NCERT Exemplar Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes 2021

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.

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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) $\frac{32}{3}\pi r^{3}$

Answer (D)

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) $27\; \text {cm}^{3}$

Answer (C)

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) $1.6\; cm$

Answer (B)
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

Answer (C)
We know that the curved surface area of a cylinder is $2\pi r h$
(Where r is radius and h is height)
Let, original radius =r
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) $2\pi r l$

Answer (B)
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) $20 : 37$

Answer (B)
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) $256\; m^{3}$

Answer (A)
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) $1840$

Answer (B)
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) $12\; m$

Answer (A)
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) $2:1$

Answer (A)
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.