# NCERT Solutions for Class 9 Science Chapter 8 Motion

NCERT solutions for class 9 science chapter 8 Motion: How do you describe the term Motion? After going through chapter 8 motion you will be able to give the answer to this question. Motion is a relative term. That is if we are traveling on a bus, a person sitting behind us is at rest with respect to us but for a person outside the bus, he is moving. We feel like trees are moving when we are traveling in a bus, this is due to relative motion. We can find such an enormous examples related to motion. By using some good examples, CBSE NCERT solutions for class 9 science chapter 8 Motion will also give you a better understanding of the concept. For example, if a person says that my home is 60 Km north of the airport. Here, the reference point is the airport. To specify the position of an object we need to choose a reference point and a direction. If in the previous case the person says that my home is 60 Km from the airport then we can go 60 Km in any direction. To know the exact point specifying direction along with reference is also important. Along with exercise solutions for NCERT class 9 science chapter 8 Motion, you will also get NCERT solutions for the questions mentioned in between the chapter. The solutions are created by our experts. Below you can read important points of the chapter which will help you to understand the chapter properly.

NCERT solutions for class 9 science chapter 8 Motion - Points to Remember for important concepts

1. Distance and Displacement: To understand the concept of distance and displacement mentioned in the chapter let's consider an ant moving along a straight line. The ant starts from point O and moves a distance 50 cm and reaches point B, then it comes back a distance of 25 cm and reaches point A. We need to calculate the distance and displacement when the ant is at B and also when the ant is at A.

When the ant is at B the distance travelled = 50cm and displacement (shortest distance from the origin) = 50 cm. That is displacement = distance. When the ant comes back to A total distance travelled = OB+BA=50+25=75 cm but displacement = OA =25 cm. Here the displacement is less than distance. The main point to remember from this session of NCERT Solutions for Class 9 is that distance is always greater than or equal to displacement. The distance (path length) travelled cannot be zero but the displacement can be zero. In the above example of the ant comes back to O then displacement=0 but distance = 50+50 =100 cm.

2. Speed and Velocity: A very important and interesting concept of the NCERT class 9 science chapter 8 Motion is speed and velocity. Certain points to remember from this topic are listed below

• $Average \ speed = \frac{distance\ travelled}{total\ time\ taken}$
• If we specify the direction along with speed the term is known as velocity. Velocity is the speed of an object in a definite direction
• $Average \ velocity = \frac{displacement}{total\ time\ taken}$
• If the object is moving with varying velocity then the

$Average \ velocity = \frac{initial\ velocity+final\ velocity}{2}$

For example, if an ant moves along a rectangle as shown in figure 2 from point A to C through B in 5 seconds then what is the average speed and average velocity of the ant?

figure 2

Solution: The distance travelled by the ant = 3+4 =7m

average speed=7/5 = 1.4 m/s

The displacement = AC

$AC=\sqrt{AB^2+BC^2}=\sqrt{3^2+4^2}=5m$

AC is obtained using Pythagoras theorem

average velocity = displacement / total time taken= 5/5 =1 m/s

The next topic of the NCERT Solutions for Class 9 Science Chapter 8 Motion is acceleration:

• Acceleration is the rate of change of velocity
• $acceleration=\frac{change\ in\ velocity }{time\ taken}$
• The SI unit of acceleration is

$ms^{-2}or\ \frac{m}{s^2}$

• The acceleration is taken to be positive if it is in the direction of velocity and negative when it is opposite to the direction of the velocity
• If an object travels in a straight line and its velocity changes by equal amounts in equal intervals of time, then the acceleration is uniform. If the change in velocity is not equal in equal interval of time then the acceleration is non-uniform

3. Distance time graph of an object moving with uniform velocity:

Figure 3

The distance-time graph for an object moving with uniform velocity is a straight line. The speed of the object can be determined as follows

Consider a small portion AB on the line. Draw a line parallel to x-axis from A and a line parallel to y-axis. Let both the lines meet at C, then and mark the corresponding points of A, B and C as shown in the above figure. The points are s1,s2,t1 and t2

$speed=\frac{BC}{AC}=\frac{s_2-s_1}{t_2-t_1}$

4. The velocity-time graph of an object moving with uniform velocity:

figure 4

The graph of an object moving with uniform velocity is a straight line parallel to the x-axis. To find the distance travelled for a given time interval (say t2-t1 as in the above figure) we have to calculate the area of the velocity-time graph. That is in the above figure the area ABCD gives the distance travelled in time t2-t1.

5. Velocity-time graph for a car moving with uniform acceleration:

Figure 5

Velocity-time graph for a car moving with uniform acceleration is a straight line and the area of the graph gives the distance travelled during a time interval. In the above graph during t2-t1 the distance travelled= area of triangle ADE+area of rectangle ABCD.

An important topic of the NCERT Solutions for Class 9 Science Chapter 8 is equations of motion for an object moving with uniform acceleration. If the initial velocity is u, the final velocity is v and t is the time taken and s is the distance travelled then the following are the three equations of motion.

$\\v=u+at\\s=ut+\frac{1}{2}at^2\\v^2-u^2=2as$

6. Uniform circular motion: Let's understand this through an example: If an ant moves along a circular path of radius 7 cm from point A to B through C and then to A through D as shown in figure 6 with a uniform speed of 1cm/sec

i)what is the distance travelled when the ant is at B. Is the distance equal to displacement?

Solution:

$distance=ACB=\frac{1}{2}circumferance=\frac{1}{2}\times2\pi r=\frac{1}{2}\times2\pi \times 7=22cm$

Displacement =AB=diameter =14cm. So distance not equal to displacement

ii)what is the total distance travelled and the final displacement?

Total distance travelled = circumference of the circle = $2\pi r=2\pi \times 7=44cm$

Displacement =0 as the initial and final position are the same

iii) What is the time taken by the ant to travel from A to B

The time = distance / speed = 22 / 1 = 22 sec

## The main topics of NCERT solutions for class 9 science chapter 8 Motion are listed below:

8.1 Describing Motion

8.1.1 Motion along a straight line

8.1.2 Uniform Motion and Nonuniform Motion

8.2 Measuring the rate of motion

8.2.1 Speed with Direction

8.3 Rate of Change of Velocity

8.4 Graphical Representation of Motion

8.4.1 Distance- Time Graphs

8.4.2 Velocity-Time Graphs

8.5 Equations of Motion by Graphical Method

8.5.1 Equation for Velocity-Time Relation

8.5.3 Equation for Position-Velocity Relation

8.6 Uniform Circular Motion

If you stuck anywhere or want to complete your homework on time, refer to the solutions mentioned below:

## Topic 8.1 Describing Motions

Yes, an object has moved through a distance can have zero displacements.

If an object moves and returns to the original position, the displacement will be zero. Consider the movement in a circular path. A man walks from point A in a circular path in a park and comes back to point A.

The distance traveled is equal to the circumference of the circular path, but displacement is zero.

Side of the square field = $10\ m$
$\therefore$ The perimeter of the square = $4\times10 = 40\ m$
According to question,
He completes 1 round in $40\ s$.
$\therefore$ Speed of the farmer = $\frac{40}{40}= 1\ m/s$
$\therefore$ Distance covered in $2\ min\ 20\ s\ (=140\ s)$ = $140 \times 1 = 140\ m$
Now,
Number of round trips completed traveling = $\frac{140}{40} = 3.5$
We know, in 3 round trips the displacement will be zero.
In $0.5$ round, the farmer will reach diametrically opposite to his initial position.
$\therefore$ Displacement = $AC = \sqrt{(AB^2 + BC^2 )} = \sqrt{(100^2+100^2)} = 10\sqrt{2}\ m$

(a) It cannot be zero.

(b) Its magnitude is greater than the distance traveled by the object.

(a) The first statement is false. Because displacement can be zero when the initial point coincides with the final point.

(b) The second statement is false. The magnitude of displacement can never be greater than the distance travelled by the object. It can be either equal or less.

## Solutions of NCERT for class 9 science chapter 8 Motion

Topic 8.2 Measuring the Rate of Motion

 Speed Velocity Speed is the distance travelled by an object in unit time Velocity is the speed of an object moving in a definite direction. Speed is a scalar quantity Velocity is a vector quantity Speed does not depend on the direction Velocity changes with change in direction Speed can never be negative Velocity can be positive, negative or zero.

When the total distance traveled by the object is equal to the displacement, the magnitude of the average velocity will be equal to the average speed. Average speed is the total distance upon the time taken, whereas average velocity is the total displacement upon time taken.

Odometer is a device that measures the total distance traveled by automobile.

An object is having a uniform motion if it covers equal distance in equal interval of time (which implies speed is constant!). So the path can be straight or curved.

For eg. Consider a circular path. For understanding purposes, divide the circumference of the circle in six equal parts each subtending $60^{\circ}$ at the centre. The object covers each equal part in equal amount of time. Hence, by definition, this object is in uniform motion.

Q 5.  During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is$3 \times 10^{8}ms^{-1}$.

Given, the signal travels at the speed of light,  $v = 3 \times 10^{8}ms^{-1}$.

Time taken by the signal =$5\ mins = 5\times60 s= 300\ s$

Let the distance of the spaceship from the ground station be $D\ m$

We know, $Speed = \frac{Distance\ travelled }{Time\ taken}$

$\\ \implies v = \frac{D}{t} \\ \\ \implies D = v\times t = 3\times10^8 \times 300$

$\\ \implies D = 9\times10^{10}\ m$

Therefore, the distance of spaceship from the ground station is $9\times10^{10}\ m$

## NCERT textbook solutions for class 9 science chapter 8 Motion

Topic 8.3 Rate of Change of Velocity

(i) uniform acceleration?

(ii) nonuniform acceleration?

(i) If the velocity of an object traveling in a straight line increases or decreases by equal amounts in equal intervals of time, then the acceleration of the object is said to be uniform. For example, An apple having a free-fall motion.

(ii) On the other hand, if the velocity of the object increases or decreases by unequal amounts in equal intervals of time, then the acceleration of the object is said to be non-uniform. For example, A car travelling along a straight road increasing its speed by unequal amounts in equal intervals of time.

(We know, $1\ km = 1000\ m ; 1\ hr = 3600\ s$)

Given, Initial speed of the bus,$u$  = $80\ kmh^{-1} = \frac{80\times1000}{3600} = \frac{200}{9} ms^{-1}$

The final speed of the bus,$v$$60\ kmh^{-1} = \frac{60\times1000}{3600} = \frac{100}{6} ms^{-1}$

Time is taken, $t= 5s$

We know, $v = u+at$

$\\ \implies \frac{100}{6} = \frac{200}{9}+a(5) \\ \implies 5a = \frac{900-1200}{6\times9} = \frac{-300}{6\times9} \\ \implies a = -\frac{100}{9} = -1.11\ ms^{-2}$

The negative sign implies retardation.

Therefore, the acceleration of the bus is $-1.11\ ms^{-2}$

Or, the retardation(de-acceleration) of the bus is $1.11 ms^{-2}$

(We know, $1\ km = 1000\ m ; 1\ hr = 3600\ s$)

Given, The train starts from rest. Hence, the initial speed of the train = $0\ kmh^{-1} = 0\ ms^{-1}$

Final speed of the train = $40\ kmh^{-1} = \frac{40\times1000}{3600} = \frac{100}{9} ms^{-1}$

Time taken, $t = 10\ min = 10\times60\ s = 600\ s$

We know, $v = u+at$

$\\ \implies \frac{100}{9} = 0+a(600) \\ \implies a = \frac{100}{9\times600} = \frac{1}{6\times9} \\ \implies a = \frac{1}{54} = 0.0185\ ms^{-2}$

Therefore, the acceleration of the train is  $0.0185\ ms^{-2}$

## Solutions for NCERT class 9 science chapter 8 Motion

Topic 8.4 Graphical Representation of Motion

Distance-time graph is the plot of distance travelled by an object along x-axis against time along y-axis.

For the uniform motion of an object, the distance-time graph is a straight line with a constant slope. (Note: However, the path may be straight or curved!)

For non-uniform motion of an object, the distance-time graph is a curved line with an increasing or decreasing slope.

If the distance-time graph of an object is a straight line parallel to the time axis, it means that the distance of the object is the same from its initial position at any point of time. This implies that the object is not moving and is at rest.

If the speed-time graph of an object is a straight line parallel to the time axis, it means that the speed of the object is not changing with time. Hence the speed of the object is constant. This also implies that the acceleration of the object is zero.

The area occupied below the velocity-time graph denotes the total distance travelled by an object in the given time frame.

We know,

$\\ Speed = \frac{Distance}{Time} \\ \implies Distance = Speed\times Time$

## Free NCERT solutions for class 9 science chapter 8 Motion

Topic 8.5 Equations of motion by graphical method

Given, The bus starts from rest. Hence, the initial speed of the bus = $0\ ms^{-1}$

Acceleration of the bus, $a$ = $0.1\ ms^{-2}$

Time is taken, $t = 2\ min = 2\times60\ s = 120\ s$

(a) We know, $v = u+at$

$\\ \implies v = 0+(0.1)(120) = 12\ ms^{-1}$

Therefore, the speed acquired by the bus is $12\ ms^{-1}$

Given, The bus starts from rest. Hence, the initial speed of the bus, u = $0\ ms^{-1}$

Acceleration of the bus, $a$ = $0.1\ ms^{-2}$

Time taken, $t = 2\ min = 2\times60\ s = 120\ s$

(b) We know, $s = ut+\frac{1}{2}at^2$

$\\ \implies s = 0(120) + \frac{1}{2}(0.1)(120)^2 \\ \implies s = \frac{1}{2}(0.1)(14400) \\ \\ \implies s = 720\ m$

Therefore, the distance travelled by bus is $720\ m$

(We know, $1\ km = 1000\ m ; 1\ hr = 3600\ s$)

Given, Initial speed of the train,$u$  = $90\ kmh^{-1} = \frac{90\times1000}{3600} = 25\ ms^{-1}$

Acceleration of the train,$a = -0.5\ ms^{-2}$   (Negative sign implies retardation)

Since, the train has to be brought to rest, final speed of the train, $v$$0\ ms^{-1}$

We know, $v^2 = u^2 + 2as$

$\\ \implies 0^2 = 25^2 + 2(-0.5)s \\ \implies 0 = 625 -s \\ \implies s = 625\ m$

Therefore, the train travels a distance of $625\ m$ before coming to rest.

Given, The trolley starts from rest. Hence, the initial speed of the trolley,$u$ = $0\ ms^{-1}$

Acceleration of the trolley, $a$ = $2\ cms^{-2}$

Time is taken, $t =3\ s$

(a) We know, $v = u+at$

$\\ \implies v = 0+(2)(3) = 6\ cms^{-1}$

Therefore, the velocity of the trolley after 3 sec is $6\ cms^{-1}$

Given, Initial speed of the racing car, u = $0\ ms^{-1}$

Acceleration of the car, $a$ = $4\ ms^{-2}$

Time taken, $t = 10\ s$

We know, $s = ut+\frac{1}{2}at^2$

$\\ \implies s = 0(10) + \frac{1}{2}(4)(10)^2 \\ \implies s = 2(100) \\ \implies s = 200\ m$

Therefore, the distance travelled by the racing car in $10\ s$ is $200\ m$

Taking upward direction as positive (+) direction:

Given,

$u = 5\ ms^{-1}$

$a = -10\ ms^{-2}$         (This is due to gravitational force!)

The stone will move up until its velocity becomes zero.

$\therefore v = 0\ ms^{-1}$

We know, $v^2 = u^2 + 2as$

$\\ \implies 0^2 = 5^2 + 2(-10)s \\ \implies 20 s = 25 \\ \implies s = 1.25\ m$

Therefore, the stone reaches to a height of $1.25\ m$

Now,

We know, $v = u + at$

$\\ \implies 0 = 5 + (-10)t \\ \implies t = 0.5\ s$

Therefore, the time taken by the stone to reach the maximum height is $0.5\ s$.

## NCERT solutions for class 9 science chapter 8 Motion- Exercise solutions

Given, Diameter of the circular track = $200\ m$

$\therefore$ The circumference of the circular track, $d = 2\pi\frac{D}{2} = 200\pi\ m$

The athlete completes one round of a circular track in 40 s.

$\therefore$ Speed of the athlete = $u = \frac{200\pi\ m}{40\ s} = 5\pi\ ms^{-1}$

In $t = 2\ min\ 20\ s = 140\ s$,

Distance travelled by the athlete = $Speed\times time = (5\pi)\times(140)$

$= 5\times\frac{22}{7}\times140 = 2200\ m$

Also, number of rounds the athlete will complete in $140\ s$ = $\frac{140}{40} = 3.5$

Therefore, the final position of the athlete after $140\ s$ will be diametrically opposite to his initial point.

(3 complete rounds and one half round.)

Hence, displacement of the athlete = magnitude of diameter of the circle =200m

## Q2.  Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging

(a) from A to B and

(b) from A to C?

Given,

(a) Distance between A and B = $300\ m$

Time taken to reach from A to B = $2\ min\ 30\ s = 150\ s$

$\therefore$ Average speed from A to B = $\frac{Distance}{Time} = \frac{300}{150} = 2\ ms^{-1}$

And, Average velocity from A to B = $\frac{Displacement}{Time} = \frac{300}{150} = 2\ ms^{-1}$

(In this case, average speed is equal to the average velocity)

(b) Distance travelled from A to reach C = $AB + BC = 300 + 100 = 400\ m$

And, Displacement from A to C = $AC= 300-100 = 200\ m$

Also, time taken to reach C from A = $2\ min\ 30\ s + 1\ min= (150+60)\ s =210\ s$

$\therefore$ Average speed from A to C = $\frac{Distance}{Time} = \frac{400}{210} = 1.90\ ms^{-1}$

And, Average velocity from A to C = $\frac{Displacement}{Time} = \frac{200}{210} = 0.95\ ms^{-1}$

(In this case, average speed is not equal to the average velocity)

Given, Average speed while going to school, $v_1 = 20\ km h^{-1}$

And Average speed while returning back from school, $v_2 = 30\ km h^{-1}$

Let the distance between starting point and school be $x\ m$

And time taken by Abdul during the two trips be $t_1\ s\ and\ t_2\ s$

We know, $Speed = \frac{Distance}{Time}$

$\therefore v_1 = \frac{x}{t_1} = 20$

And, $\therefore v_2 = \frac{x}{t_2} = 30$             -(i)

Now, Total distance that Abdul covers = $x +x = 2x$

And total time Abdul takes to cover this distance =$t_1 + t_2$

$\\ \therefore v_{avg} = \frac{2x}{t_1+t_2} \\ = \frac{2x}{\frac{x}{20}+\frac{x}{30}} \\ \\ = \frac{60\times2x}{5x} \\ \\ = 24\ ms^{-1}$

Therefore, the average speed for Abdul's trip is $24\ ms^{-1}$

(Note: $\frac{20+30}{2} =25\ ms^{-1} \neq 24\ ms^{-1}$)

Given, The motorboat starts from rest. Hence, initial speed of the motorboat, u = $0\ ms^{-1}$

Acceleration of the motorboat, $a$ = $3.0\ m s^{-2}$

Time taken, $t = 8\ s$

We know, $s = ut+\frac{1}{2}at^2$

$\\ \implies s = 0(8) + \frac{1}{2}(3)(8)^2 \\ \implies s = \frac{1}{2}(3)(64) \\ \\ \implies s = 96\ m$

Therefore, the distance travelled by the motorboat is $96\ m$

The initial speed  $=52\times \frac{5}{18}=14.4 \frac{m}{s}$

After 5 sec the car stops

The graph is represented by the blue line ( x-axis is time and the y-axis is speed)

For the car with 3Kmh-1. Initial speed $=3\times\frac{5}{18}=0.833\frac{m}{s}$  .  The graph which is represented by the golden line

The area covered by the blue graph is greater than the golden graph so the car with 15 m/s initial velocity travells large distance.

Given is a distance-time graph. The slope of this graph gives us speed. Hence, the graph with the highest slope will have the highest speed.

Since B has the highest slope(inclination), it travels the fastest.

Given is a distance-time graph. Any point on the curve will give the distance of object from O. Since there is no intersection point of all the three graphs, they never meet at the same point on the road.

(Although any two of them do meet at some point on the road!)

Given is a distance-time graph. Any point on the curve will give the distance of object from O. To find how far C has travelled when B passes A, draw a perpendicular from the intersection point of A and B on time axis. The point where it intersects on the C graph, from C draw a perpandicular to y axis . Therefore, distance travelled by C will be (Final distance from O - Initial distance from O)

Therefore, C has traveled 6.5 km when B passes A.

Given is a distance-time graph. The graph of B and C intersect at a point whose y-coordinate is 5. Hence, B has travelled $5\ km$ by the time it passes C.

Considering downward direction as positive direction.

Given, Height from which ball is dropped, $s = 20\ m$

Acceleration of the ball, $a$ = $10\ ms^{-2}$

Initial velocity, $u = 0\ ms^{-1}$

(i) We know, $v^2 = u^2 + 2as$

$\\ \implies v^2 = 0^2 + 2(10)(20) \\ \implies v^2 = 400 \\ \implies v = 20\ ms^{-1}$(In downward direction)

Therefore, the ball will strike the ground with a velocity of $20\ ms^{-1}$

(ii) Now, we know, $v = u + at$

$\\ \implies 20 = 0 + 10t \\ \implies t = 2\ s$

Therefore, the ball reaches the ground in $2\ s$.

Note: $v = -20\ ms^{-1}$ was rejected because in this case, the negative sign implies the velocity in upward direction, which is opposite to the direction of the motion of the ball(before collision).

Given is a speed-time graph. The area under the curve will give the distance travelled by the car.

In time $t= 4\ s$, the distance travelled by the car will be equal to the area under the curve from $x = 0\ to\ x=4$

Considering this part of the graph as a quarter of a circle whose radius = 4 unit.

Therefore, required area = $\frac{1}{4}\pi r^2 = \frac{1}{4}\pi (4)^2 = 12.56\ m$

Therefore, distance the car travelled in the first 4 seconds is $12.56\ m$

Which part of the graph represents uniform motion of the car?

In uniform motion, the speed of car will become constant which is represented by line parallel to the time axis. In the given figure, the straight line graph from $t = 6\ s\ to\ t= 10\ s$ represents the uniform motion of the car.

(a) The given situation is possible.

When an object is thrown upwards (under gravity only), it reaches to a maximum height where its velocity becomes zero. However, it still has an acceleration acting in the downward direction (acceleration due to gravity).

Note: This is possible for a given point of time, however, it is not possible for a period of time.

(b) The given situation is possible.

An object moving in a circular path with uniform speed, i.e covering equal distance in equal amount of time is still under acceleration. Because, the velocity keeps on changing due to continuous change in the direction of motion. Therefore, circular motion is an example of an object moving with an acceleration but with uniform speed.

(c) The given situation is possible.

For an object moving in a circular path with constant speed, the direction of its velocity at any point will be tangential to that point. However, its acceleration will be directed radially inwards. (Constant speed but still having an acceleration - Due to continuous change in direction.)

Given, Radius of the circular orbit = $42250\ km$

$\therefore$ Circumference of the orbit = $2\pi r = 2.\frac{22}{7}. (42250) = 265571.4\ km$

The satellite takes 24 hours to revolve around the earth.

We know, $Speed = \frac{Distance}{Time}$

$\\ = \frac{265571.4}{24} \\ \\ = 11065.4\ kmh^{-1} \\ \\ = \frac{11065.4}{3600}\ kms^{-1} \\ \\ = 3.07\ kms^{-1}$

Therefore, the speed of the artificial satellite is $3.07\ kms^{-1}$

## NCERT Solutions for Class 9 Science- Chapter Wise

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

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