinder is $1.131\text{ }{{\text{m}}^{3}}$.

8. A spherical glass vessel has a cylindrical neck 8cm long, $2\text{ cm}$ in diameter; the diameter of the spherical part is $8.5\text{ cm}$. By measuring the amount of water it holds, a child find its volume to be $345\text{ c}{{\text{m}}^{3}}$. Check whether she is correct, taking the above as the inside measurement and $\pi =3.14$

Ans:

Given that,

Height of the cylindrical part $=8\text{ cm}$

Diameter of cylindrical neck $=2\text{ cm}$

Diameter of spherical glass vessel $=8.5\text{ cm}$

Volume of the water that the vessel holds $=345\text{ c}{{\text{m}}^{3}}$

To find it the above given volume is correct.

Volume of the vessel = Volume of sphere + Volume of cylinder

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

$=\frac{4}{3}\times 3.14\times {{\left( \frac{8.5}{2} \right)}^{3}}$

$=321.392\text{ c}{{\text{m}}^{3}}$

Volume of cylinder $=\pi {{r}^{2}}h$

$=3.14\times {{\left( 1 \right)}^{2}}\times 8$

$=25.12\text{ c}{{\text{m}}^{3}}$

Volume of the vessel $=321.392+25.12$

$=346.51\text{ c}{{\text{m}}^{3}}$

$\therefore$The volume of the vessel is $346.51\text{ c}{{\text{m}}^{3}}$ and hence the child is wrong.

## Overview of Deleted Syllabus for CBSE Class 10 Maths Surface Area and Volume

 Chapter Dropped Topics Surface Area and Volume 12.4 : Conversion of solid from one shape to another 12.5 : Frustum of a cone

## Class 10 Maths Chapter 12 : Exercises Breakdown

 Exercise Number of Questions Exercise 12.1 9 Questions with Solutions (7 Long, 2 Short questions) Exercise 12.2 8 Questions with Solutions (7 Long and 1 Short questions)

## Conclusion

The NCERT Solutions for Class 10 Maths Chapter 12 Surface Area and Volume provided by Vedantu offer comprehensive explanations and step-by-step solutions to the problems in this chapter. It's crucial to focus on understanding the formulas for the surface area and volume of different shapes like cylinders, cones, spheres, and hemispheres. This chapter is important for building a solid foundation in geometry. In the previous year's question papers, about 3-4 questions were typically asked from this chapter, highlighting its significance in exams. Make sure to practice these problems thoroughly to score well.

## Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths Chapter 12. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 10 Maths Chapter 12 Surface Area and Volume

1. List out all the formulas of Class 9 Maths Chapter 13 Surface Area and Volume

All the important formulas that are included in Class 9 Maths Chapter 13 Surface Area and Volume are listed in the table below:

 NCERT Class 9 Maths Chapter 13 Surface Area and Volume Formulas Objects Total Surface Area Curved Surface Area Volume Cube 6a2 4a2 a2 Cuboid 2(lb+bh+hl) 2(l+b)h lbh Right-circular Cone πr(l+r) πrl ⅓πr2h Right-circular Cylinder 2πr(h+r) 2πrh π2h Sphere 4πr2 4πr2 4/3πr3 Hollow Hemisphere 2πr2 2πr2 2/3πr3 Solid Hemisphere 3πr2 2πr2 2/3πr3

2. What are the important topics that come under NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes?

The important topics that are included under NCERT Class 9 Maths Chapter 13 Surface Areas and Volumes are mentioned below:

• Curved surface area, total surface area and volume of cubes and cuboids

• Surface area and volume of a right circular cone and right-circular cylinder.

• The curved surface area of a sphere and a hemisphere

• Surface area and volume of a sphere, hollow hemisphere and solid hemisphere

• Curved surface area and volume of a right circular cylinder and cone.

All the above-mentioned topics are very important for the CBSE Class 9 final examination. Students must practise these topics and formulas regularly for better performance.

You can download free PDFs of NCERT Solutions for Class 9 Maths Chapter 13 by registering yourself on the Vedantu app. We provide free PDF downloads of solved NCERT Solutions which are created by expert teachers in a very simple and step-by-step manner.

4. How can Vedantu help me in scoring good grades in Class 10 Maths?

Class 10 holds an important place in the life of every student. Securing good grades on 10th boards is what every student wishes for. Therefore, it is important to have the best study partner for exam preparations. Given that, Vedantu comes with exclusive study materials developed by a team of experts. Also, you will get NCERT Solutions for Class 10 Maths, mock tests, and previous year papers for Class 10 Maths.

5. Is Class 10 Maths Chapter 13 important for Boards?

Class 10 Maths Chapter 13 is essential for Boards. This chapter may present you with long word-problem-type questions. Since it is fairly interesting and fun to solve, all you need to do is recall the key formulas of surface areas and volumes and their applications. You can use Vedantu's study materials to help you understand all of the ideas. You will be given a series of questions to aid you with quality practices.

6. How is Chapter 13 ‘Surface area and volume’ helpful in real life?

Chapter 13 ‘Surface area and volume’ is not just a chapter that teaches you a few formulas but if you understand the concepts thoroughly, you can apply them in real-world situations as well. In our daily life, we apply them unknowingly. For example, filling oil in a can, covering the floor with tiles, making a box out of cardboard, or wrapping a box with gift paper, all come under surface area and volume.

7. Are PDFs of NCERT Solutions for Class 10 Maths Chapter 13 available online?

Yes, PDFs of NCERT Solutions for Class 10 Maths Chapter 13 are available on the online platform of Vedantu (vedantu.com). It is freely accessible for download from the website as well as the Vedantu app. The NCERT solutions are offered in an easy-to-understand and detailed format. Subject matter experts have ensured that the finest answers are curated for students. You can browse over these solutions and get clarity on any sort of query or doubt.

8. How should I prepare myself for Class 10 CBSE Boards?

The Class 10 boards are essential in determining your specializing subject for later. Hence, we recommend that you take the following steps to ensure good scores in the Class 10 Boards:

•  Choose books wisely.

•  Focus on important chapters and topics.

•  Once the subject is prepared, revise and practice on a regular basis.

•  Solve the sample papers and also previous years' exam papers.

9. How to avoid mistakes in surface area and volume in Chapter 12 maths class 10?

Understanding what surface area and volume mean is key. Imagine the difference between a box's painted surface and the amount of air it can hold. When solving problems, use the correct formula for the shape (cube, cylinder, etc.) and plug in the right measurements (length, radius, etc.). Always double-check your calculations and make sure your answer has the correct unit (square centimetres for area, cubic centimetres for volume). By following these simple tips, you'll be a surface area and volume pro in no time!

10. Why is surface area and volume important in class 10 maths surface area and volume ?

These concepts are crucial in real-world applications! Surface area helps calculate paint needed, material costs, or heat transfer in objects. Volume helps determine how much liquid a container can hold or the amount of material needed to build something.

11. What is the key concept of surface area and volume class 10?

Surface area and volume are like two sides of the same coin; they both deal with 3D shapes, but in different ways. Surface area focuses on the total "outer covering" of a shape, like the amount of paint you'd need to coat a box. Volume, on the other hand, deals with the "inner space" a shape can hold, like how much juice a cylindrical glass can contain. Mastering these concepts is like having a 3D measuring toolkit to solve problems involving boxes, cans, balls, and more!