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

Question:1
The radius of a sphere is 2r, then its volume will be
(A) 4/3 πr³
(B) 4 πr³
(C) 8/3 πr³
(D) 32/3 πr³

Answer (D)

We know that the volume of a sphere is given as 4/3 πr³, where r is the radius of the sphere.

In this question, it is given that the Radius of the sphere is 2r.
So, the Volume of the sphere will be equal to:
= 4/3 π(2r)³
= (4×8)/3 π(r)³
= 32/3 πr³
So, option (D) is the correct.

Question:2

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

Answer (C)

We know that the total surface area of a cube is given as 6a² (where a is the side length of the cube)
Here we have been given that total surface area =96$\text{cm}^2$
So, 6a² = 96 $\text{cm}^2$
a²= $\frac{96}6=16$
a=4cm
Now we know that the volume of a cube =(sidelength)3
Volume =a³=(4)³=64 $\text{cm}^3$
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) 4.2cm
(B) 2.1cm
(C) 2.4cm
(D) 1.6cm

Answer (B)

In the question, it is given that
Height of the cone, $h=8.4 \mathrm{~cm}$
Radius of the cone, $r=2.1 \mathrm{~cm}$

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 the sphere, $r$ is the radius of the cone, and $h$ is the height of the cone)

$
\begin{aligned}
4 R^3 & =r^2 h \\
R^3 & =\frac{r^2 h}{4}
\end{aligned}
$
Putting the given values,

$
\begin{gathered}
R^3=\frac{(2.1)^2 \times 8.4}{4} \\
⇒R^3=(2.1)^2 \times 2.1 \\
\end{gathered}$
$⇒R^3=(2.1)^3$
$⇒R=2.1 \mathrm{~cm}$
Hence, option (B) is the correct answer.

Question:4

In a cylinder, radius is doubled and the height is halved, the curved surface area will be
(A) halved
(B) doubled
(C) same
(D) four times

Answer (C)

We know that the curved surface area of a cylinder is

$
\text { Curved Surface Area }=2 \pi r h
$

(where $r$ is the radius and $h$ is the height)
Let,
Original radius $=r$
Original height $=h$
Original curved surface area,

$
C_1=2 \pi r h
$
It is given that the radius is doubled and the height is halved.
New radius,

$
R=2 r
$
New height,

$
H=\frac{h}{2}
$

New curved surface area,

$
C_2=2 \pi R H
$
Substituting the values,

$
\begin{gathered}
C_2=2 \pi(2 r)\left(\frac{h}{2}\right) \\
C_2=2 \pi r h
\end{gathered}
$
Thus,

$
C_2=C_1
$
So, the curved surface area remains unchanged.

Hence, option (C) is the correct answer.

Question:5

The total surface area of a cone whose radius is r2 and slant height 2l is
(A) 2πr(l+r)
(B) πr(l+r4)
(C) πr(l+r)
(D) 2πrl

Answer (B)

We know that the total surface area of a cone is given by

$
\text { Total Surface Area }=\pi R L+\pi R^2
$

(where $R$ is the radius of the base and $L$ is the slant height)
Now, it is given that:
Radius of the cone,

$
R=\frac{r}{2}
$
Slant height,

$
L=2 l
$
Substituting these values,

$
\begin{gathered}
\text { Total Surface Area }=\pi\left(\frac{r}{2}\right)(2 l)+\pi\left(\frac{r}{2}\right)^2 \\
=\pi r l+\pi \frac{r^2}{4} \\
=\pi r\left(l+\frac{r}{4}\right)
\end{gathered}
$
Hence, 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

Answer (B)

We know that the volume of a cylinder is given by

$
\text { Volume }=\pi r^2 h
$

where $r$ is the radius of the base and $h$ is the height.
It is given that:

$
\begin{aligned}
& r_1: r_2=2: 3 \\
& h_1: h_2=5: 3
\end{aligned}
$
Required ratio of volumes,

$
\begin{gathered}
V_1: V_2=\pi r_1^2 h_1: \pi r_2^2 h_2 \\
=r_1^2 h_1: r_2^2 h_2 \\
=\left(r_1: r_2\right)^2 \times\left(h_1: h_2\right) \\
=(2: 3)^2 \times(5: 3) \\
=4: 9 \times 5: 3 \\
=20: 27
\end{gathered}
$
Hence, option (B) is the correct answer.

Question:7

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

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

Question:8

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

Answer (B)
We can see that both the plank and pit will be in the form of a cuboid.
Volume of a cuboid =l×b×h
Where l is its length, b is its breadth, and h is its height.
Given dimensions of plank =(4m×50cm×20cm)
We know that, 1m=100cm
So, Dimension of plank =(4m×0.5m×0.2m)
Volume of plank=4m×0.5m×0.2m=4$\text{m}^3$
Now, Dimensions of pit =(16m×12m×40m)
Volume of pit=16m×12m×40m=7680$\text{m}^3$
Thus number of planks that can be fitted into the pit = Volume of Pit / 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 (10m×10m×5m) is
(A) 15m
(B) 16m
(C) 10m
(D) 12m

Answer (A)

Dimensions of the room are

$
\begin{aligned}
& \text { Length }=10 \mathrm{~m} \\
& \text { Breadth }=10 \mathrm{~m} \\
& \text { Height }=5 \mathrm{~m}
\end{aligned}
$
The room is in the shape of a cuboid.
The longest rod (pole) that can be placed inside a cuboid is along its space diagonal.
Length of the diagonal of a cuboid is given by

$
\sqrt{l^2+b^2+h^2}
$
Substituting the given values,

$
\begin{gathered}
\sqrt{10^2+10^2+5^2} \\
=\sqrt{100+100+25} \\
=\sqrt{225} \\
=15 \mathrm{~m}
\end{gathered}
$
Hence, option (A) is the correct answer.

Question:10

The radius of a hemispherical balloon increases from 6cmto12cm 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

Answer (A)

We know that for a hemisphere,
Total surface area

$
=2 \pi r^2+\pi r^2=3 \pi r^2
$

(where $r$ is the radius)
It is given that the radius of the hemispherical balloon increases from 6 cm to 12 cm.
When $r=6 \mathrm{~cm}$,

$
S_1=3 \pi(6)^2=3 \pi \times 36
$
When $r=12 \mathrm{~cm}$,

$
S_2=3 \pi(12)^2=3 \pi \times 144
$
Required ratio of surface areas,

$
\begin{aligned}
S_1: S_2= & 3 \pi \times 36: 3 \pi \times 144 \\
= & 36: 144 \\
= & 1: 4
\end{aligned}
$
Hence, option (A) is the correct answer.

Question:1

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

Answer: True

We know that the volume of a sphere is $\frac{4}{3} \pi r^3$

(where $r$ is the radius)

Volume of a cylinder is $\pi R^2 h$

(where $R$ is the radius and $h$ is the height)

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

So,
Radius of the cylinder, $R=r$
Height of the cylinder, $h=2 r$

Volume of the cylinder,

$
=\pi r^2 \times 2 r=2 \pi r^3
$
Volume of the sphere,

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

$
\begin{gathered}
\frac{\text { Volume of sphere }}{\text { Volume of cylinder }}=\frac{\frac{4}{3} \pi r^3}{2 \pi r^3}=\frac{2}{3} \\
\text { Volume of sphere }=\frac{2}{3} \times \text { Volume of cylinder }
\end{gathered}
$
Therefore, the given statement is True.

Question:2

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

Answer: False

We know that the volume of a cone is

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

(where $r$ is the radius and $h$ is the height)
It is given that the radius of a right circular cone is halved and the height is doubled.
New radius,

$r^{\prime}=\frac{r}{2}$
New height,

$h^{\prime}=2 h$
New volume of the cone,

$
\begin{gathered}
=\frac{1}{3} \pi\left(r^{\prime}\right)^2 h^{\prime} \\
=\frac{1}{3} \pi\left(\frac{r}{2}\right)^2(2 h) \\
=\frac{1}{3} \pi \cdot \frac{r^2}{4} \cdot 2 h \\
=\frac{1}{2}\left(\frac{1}{3} \pi r^2 h\right) \\
=\frac{1}{2} \times(\text { original volume })
\end{gathered}
$
Thus, the new volume is half of the original volume.
Therefore, the given statement is false.

Question:3

Directions: 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.

Answer: 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:
(hypotenuse²=base²+height²)
l²=h²+r²
Thus, the statement is False.

Question:4

Directions: 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.

Answer: True
We know that the curved surface area of the cylinder is given as 2πrh
(where r is the radius and h is the height)
Given that the radius of a cylinder is doubled, and its curved surface area is not changed
For the new cylinder,
New radius r₁=2r
Let the new height be h₁ cm
Now, according to the question –
2πrh=2πr₁h₁
2πrh=2π.2r.h₁
⇒h₁=h₂
Hence, the height is halved.
Thus, the statement is true.

Question:5

Directions: 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.

Answer: True

We know that
Volume of a cone is $\frac{1}{3} \pi r^2 h$

(where $r$ is the radius and $h$ is the height)
Volume of a hemisphere is $\frac{2}{3} \pi r^3$

(where $r$ is the radius)

For the largest right circular cone that can be fitted in a cube of edge $2 r$ :
Height of the cone, $h=2 r$

Diameter of the base of the cone, $=2 r$

So, radius of the base, $=r$
Volume of the cone,

$
\begin{gathered}
=\frac{1}{3} \pi r^2(2 r) \\
=\frac{2}{3} \pi r^3
\end{gathered}
$
This is equal to the volume of the hemisphere.
Thus, the given statement is true.

Question:6

Directions: 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.

Answer: True

We know that
Volume of a cylinder is $\pi R^2 H$

(where $R$ is the radius and $H$ is the height)
Volume of a cone is $\frac{1}{3} \pi r^2 h$

(where $r$ is the radius and $h$ is the height)
It is given that a cylinder and a right circular cone have the same base and the same height.
So, let
Radius of both $=r$
Height of both $=h$
Volume of the cylinder,

$V_1=\pi r^2 h$

Volume of the cone, $V_2=\frac{1}{3} \pi r^2 h$
Comparing the two volumes,

$
\begin{aligned}
V_2 & =\frac{1}{3} V_1 \\
V_1 & =3 V_2
\end{aligned}
$
Hence, the volume of the cylinder is three times the volume of the cone.
Therefore, the given statement is true.

Question:7

Directions: 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.

Answer: True

It is given that a cone, a hemisphere, and a cylinder stand on equal bases and have the same height.
Let the radius of the base be $r$ and the height be $h$.
Since the height of a hemisphere is equal to its radius, $h=r$
Now, we know that:

Volume of the cone,

$V_{\text {cone }}=\frac{1}{3} \pi r^2 h$
Volume of the hemisphere,

$V_{\text {hemisphere }}=\frac{2}{3} \pi r^3=\frac{2}{3} \pi r^2 h$
Volume of the cylinder,

$V_{\text {cylinder }}=\pi r^2 h$
From equations (i), (ii), and (iii),

$V_{\text {cone }}: V_{\text {hemisphere }}: V_{\text {cylinder }}=\frac{1}{3} \pi r^2 h: \frac{2}{3} \pi r^2 h: \pi r^2 h$
Cancelling common terms,

$=\frac{1}{3}: \frac{2}{3}: 1$
Multiplying each term by 3,

$=1: 2: 3$
Thus, the given statement is true.

Question:8

Directions: Write True or False and justify your answer in each of the following :
If the length of the diagonal of a cube is 63 cm, then the length of the edge of the cube is 3 cm.

Answer: False
We know that the diagonal of the cube with side a is given as a³
Now in the question, the diagonal of the cube is 6³ cm
So we can write
⇒ a³ = 6³
Which gives,
a = 6 cm
So the side of the cube is 6 cm.
Thus, the statement is false.

Question:9

Directions: 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:π.

Answer: True

Explanation:
Volume of the cube,

$
V_1=(\text { side })^3
$
Radius of the sphere,

$
r=\frac{\text { side }}{2}
$
Volume of the sphere,

$
V_2=\frac{4}{3} \pi r^3
$
Substituting the value of $r$,

$
\begin{aligned}
V_2 & =\frac{4}{3} \pi\left(\frac{\text { side }}{2}\right)^3 \\
& =\frac{4}{3} \pi \cdot \frac{(\text { side })^3}{8} \\
& =\frac{\pi(\text { side })^3}{6}
\end{aligned}
$

Now, ratio of volumes,

$
V_1: V_2=(\text { side })^3: \frac{\pi(\text { side })^3}{6}
$
Cancelling (side) ${ }^3$,

$
\begin{aligned}
& =1: \frac{\pi}{6} \\
& =6: \pi
\end{aligned}
$
Hence,

$
V_1: V_2=6: \pi
$

Question:10

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

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

We know that the volume of a cylinder is $V=\pi r^2 h$

where $r$ is the radius and $h$ is the height.
Let the original volume be, $V_1=\pi r^2 h$

Now, the radius of the cylinder is doubled and the height is halved.
New radius, $R=2 r$
New height, $H=\frac{h}{2}$
New volume of the cylinder, $V_2=\pi R^2 H$
Substituting the values,

$
\begin{gathered}
V_2=\pi(2 r)^2\left(\frac{h}{2}\right) \\
=\pi \cdot 4 r^2 \cdot \frac{h}{2} \\
=2 \pi r^2 h \\
=2 V_1
\end{gathered}
$
Hence, the given statement is true.

Question:1

Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16cm×8cm×8cm. When 16 spheres are packed the box is filled with preservative liquid. Find the volume of this liquid. Give your answer to the nearest integer. [Useπ=3.14]

Answer: 488

We know that the volume of a sphere is given by $V=\frac{4}{3} \pi r^3$

where $r$ is the radius.
Given that the radius of each sphere $r=2 \mathrm{~cm}$,

$V_{\text {sphere }}=\frac{4}{3} \pi(2)^3=\frac{32}{3} \pi \approx 33.51 \mathrm{~cm}^3$

Volume of 16 spheres,

$V_{16 \text { spheres }}=16 \times 33.51 \approx 535.89 \mathrm{~cm}^3$
Internal dimensions of the given rectangular box are $16 \times 8 \times 8 \mathrm{~cm}$.
Volume of the rectangular box,

$V_{\text {box }}=l \times b \times h=16 \times 8 \times 8=1024 \mathrm{~cm}^3$
When 16 spheres are packed, the box is filled with preservative liquid.

$
\begin{aligned}
& \text { Volume of preservative liquid }=V_{\text {box }}-V_{16 \text { spheres }} \\
& \qquad=1024-535.89 \approx 488.11 \mathrm{~cm}^3
\end{aligned}
$
Hence, the volume of preservative liquid is $488.11 \mathrm{~cm}^3$.

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 $\text{m}^3$. If the present depth of water is 1.3m, find the volume of water already used from the tank.

Answer: 8.125 $\text{m}^3$

In the question, it is given that:
Volume of water in the tank $=15.625 \mathrm{~m}^3$
Depth of water $=1.3 \mathrm{~m}$

We know that the volume of a cube is given by

$V=a^3$, where $a$ is the side length.

$
\begin{gathered}
a^3=15.625 \\
⇒ a=\sqrt[3]{15.625}=2.5 \mathrm{~m}
\end{gathered}
$
The present depth of water volume is given as 1.3 m.
Height of water used $=1.2 \mathrm{~m}$
Volume of water already used from the tank,

$
\begin{aligned}
& V_{\text {used }}=\text { area of base × height of water used } \\
& =a^2 \times 1.2 \\
& =(2.5)^2 \times 1.2 \\
& =6.25 \times 1.2 \\
& =7.5 \mathrm{~m}^3
\end{aligned}
$
Hence, the volume of water already used from the tank is $7.5 \mathrm{~m}^3$.

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 c$\text{m}^3$

Given that a solid spherical ball has a diameter $d=4.2 \mathrm{~cm}$.
Radius of the ball, $r=\frac{d}{2}=\frac{4.2}{2}=2.1 \mathrm{~cm}$
We know that the volume of a sphere is

$
\begin{gathered}
V=\frac{4}{3} \pi r^3 \\
=\frac{4}{3} \pi(2.1)^3 \\
=\frac{4}{3} \pi(9.261) \\
=\frac{4}{3} \times \frac{22}{7} \times 9.261 \\
\approx 38.808 \mathrm{~cm}^3
\end{gathered}
$
Hence, the volume of the spherical ball is $38.808 \mathrm{~cm}^3$.
The amount of water displaced is equal to the volume of the ball, so the amount of water displaced = $38.808 \mathrm{~cm}^3$.

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 $\text{m}^2$

We have to find the area of canvas required for the given conical tent.
Area of canvas required $=$ Curved surface area of the conical tent.

We know that the curved surface area of a cone is

$\mathrm{CSA}=\pi r l$

where $r$ is the radius of the base and $l$ is the slant height.
Slant height of the cone,

$l=\sqrt{r^2+h^2}$
Given:
Height $h=3.5 \mathrm{~m}$
Radius $r=12 \mathrm{~m}$

$
l=\sqrt{12^2+3.5^2}=\sqrt{144+12.25}=\sqrt{156.25}=12.5 \mathrm{~m}
$
Curved surface area,

$\mathrm{CSA}=\frac{22}{7} \times 12 \times 12.5 \approx 471.43 \mathrm{~m}^2$
Hence, the area of the canvas required is $471.43 \mathrm{~m}^2$.

Question:5

Two solid spheres made of the same metal have weights of 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

The density $D$ of an object is mass $m$ per unit volume $V$: $D=\frac{m}{V}$
For the same metal, density remains constant, so mass is directly proportional to volume.
For solid 1:

$\text { Mass }=M_1, \quad \text { Volume }=V_1$
For solid 2:

$
\begin{gathered}
\text { Mass }=M_2, \quad \text { Volume }=V_2 \\
\frac{M_1}{M_2}=\frac{V_1}{V_2}
\end{gathered}
$
In the question, the objects are spheres.
Volume of a sphere: $V=\frac{4}{3} \pi R^3$

So volume is directly proportional to $R^3$.

$
\frac{M_1}{M_2}=\frac{V_1}{V_2}=\frac{R_1^3}{R_2^3}
$
Given $M_1=5920 \mathrm{~g}$ and $M_2=740 \mathrm{~g}$ :

$
\begin{gathered}
\frac{5920}{740}=\frac{R_1^3}{R_2^3} \\
R_1^3 / R_2^3=8 \\
\frac{R_1}{R_2}=2
\end{gathered}
$
The diameter of the smaller sphere is 5 cm, so

$
\begin{gathered}
R_2=\frac{5}{2}=2.5 \mathrm{~cm} \\
R_1=2 \times 2.5=5 \mathrm{~cm}
\end{gathered}
$
Hence, the radius of the 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 litres of milk

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

$r=\frac{d}{2}=\frac{7}{2}=3.5 \mathrm{~cm}$
Height of milk in each glass,

$h=12 \mathrm{~cm}$
Volume of milk in one cylindrical glass,

$
\begin{gathered}
V=\pi r^2 h \\
=\frac{22}{7} \times(3.5)^2 \times 12 \\
\approx 461.58 \mathrm{~cm}^3
\end{gathered}
$
Number of students $=1600$, so total volume of 1600 glasses:

$V_{\text {total }}=1600 \times 461.58 \approx 738528 \mathrm{~cm}^3$
Since 1 litre $=1000 \mathrm{~cm}^3$,

$V_{\text {total }}=\frac{738528}{1000} \approx 738.528 \text { litres }$
Hence, to serve 1600 students, we need 738.528 litres of milk.

Question:7

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

Answer: 200 revolutions

It is given that a cylindrical roller covers an area of $5500 \mathrm{~m}^2$ when rolled on the road.

The area covered by the roller in one revolution = Curved surface area of the cylinder.

$\text { CSA of cylinder }=2 \pi r h$

where $r$ is the radius and $h$ is the height (length) of the cylinder.

Given:
Length of roller $h=2.5 \mathrm{~m}$
Radius of roller $r=1.75 \mathrm{~m}$

$
\begin{aligned}
\mathrm{CSA}= & 2 \times 3.14 \times 1.75 \times 2.5 \\
= & 6.28 \times 4.375 \\
& \approx 27.475 \mathrm{~m}^2
\end{aligned}
$
Total area to be covered $=5500 \mathrm{~m}^2$

Number of revolutions:

$
\begin{aligned}
& \text { Number of revolutions }=\frac{\text { Total area }}{\text { Area covered in one revolution }} \\
&=\frac{5500}{27.475} \approx 200.18 \approx 200
\end{aligned}
$
Hence, the 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 40m×25m×15m. For how many days will the water of this tank last?

Answer: 40 days

Given, dimensions of the overhead tank: $40 \mathrm{~m} \times 25 \mathrm{~m} \times 15 \mathrm{~m}$
The tank is in the shape of a cuboid, so its volume:

$\text { Volume }=l \times b \times h=40 \times 25 \times 15=15000 \mathrm{~m}^3$
Population of village $=5000$
Daily water requirement for 1 person $=75$ litres
Daily water requirement for 5000 people:

$75 \times 5000=375000 \text { litres }$
Since 1 litre $=0.001 \mathrm{~m}^3$,

$\text { Daily water requirement }=375000 \times 0.001=375 \mathrm{~m}^3$
Number of days the water in the tank will last:

$\text { Number of days }=\frac{\text { Volume of tank }}{\text { Daily water requirement }}=\frac{15000}{375}=40$
Hence, the water in this tank will last for $\mathbf{4 0}$ days.

Question:9

A shopkeeper has one spherical laddoo of radius 5cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made?

Answer: 8 laddoos

Given:
Larger laddoo with radius $R=5 \mathrm{~cm}$
Smaller laddoo with radius $r=2.5 \mathrm{~cm}$

Volume of a sphere:

$V=\frac{4}{3} \pi R^3$
Volume of 1 larger laddoo:

$V_{\text {large }}=\frac{4}{3} \pi(5)^3$
Volume of 1 smaller laddoo:

$V_{\mathrm{small}}=\frac{4}{3} \pi(2.5)^3$
Let the number of smaller laddoos be $n$.

$
\begin{aligned}
V_{\text {large }} & =n \times V_{\text {small }} \\
\frac{4}{3} \pi(5)^3 & =n \times \frac{4}{3} \pi(2.5)^3 \\
125 & =n \times 15.625 \\
n & =8
\end{aligned}
$
Hence, 8 smaller laddoos can be formed from one larger laddoo.

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:

A right triangle with sides $6 \mathrm{~cm}, 8 \mathrm{~cm}$ and 10 cm is revolved about the side 8 cm.
On revolving, we get a right circular cone with:
Radius $r=6 \mathrm{~cm}$, Height $h=8 \mathrm{~cm}$, Slant height $l=10 \mathrm{~cm}$
Volume of the cone:

$
\begin{gathered}
V=\frac{1}{3} \pi r^2 h \\
=\frac{1}{3} \times 3.14 \times 6^2 \times 8 \\
=\frac{1}{3} \times 3.14 \times 36 \times 8 \\
\approx 301.44 \mathrm{~cm}^3
\end{gathered}
$

Curved surface area of the cone:

$
\begin{gathered}
\text { CSA }=\pi r l \\
=\frac{22}{7} \times 6 \times 10 \\
\approx 188.57 \mathrm{~cm}^2
\end{gathered}
$
Hence, Volume $=301.44 \mathrm{~cm}^3$ and Curved Surface Area $=188.57 \mathrm{~cm}^2$.

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 the volume of a cylinder is given by $V=\pi r^2 h$

(where $r$ is the radius and $h$ is the height)
Here, the length of the cylinder is taken as its height. $h=100 \mathrm{~cm}$
Outer diameter of the cylinder, $d=16 \mathrm{~cm}$
So, outer radius,

$r=\frac{16}{2}=8 \mathrm{~cm}$
Volume of the outer cylinder,

$
\begin{gathered}
=\pi r^2 h \\
=3.14 \times(8)^2 \times 100 \\
=20096 \mathrm{~cm}^3
\end{gathered}
$
Thickness of the iron sheet $=2 \mathrm{~cm}$.
Inner diameter,
$
=16-(2 \times 2)=12 \mathrm{~cm}
$
So, inner radius,

$
R=\frac{12}{2}=6 \mathrm{~cm}
$
Volume of the inner cylinder,

$
\begin{gathered}
=\pi R^2 h \\
=3.14 \times(6)^2 \times 100 \\
=11304 \mathrm{~cm}^3
\end{gathered}
$
Volume of iron used,

$
\begin{aligned}
& =\text { Volume of outer cylinder }- \text { Volume of inner cylinder } \\
& =20096-11304 \\
& =8792 \mathrm{~cm}^3
\end{aligned}
$

Hence, the answer is 8792 $\text{cm}^3$.

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 c$\text{m}^3$
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 =28cm
So, radius of semi-circular sheet (R)=14cm
Slant height of the conical cup (l)=14cm
Now,
Let the radius of the base of the cup = r cm.

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

$
\pi R=2 \pi r
$
Given, $R =14 \mathrm{~cm}$

$
\begin{aligned}
\frac{22}{7} \times 14 & =2 \times \frac{22}{7} \times r \\
⇒r & =7 \mathrm{~cm}
\end{aligned}
$
Let the height of the conical cup be $h \mathrm{~cm}$.
For a right circular cone, $l^2=r^2+h^2$

Given slant height, $l=14 \mathrm{~cm}$
$
\begin{aligned}
\frac{22}{7} \times 14 & =2 \times \frac{22}{7} \times r \\
r & =7 \mathrm{~cm}
\end{aligned}
$

Let the height of the conical cup be $h \mathrm{~cm}$.
For a right circular cone,

$
l^2=r^2+h^2
$
Given slant height,

$
\begin{gathered}
l=14 \mathrm{~cm} \\
(14)^2=(7)^2+h^2 \\
196=49+h^2 \\
h^2=147 \\
h=7 \sqrt{3} \mathrm{~cm}
\end{gathered}
$
Now, the capacity of the cup
= Volume of the cone

$
\begin{gathered}
V=\frac{1}{3} \pi r^2 h \\
=\frac{1}{3} \times \frac{22}{7} \times(7)^2 \times 7 \sqrt{3} \\
=\frac{1078 \sqrt{3}}{3} \\
=359.33 \sqrt{3} \\
\approx 622.38 \mathrm{~cm}^3
\end{gathered}
$

Question:3

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

1. How many students can sit in the tent if a student, on average, occupies 57 $\text{m}^2$ on the ground?
2. Find the volume of the cone.

Answer: (i) 110 students
(ii) 241.73 $\text{m}^3$

We have, Area of cloth $=165 \mathrm{~m}^2$.
This cloth is shaped in the form of a conical tent with radius $r=5 \mathrm{~m}$.
Curved surface area of cone $=\pi r l$

$
\begin{gathered}
165=\frac{22}{7} \times 5 \times l \\
\Rightarrow l=10.5 \mathrm{~m}
\end{gathered}
$

(i)Area covered by 1 student $=57 \mathrm{~m}^2$
Number of students $=\frac{\text { Area of base }}{\text { Area per student }}$

$=\frac{\pi r^2}{57}=\frac{\frac{22}{7} \times 25}{57} \approx 110$

So, 110 students can sit in the tent.

(ii) Volume of cone $=\frac{1}{3} \pi r^2 h$
For right circular cone,

$
\begin{gathered}
r^2+h^2=l^2 \\
5^2+h^2=10.5^2 \\
h=\sqrt{110.25-25} \approx 9.23 \mathrm{~m}
\end{gathered}
$

Volume,

$V=\frac{1}{3} \times \frac{22}{7} \times 25 \times 9.23 \approx 241.73 \mathrm{~m}^3$

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 $\text{m}^3$

We know that the volume of a hemispherical tank is

$V=\frac{2}{3} \pi r^3$

where $r$ is the radius of the hemispherical tank.
The internal diameter of the tank is 14 m, so the internal radius is

$
r=\frac{14}{2}=7 \mathrm{~m}
$
The tank already contains 50 kL of water. Since

$
1000 \mathrm{~L}=1 \mathrm{~m}^3 \Rightarrow 1 \mathrm{~kL}=1 \mathrm{~m}^3
$

the tank contains $50 \mathrm{~m}^3$ of water.

Total volume of the hemispherical tank,

$V=\frac{2}{3} \times \frac{22}{7} \times 7^3=718.66 \mathrm{~m}^3$
Volume of water to be pumped into the tank,

$
\begin{aligned}
V_{\text {pumped }}= & \text { Total volume }- \text { Already present volume } \\
& =718.66-50=668.66 \mathrm{~m}^3
\end{aligned}
$
Hence, the answer is $668.66 \mathrm{~m}^3$.

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 spheres is in the ratio
$V_1: V_2=64: 27$
We know that:
Volume of a sphere $=\frac{4}{3} \pi r^3$
Surface area of a sphere $=4 \pi r^2$

Let the radius of sphere 1 be $r_1$ and of sphere 2 be $r_2$.

$
\begin{gathered}
\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}=\frac{64}{27} \\
\frac{r_1}{r_2}=\frac{4}{3}
\end{gathered}
$
Now, the ratio of the surface areas of the spheres,

$
\begin{gathered}
\frac{\text { Area of sphere } 1}{\text { Area of sphere } 2}=\frac{4 \pi r_1^2}{4 \pi r_2^2}=\left(\frac{r_1}{r_2}\right)^2 \\
=\left(\frac{4}{3}\right)^2=\frac{16}{9}
\end{gathered}
$
Hence, the required ratio of surface areas 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 $\text{cm}^3$

Volume of the cube is

$V_{\text {cube }}=(\text { side length })^3=(4 \mathrm{~cm})^3=64 \mathrm{~cm}^3$

As the cube contains a sphere,
Diameter of the sphere $=$ side length of the cube $=4 \mathrm{~cm}$
Radius of the sphere, $r=\frac{4}{2}=2 \mathrm{~cm}$
Volume of the sphere,

$V_{\text {sphere }}=\frac{4}{3} \pi r^3=\frac{4}{3} \times \frac{22}{7} \times 2^3=33.51 \mathrm{~cm}^3$
Volume of the gap in between,

$
\begin{gathered}
V_{\text {gap }}=V_{\text {cube }}-V_{\text {sphere }}=64-33.51 \\
\approx 30.49 \mathrm{~cm}^3
\end{gathered}
$
Hence, the required answer is $30.49 \mathrm{~cm}^3$.

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 exceed its height?

Answer: 50%

Let the radius of the sphere be $r$, the radius of the cylinder be $R$, and the height of the cylinder be $H$.
Volume of the sphere $=\frac{4}{3} \pi r^3$
Volume of the cylinder $=\pi R^2 H$
It is given that $r=R$, so

$
\begin{aligned}
\frac{4}{3} \pi r^3 & =\pi r^2 H \\
\frac{4}{3} r^3 & =r^2 H \\
H & =\frac{4}{3} r
\end{aligned}
$
Diameter of the sphere, $d=2 r$

$
\begin{gathered}
d=2 r=2 \times \frac{3}{4} H=\frac{3}{2} H \\
d=1.5 H=(1+0.5) H \\
d=\left(1+\frac{50}{100}\right) H
\end{gathered}
$
Hence, the diameter of the sphere exceeds the height of the cylinder by $50 \%$.

Question: 8(i)

(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 3 cm are placed one above the another to form a cylindrical solid. Find the volume of the cylinder so formed.

(i) Answer: 9152 $\text{cm}^2$

It is given that 30 circular plates are placed one above the other to form a cylindrical solid.
Height of the cylindrical solid = total thickness of 30 plates

$\text { Height, } h=30 \times 3=90 \mathrm{~cm}$

Radius of each plate, $r=14 \mathrm{~cm}$

The total surface area of a cylinder is given by

$\mathrm{TSA}=2 \pi r(h+r)$
Substituting the values,

$
\begin{aligned}
\mathrm{TSA}=2 \times & \frac{22}{7} \times 14 \times(90+14) \\
= & 44 \times 104 \\
= & 9152 \mathrm{~cm}^2
\end{aligned}
$
Hence, the total surface area of the cylindrical solid is $9152 \mathrm{~cm}^2$.

(ii)
Answer: 55440 $\text{cm}^3$

It is given that 30 circular plates are placed one above the other to form a cylindrical solid.
Height of the cylindrical solid $=$ total thickness of 30 plates

$h=30 \times 3=90 \mathrm{~cm}$
Radius of each plate,

$r=14 \mathrm{~cm}$
The volume of a cylinder is given by

$V=\pi r^2 h$
Substituting the values,

$
\begin{gathered}
V=\frac{22}{7} \times 14^2 \times 90 \\
=\frac{22}{7} \times 14 \times 14 \times 90 \\
=55440 \mathrm{~cm}^3
\end{gathered}
$
Hence, the volume of the cylindrical solid is $55440 \mathrm{~cm}^3$.