NCERT Exemplar Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes

# NCERT Exemplar Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes

Edited By Safeer PP | Updated on Aug 31, 2022 03:29 PM IST

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.

### Question:1

The radius of a sphere is $2r$, then its volume will be
(A) $\frac{4}{3}\pi r^{3}$
(B) $4\pi r^{3}$
(C) $\frac{8\pi r^{3}}{3}$
(D) $\frac{32}{3}\pi r^{3}$

We know that volume of a sphere is given as $\frac{4}{3}\pi r^{3}$ where r is the radius of the sphere.

In this question it is given that the Radius of sphere is 2r
So, Volume of sphere will be equal to:
$=\frac{4}{3}\pi \left ( 2r \right )^{3}$
$=\frac{4\times 8}{3}\pi \left ( r \right )^{3}$
$=\frac{32}{3}\pi r^{3}$
So, option (D) is the correct.

### Question:2

The total surface area of a cube is $96\; \text {cm}^{3}$. The volume of the cube is:
(A) $8\; \text {cm}^{3}$
(B) $512\; \text {cm}^{3}$
(C) $64\; \text {cm}^{3}$
(D) $27\; \text {cm}^{3}$

We know that the total surface area of cube is given as $6a^{2}$ (where a is the side length of the cube)
Here we have been given that total surface area $=96\; cm^{2}$
So, $6a^{2}=96\; cm^{2}$
$a^{2}=\frac{96}{6}=16$
$a=4\; cm$
Now we know that volume of a cube $=\left ( side\; length \right )^{3}$
Volume $=a^{3}=\left ( 4 \right )^{3}=64\; cm^{3}$
So, option (C) is the correct answer.

### Question:3

A cone is $8.4\; cm$ high and the radius of its base is $2.1\; cm$. It is melted and recast into a sphere. The radius of the sphere is :
(A) $4.2\; cm$
(B) $2.1\; cm$
(C) $2.4\; cm$
(D) $1.6\; cm$

In the question it i given that the height of cone $=8.4\; cm$
And, Radius of cone $=2.1\; cm$
The cone is melted and recast into a sphere
So,
Volume of sphere = Volume of cone
$\frac{4}{3}\pi R^{3}=\frac{1}{3}\pi r^{2}h$
(where R is the radius of sphere, r is the radius of cone, h is the height of cone)
$R^{3}=\frac{r^{2}h}{4}$
Putting the given values,
$R^{3}=\frac{\left ( 2.1 \right )^{2}\times\left ( 8.4 \right )}{4}=\left ( 2.1 \right )^{2}\times 2.1$
$R^{3}=\left ( 2.1 \right )^{3}$
$R=2.1 \; cm$
Hence, option (B) is the correct answer.

### Question:4

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 $2\pi r h$
(Where r is radius and h is height)
Original heigth = h
Original curved surface area $\left ( C_{1} \right )=2 \pi r h$
Now it is given that radius is doubled and height is halved
New radius (R)$=2r$
New height (H) $=\frac{h}{2}$
So, new curved surface area $\left ( C_{2} \right )=2 \pi RH$
Now, $C_{2} =2 \pi RH$
$C_{2} =2\pi \left ( 2r \right )\times \frac{h}{2}$
$=2 \pi rh$
$C_{2}=2 \pi rh$
then
$C_{2}=C_{1}$
So, option (C) is the correct answer.

### Question:5

The total surface area of a cone whose radius is $\frac{r}{2}$ and slant height $2l$ is
(A) $2\pi r\left ( l+r \right )$
(B) $\pi r\left ( l+\frac{r}{4} \right )$
(C) $\pi r\left ( l+r \right )$
(D) $2\pi r l$

We know that total surface area of cone is given as $=\pi .R\; L+\pi R^{2}$
Where R is the radius of its base and L is the slant height.
Now, it is given that :
Radius of cone $\left ( R \right )=\frac{r}{2}$
Slant height (L) $=2l$
So total surface area $=\pi \left ( \frac{r}{2} \right )\left ( 2l \right )+\pi \left ( \frac{r}{2} \right )^{2}$
$=\pi r l+\frac{\pi r^{2}}{4}$
$=\pi r\left ( l+\frac{r}{4} \right )$
So, option (B) is the correct answer.

### Question:6

The radii of two cylinders are in the ratio of $2:3$ and their heights are in the ratio of $5:3$. The ratio of their volumes is:
(A) $10 : 17$
(B) $20 : 27$
(C) $17 : 27$
(D) $20 : 37$

We know that volume of a cylinder is given as $\pi r^{2}h$
Where r is the radius of its base and h is the height.
Now it is given that ratio of Radius $=r_{1}:r_{2}=2:3$
And, Ratio of Heights $=h_{1}:h_{2}=5:3$
So, Ratio of volumes,
$\pi r_{1}^{2}h_{1}:\pi r_{2}^{2}h_{2}$
$=\frac{\pi r_{1}^{2}h_{1}}{\pi r_{2}^{2}h_{2}}$
$=\left ( \frac{r_{1}}{r_{2}} \right )^{2}\left ( \frac{h_{1}}{h_{2}} \right )$
$=\left ( \frac{2}{3} \right )^{2}\left ( \frac{5}{3} \right )$
$=\left ( \frac{4}{9} \right )\left ( \frac{5}{3} \right )$
$= \frac{20}{27}$
$= 20:27$
So, option (B) is the correct answer.

### Question:7

The lateral surface area of a cube is $256\; m^{2}$. The volume of the cube is
(A) $512\; m^{3}$
(B) $64\; m^{3}$
(C) $216\; m^{3}$
(D) $256\; m^{3}$

We know that Lateral surface area of cube is given as: $4\left ( a \right )^{2}$
Where 'a' is the edge (side) length of the cube.
Given that lateral surface area of cube $=256\; m^{2}$
Now,
$4\left ( a \right )^{2}=256$
$\left ( a \right )^{2}=64$
$a=8m$
Then
$\text {Volume of cube = (edge length)}^{3}$
$\text {Volume of cube = (a)}^{3}$
$\text {Volume of cube = (8)}^{3}$
$\text {Volume of cube = 512 m}^{3}$
So, option (A) is the correct answer.

### Question:8

The number of planks of dimensions $\left ( 4\; m\times 50\; cm \times 20\; cm \right )$ that can be stored in a pit which is $16\; m$ long, $12\; m$ wide and $4\; m$ deep is
(A) $1900$
(B) $1920$
(C) $1800$
(D) $1840$

We can see that both the plank and pit will be in the form of a cuboid.
Volume of a cuboid $=l \times b \times h$
Where l is its length, b is breadth and h is height.
Given dimensions of plank $=\left ( 4\; m\times 50\; cm \times 20\; cm \right )$
We know that, $1m=100 cm$
So, Dimension of plank $=\left ( 4\; m\times 0.5\; m \times 0.2\; m \right )$
Volume of plank$=4\; m\times 0.5\; m \times 0.2\; m=4\; m^{3}$
Now, Dimensions of pit $=\left ( 16\; m\times 12\; m \times 40\; m \right )$
Volume of pit$=16\; m\times 12\; m \times 40\; m =7680\; m^{3}$
Thus number of planks that can be fitted into the pit $=\frac{\text {Volume of Pit}}{\text {Volume of plank}}$
$=\frac{7680}{4}$
$=1920$
This is option (B) is the correct.

### Question:9

The length of the longest pole that can be put in a room of dimensions $\left ( 10\; m\times\; 10\; m \times 5\; m \right )$ is
(A) $15\; m$
(B) $16\; m$
(C) $10\; m$
(D) $12\; m$

Dimensions of room $=10\; m \times \; 10\; m\; \times 5\; m$
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 $=\sqrt{l^{2}+b^{2}+h^{2}}$
$=\sqrt{10^{2}+10^{2}+5^{2}}$
$=\sqrt{225}$
$=15\; m$
So, option (A) is the correct answer.

### Question:10

The radius of a hemispherical balloon increases from $6 cm \; to\; 12 cm$ as air is being pumped into it. The ratios of the surface areas of the balloon in the two cases is:
(A) $1:4$
(B) $1:3$
(C) $2:3$
(D) $2:1$

We know that for a hemisphere
Total surface area of $=2\pi r^{2}+\pi r^{2}=3 \pi r^{2}$ ( Where r is the radius)

Now it is given that the radius of a hemispherical balloon increases from 6 cm to 12 cm
So,
When $r=6$, total surface area $\left ( S_{1} \right )=3 \pi\left ( 6 \right )^{2}=36$
When $r=12$, total surface area $\left ( S_{2} \right )=3 \pi\left ( 12 \right )^{2}=144$
Hence, required ratio$=\left ( S_{1} \right ):\left ( S_{2} \right )$
$=36 :144$
$=1 : 4$
So, option (A) is the correct answer.

### Question:1

Direction: Write True or False and justify your answer in each of the following :
The volume of a sphere is equal to two-third of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere.

We know that volume of the sphere $=\frac{4}{3}\pi r^{3}$ (where r is the radius)
And Volume of cylinder $=\pi R^{2}h$ (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 $=\pi r^{2}.2r$
Volume of sphere$=\frac{4}{3}\pi r^{3}=\frac{2}{3}\pi r^{2}\left ( 2r \right )=\frac{2}{3}$ volume of cylinder
Therefore the given statement is True.

### Question:2

Direction: Write True or False and justify your answer in each of the following :
If the radius of a right circular cone is halved and height is doubled, the volume will remain unchanged.

We know that Volume of a cone $=\frac{1}{3}\pi r^{2}h$ (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$=\frac{1}{3}\pi \left ( \frac{r}{2} \right )^{2}.2h=\frac{1}{2}\left ( \frac{1}{3}\pi r^{2}h \right )$
$=\frac{1}{2}$ (original volume of cone )
Thus the statement is false.

### Question:3

Direction: Write True or False and justify your answer in each of the following :
In a right circular cone, height, radius and slant height do not always be sides of a right triangle.

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:
$\left ( \text {hypotenuse} ^{2}=\text {base}^{2}+\text {height}^{2}\right )$
$l^{2}=h^{2}+r^{2}$
Thus the statement is False.

### Question:4

Direction: Write True or False and justify your answer in each of the following :
If the radius of a cylinder is doubled and its curved surface area is not changed, the height must be halved.

We know that curved surface area of the cylinder is given as $2\pi rh$
(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 $r_{1}=2r$
Let new height be$=h_{1}\; cm$
Now, according to the question –
$2\; \pi \; r\; h=2\; \pi \; r_{1}h_{1}$
$2\; \pi \; r\; h=2\; \pi. \;2 r.h_{1}$
$\Rightarrow h_{1}=\frac{h}{2}$
Hence the height is halved.
Thus, the statement is true.

### Question:5

Direction: Write True or False and justify your answer in each of the following :
The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals to the volume of a hemisphere of radius r.

We know that
Volume of cone $=\frac{1}{3}\pi r^{2}h$ (where r is the radius and h is height)
Volume of the hemisphere $=\frac{2}{3}\pi r^{3}$ (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 $=\frac{1}{3}\pi r^{2}\left ( 2r \right )=\frac{2}{3}\pi r^{3}$
= volume of hemisphere
Thus, the statement is true.

### Question:6

Direction: Write True or False and justify your answer in each of the following :
A cylinder and a right circular cone are having the same base and same height. The volume of the cylinder is three times the volume of the cone.

We know that
Volume of cylinder $=\pi R^{2}H$ (where R is the radius and H is height)
Volume of cone $=\frac{1}{3}\pi r^{2}h$ (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 , $V_{1}=\pi r^{2}h$
and Volume of cone will be, $V_{2}=\frac{1}{3}\pi r^{2}h$
So we get:
$\therefore V_{2}=\frac{1}{3}V_{1}$
Therefore, $V_{1}=3V_{2}$
Hence, the statement is true.

### Question:7

Direction: Write True or False and justify your answer in each of the following :
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is $1 : 2 : 3$.

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, $V_{Cone}=\frac{1}{3}\pi r^{2}h\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(i)$
Volume of hemisphere, $V_{Hemisphere}=\frac{2}{3}\pi r^{2}h\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(ii)$
Volume of cylinder, $V_{Cylinder}=\pi r^{2}h\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(iii)$
From equations (i) (ii) and (iii)
$V_{Cone}:V_{Hemisphere}:V_{Cylinder}$
$=\frac{1}{3}\pi r^{2}h:\frac{2}{3}\pi r^{2}h:\pi r^{2}h$
$=\frac{1}{3}:\frac{2}{3}:1$
$=1:2:3$
Thus the statement is true.

### Question:8

Direction: Write True or False and justify your answer in each of the following :
If the length of the diagonal of a cube is $6\sqrt{3}$ cm, then the length of the edge of the cube is 3 cm.

We know that the diagonal of cube with side a is given as $a\sqrt{3}$
Now in the question, the diagonal of the cube is $6\sqrt{3}$ cm
So we can write
$\Rightarrow a\sqrt{3}=6\sqrt{3}$
Which gives,
$a=6cm$
So the side of cube is 6 cm
Thus, the statement is false.

### Question:9

Direction: Write True or False and justify your answer in each of the following :
If a sphere is inscribed in a cube, then the ratio of the volume of the cube to the volume of the sphere will be $6:\pi$.

Explaination:
$\\ Volume of cube \left(V_{1}\right)=(\text { Side })^{3}\\ Radius of sphere (r)=\frac{\text { Side }}{2}\\ Volume of sphere \left(V_{2}\right)=\frac{4}{3} \pi\left(r^{3}\right) \\ =\frac{4}{3} \pi\left(\frac{\text { Side }}{2}\right)^{3}\\=\frac{4}{3} \pi \times \frac{(\text { Side })^{3}}{8} \\ \frac{V_{1}}{V_{2}} =\frac{(\text { side })^{3}}{\frac{4}{3} \pi \frac{(\text { side })^{3}}{8}}=\frac{6}{\pi} \\\\ V_{1}:V_{2}=6: \pi$

### Question:10

Direction: Write True or False and justify your answer in each of the following :
If the radius of a cylinder is doubled and height is halved, the volume will be doubled.

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 $\pi r^{2}h$
where the radius of cylinder is r and height is h.
Let original volume be,
$V_{1}=\pi r^{2}h$
Now the radius of a cylinder is doubled and height is halved,
So,
New Height (H)$=\frac{h}{2}$
Then, new Volume of cylinder
$V_{2}=\pi \left ( 2r \right )^{2}\frac{h}{2}$
$V_{2}=2\pi r^{2}h=2 \times V_{1}$
Hence, the given statement is true.

### Question:1

We know that, Volume of a sphere is given as : $\frac{4}{3}\pi r^{3}$ (where r is the radius)
Given that radius of each sphere $\left ( r \right )=2\; cm$
So, Volume of sphere$=\frac{4}{3}\pi r^{3}$
$=\frac{4}{3}\times \pi \times \left ( 2 \right )^{3}$
$=\frac{32}{3}\times 3.14$
Volume of 16 spheres $=16\times \frac{32}{3}\times3.14$
$=535.89\; cm^{3}$
Internal dimensions of given rectangular box $=16 \times 8 \times 8\; cm$
The box is in the form of a cuboid.
Volume of rectangular box $=l \times b \times h$
$=16\; cm \times 8\; cm \times 8\; cm$
$=1024\; cm^{3}$
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 $=\left ( 1024-535.89 \right )cm^{3}$
$=488.11\; cm^{3}$
Thus, the volume of preservative liquid is $488.11\; cm^{3}$

### Question:2

A storage tank is in the form of a cube. When it is full of water, the volume of water is $15.625\; m^{3}$. If the present depth of water is $1.3\; m$, find the volume of water already used from the tank.

Answer: $8.125\; m^{3}$
In the question, it is given that:
Volume of water in tank$=15.625\; m^{3}$
Depth of water $=1.3 \; m$
Now we know that,
Volume of cube is given as $a^{3}$ (where a is the side length)
$a^{3}=15.625$
$a=\sqrt[3]{15.625}$
$a=2.5\; m$
The present depth of water volume is given as $1.3\; m$.
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
$=a^{2}\times 1.2$
$=\left ( 2.5 \right )^{2}\times 1.2$
$=6.25 \times 1.2$
$=7.5\; m^{3}$
Hence volume of water already used from the tank is $7.5\; m^{3}$.

### Question:3

Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water.

Answer: $38.808\; cm^{3}$
Given that a solid spherical ball of diameter (d) 4.2 cm is completely immersed in water.
Radius of ball$=\frac{d}{2}$
$=\frac{4.2}{2}=2.1\; cm$
Now we know that,
Volume of spherical ball $=\frac{4}{3}\pi r^{3}$ (where r is the radius)
$=\frac{4}{3} \times \pi \times \left ( 2.1 \right )^{3}$
$=\frac{4}{3} \times \pi \times \left ( 9.261 \right )$
$=\frac{4}{3} \times \frac{22}{7} \times 9.261$
$=38.808\; cm^{3}$
Hence volume of spherical ball$=38.808\; cm^{3}$
Amount of water displaced is same as the volume of spherical ball.
So we have Amount of water displaced $=38.808\; cm^{3}$

### Question:4

How many square metres of canvas is required for a conical tent whose height is 3.5 m and the radius of the base is 12 m?

Answer:$471.43\; m^{2}$
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$=\pi r \sqrt{r^{2}+h^{2}}$
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

On putting the values,
Curved surface area $=\frac{22}{7}\times 12\sqrt{\left ( 12 \right )^{2}+\left ( 3.5 \right )^{2}}$
$=\frac{22}{7}\times 12 \times \sqrt{156.25}$
$=\frac{22}{7}\times 12 \times 12.5$
$=471.43\; m^{2}$
So, the area of the canvas $=471.43\; m^{2}$
Therefore, the area of canvas required is $471.43\; m^{2}$.

### Question:5

Two solid spheres made of the same metal have weights 5920 g and 740 g, respectively. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.

Answer : $5\; cm$
Density (D) of an object is mass (m) per unit volume (v).

$D=\frac{m}{v}$
So mass is directly proportional to volume for same metal (as density remains same)
For Solid 1
Let Mass = $M_{1}$
Volume = $V_{1}$
For solid 2
Mass = $M_{2}$
Volume = $V_{2}$
Now, $\frac{M_{1}}{M_{2}}=\frac{V_{1}}{V_{2}}$
In the question, the objects are spheres
Volume of sphere $=\frac{4}{3}\pi R^{3}$ (where R is radius)
So, volume is directly proportional to $R^{3}$
Hence, $\frac{M_{1}}{M_{2}}=\frac{V_{1}}{V_{2}}=\frac{R_{1}^{3}}{R_{2}^{3}}$
Putting the given values $\left ( 5920\; g \; and\; 740\; g \right )$
$\Rightarrow \frac{5920}{740}=\frac{R_{1}^{3}}{R_{2}^{3}}$
$\Rightarrow \frac{R_{1}^{3}}{R_{2}^{3}}=8$
$\Rightarrow \frac{R_{1}}{R_{2}}=\sqrt[3]{8}$
$\Rightarrow \frac{R_{1}}{R_{2}}=2$
So, $R_{1}=R_{2}\times 2$
Diameter of the smaller one is 5 cm
So, $R_{2}=\frac{5}{2}=2.5\; cm$
$R_{1}=2.5 \times 2$
Hence, $R_{1}=5\; cm$
Therefore radius of larger sphere is 5 cm.

### Question:6

A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. If the glass is filled with milk up to an height of 12 cm, find how many litres of milk is needed to serve 1600 students.

Answer: $738.528\; \text {litre of milk}$
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 $\left ( r \right )=\frac{d}{2}$
$=\frac{7}{2}$
$=3.5\; cm$
Now milk is filled in this cylindrical glass.
Height of milk up to which the glass is filled $(h) = 12 cm$
So, volume of milk filled in the cylindrical glass
$V=\pi r^{2}h$ (r is the radius, h is the height up to which the milk is filled)
Putting the values,
$V=\frac{22}{7}\times 3.5 \times 3.5 \times 12$
$\Rightarrow 461.58\; cm^{3}$
Now, number of students=1600=Total number of glasses required
Volume of 1600 cylindrical glasses = 1600 (Volume of one glass)
$=461.58 \times 1600$
$=738528\; cm^{3}$
Now we know that 1 litre $=1000\; cm^{3}$
Volume of 1600 cylindrical glasses $=\frac{738528}{1000}$
$=738.528\; litre$
Hence, to serve 1600 students, we need $738.528\; \text {litre of milk}$.

### Question:7

A cylindrical roller $2.5\; m$ in length, $1.75\; m$ in radius when rolled on a road was found to cover the area of $5500\; m^{2}$. How many revolutions did it make?

It is given that when cylindrical roller is rolled on the road its covers $5500\; m^{2}$ area
Now,
Area covered by cylindrical roller in one revolution = curved surface area of cylinder
And we know that,
Curved surface area of cylinder $=2\pi rh$ (where r is the radius and h is its height)
Given length of roller $=2.5\; m$ (This will be taken as height of the roller)
Cylindrical roller's radius = 1.75 m
So, Curved surface area $=2 \times 3.14 \times 1.75 \times 2.5$
$= 6.28 \times 4.375\; m^{2}$
$= 27.475\; m^{2}$
Total Area covered in one revolution $= 27.475\; m^{2}$
Now total area covered $= 5500\; m^{2}$
So, Number of revolutions $=\frac{\text {Total Area}}{\text {area covered in one revolution}}$
$=\frac{5500}{27.475}$
$=200.181\cong 200$
Hence, cylindrical roller makes 200 revolutions.

### Question:8

A small village, having a population of 5000, requires 75 litres of water per head per day. The village has got an overhead tank of measurement $40\; m \times 25\; m \times 15\; m$. For how many days will the water of this tank last?

Given, dimension of overhead tank $=40\; m \times 25 \; m \times 15\; m$
Now the tank is in the shape of a cuboid, so we have volume $=l\; b\; h$
$\Rightarrow$ Volume of tank $=40\; m \times 25 \; m \times 15\; m=15000\; m^{3}$
Population of village $=5000$
Daily water requirement for 1 person = 75 litre
Then daily water requirement for 5000 people $=75 \times 5000=375000\; l$
Now we know that 1 litre $=0.001\; m^{3}$
So, daily water requirement for 5000 people $=375000 \times 0.001\; m^{3}=375\; m^{3}$
Number of Days till the water in this tank will last
$=\frac{\text {Volume of tank}}{\text {daily water requirement for 5000 people}}$
$=\frac{15000\; m^{3}}{375\; m^{3}}=40$
Hence, the water of this tank will last for 40 days.

### Question:9

Given that a shopkeeper has:
Volume of sphere $=\frac{4}{3}\pi R^{3}$ (where R is the radius)
Volume of 1 larger laddoo $=\frac{4}{3}\pi \left ( 5 \right )^{3}$
Volume of 1 smaller laddoo $=\frac{4}{3}\pi \left ( 2.5 \right )^{3}$
Volume of 1 larger laddoo = Total volume of all smaller laddoos formed
Let number of smaller laddoos be n.
Volume of n smaller laddoo $=n\left ( \frac{4}{3}\pi \left ( 2.5 \right )^{3} \right )$
So,
$\Rightarrow \frac{4}{3}\pi \times 5^{3}=n\times\frac{4}{3}\pi \times \left ( 2.5 \right )^{3}$
$\Rightarrow 125=n \times \left ( \frac{25}{10} \right )^{3}$
$\Rightarrow n=8 \; laddoos$

### Question:10

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the solid so formed.

Answer : Volume $=301.44\; cm^{3}$ and Curved surface area $=188.57\; cm^{2}$
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$=\frac{1}{3}\pi r^{2}h$ (where r is radius and h is height)
$V=\frac{1}{3}\times 3.14 \times \left ( 6 \right )^{2}\times 8$
$=\frac{1}{3}\times 3.14 \times 36 \times 8$
$=\frac{1}{3}\times 12 \times 8$
$=301.44\; cm^{3}$
Curved surface area $=\pi r l$ (where r is radius and l is slant height)
$=\frac{22}{7}\times 6\times 10$
$=\frac{1320}{7}$
$=188.57\; cm^{2}$
Hence Volume $=301.44\; cm^{3}$ and curved surface area $=188.57\; cm^{2}$

### Question:1

A cylindrical tube opened at both the ends is made of iron sheet which is 2 cm thick. If the outer diameter is 16 cm and its length is 100 cm, find how many cubic centimeters of iron has been used in making the tube?

Answer : $8792\; \text {cm}^{3}$
We know that
Volume of cylinder is given as $\pi r^{2}h$ (where r = outer radius and h is the height)
Here, length of cylinder will be considered as its height
$\left ( \text {Height=length=100 cm} \right )$
Outer diameter is given as, $d=16\; cm$
So, outer radius $r=\frac{16}{2}=8 \; cm$
Thus volume of cylinder $=\pi r^{2}h$
$=3.14 \times \left ( 8 \right )^{2}\times 100$
$=20096\; cm^{3}$
Thickness of iron sheet = 2 cm
Now, inner diameter = outer diameter – 2 (Thickness of iron sheet)
Inner diameter $=16-\left ( 2 \times 2 \right )=12\; cm$
So, inner radius $\left ( R \right )=\frac{12}{2}=6 \; cm$
Now, Volume of inner cylinder $=\pi r^{2}h$, (where R = inner radius)
$=\pi R^{2}h$
$=3.14 \times \left ( 6 \right )^{2}\times 100$
$=11304\; cm^{3}$
Thus, Volume of iron used = Volume of outer cylinder - Volume of inner cylinder
$=\left ( 20096-11304 \right )cm^{3}$
$=8792\; cm^{3}$
Hence the answer is $8792\; cm^{3}$

### Question:2

A semi-circular sheet of metal of diameter 28cm is bent to form an open conical cup. Find the capacity of the cup.

Answer : $622.38\; cm^{3}$
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 $=28\; cm$
So, radius of semi-circular sheet $\left ( R \right )=14\; cm$
Slant height of the conical cup $\left ( l \right )=14\; cm$
Now,
Let the radius of base of the cup = r cm.
Circumference of semi-circular sheet = Circumference of base of conical cup
$\pi R=2\; \pi\; r$
$\Rightarrow \frac{22}{7}\times 14=2.\frac{22}{7}.r$
$\Rightarrow r=7\; cm$
Let the height of cup be h cm.
Then we know that for a right circular cone,
$l^{2}=r^{2}+h^{2}$
$\Rightarrow \left ( 14 \right )^{2}=\left ( 7 \right )^{2}+h^{2}$
$\Rightarrow 196-49=h^{2}$
$\Rightarrow h=7\sqrt{3}\; cm$
$\therefore$ Capacity of the cup = Volume of a cone $=\frac{1}{3}\pi r^{2}h$
$=\frac{1}{3}.\frac{22}{7}.\left ( 7 \right )^{2}.7\sqrt{3}$
$=\frac{1078}{3}\sqrt{3}$
$=\frac{1078}{3}\times 1.732$
$=622.38\; cm^{3}$

### Question:3

A cloth having an area of $165\; m^{2}$ is shaped into the form of a conical tent of radius $5\; m$

1. How many students can sit in the tent if a student, on an average, occupies $\frac{5}{7}m^{2}$ on the ground?
2. Find the volume of the cone.
(ii) $241.73\; m^{3}$
We have, Area of cloth $=165\; m^{2}$
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 $\pi rl$ Where, r = radius of a cone
l = slant height of a cone
So curved surface area of conical tent,
$165=\frac{22}{7}\times 5 \times l$
$\Rightarrow l=\frac{165 \times 7}{22 \times 5}=\frac{21}{2}=10.5\; m$
(i) Now it is given that area covered by 1 student $=\frac{5}{7}m^{2}$
$\text {So the number of students}=\frac{\text {Area of a circular base of a cone}}{\text {Area covered by 1 student}}$
$\Rightarrow \text {Number of student}=\frac{\pi r^{2}}{\frac{5}{7}}=\frac{\left ( \frac{22}{7}\times 5^{2} \right )}{\frac{5}{7}}=110$
So 110 students can sit in the tent.
(ii) Volume of cone : $\frac{1}{3}\pi r^{2}h$
For a right circular cone, we have
$r^{2}+h^{2}=l^{2}$
$\Rightarrow 5^{2}+h^{2}=\left ( 10.5 \right )^{2}$
$\Rightarrow 25+h^{2}=110.25$
$\Rightarrow h^{2}=110.25-25=85.25$
$h=\sqrt{85.25}=9.23\; m$
So, volume $=\frac{1}{3}\times \frac{22}{7}\times 5^{2}\times 9.23$
$=\frac{5076.5}{21}=241.73\; m^{3}$

### Question:4

The water for a factory is stored in a hemispherical tank whose internal diameter is 14 m. The tank contains 50 kilolitres of water. Water is pumped into the tank to fill to its capacity. Calculate the volume of water pumped into the tank.

Answer : $668.66 \; m^{3}$
We know that Volume of hemispherical tank$=\frac{2}{3}\pi r^{3}$
Where, r = radius of hemispherical tank
Now, internal diameter of hemispherical tank is given as 14 m
So, internal radius of hemispherical tank$=\frac{14}{2}=7\; m$
Tank contains 50 kL of water
We have, $1000 \; L=1\; m^{3}$
So, $1\; kL=1\; m^{3}$
This means that tank contains $50\; m^{3}$ of water
$\Rightarrow$ Volume of hemispherical tank $=\frac{2}{3}\times\frac{22}{7}\times \left ( 7 \right )^{3}=\frac{2156}{3}=718.66\; m^{3}$
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 $-50\; m^{3}$
Volume of water pumped into the tank $=\left ( 718.66-50 \right )m^{3}=668.66\; m^{3}$
Hence the answer is $668.66\; m^{3}$

### Question:5

The volumes of the two spheres are in the ratio $64:27$. Find the ratio of their surface areas.

Answer : $16:9$
We have been given that the volume of two sphere is in the ratio $=64:27$
We know that, Volume of sphere is given as $\frac{4}{3}\pi r^{3}$ (where r is its radius)
And surface area is given as $4\pi r^{2}$
Let volume of sphere $1=V_{1}$ (radius $r_{1}$)
And, volume of sphere $2=V_{2}$ (radius $r_{2}$)
Then, $\frac{V_{1}}{V_{2}}=\frac{64}{27}$
$\frac{\frac{4}{3}\pi r_{1}^{3}}{\frac{4}{3}\pi r_{2}^{3}}=\frac{64}{27}$
$\frac{r_{1}^{3}}{r_{2}^{3}}=\left ( \frac{4}{3} \right )^{3}$
$\frac{r_{1}}{r_{2}}=\frac{4}{3}$
Then, ratio of areas of both spheres
$\frac{\text {area of sphere (1))}}{\text {area of sphere }(2)}=\frac{4\pi r_{1}^{2}}{4\pi r_{2}^{2}}$
$=\frac{r_{1}^{2}}{r_{2}^{2}}=\left ( \frac{r_{1}}{r_{2}} \right )^{2}=\left ( \frac{4}{3} \right )^{2}=\frac{16}{9}$
Hence the required ratio is $16:9$

### Question:6

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

Answer : $30.47\; cm^{3}$
Volume of cube is given as $(\text {side length})^{3} = (4\; \text {cm})^{3} = 64 \text {cm}^{3}$
As the cube contains the sphere,
Diameter of sphere = side length of the cube = 4 cm
Radius of sphere $(r)=\frac{4}{2}cm=2cm$
We know that volume of a sphere is given as $\frac{4}{3}\pi r^{3}$ (where r is the radius)
Volume $=\frac{4}{3}.\frac{22}{7}.2.2.2=\frac{704}{21}cm^{3}$
$\therefore \text {Volume of the gap in between}=\text {Volume of cube}-\text {volume of sphere}$
$=\left ( 64-\frac{704}{21} \right )cm^{3}$
$=\frac{640}{21}cm^{3}$
$=30.47\; cm^{3}$
Hence the required answer is $30.47\; cm^{3}$

### Question:7

A sphere and a right circular cylinder of the same radius have equal volumes. By what percentage does the diameter of the cylinder exceeds its height ?

Let radius of sphere be r, radius of cylinder be R, height of cylinder be H
Volume of sphere $\left ( \frac{4}{3}\pi r^{3} \right )$ = Volume of cylinder $\left ( \pi R^{2}H \right )$
Now, it is given that r = R
$\left ( \frac{4}{3}\pi r^{3} \right )=\left ( \pi r^{2}H \right )$
$\frac{4}{3}r=h$
$4r=3h$
$2(2r)=3h$ $\left ( \therefore d=2r \right )$
$\Rightarrow 2d=3h$
$d=1.5 h=\left ( 1+0.5 \right )h$
$d=\left ( 1+\frac{50}{100} \right )h$
The diameter exceeds the height by 50%.

### Question:8

(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.

(i) Answer : $9152\; cm^{2}$
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 $=30 \times 3 = 90\; cm$
Now, radius of circular plate, r = 14 cm
We know that
Total surface area of the cylindrical solid of formed will be equal to $2\pi r\left ( h+r \right )$
Where r is its radius and h is the height. So, putting the above calculated values:
Total surface area $=2 \times \frac{22}{7}\times 14\left ( 90+14 \right )$
$=44 \times 2\times 104$
$=9152\; cm^{2}$
Hence, the total surface area of the cylindrical solid is $9152\; cm^{2}$.
(ii)
Answer : $55440\; cm^{3}$
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 $=30 \times 3=90\; cm$
Now, radius of circular plate, r = 14 cm
We know that
Volume of the cylinder so formed will be equal to $\pi r^{2}h$
Where r is its radius and h is the height. So, putting the above calculated values:
Volume of the cylinder $=\frac{22}{7}\times (14)^{2}\times 90$
$=\frac{22}{7}\times 14\times 14 \times 90$
$=55440\; cm^{3}$
Hence, the volume of the cylindrical solid is $55440\; cm^{3}$

### Question:9

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.

Answer : $55440\; cm^{3}$
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 $=30 \times 3=90\; cm$
Now, radius of circular plate, r = 14 cm
We know that
Volume of the cylinder so formed will be equal to $\pi r^{2}h$
Where r is its radius and h is the height. So, putting the above calculated values:
Volume of the cylinder $=\frac{22}{7}\times (14)^{2}\times 90$
$=\frac{22}{7}\times 14\times 14 \times 90$
$=55440\; cm^{3}$
Hence, the volume of the cylindrical solid is $55440\; cm^{3}$

### 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.

Answer: (A) a cone and a cylinder
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.

### 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.

(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.

### 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

Answer (B) a hemisphere and a cone
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.

### 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

Answer: (B) frustum of a cone
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.

### Question:

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

Answer: (C) two cones and a cylinder
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.

## Important Topics for NCERT Exemplar Solutions Class 9 Maths Chapter 13:

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.

## NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

 Chapter 1 Number System Chapter 2 Polynomials Chapter 3 Coordinate geometry Chapter 4 Linear equations in Two Variable Chapter 5 Introduction to Euclid’s Geometry Chapter 6 Lines and Angles Chapter 7 Triangles Chapter 8 Quadrilaterals Chapter 9 Area of Parallelograms and Triangles Chapter 10 Circles Chapter 11 Constructions Chapter 12 Heron’s Formula Chapter 14 Statistics and Probability

## Features of NCERT Exemplar Class 9 Maths Solutions Chapter 13:

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.

### Check the Solutions of Questions Given in the Book

 Chapter No. Chapter Name Chapter 1 Number Systems Chapter 2 Polynomials Chapter 3 Coordinate Geometry Chapter 4 Linear Equations In Two Variables Chapter 5 Introduction to Euclid's Geometry Chapter 6 Lines And Angles Chapter 7 Triangles Chapter 8 Quadrilaterals Chapter 9 Areas of Parallelograms and Triangles Chapter 10 Circles Chapter 11 Constructions Chapter 12 Heron’s Formula Chapter 13 Surface Area and Volumes Chapter 14 Statistics Chapter 15 Probability

### Also Check NCERT Books and NCERT Syllabus here

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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