# NCERT Exemplar Class 10 Maths Solutions Chapter 12 Surface Areas and Volumes

NCERT Exemplar class 10 maths solutions chapter 12 discusses the method to determine the surface area and volume of various three-dimensional objects. These three-dimensional objects can be cones, spheres, or cylinders. The NCERT Exemplar class 10 maths solutions chapter 12 consist of detailed solutions to study and understand the NCERT class 10 Maths. These NCERT exemplar class 10 maths chapter 12 solutions are prepared by our skilled subject matter experts with 10+ years of teaching and content development experience. The CBSE 10th maths syllabus acts as the outline of these NCERT exemplar class 10 maths solutions chapter 12.

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### Question:1

A cylindrical pencil sharpened at one edge is the combination of

(A) a cone and a cylinder (B) frustum of a cone and a cylinder

(C) a hemisphere and a cylinder (D) two cylinders.

Solution

(A) A cone –A cone is a three-dimensional geometric shape that tapers smoothly from a flat base to a point called the apex or vertex.

A cylinder – A cylinder is a three dimensional solid that holds two parallel bases joined by a curve surface at a field distance.

(B) Frustum of a cone – It is the portion of a solid that lies between one or two parallel planes cutting it.

A cylinder – A cylinder is a three dimensional solid that holds two parallel bases joined by a curve surface at a fixed distance.

(C) Hemisphere – In geometry it is an exact half of a sphere and it is a three dimensional geometric.

A cylinder – A cylinder is a three dimensional solid that holds two parallel bases joined by a curve surface at a fixed distance.

(D) Cylinder – A cylinder is a three dimensional solid that holds two parallel bases joined by a curve surface at a fixed distance.

(Combination of two cylinders)

Hence form the above diagrams we conclude that option (A) is correct.

Therefore, A cylindrical pencil sharped at one edge is the combination of a cone and a cylinder.

### Question:2

A Surahi is the combination of

(A) a sphere and a cylinder (B) a hemisphere and a cylinder

(C) two hemispheres (D) a cylinder and a cone.

Solution

(A) A Sphere – The set of all points in three dimensions space lying the same distance from a given point.

A cylinder – A cylinder is a three dimensional solid that holds two parallel bases joined by a curve surface at a fixed distance.

### Question:3

A plumbline (sahul) is the combination of

(A) a cone and a cylinder (B) a hemisphere and a cone

(C) frustum of a cone and a cylinder (D) sphere and cylinder

### Question:4

The shape of a glass (tumbler) (see figure) is usually in the form of

(A) a cone (B) frustum of a cone

(C) a cylinder (D) a sphere

Solution

A cone – A cone is a three-dimensional geometrical shape that tapers smoothly from a flat base to a point called the apex or vertex.

(B) Frustum of a cone – It is a portion of a solid that lies between one or two parallel planes cutting it.

(C) A cylinder – A cylinder is a three dimensional solid that has two parallel based joined by a curved surface at a fixed distance.

(D) A sphere – The set of all points in three-dimensional space lying the same distance from a given point.

Hence the shape of glass is usually in the form of the frustum of a cone.

### Question:5

The shape of a gilli, in the gilli-danda game (See figure), is a combination of

(A) two cylinders (B) a cone and a cylinder

(C) two cones and a cylinder (D) two cylinders and a cone

### Question:6

A shuttle cock used for playing badminton has the shape of the combination of

(A) a cylinder and a sphere

(B) a cylinder and a hemisphere

(C) a sphere and a cone

(D) frustum of a cone and a hemisphere

Solution

Cylinder – A cylinder is a three dimensional solid that has two parallel based joined by a curved surface at a fixed distance.

Sphere – The set of all points in thee dimensional space lying the same distance from a given point.

Similarly hemisphere

Cone – A cone is a three dimensional geometrical shape that tappers smoothly from a flat base to a point called the apex or vertex.

Similarly frustum of cone

Hence we conclude that shuttle cock which is used for playing badminton has the shape of combination of frustum of a cone and hemisphere.

### Question:7

A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the otherside of the plane is called

(A) a frustum of a cone (B) cone

(C) cylinder (D) sphere

Solution

Cone – A cone is a three dimensional geometrical shape that tapers smoothly from a flat base to a point called the apex or vertex.

Frustum of a cone – It is portion of a solid that lies between one or two parallel planes cutting it

Cylinder – A cylinder is a three dimensional solid that has two parallel based joined by a curved surface at a fixed distance.

Sphere – The set of all points in three dimensional space lying the same distance from a given point.

Thus A cone which is cut through a plane parallel to its base and then the cone that is formed one side of that plane is removed then the new part which is formed is called frustum of a cone.

### Question:8

A hollow cube of internal edge 22cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is

(A) 142296 (B) 142396 (C) 142496 (D) 142596

Solution

Diameter of marble = 0.5 cm

Radius

Volume of marble

Edge of cube = 22 cm

Volume (V)

Space occupied by marble = total volume part of volume

Number of marble

### Question:10

A solid piece of iron in the form of a cuboid of dimensions , is moulded to form a solid sphere. The radius of the sphere is

(A) 21cm (B) 23cm (C) 25cm (D) 19cm

Solution

Length of cuboid = 49 cm

Breadth of cuboid = 33 cm

Height of cuboid = 24 cm

Volume

We know that volume of sphere

Also volume of sphere 38808 (given)

Hence radius of sphere = 21 cm

### Question:11

A mason constructs a wall of dimensions with the bricks each of size and it is assumed that space is covered by the mortar. Then the number of bricks used to construct the wall is

(A) 11100 (B) 11200 (C) 11000 (D) 11300

Solution

Length of wall = 270 cm

Breadth = 300 cm

Height = 350 cm

Volume

Length of brick = 22.5 cm

Breadth = 11.25 cm

Height = 8.75 cm

Volume

Space is covered by mortar (given)

Remaining space

Surface constructed

Number of bricks used

### Question:12

Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is

(A) 4 cm (B) 3 cm (C) 2 cm (D) 6 cm

Solution

Diameter of metallic cylinder = 2 cm

Radius

Height = 16cm

Volume

We know that twelve solid sphere are made by melting of solid metallic cylinder

Volume of sphere

Hence volume of 12 spheres

Radius = 1 cm

Diameter

### Question:13

The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm, respectively. The curved surface area of the bucket is

(A) 4950 cm^{2} (B) 4951 cm^{2} (C) 4952 cm^{2} (D) 4953 cm^{2}

^{2}

Solution

Slant height of a bucket = 45 cm

Top radius

Bottom radius

Curved surface area of bucket is

### Question:14

A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. The capacity of the capsule is

(A) 0.36 cm^{3} (B) 0.35 cm^{3} (C) 0.34 cm^{3} (D) 0.33 cm^{3}

### Answer:

Diameter of hemisphere = 0.5 cm

Similarly radius of cylinder = 0.25

Height = 2- 0.25 -0.25

= 2- 0.5

=1.5 cm

The total volume of capsule = volume of two hemispheres + volume of the cylinder

= 0.359cm^{3}

=0.36cm^{3} (approximate)

### Question:15

If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is

(A) (B) (C) (D)

The radius of hemisphere = r

The curved surface area of two solid hemisphere

### Question:16

A right circular cylinder of radius r cm and height h cm (h > 2r) just encloses a sphere of diameter

(A) r cm (B) 2r cm (C) h cm (D) 2h cm

Solution

Here we found that the right circular cylinder of radius r cm and height h cm (h > 2r) can enclosed a sphere of radius upto r cm.

The required sphere is of diameter 2r.

### Question:17

During conversion of a solid from one shape to another, the volume of the new shape will

(A) increase (B) decrease

(C) remain unaltered (D) be doubled

Solution

Volume – Volume is defined as the amount of space the object takes up that is the amount of fluid that the container could hold.

During conversion of a solid from one shape to another, the volume of the new shape remains unchanged.

That is when you convert one solid shape to another then the volume of the original as well as the new solid remains the same.

### Question:18

The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is

(A) 32.7 litres (B) 33.7 litres (C) 34.7 litres (D) 31.7 litres

Solution

Volume of frustum of cone

Diameter of first end = 44 cm

Radius

Diameter of second end = 24 cm

Radius

Height (h) = 35cm

Volume

We know that 1 liter = 1000 cm

^{3}

### Question:19

In a right circular cone, the cross-section made by a plane parallel to the base is a

(A) circle (B) frustum of a cone

(C) sphere (D) hemisphere

### Answer:

According to the question if a right circular cone is cut by a plane parallel to its base the figure formed isHere BECD is not a circle, not a sphere not a hemisphere but it is a frustum of a cone.

Hence in a right circular cone, the cross-section made by a plane parallel to the base is a frustum of a cone.

### Question:20

Volumes of two spheres are in the ratio 64:27. The ratio of their surface areas is

(A) 3 : 4 (B) 4 : 3 (C) 9 : 16 (D) 16 : 9

Let two sphere having radius and

According to question

Ratio of their surface area is

Hence required ratio is 16 : 9

### Question:1

Write ‘True’ or ‘False’ and justify your answer in the following:

Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is .

### Answer:

It is given that there is two hemispheres of radius r.Let A and B are two hemispheres of radius r.

Join A and B along with their bases

Now it is a full sphere of radius r

The total surface area of a sphere

Here we found that the total surface area of the combination is , but not

### Question:2

Write ‘True’ or ‘False’ and justify your answer in the following :

A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is

### Answer:

It is given that there is two cylinders of height h and radius r.Where one is placed on other then the shape formed is

The total surface area of the shape formed

So the surface area is not equal to

### Question:3

Write ‘True’ or ‘False’ and justify your answer in the following :

A solid cone of radius r and height h is placed over a solid cylinder having same base radius and height as that of a cone. The total surface area of the combined solid is

### Answer:

It is given that there is a cone of radius r and height h and a cylinder of height h and radius r.Where A is placed on B

Total surface area = Surface area of cone + Total surface area of cylinder – Surface area of part I – the surface area of part II

Total surface area is not equal to

### Question:4

Write ‘True’ or ‘False’ and justify your answer in the following :

A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is

### Answer:

It is given that the ball is exactly filled inside the cubical box of side a.Hence the diameter of the sphere = a

Hence the given statement is False.

### Question:5

Write ‘True’ or ‘False’ and justify your answer in the following:

The volume of the frustum of a cone is where h is vertical height of the frustum and are the radii of the ends.

According to question

In this figure ABCE is a frustum of cone ABD, h is the height of frustum and are the radii of the frustum.

From the figure

Volume of frustum ABCE = volume of ABD – volume of ECD

[ using (1) ]

Hence the volume of frustum of cone is not equal to

Hence the given statement is false.

### Question:6

Write ‘True’ or ‘False’ and justify your answer in the following:

The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Figure is

Solution

It is given that a cylindrical vessel of height h and radius r is raised upward with a hemispherical portion.

From the figure radius of hemisphere = r cm

The volume of the figure = volume of the cylinder – the volume of the hemisphere

So the given statement is True.

### Question:7

Write ‘True’ or ‘False’ and justify your answer in the following:

An open metallic bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to curved surface area of frustum of a cone + area of circular base + curved surfacearea of cylinder

Solution

According to question, here is a metallic bucket in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet.

The surface area of figure (A) is the surface area of (a), (b), (c)

(a) = curved surface area of frustum

(b) = area of circular base

(c) = curved area of cylinder

Total surface area = curved surface area of frustum of cone + area of circular base + curved surface are of cylinder.

So the given statement is True.

### Question:1

Answer 6 cmSolution

It is given that there are three cubes of sides

The volume of the cube formed by melting these three cubes = volume of first + volume of second + volume of the third.

Let the edge of the cube formed by melting these three cubes is .

Hence the edge of the cube formed is 6 cm.

### Question:2

How many shots each having a diameter of 3 cm can be made from a cuboidal lead solid of dimensions

Answer 84Solution

Here the dimensions of a cuboid are

Volume of cuboid

The diameter of the shot is 3 cm

Radius of shot

The volume of 1 shot

Number of shots

### Question:3

Answer15 cmSolution

Given :

Volume of bucket = 28.490 liters

Radii of the top

Radii of the bottom

Volume of bucket = 28.490 liters

Or

Volume of bucket = 28.490 liters × 1000

=28490 cm

^{3}

(Q volume of frustum of cone )

h = 15 cm

Hence the height of the bucket is 15 cm.

### Question:4

Answer 1:7Solution

When a cone is divided into two parts by a plane through the mid-point the image formed is

In figure is common angle

So the corresponding sides are in equal ratio.

FC = 4 cm

Volume of cone

Volume of frustum of cone

Volume of cone EDC : volume of ABCD

100.48 : 703.36

1 : 7

### Question:5

### Answer:

It is given that volume of cube = 64 cm^{3}

( Because the volume of cube = a

^{3})

So the side of the two cubes are 4 cm

The cuboid formed by joining two cubes.

The surface area of the cuboid

### Question:6

Answer 277 cm^{3}

Solution

The figure formed when a conical cavity is cut out from a cube.

Volume of cube

The volume of the conical cavity

The volume of remaining solid = volume of the cube – the volume of the conical cavity

### Question:7

Answer 854 cm^{2}

Solution

According to question

Here are two cones joined together along their bases

Height of both cone = 15 cm

Base radius of both cone = 8 cm

Surface area of combination = 2(surface area of one cone)

(Q both cones are same)

Hence the surface area of the combination is 854 cm

^{2}

### Question:8

Answer 396.18 cm^{3 }

Solution

Height of the tube = 21 cm

Base radius of the tube =3cm

Volume of tube

Let the height of cone A is h cm

Height of cone

Base radius of both A and B =3 cm

Volume of cone

Volume of cone

It is given that the ratio of the volume is 2: 1

Height of cone A = 14 cm

Height of cone B =21-4=7 cm

Volume of cone

Volume of cone

The volume of remaining portion = Volume of the tube – the volume of cone A – the volume of cone B

= 396.18 cm

^{3}

Volume of remaining portion = 396.18 cm

^{3}

### Question:9

### Answer:

In this figure, there is a hemisphere of radius of 5 cmAnd a cone of radius 5 cm and of height

Volume of cone

Volume of hemisphere

The volume of complete figure = volume of cone + volume of the hemisphere

Volume of ice cream = volume of complete figure - volume of unfilled part

Hence the volume of ice cream is 327.08 cm

^{3}

### Question:10

Answer 150Solution

Given:- Diameter of marble = 1.4 cm

Diameter of beaker = 7cm

Diameter of marble = 1.4 cm

The volume of 1 marble

Diameter of beaker = 7 cm

Water level rises(h) = 5.6cm

Volume of water

Hence 150 marbles should be dropped into the beaker.

So that the water level rises by 5.6 cm.

### Question:11

Answer 1501Solution

It is given that the length, breadth and height of rectangular solid is 66cm, 42cm and 21cm respectively.

Volume of solid rectangular lead piece =

The diameter of spherical lead shot =4.2

The radius of spherical lead shot

Hence 1501 lead shot can be obtained from the lead piece of dimensions 66cm, 42cm and 21cm

### Question:12

Answer 2542Solution

Given : Diameter of spherical lead shot = 4 cm

Edge of cube = 44 cm

Volume of cube =a

^{3}

=44

^{3}(=44 cm)

=85184cm

^{3}

Radius of spherical lead shot =2cm

Hence 2542 lead shots can be made out of a cube of lead whose edge measures 44 cm.

### Question:13

Answer 12960Solution

Length of wall ( because 1m = 100cm )

Breadth of wall

Height of wall

The volume of wall = length × breadth × height

Remaining volume = 57600000 – 5760000 = 51840000 cm

^{3}

Length of brick = 25 cm

Breadth of brick = 16cm

Height of brick = 10cm

Volume of brick = length × breadth × height

= 25 × 16 × 10 = 4000 cm

^{3}

Hence the number of bricks is 12960

### Question:14

Answer 450Solution

Base diameter of disc = 1.5 cm

Radius of disc

Height of disc = 0.2 cm

Volume of disc

Height of required cylinder = 10 cm

Diameter of required cylinder = 4.5cm

Radius of required cylinder

Volume of required cylinder

Number of disc

Hence 450 metallic disc are required to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

### Question:1

Answer h = 28.44 cmSolution

Radius of hemisphere

Volume

Radius of cone = 6 cm

Let height = h

Volume =

If hemisphere is melted and recast into a right circular cone.

Then, the volume of hemisphere = volume of the cone

28.44 = h

### Question:2

Answer h = 8.579mSolution

Length of cuboid = 11m

Breadth = 6m

Height = 5m

Radius of cylindrical tank = 3.5m

Let height = h

To find the height of water level

Volume of cuboid = volume of cylindrical tank

### Question:3

Answer 37867.5g or 37.867 kgSolution

External Length = 36 cm

Breadth = 25 cm

Height = 16.5 cm

Volume

= 36 × 25 × 16.5

Thickness of iron = 1.5 cm

Internal length {subtract border from both sides}

= 33 cm

Breadth

Height

Internal volume

Volume of iron = external volume – internal volume

Weight of one cubic cm of iron = 7.5 g

### Question:4

The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen is used upon writing 3300 words on average. How many words can be written in a bottle of ink containing one-fifth of a litre?

3300 words can be written with 0.001375 L of ink

So with 1L of ink 3300/0.001375 words can be written=2400000

So with 1/5th of a litter 2400000/5 words can be written=48000

### Answer:

Mastermind of math class 10th### Question:5

### Answer:

Diameter of water that flows out of cylinder = 5mm

(Because 1 cm = 10mm)

( Because 1m =100cm)

### Question:6

Answer 74.18 m3, 67.11 m2Solution

Here l is slant height

Curved surface area is

Hence 67.11m2 canvas cloth is required to just cover the heap.

### Question:7

Answer Rs. 2250Solution

Given

Cost of coloring 1dm2 in one day =0.05 Rs

Cost of coloring 45000 dm

^{2}=0.05 x 45000

= 2250 Rs.

### Question:8

Answer 2 hoursSolution

Length of cuboidal pond = 50m

Breadth = 44 m

Speed of water = 15 km per hour = 15000 m

### Question:9

### Answer:

Answer 112mSolution

Length of cuboidal block = 4.4m

Breadth = 2.6 m

Height = 1 m

Volume

Radius of cylindrical pipe r

_{1}= 30cm = 0.3m

r

_{2}= 30+5 = 35cm = 0.35 m

Let, Height h

_{1}

Volume

The volume of cuboid = volume of the cylindrical pipe

### Question:10

### Answer:

Length of cuboidal pond = 80mBreadth = 50 m

Let, Height = h

Average water of one person =0.04m

^{3}

Average water of 500 persons =

According to question

The volume of cuboidal pond = Average water of 500 persons

### Question:11

Answer 487.6 cm^{3}

Solution

Length of cuboidal box = 16 cm

Breadth = 8cm

Height = 8 cm

Radius of sphere = 2 cm

The volume of liquid = volume of the cuboidal box – the volume of 16 spheres

### Question:12

Answer 230.10 Rs.Solution

Upper radius of frustum of cone (R) = 20cm

Lower radius of frustum of cone r = 8cm

Height (h)= 16cm

Volume

= 10.459 liter

Cost of 1 liter milk = 22 Rs.

Cost of 10.459 liter milk

= 2301.10 Rs

### Question:13

### Answer:

r = 36 cm,Height of cylindrical bucket (h

_{1}) = 32 cm

Radius (r

_{1}) = 18 cm

Volume =

Height of conical heap (h

_{2}) = 24 cm

Let radius = r

_{2}

Volume

According to question

Volume of cylindrical bucket = Volume of conical heap

Slant height

### Question:14

### Answer:

Diameter of cylinder = 6 cmRadius = 3 cm

Height = 12 cm

Similarly radius of circle = 3 cm

Slant height of cone (l)=5cm

Radius = 3 cm

Total surface area = area of cylinder + area of circle + area of the cone

Slant height (l) = 5cm

We know that,

### Question:16

### Answer:

The radius of hemispherical bowl = 9 cmRadius of cylindrical bottle = 1.5 cm

Height = 4cm

_________________

### Question:17

### Answer:

Height of cone = 120cmRadius =60cm

Height of cylinder = 180

Radius = 60 cm (given that radius of the cylinder is equal to the radius of the cone)

Volume of water left in the cylinder = volume of cylinder – volume of cone

2036571.43 – 452571.43

= 1584000 cm3

Or 1.584 m3 [Q 1m = 100cm]

2m

^{3}(approximate)

### Question:18

### Answer:

Radius of cylindrical pipe = 1 cmHeight = 80 cm

In half an hour volume of water is

Radius of cylindrical tank = 40 cm

Let height = h

Volume

According to question

The volume of cylindrical pipe = volume of the cylindrical tank

h =89.99

h = 90 cm (approximate)

### Question:19

### Answer:

The radius of the cylindrical vesselHeight = 3.5 cm

Let the height of rainfall = x

Length = 22m

Breadth = 20m

Rainfall = volume of water = volume of the cylindrical vessel

x=0.025 m

Or x= 2.5cm [Q 1 m = 100 cm]

Hence the rainfall is in 2.5 cm

### Question:20

### Answer:

Radius of conical depression = 0.5 cmDepth = 2.1 cm

the volume of 4 cones

Edge of cube = 3

The volume of cube =3

^{3}( Because the volume of cube = a

^{3})

Length of cuboid = 10 cm

Breadth = 5 cm

Height = 4 cm

Volume of wood = volume of cuboid – volume of cube - volume of 4 cones

=200 - 2.2 -27 = 170.8 cm

^{3}

## NCERT Exemplar Solutions Class 10 Maths Chapter 12 Important Topics:

NCERT exemplar class 10 chapter 12 maths solutions covers the following topics:

◊ Surface area and Volume up of frustum.

◊ Area and volume of multiple composite bodies made by known structures.

◊ NCERT Exemplar class 10 maths solutions chapter 12 discusses the method to find out curved surface area and total surface area.

## NCERT Class 10 Solutions for Other Subjects:

NCERT Exemplar Class 10 Maths solutions

NCERT Exemplar Class 10 Science solutions

## NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

Chapter | Link |

NCERT exemplar solution for class 10 Maths Chapter 1 Real Numbers | Click here |

NCERT exemplar solutions for class 10 Maths Chapter 2 Polynomials | Click here |

NCERT exemplar solution for class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables | Click here |

NCERT exemplar solution for class 10 Maths Chapter 4 Quadratic Equations | Click here |

NCERT exemplar solution for class 10 Maths Chapter 5 Arithmetic Progressions | Click here |

NCERT exemplar solution for class 10 Maths Chapter 6 Triangles | Click here |

NCERT exemplar solution for class 10 Maths Chapter 7 Coordinate Geometry | Click here |

NCERT exemplar solution for class 10 Maths Chapter 8 Introduction to Trigonometry & Its Equations | Click here |

NCERT exemplar solution for class 10 Maths Chapter 9 Circles | Click here |

NCERT exemplar solution for class 10 Maths Chapter 10 Constructions | Click here |

NCERT exemplar solution for class 10 Maths Chapter 11 Areas related to Circles | Click here |

NCERT exemplar solution for class 10 Maths Chapter 12 Surface Areas and Volumes | Click here |

NCERT exemplar solution for class 10 Maths Chapter 13 Statistics and Probability | Click here |

## Features of NCERT exemplar class 10 maths solutions chapter 12:

These class 10 maths NCERT exemplar chapter 12 solutions emphasise finding out curved surface area, total surface area, and volume of various three-dimensional objects such as cones, cylinders, and spheres. These calculations and learning will be helpful in physics and maths of higher classes and are essential for both medical and engineering aspirants. The expressive nature of these solutions provides the students a perfect environment to attempt and practice Surface Areas and Volumes based practice problems. The class 10 maths NCERT exemplar solutions chapter 12 Surface Areas and Volumes consist of plenty of practice problems, making the student prepared for NCERT class 10 maths, RD Sharma class 10 maths, RS Aggarwal class 10 maths et cetera.

NCERT exemplar class 10 maths solutions chapter 12 pdf download is a free-to-use feature that provides the students with the facility to study the NCERT exemplar Class 10 maths chapter 12 in an offline environment.

## Frequently Asked Question (FAQs) - NCERT Exemplar Class 10 Maths Solutions Chapter 12 Surface Areas and Volumes

**Question: **How to find out the volume of an ice cream cone?

**Answer: **

Ice cream cones can be seen as a combination of a hemisphere and a cone in simplified form. We know the volume of cone and volume of the hemisphere, so we can find out its complete volume.

**Question: **How to find out the volume of a sharpened pencil?

**Answer: **

We can find out the volume of sharpened pencil by assuming it as a combination of cylinder and a cone.

**Question: **Is the chapter Surface Areas and Volumes important for Board examinations?

**Answer: **

The chapter Surface Areas and Volumes is vital for Board examinations as it holds around 8-10% weightage of the whole paper.

**Question: **What are the critical topics of Surface Areas and Volumes from a board examination perspective?

**Answer: **

It is highly suggested that students practice and study every topic covered in NCERT Exemplar class 10 maths solutions chapter 12 to score high in Surface Areas and Volumes.

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## Questions related to CBSE Class 10th

### what is the timmg of result declaration of class 10 cbse

Hello,

Central Board of Secondary Education may release the CBSE Class 10 result 2021 by July 20, 2021. The officials have not declared any timings exactly when the results will be out. The results will out at any point of the day. Keep an eye on the official website for the latest information. The board publishes CBSE 10th result 2021 online on cbse.nic.in, cbse.gov.in and cbseresults.nic.in. To check the CBSE Class 10th result 2021, students need their roll number, date of birth and other details.

### How to Download MP Board /CBSC 10 th class books PDF 2021-2022

Hy,

If you want the soft copies of the books I suggest you to join some telegram groups which provides books, PDFs , question papers etc . It will be beneficial for you .

Secondly you can download NCERT book app from Google play store for Android where you will get every books of NCERT from class 1 to class 12 for every stream.

Thirdly visit the websites given below

https://ncert.nic.in

https://www.ncertbooks.guru

Hopes this helps.

Wish you luck!

### After class 10 result ,how will subjects distribute to students, on the basis of result!?

It depends on the school in which you want admission.

Some schools have the rule that

i. 75%+ can take science.

ii.60% - 75% can take commerce.

iii.Below 60% will take humanities.

The choice of choosing stream totally depends on students.

And some schools also check your marks in the subject you want to take admission.

Example- If you want to take admission in maths stream they will check your maths and science marks.

If you want to take admission in humanities they will check your social science marks.

Some schools also conduct an entrance exam followed by an interview.

Refer to this link if you are confused about which stream to choose after Class 10th.

https://school.careers360.com/articles/which-stream-select-after-class-x

Feel free to ask if you have any more queries.

All the best!

### Board cbse class 10 Is cancelled and promote all student please sir because of corona do not study

Hello aspirant,

No CBSE and ICSE board didn't not cancelled the 10th and 12th exam.

As they had declared that 10th and 12th class exam will not be cancelled and students will have to give it in offline mode in this pandemic also.

Hope this helps you

All the best for your future

### CBSE class 10th malayalam can someone please tell me a good one

Adamas World school is the best CBSE school in Kolkata. To get admission please visit https://www.adamasworldschool.org/