NCERT Exemplar Class 9 Science Solutions chapter 8 Motion

# NCERT Exemplar Class 9 Science Solutions chapter 8 Motion

Edited By Safeer PP | Updated on Sep 01, 2022 04:09 PM IST

NCERT exemplar Class 9 Science solutions chapter 8 involves the understanding of a key concept of physics: Motion. Any motion can be defined with the help of four basic properties: position, time, velocity, and acceleration. The knowledge of motion or study of motion is also known as kinematics. These NCERT exemplar Class 9 Science chapter 8 solutions are prepared by our highly experienced physics teachers. The intent of these NCERT exemplar Class 9 Science solutions chapter 8 is to provide exhaustive solutions with a step-by-step guide to understand and build the concepts of NCERT Class 9 Science. These NCERT exemplar Class 9 Science chapter 8 solutions explore all the possible questions based on the concept of Motion and are in accordance with the CBSE Syllabus for Class 9.

## NCERT Exemplar Class 9 Science Solutions Chapter 8-MCQ

Question:1

A particle is moving in a circular path of radius r. The displacement after half a circle would be:
(a) Zero
(b) $\pi r$
(c) 2 r
(d) 2 $\pi r$

Displacement is the direct separation length from initial point to final point.
Distance is the total path length travelled by the particle.
As the particle covers half circle it will reach at diametrically opposite point.
If someone ask, what is the distance travelled, answer will be half of Perimeter.
But displacement will be the direct separation between these two diametric points and it would be equal to Diameter.
Hence the correct answer is C which is 2r.

Question:2

A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a) $\frac{u}{g}$
(b) $\frac{u^{2}}{2g}$
(c) $\frac{u^{2}}{g}$
(d) $\frac{u}{2g}$

The body is thrown upwards with some initial velocity and acceleration due to gravity is constant, which will retard the body.
As the acceleration is constant we can use equation of motion.
By using third equation of motion and putting proper values as-
$v^{2}=u^{2}+2as$
where v = 0 and a= -g
$0^{2}=u^{2}+2(-g)h$
$h=\frac{u^{2}}{2g}$
Hence, the correct answer is option B

Question:3

The numerical ratio of displacement to distance for a moving object is
(a) always less than 1
(b) always equal to 1
(c) always more than 1
(d) equal or less than 1

The shortest distance between initial point and final point is called displacement and total path length is called distance.
If a particle travels in a straight line without turning back, then its distance and displacement will be equal. In all other cases distance will be more than or equal to displacement.
Therefore, the ratio of displacement and distance will be equal or less than 1
Hence the correct option is D.

Question:4

If the displacement of an object is proportional to square of time, then the object moves with
(a) uniform velocity
(b) uniform acceleration
(c) increasing acceleration
(d) decreasing acceleration

If the object moves with uniform velocity the displacement increases linearly with time. That means displacement will be proportional to time.
In case of uniform acceleration, we can use equation of motion.
$v = u + at$
$s = ut + \frac{1}{2}$
$v^{2}=u^2 + 2as$
If we observe second equation of motion carefully for zero initial velocity, We will observe that displacement is proportional to square of time.
$s = \frac{1}{2}at^2$
In case of variable acceleration, whether it is increasing or decreasing, the displacement will not be proportional to square of time.
Hence the correct option is B.

Question:5

From the given v – t graph (Figure), it can be inferred that the object is
(a) in uniform motion
(b) at rest
(c) in non-uniform motion
(d) moving with uniform acceleration

If we observe the graph carefully we can see that at every time, the value of velocity remains constant.
Hence the acceleration or retardation is zero as a speed is not changing.
The velocity is constant but it is non-0 hence particle must be moving with uniform velocity.
The motion of such type is called uniform motion.
hence the correct option is option A

Question:6

Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 ms–1. It implies that the boy is:
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity

Velocity is a vector quantity, which has direction as well as magnitude.
In merry-go-round, even the boy is moving with constant speed, the direction of velocity keep on changing.
Hence, we can say velocity of the boy is changing.
The quantity, which changes the velocity is called acceleration.
Therefore, the boy must be in accelerated motion.
Hence the correct option of this question is C.

Question:7

Area under a v – t graph represents a physical quantity which has the unit
(a) $m^2$
(b) m
(c) $m^3$
(d) $ms^{-1}$

Area under any x-y Graph will have unit as product of unit of x and unit of y.
That means area under the Graph of velocity and time will have unit equal to Product of unit of velocity (m/s) and unit of time (s).
That would be meter.
For the information, the area under the v-t graph gives displacement.
Hence, the correct option is option B

Question:8

Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in Figure. Choose the correct statement
(a) Car A is faster than car D.
(b) Car B is the slowest.
(c) Car D is faster than car C.
(d) Car C is the slowest.

If we closely observe the graph, we will notice:
For any time interval distance travelled by car C is maximum followed by D, A and B.
Hence, car C is the fastest car and car B is slowest car.
In distance time graph, if the graph is a straight line then the angle between the straight line and time axis represents the slope or velocity of the car.
Larger the angle implies larger the slope means larger velocity.
Hence the correct option is B.

Question:9

Uniform motion means the body is moving with constant velocity.
Hence, in every second it will cover same distance.
Therefore, for uniform motion distance time graph will be a straight line.
In the graph shown in option B, the distances travelled in consecutive seconds are increasing hence speed is increasing.
In the graph shown in option C, the distances travelled in consecutive seconds are decreasing hence speed is decreasing.
In option D distance is not changing hence the object is not moving.
Therefore, the correct option is option A.

Question:10

Slope of a velocity – time graph gives
(a) the distance
(b) the displacement
(c) the acceleration
(d) the speed

If we observe velocity time graph, we can calculate displacement as well as acceleration.
The area under the velocity time graph, gives the displacement and slope of velocity time graph gives the acceleration.
Qualitatively, the slope of line represents angle made by straight line from time axis.
Larger the angle implies larger the slope.
Hence the correct option is option C

Question:11

In which of the following cases of motions, the distance moved and the magnitude of displacement are equal?
(a) If the car is moving on straight road
(b) If the car is moving in circular path
(c) The pendulum is moving to and fro
(d) The earth is revolving around the Sun

Sol.
The distance is total path travelled by anybody. The displacement is shortest distance between initial and final point of the motion.
Only in the case when the body moves in a straight line without turning, the distance will be equal to displacement.
In all other cases distance will be more than displacement.
Hence the correct option of this question is option A.

## NCERT Exemplar Class 9 Science Solutions Chapter 8-Short Answer

Question:12

The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Justify you answer.

The displacement is the shortest distance between initial point and final point in any motion.
The distance is total path length of the motion.
If anybody starts from a point and comes back to the same point. The displacement will be zero but distance will be non-zero.
For example, if you start your scooter from your home and after some time you park the scooter at the same place. The displacement will be zero but distance is non-zero.
The reading of scooter-meter will show the distance travelled.
Hence, zero displacement does not guarantee zero distance.

Question:13

How will the equations of motion for an object moving with a uniform velocity change?

If a particle is moving with uniform velocity, it means its acceleration is zero.
Hence, its speed remains constant at all times therefore 1st equation looks as:
v = u
The $2^{nd}$ equation of motion looks as:
s = ut
and the 3rd equation looks as:
$v^{2}=u^{2}$

Question:14

If we observe the displacement time graph, for the part of graph, where distance is increasing the graph is a straight line. It means that its slope is constant. Therefore, we can say its velocity will be constant for that time interval.
Later, its distance is decreasing with constant slope. It means that it is moving with constant velocity in opposite direction.
Mathematically, opposite direction of velocity will be negative velocity.
Hence, its v-t graph will look as shown below:

Question:15

A car starts from rest and moves along the x-axis with constant acceleration $5ms^{-2}$ for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?

For first eight seconds the car is moving with constant acceleration with zero initial velocity.
Hence, we can use equations of motion to find out distance and final velocity after that time interval of eight seconds.
$s =ut +\frac{1}{2}at^{2}$
$v = u + at$
where u=0 , a =5 , t=8
By putting values
s = 160 m
v = 40 m/s
Now for the next four seconds the car will move with constant speed of 40 m/s. Hence, in this time interval it will travel 160 m.
There for total distance travelled will be 320 m.

Question:16

A motorcyclist drives from A to B with a uniform speed of 30 $km h^{-1}$ and returns back with a speed of 20 $km h^{-1}$. Find its average speed.

The average speed is total distance travelled divided by total time taken.
Let the distance between A to B is X kilometre.
As it is moving with uniform speed, time taken from A to B will be $\frac{X}{30}$.
Similarly time taken from B to A will be $\frac{X}{20}$.
Hence total time of motion will be $\left ( \frac{X}{30} \right ) +\left (\frac{X}{20} \right )= \frac{X}{12}$.
Total distance travelled will be equal to 2X.
Therefore, by dividing total distance by total time we can find average speed that would be 24 km/h.

Question:17

(i) As there is no change in velocity, we can say that its acceleration is zero all the time.
(ii) If we observe velocity time graph, we can easily deduce that velocity is constant and equal to 20 m/s during complete motion.
(iii) The distance travelled by the cyclist can calculated by the formula:
$s = v t = (20) (15) m = 300 m$
Hence the distance travelled will be 300 m.

Question:18

Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.

When a stone is thrown upwards, it will have initial positive velocity.
This velocity will decrease uniformly due to gravity and become zero at the highest point.
After that its velocity will increase its magnitude but in downward direction.
Hence the correct graph of velocity will be as shown:

If we try to draw the graph of speed versus time, it would look like as shown:

## NCERT Exemplar Class 9 Science Solutions Chapter 8-Long Answer

Question:19

An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time?

As the objects are dropped from rest their initial velocity is zero.
The acceleration will be acceleration due to gravity, which will be equal to 10 ms-2.
The height of the particle is written as:
$D = H-S$
Where H is initial height and S is displacement.
If we write heights of both the particles by using second equation of motion: $s = ut + \frac{1}{2}at^2$
$D_{1}=H_{1}- \left ( ut + \frac{1}{2}at^{2} \right )$
$D_{2}=H_{2}- \left ( ut + \frac{1}{2}at^{2} \right )$
Hence,
$D_{1}-D_{2}=H_{1}-H_{2}=50m$
The difference in height at any time is coming out to be 50 m.
So even, after two seconds, the difference will be 50 m and it will not vary with time.

Question:20

An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start.

We can observe that the distance travelled in consecutive seconds is increasing hence we can conclude that object is accelerating.
Let its initial speed is u and acceleration is a.
Hence, by the 2nd equation of motion we can write:
$s= ut + \frac{1}{2}at^{2}$
$20= u(2) + \frac{1}{2}a(2)^{2}$
$\Rightarrow 20= 2u +2a$
$\Rightarrow 10= u +a$ ..................(i)
and
$180 =u (6)+ \frac{1}{2}a(6)^{2}$
$180 =6u+ 18 a$
$\Rightarrow 30 =u+ 3 a$ ............(ii)
Solving (i) and (ii)
u = 0
$a= 10m /s^{2}$
Since, acceleration is same, the velocity after 7 s from the start will be $v' = 0+(10\times 7 ) = 70 \;ms^{-1}$

Question:21

Using following data, draw time displacement graph for a moving object:
Time (s) 0 2 4 6 8 10 12 14 16
Displacement (m) 0 2 4 4 4 6 4 2 0
Use this graph to find average velocity for first 4 s, for next 4 s and for last 6 s.

By using the data given in the table, if we draw the displacement time graph, it will look as shown:

Average velocity is displacement divided by time in any time interval.
In first four seconds, it has travelled four meter hence average velocity will be 1 m/s.
For the next four seconds, its position is not changing. This means it is not moving. Hence, its average velocity will be zero.
In last six seconds, from t = 10 to t = 16 sec, it has changed its position from 6 m to 0 m.
Hence, in this time interval, displacement is 6 m and time is also 6 seconds.
Therefore, average velocity will be 1 m/s.

Question:22

An electron moving with a velocity of $5 \times 10^4 ms^{-1 }$enters into a uniform electric field and acquires a uniform acceleration of $10^4 ms^{-2 }$ in the direction of its initial motion.
(i) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
(ii) How much distance the electron would cover in this time?

(i) Given that, Initial velocity of the electron is u =$5 \times 10^4 ms^{-1 }$
Acceleration of the electron is a = $10^4 ms^{-2 }$
We know that first equation of motion is,
v = u + at
2u = u + at
$t = \frac{u}{a} =\frac{ (5 \times 10^4) }{ (10^4) }= 5 seconds$
(ii) Now we know that,
$s = ut + \frac{1}{2}at^{2}$
$s =( 5 \times 10^{4})5+ \frac{1}{2}(10)^{4}(5)^{2}$
$s = 37.5 \times 10^{4}m$

Question:23

Obtain a relation for the distance travelled by an object moving with uniform acceleration in the interval between 4th and 5th seconds.

Distance travelled in nth second by any particle is equal to the difference of distances travelled in n seconds and n -1 seconds.

For example. distance travelled in the fourth second is the difference of distance travel in four seconds and three seconds.

Hence, by using the second equation of motion we can find out the distance travelled in nth second.

If the initial velocity is zero,

In that case, the distance travelled in the fourth second and fifth second will be in the ratio of 7:9.

Question:24

Two stones are thrown vertically upwards simultaneously with their initial velocities u1and u2 respectively. Prove that the heights reached by them would be in the ratio of u12 : u22

(Assume upward acceleration is –g and downward acceleration to be +g ).

The height reached by any stone can be calculated by using the third equation of motion.

$v^2=u^2+2as$

We can use the fact, that at the highest point the velocity will be zero. The acceleration due to gravity will retard the particles.

Therefore, the height raised for any particle will be given as:

$\\0^2=u^2+2as\\\Rightarrow h=\frac{u^2}{2g}$

Therefore

$\frac{h_1}{h_2}=\frac{u_1^2}{u_2^2}\\$

## NCERT Exemplar Class 9 Science Solutions Chapter 8 Important Topics:

The key topics covered in the chapter on Motion through NCERT exemplar Class 9 Science solutions chapter 8 are:

• Basics of motion and relations among kinematic quantities.
• In this chapter, the fundamentals of graphs are discussed for different types of physical quantities involved in motion.
• We will learn in this chapter that how to find out acceleration in a speed-time graph or how to find distance or displacement in the same graph.
• NCERT exemplar Class 9 Science solutions chapter 8 discusses the importance of slope and area under the curve is explained by the help of relations in kinematical variables

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

 Chapter wise solutions Chapter 1 Matter in our Surroundings Chapter 2 Is Matter Around Us Pure? Chapter 3 Atoms and Molecules Chapter 4 Structure of the Atom Chapter 5 The Fundamental Unit of Life Chapter 6 Tissues Chapter 7 Diversity in Living Organisms Chapter 9 Forces and Laws of Motion Chapter 10 Gravitation Chapter 11 Work and Energy Chapter 12 Sound Chapter 13 Why do We Fall ill? Chapter 14 Natural Resources Chapter 15 Improvement in Food Resources

## Features of NCERT exemplar class 9 science solutions chapter 8:

• These Class 9 Science NCERT exemplar chapter 8 solutions provide a basic understanding of motion.

• The knowledge of this chapter is very important for all students who will pursue science in higher Classes.

• Whether you want to be an engineer or a doctor, you have to study physics. Motion can be treated as the first fundamental chapter of physics in higher classes.

• The NCERT exemplar Class 9 Science solutions chapter 8 are appropriate for students planning to attempt other books such as NCERT Class 9 Science textbook, Physics Question Bank by Oswaal Publications, Lakhmir Singh and Manjit Kaur by S. Chand et cetera.

Also read - NCERT Solutions for Class 9

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

 Chapter No. Chapter Name Chapter 1 Matter in Our Surroundings Chapter 2 Is Matter Around Us Pure Chapter 3 Atoms and Molecules Chapter 4 Structure of The Atom Chapter 5 The Fundamental Unit of Life Chapter 6 Tissues Chapter 7 Diversity in Living Organisms Chapter 8 Motion Chapter 9 Force and Laws of Motion Chapter 10 Gravitation Chapter 11 Work and Energy Chapter 12 Sound Chapter 13 Why Do We Fall ill? Chapter 14 Natural Resources Chapter 15 Improvement in Food Resources

### Frequently Asked Question (FAQs)

1. Q1. Are graphs important in understanding motion?

A1. Graphs are important to analyze the behavior of any phenomenon. In motion, graphs are very important and with the help of a graph we can easily find out one variable from another variable such as displacement, distance, or acceleration from the speed-time graph.

2. Q2. “Motion or rest “can we absolutely determine?

A2. No motion or rest is totally relative. Whether a body is in motion or at rest will depend on the observer who is observing that body. A man sitting on the train will find that everybody else as the fellow passengers are at rest but they are moving with respect to the man on the ground.

3. Q3. How to find displacement from the speed-time graph?

A3. The area under the speed-time graph will give displacement or distance of any moving body.

4. Q4. Is Motion an important topic for competitive examination?

A4. Motion is one of the building blocks of physics and is an extremely important topic from the perspective of competitive exams such as JEE Advanced and NEET.

5. Q5. Will I be able to download these solutions and view them in an offline mode?

A5. Certainly, NCERT exemplar Class 9 Science solutions chapter 8 pdf download is the feature provided to students to view/download these solutions for use in offline mode and attempt NCERT exemplar Class 9 science chapter 8. NCERT exemplar Class 9 science solutions chapter 7 are elaborate in nature and judicious use can help the students score well in the examinations.

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