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NCERT exemplar Class 9 Maths solutions chapter 13 go through the questions related to surface areas of cube, cuboid, etc. Our highly experienced team at Careers 360 has curated these NCERT exemplar Class 9 Maths chapter 13 solutions to provide elaborate and accurate answers to the students practicing the NCERT Class 9 Maths Book. These NCERT exemplar Class 9 Maths chapter 13 solutions build a better understanding of calculations based on surface area and volume as they are highly intricate. The exemplar follows the CBSE Prescribed Syllabus for Class 9 and the same is incorporated in the NCERT exemplar Class 9 Maths solutions chapter 13.

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The radius of a sphere is , then its volume will be

(A)

(B)

(C)

(D)

Answer (D)

We know that volume of a sphere is given as where r is the radius of the sphere.

In this question it is given that the Radius of sphere is 2rSo, Volume of sphere will be equal to:

So, option (D) is the correct.

The total surface area of a cube is . The volume of the cube is:

(A)

(B)

(C)

(D)

Answer (C)

We know that the total surface area of cube is given as (where a is the side length of the cube)Here we have been given that total surface area

So,

Now we know that volume of a cube

Volume

So, option (C) is the correct answer.

A cone is high and the radius of its base is . It is melted and recast into a sphere. The radius of the sphere is :

(A)

(B)

(C)

(D)

In the question it i given that the height of cone

And, Radius of cone

The cone is melted and recast into a sphere

So,

Volume of sphere = Volume of cone

(where R is the radius of sphere, r is the radius of cone, h is the height of cone)

Putting the given values,

Hence, option (B) is the correct answer.

In a cylinder, radius is doubled and height is halved, curved surface area will be

(A) halved

(B) doubled

(C) same

(D) four times

We know that the curved surface area of a cylinder is

(Where r is radius and h is height)

Let, original radius =r

Original heigth = h

Original curved surface area

Now it is given that radius is doubled and height is halved

New radius (R)

New height (H)

So, new curved surface area

Now,

then

So, option (C) is the correct answer.

The total surface area of a cone whose radius is and slant height is

(A)

(B)

(C)

(D)

We know that total surface area of cone is given as

Where R is the radius of its base and L is the slant height.

Now, it is given that :

Radius of cone

Slant height (L)

So total surface area

So, option (B) is the correct answer.

The radii of two cylinders are in the ratio of and their heights are in the ratio of . The ratio of their volumes is:

(A)

(B)

(C)

(D)

We know that volume of a cylinder is given as

Where r is the radius of its base and h is the height.

Now it is given that ratio of Radius

And, Ratio of Heights

So, Ratio of volumes,

So, option (B) is the correct answer.

The lateral surface area of a cube is . The volume of the cube is

(A)

(B)

(C)

(D)

We know that Lateral surface area of cube is given as:

Where 'a' is the edge (side) length of the cube.

Given that lateral surface area of cube

Now,

Then

So, option (A) is the correct answer.

The number of planks of dimensions that can be stored in a pit which is long, wide and deep is

(A)

(B)

(C)

(D)

We can see that both the plank and pit will be in the form of a cuboid.

Volume of a cuboid

Where l is its length, b is breadth and h is height.

Given dimensions of plank

We know that,

So, Dimension of plank

Volume of plank

Now, Dimensions of pit

Volume of pit

Thus number of planks that can be fitted into the pit

This is option (B) is the correct.

The length of the longest pole that can be put in a room of dimensions is

(A)

(B)

(C)

(D)

Dimensions of room

We can see that the room is in form of a cuboid.

Longest side of a cuboid is its diagonal.

So, we can say that the longest pole will be faced along its diagonal.

Diagonal of cuboid

So, option (A) is the correct answer.

We know that volume of the sphere (where r is the radius)

And Volume of cylinder (where R is the radius and h is height)

It is given that height and diameter of cylinder are equal to the diameter of the sphere

(So, radius of cylinder = r and h = 2r)

So, Volume of cylinder

Volume of sphere volume of cylinder

Therefore the given statement is True.

We know that Volume of a cone (where r is the radius and h is height)

Now as radius of a right circular cone is halved and height is doubled,

Then new volume

(original volume of cone )

Thus the statement is false.

Consider a right circular cone, with height = h, radius = r, slant height = l

We know that, in a right triangle, one angle is equal to 90° Using Pythagoras theorem, we know that:

Thus the statement is False.

We know that curved surface area of the cylinder is given as

(where r is the radius and h is height)

Given that the radius of a cylinder is doubled and it’s curved surface area is not changed

For new cylinder,

New radius

Let new height be

Now, according to the question –

Hence the height is halved.

Thus, the statement is true.

We know that

Volume of cone (where r is the radius and h is height)

Volume of the hemisphere (where r is the radius)

For largest right circular cone that can be fitted in a cube whose edge is 2r, the height of cone will be 2r and its diameter will also be equal to 2r. So, radius of its base is equal to r.

So, volume of cone

= volume of hemisphere

Thus, the statement is true.

We know that

Volume of cylinder (where R is the radius and H is height)

Volume of cone (where r is the radius and h is height)

Given that a cylinder and a right circular cone are having the same base and same height.

So let r be the radius and h be the height of both.

Now, Volume of cylinder will be ,

and Volume of cone will be,

So we get:

Therefore,

Hence, the statement is true.

We have been given that a cone, a hemisphere and a cylinder stand on equal bases and have the same height.

Let radius of base is r and height is h.

Now we know that

Volume of cone,

Volume of hemisphere,

Volume of cylinder,

From equations (i) (ii) and (iii)

Thus the statement is true.

We know that the diagonal of cube with side a is given as

Now in the question, the diagonal of the cube is cm

So we can write

Which gives,

So the side of cube is 6 cm

Thus, the statement is false.

The given statement is: If the radius of a cylinder is doubled and height is halved, the volume will be doubled.

We know that the volume of cylinder is given as

where the radius of cylinder is r and height is h.

Let original volume be,

Now the radius of a cylinder is doubled and height is halved,

So,

New Radius (R) = 2r

New Height (H)

Then, new Volume of cylinder

Hence, the given statement is true.

We know that, Volume of a sphere is given as : (where r is the radius)

Given that radius of each sphere

So, Volume of sphere

Volume of 16 spheres

Internal dimensions of given rectangular box

The box is in the form of a cuboid.

Volume of rectangular box

Now it is given that when 16 spheres are packed, the box is filled with preservative liquid. So,

The total volume of rectangular box = Volume of 16 spheres - Volume of preservative liquid

The volume of preservative liquid

Thus, the volume of preservative liquid is

In the question, it is given that:

Volume of water in tank

Depth of water

Now we know that,

Volume of cube is given as (where a is the side length)

The present depth of water volume is given as .

Height of water used = 1.2 m

Volume of water already used from the tank

= area of base of the cube × Height of water used

Hence volume of water already used from the tank is .

Given that a solid spherical ball of diameter (d) 4.2 cm is completely immersed in water.

Radius of ball

Now we know that,

Volume of spherical ball (where r is the radius)

Hence volume of spherical ball

Amount of water displaced is same as the volume of spherical ball.

So we have Amount of water displaced

We have to find the square metres of canvas required for the given conical tent.

Area of canvas required = Curved surface area of conical text.

We know that,

Curved surface area of cone

Where r is the radius of its base and height is h.

As given in the question, dimensions of conical tent are:

Height = 3.5 m

Radius = 12 m

On putting the values,

Curved surface area

So, the area of the canvas

Therefore, the area of canvas required is .

Density (D) of an object is mass (m) per unit volume (v).

So mass is directly proportional to volume for same metal (as density remains same)

For Solid 1

Let Mass =

Volume =

For solid 2

Mass =

Volume =

Now,

In the question, the objects are spheres

Volume of sphere (where R is radius)

So, volume is directly proportional to

Hence,

Putting the given values

So,

Diameter of the smaller one is 5 cm

So,

Hence,

Therefore radius of larger sphere is 5 cm.

Given that a school provides milk to the students daily in cylindrical glasses of diameter 7 cm

So glass is a cylinder with diameter = 7 cm

Radius of cylinder

Now milk is filled in this cylindrical glass.

Height of milk up to which the glass is filled

So, volume of milk filled in the cylindrical glass

(r is the radius, h is the height up to which the milk is filled)

Putting the values,

Now, number of students=1600=Total number of glasses required

Volume of 1600 cylindrical glasses = 1600 (Volume of one glass)

Now we know that 1 litre

Volume of 1600 cylindrical glasses

Hence, to serve 1600 students, we need .

It is given that when cylindrical roller is rolled on the road its covers area

Now,

Area covered by cylindrical roller in one revolution = curved surface area of cylinder

And we know that,

Curved surface area of cylinder (where r is the radius and h is its height)

Given length of roller (This will be taken as height of the roller)

Cylindrical roller's radius = 1.75 m

So, Curved surface area

Total Area covered in one revolution

Now total area covered

So, Number of revolutions

Hence, cylindrical roller makes 200 revolutions.

Given, dimension of overhead tank

Now the tank is in the shape of a cuboid, so we have volume

Volume of tank

Population of village

Daily water requirement for 1 person = 75 litre

Then daily water requirement for 5000 people

Now we know that 1 litre

So, daily water requirement for 5000 people

Number of Days till the water in this tank will last

Hence, the water of this tank will last for 40 days.

Given that a shopkeeper has:

Larger laddoo with radius = 5 cm

Smaller laddoo with radius = 2.5 cm

Volume of sphere (where R is the radius)

Volume of 1 larger laddoo

Volume of 1 smaller laddoo

Volume of 1 larger laddoo = Total volume of all smaller laddoos formed

Let number of smaller laddoos be n.

Volume of n smaller laddoo

So,

Hence 8 smaller laddoos can be formed by the larger laddoo.

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm.

So on revolving we get a right circular cone as shown in the figure.

Radius = 6 cm, Height = 8 cm, Slant Height = 10 cm

Now, we know that Volume of cone (where r is radius and h is height)

Curved surface area (where r is radius and l is slant height)

Hence Volume and curved surface area

We know that

Volume of cylinder is given as (where r = outer radius and h is the height)

Here, length of cylinder will be considered as its height

Outer diameter is given as,

So, outer radius

Thus volume of cylinder

Thickness of iron sheet = 2 cm

Now, inner diameter = outer diameter – 2 (Thickness of iron sheet)

Inner diameter

So, inner radius

Now, Volume of inner cylinder , (where R = inner radius)

Thus, Volume of iron used = Volume of outer cylinder - Volume of inner cylinder

Hence the answer is

It is given that a semi-circular sheet is bent to form an open conical cup.

Now, the radius of the sheet becomes the slant height of the cup and the circumference of the sheet becomes the circumference of the base of the cone.

Diameter of semi-circular sheet

So, radius of semi-circular sheet

Slant height of the conical cup

Now,

Let the radius of base of the cup = r cm.

Circumference of semi-circular sheet = Circumference of base of conical cup

Let the height of cup be h cm.

Then we know that for a right circular cone,

Capacity of the cup = Volume of a cone

A cloth having an area of is shaped into the form of a conical tent of radius

- How many students can sit in the tent if a student, on an average, occupies on the ground?
- Find the volume of the cone.

(ii)

We have, Area of cloth

This cloth is shaped in the form of a conical tent.

Radius of conical tent = 5 cm

So, we have:

Area of cloth = Curved surface area of cone

Curved surface area of cone is given as Where, r = radius of a cone

l = slant height of a cone

So curved surface area of conical tent,

(i) Now it is given that area covered by 1 student

So 110 students can sit in the tent.

(ii) Volume of cone :

For a right circular cone, we have

So, volume

We know that Volume of hemispherical tank

Where, r = radius of hemispherical tank

Now, internal diameter of hemispherical tank is given as 14 m

So, internal radius of hemispherical tank

Tank contains 50 kL of water

We have,

So,

This means that tank contains of water

Volume of hemispherical tank

Now, as we are given that the already tank contains 50 kilolitres of water

Volume of water pumped into the tank = Total Volume of hemispherical tank

Volume of water pumped into the tank

Hence the answer is

The volumes of the two spheres are in the ratio . Find the ratio of their surface areas.

Answer :We have been given that the volume of two sphere is in the ratio

We know that, Volume of sphere is given as (where r is its radius)

And surface area is given as

Let volume of sphere (radius )

And, volume of sphere (radius )

Then,

Then, ratio of areas of both spheres

Hence the required ratio is

A cube of side 4 cm contains a sphere touching its sides. Find the volume of the gap in between.

Answer :Volume of cube is given as

As the cube contains the sphere,

Diameter of sphere = side length of the cube = 4 cm

Radius of sphere

We know that volume of a sphere is given as (where r is the radius)

Volume

Hence the required answer is

Let radius of sphere be r, radius of cylinder be R, height of cylinder be H

Volume of sphere = Volume of cylinder

Now, it is given that r = R

The diameter exceeds the height by 50%.

(i) 30 circular plates, each of radius 14 cm and thickness 3cm are placed one above the another to form a cylindrical solid. Find : the total surface area

(ii) 30 circular plates, each of radius 14 cm and thickness 3cm are placed one above the another to form a cylindrical solid. Find volume of the cylinder so formed.

It is given that 30 circular plates are placed one above the other to form a cylindrical solid.

So we can see that the height of this cylindrical solid = total thickness of 30 circular plates placed one over the other

Given, thickness of one circular plate

Thickness of 30 circular plates

Now, radius of circular plate, r = 14 cm

We know that

Total surface area of the cylindrical solid of formed will be equal to

Where r is its radius and h is the height. So, putting the above calculated values:

Total surface area

Hence, the total surface area of the cylindrical solid is .

(ii)

Answer :

It is given that 30 circular plates are placed one above the other to form a cylindrical solid.

So we can see that the height of this cylindrical solid = total thickness of 30 circular plates placed one over the other

Given, thickness of one circular plate = 3 cm

Thickness of 30 circular plates

Now, radius of circular plate, r = 14 cm

We know that

Volume of the cylinder so formed will be equal to

Where r is its radius and h is the height. So, putting the above calculated values:

Volume of the cylinder

Hence, the volume of the cylindrical solid is

It is given that 30 circular plates are placed one above the other to form a cylindrical solid.

So we can see that the height of this cylindrical solid = total thickness of 30 circular plates placed one over the other

Given, thickness of one circular plate = 3 cm

Thickness of 30 circular plates

Now, radius of circular plate, r = 14 cm

We know that

Volume of the cylinder so formed will be equal to

Where r is its radius and h is the height. So, putting the above calculated values:

Volume of the cylinder

Hence, the volume of the cylindrical solid is

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.

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.

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.

(B) A 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.

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

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

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

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

Therefore a surahi is the combination of a sphere and a cylinder.

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

Solution.

(A) A cone – A cone is a three dimensional geometric shape that tapers smoothly from 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 fixed distance.

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

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.

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

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

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

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

Hence from the above figures we conclude that the given figure of plumbline is the combination of a hemisphere and a cone.

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) A cone – A cone is a three dimensional geometrical shape that topers 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 a glass is usually in the form of frustum of a cone.

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

Solution.

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

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

(A) Two cylinders

(B) A cone and a cylinder

(C) Two cones and a cylinder

(D) Two cylinders and a cone

Hence the shape of a gilli, in the gilli-danda game is a combination of two cone and a cylinder.

NCERT exemplar Class 9 Maths solutions chapter 13 deals with the comprehension of the following topics:

- Surface area of cuboid and cubes which can be seen as the sum of the surface area of six surfaces.
- The volume of cube or cuboid is given as the product of three mutually perpendicular sides.
- NCERT exemplar Class 9 Maths solutions chapter 13 discusses the curved and total surface area of cylinder and cone.
- The surface area of any sphere or hemisphere is discussed in both cases whether it is solid or hollow.
- The volume of cone, cylinder, and sphere.

These Class 9 Maths NCERT exemplar chapter 13 solutions provide a basic knowledge of how to calculate surface areas and volumes of three-dimensional objects. Surface area and volume Calculation has great importance. Surface area of any such object can be seen as the external area which can be painted and the volume of any such object can be seen as the space occupied by them. Students can clarify their doubts and use these solutions as reference material to better understand the concepts of Surface areas and Volumes related practice problems. The Class 9 Maths NCERT exemplar solutions chapter 13 Surface areas and Volumes builds the concepts of this chapter in an organised manner, the learnings from same are sufficient to solve other books such as NCERT Class 9 Maths, RD Sharma Class 9 Maths, RS Aggarwal Class 9 Maths, Mathematics Pearson Class 9, et cetera.

NCERT exemplar Class 9 Maths solutions chapter 13 pdf download is provided by Careers 360 so that students can view/download the solutions and use them while in an offline environment. This is a free to use feature and will help resolve the queries faced while solving the NCERT exemplar Class 9 Maths chapter 13.

Chapter No. | Chapter Name |

Chapter 1 | |

Chapter 2 | |

Chapter 3 | |

Chapter 4 | |

Chapter 5 | |

Chapter 6 | |

Chapter 7 | |

Chapter 8 | |

Chapter 9 | |

Chapter 10 | |

Chapter 11 | |

Chapter 12 | |

Chapter 13 | |

Chapter 14 | |

Chapter 15 |

1. What is the slant length in any cone?

It is also known as the slant height of the cone. It is the distance between the apex of the cone to any point on the perimeter of the circular base. If the radius and height of a cone is known we can find out the slant height by Pythagoras theorem.

2. If a cone and cylinder have the same base area and same height what is the ratio of their volume?

If any cylinder and cone have same base area and height then the volume of cylinder will be three times of the volume of the cone

3. If we melt a cylinder and make cones of the same height and same radius of base. How many such cones can be formed?

For the same radius and height, the volume of the cone is one third the volume of the cylinder. If we melt the cylinder its volume will not change therefore, three such cones can be formed.

4. A solid sphere is cut into equal pieces. What will be the increase in surface area?

When a solid sphere is cut in two equal pieces, two plane surfaces will be created along with the curved surface. The old surface area will be 4π2 and the new surface area will be 6π2. Hence the area will be increased by 150%.

5. Is the chapter on Surface areas and volumes very important for the final examination?

Students should consider each and every chapter as an important chapter because these are the building blocks for your future learning. One can only understand and score well if all the chapters are taken very sternly. NCERT exemplar Class 9 Maths solutions chapter 13 makes your learning experience very smooth and can be referred to score well in the final examination.

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