NCERT Exemplar Class 9 Science Solutions Chapter 11 Work and Energy
It is of utmost importance that we learn the role of work, energy, and power in our daily activities. Be it when you are riding a bike up the mountain, when you are picking up your school bag, or when you are running in a contest, regardless of the task, every activity requires energy transfer and utilisation. These are the practical concepts which comprise Class 9 Science Chapter 11 - Work, Energy, and Power, which is discussed in detail by the NCERT Exemplar Solutions.
This Story also Contains
- NCERT Exemplar Class 9 Science Solutions Chapter 11-MCQ
- Question:1
- NCERT Exemplar Class 9 Science Solutions Chapter 11-Short Answer
- NCERT Exemplar Class 9 Science Solutions Chapter 11-Long Answer
- NCERT Exemplar Solutions Class 9 Science Chapter 11: Important Concepts and Formulas
- Advantages of NCERT Exemplar Class 9 Science Chapter 11 Work, Energy and Power solutions
- NCERT Class 9 Science Exemplar Solutions for Other Chapters
- Features of NCERT Exemplar Class 9 Science Solutions Chapter 11
The NCERT Exemplar Class 9 Science Solutions Chapter 11 Work, Energy and Power have been designed with great care by subject experts to deliver clarity and accuracy. All the solutions are given in a step-wise format, enabling students to develop a solid conceptual understanding as they learn to approach questions in a logical manner. These NCERT Exemplar Class 9 Solutions Science Chapter 11 Work, Energy and Power are in alignment with the current CBSE Class 9 Science syllabus and thus, revision is more relaxed, and confidence is boosted. They assist students to achieve better marks by enhancing their problem-solving skills and raising the level of exam preparedness. There is also the fact that solving these exemplar questions on a regular basis also builds up analytical reasoning and command over key formulas and concepts. These NCERT Exemplar solutions can be quickly revised with a simple and precise explanation, which means that students can have the feeling that they are ready and can perform optimally during the exams.
NCERT Exemplar Class 9 Science Solutions Chapter 11-MCQ
The NCERT Exemplar Class 9 Science Solutions Chapter 11 Work, Energy and Power has several multiple-choice questions that are aimed at evaluating the main concepts in an objective format. These MCQs allow students to train on the most critical problem-solving skills and consolidate their knowledge on formulae and definitions. These questions should be practised regularly in order to increase accuracy, speed, and readiness to take exams.
Question:1
When a body falls freely towards the earth, then its total energy
(a) increases
(b) decreases
(c) remains constant
(d) first increases and then decreases
Solution.
The total energy of any body is the sum of kinetic energy and potential energy.
$TE =KE + PE$
When a body falls freely towards the earth, the kinetic energy of the body increases, and the potential energy of the body decreases.
However, by the law of conservation of energy total energy of the body remains constant.
The correct answer to this question is option C.
Question:2
A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car
(a) does not change
(b) becomes twice that of initial
(c) becomes 4 times that of initial
(d) becomes 16 times that of initial
Solution.
The kinetic energy of a body increases with increasing speed.
$KE=\frac{1}{2}mv^{2}$
The potential energy of the body increases with increasing height above the surface of Earth.
$PE=mgh$
In the given question, the car is increasing speed on a level road. Hence, its kinetic energy must be increasing.
The kinetic energy will be 16 times the initial kinetic energy as the velocity becomes four times.
The car's height from the ground is not changing as it is moving on a level road.
Therefore, we can say that the potential energy of the car is not changing.
Hence, the correct answer to this question is option A.
Question:3
In case of negative work the angle between the force and displacement is
(a) $0^{\circ}$ (b) $45^{\circ}$ (c) $90^{\circ}$ (d ) $180^{\circ}$
Solution.
Work done by any force is defined as:
$W=Fd\cos \theta$
This work done can be positive, negative or zero depending on the angle between force and displacement.
If the angle between them is acute, then the work done will be positive because $\cos \theta$ will be positive.
If the angle between them is $90^{\circ}$, then the work done will be zero.
If the angle between them is obtuse, then the work done will be negative.
$W=Fd(\cos 180^{\circ})=-Fd$
Hence, the correct answer is option D.
Question:4
An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same
(a) acceleration
(b) momenta
(c) potential energy
(d) Kinetic energy
Solution.
When a body falls under gravity, the body’s acceleration does not depend on the mass of the body.
Therefore, the acceleration for both spheres will be the same at that instant.
Their velocities will be the same at that instant, too.
$v = u + at$
Momentum and kinetic energy depend on speed as well as mass.
$KE=\frac{1}{2}mv^{2}$
$P=mv$
As the masses of both spheres are different, however, speed is the same; hence, their momentum and kinetic energy will be different.
Gravitational potential energy is defined as:
$PE=mgh$
Therefore, it would be different for the spheres of different masses at the same height.
Hence, the correct answer is option A.
Question:5
A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be $(g =10 m s^{-2})$
(a) $6 \times 10^3 J$
(b) $6 J$
(c) $0.6 J$
(d) zero
Solution.
We know that work is defined as:
$W=Fd\cos \theta$
As the girl is carrying a schoolbag of 3 kg, the girl must be applying force in the vertically upward direction against gravity.
The displacement of the girl is in the horizontal direction, as it is moving on a level road.
The angle between force and displacement is $90^{\circ}$.
Therefore, by the formula, the work done by the girl against gravity will be zero.
$W=Fd\cos 90^{\circ}=0$
Hence, the correct answer for this question is option D.
Question:6
Which one of the following is not the unit of energy?
(a) joule
(b) newton metre
(c) kilowatt
(d) kilowatt hour
Solution.
We know that work done and energy have the same unit.
The SI unit of energy is the Joule.
Work done is defined as the product of force and displacement; therefore, its unit will be newton-meters.
$W=Fd\cos \theta$
Power is defined as work done per unit time, and its unit will be watts or kilowatts.
If we multiply power by time, we will get energy or work done.
Therefore, kilowatt-hour will be the unit of energy.
[Only kilowatt is the unit of power, and that is not a unit of energy.]
The correct answer to this question is option c.
Question:7
The work done on an object does not depend upon the
(a) displacement
(b) force applied
(c) angle between force and displacement
(d) initial velocity of the object
Solution.
We know that the work done is defined as:
$W=Fd\cos \theta$
It means work done depends on force, displacement and the angle between force and displacement.
Work done does not depend on any other property other than the three mentioned above.
Therefore, we can say the work done does not depend on the initial velocity of the object.
Hence, the correct option for this question is option d.
Question:8
Water stored in a dam possesses
(a) no energy
(b) electrical energy
(c) kinetic energy
(d) potential energy
Answer: [d]
Solution.Water stored in a dam remains at rest.
The kinetic energy of the water stored will be zero as its speed is zero.
$KE=\frac{1}{2}mv^{2}$
Electrical energy is not possessed by stored water.
The water has mass, and it occupies the space above the surface of the ground.
Therefore, the water will have gravitational potential energy.
$PE=mgh$
Hence, the correct answer to this question is option d.
Question:9
A body is falling from a height h. After it has fallen a height h/2, it will possess
(a) only potential energy
(b) only kinetic energy
(c) half potential and half kinetic energy
(d) more kinetic and less potential energy
Solution.
When a body is dropped from a height, it gets acceleration due to gravity.
Acceleration increases its speed; hence, its kinetic energy keeps increasing.
$KE=\frac{1}{2}mv^{2}$
The ball is dropped from a height H, and we are analysing the ball at height H/2.
The total energy of the ball will not change in free fall under gravity.
As the height becomes half of the initial height, the potential energy will become half.
$PE=mgh$
This decrease in potential energy will come in the form of kinetic energy because total energy is conserved.
$TE = PE +KE$
Therefore, the kinetic energy of the ball will be half of the initial potential energy.
Hence, the correct answer to this question is option C.
NCERT Exemplar Class 9 Science Solutions Chapter 11-Short Answer
The NCERT Exemplar Class 9 Science Chapter 11 Short Answer questions are used to reinforce conceptual clarity of the students on topics such as work, energy, and power. All these questions stimulate analytical and logical thinking, which would allow the students to solve both numerical and theoretical problems successfully.
Question:10
Answer:
We know that kinetic energy is defined as:$KE=\frac{1}{2}mv^{2}$
We can see that the kinetic energy will be proportional to the square of speed for a given body of constant mass.
As the rocket's speed is suddenly tripled, the rocket's kinetic energy will be nine times.
$KE=\frac{1}{2}mv^{2}$
$\frac{KE_{i}}{KE_{f}}=\frac{V^{2}_{i}}{V^{2}_{f}}$
Question:11
Answer:
The power of anything is defined as the rate of work done by that thing.Mathematically, power is defined as the product of force and velocity.
Power =Fv
If we calculate the power of Avinash and Kapil, by the given information:
$Power_{Avinash}=10\times 8=80watt$
$Power_{Kapil}=25\times 3=75watt$
Therefore, it is quite visible that the power of Avinash is more than the power of Kapil.
Question:12
A boy is moving on a straight road against a frictional force of 5 N. After travelling a distance of 1.5 km he forgot the correct path at a round-about of radius 100 m. However, he moves on the circular path for one and half cycle and then he moves forward upto$2.0km.$ Calculate the work done by him.
Answer:
The boy has to apply force in the direction of his motion against the frictional force.Therefore, the angle between force and displacement will always be $0^{\circ}$.
For calculating the work done, we have to multiply the force by the distance travelled by the boy.
The magnitude of the force is always 5 newtons.
Total distance travelled will be equal to the sum of 3/2 times the perimeter of the circular path and the distance travelled along a straight track.
$d=1500m+(3/2)(2\pi \times 100)m+2000m$
$d=4442m$
Hence, work done:
$W=5\times 4442=22210J$
Question:13
Can any object have mechanical energy even if its momentum is zero? Explain.
Answer:
Mechanical energy is the sum of kinetic energy and potential energy. It is also known as total energy.Kinetic energy is the virtue of motion, which comes with speed.
$KE=\frac{1}{2}mv^{2}$
Potential energy is the virtue of position, which comes with a height above the ground surface.
$PE=mgh$
As the object has zero momentum, it means it is not moving.
Momentum is defined as the product of mass and speed.
P = mv
A body at rest and some height from the ground will have potential energy or mechanical energy.
Question:14
Can any object have mechanical energy even if its momentum is zero? Explain.
Answer:
Mechanical energy consists of both potential energy and kinetic energy. Momentum is said to be zero when the velocity is zero. Hence, there is no kinetic energy, but the object may have potential energy.Question:15
The power of a motor pump is 2 kW. How much water per minute the pump can raise to a height of 10 m? (Given $g = 10 m s^{-2}$)
Answer:
The power of the pump is 2 kW, which means it can give energy equal to 2000 joules in one second.Total time given is one minute; therefore, the motor pump's total energy will be equal to 1,20,000 joules.
$Energy=power\times time$
This energy will be equal to the potential energy of the water raised to 10 m.
Let the mass of the water raised be M kg.
Therefore, its potential energy at height $H =10 m$ will be:
$PE=mgh$
$=M\times 10\times 10$
$\Rightarrow 120000=100M$
$\Rightarrow M=1200kg$
It means 1200 kg of water can be raised to a height of 10 m in one minute by this motor pump.
Question:16
Answer:
We will assume that the energy level (power) of a person does not change on changing planets.The energy used by the person will increase their potential energy.
Therefore, potential energy at the end will be the same on planet A and Earth.
We know the formula of potential energy is given as:
$PE=mgh$
As the acceleration due to gravity is half that on planet A. Therefore, the height raised will be twice on planet A.
Hence, height raised on planet A will be $0.8m$.
Question:17
Answer:
By the equation of motion, we can say:$v^{2}=u^{2}+2as$
$\Rightarrow s=\frac{v^{2}-u^{2}}{2a}$
By Newton’s second law, we can say:
F= ma
We know that work done is defined as the product of force and displacement if the body is moving on a straight track.
$W=Fs$
$\Rightarrow W=ma\times \frac{v^{2}-u^{2}}{2a}$
$\Rightarrow W=m\times \frac{v^{2}-u^{2}}{2}$
$\Rightarrow W=\frac{mv^{2}}{2}-\frac{mu^{2}}{2}$
$\Rightarrow W=KE_{f}-KE_{i}$
Hence, work done is equal to the change in kinetic energy.
Question:18
Answer:
Yes, it is possible. A body moving in a uniform circular motion is an example of accelerated motion. Consider the motion of the Earth around the Sun. The Earth is constantly moving in a circular orbit in a direction perpendicular to the gravitational pull of the Sun. So, the work done by the gravitational force is zero. Thus, the work done can be zero for an accelerated body.Question:19
A ball is dropped from a height of 10 m. If the energy of the ball reduces by $40\%$ after striking the ground, how much high can the ball bounce back?$(g = 10 m s^{-2})$
Answer:
When the ball is dropped from 10 meters, the total energy will be in the form of potential energy.Potential energy can be calculated as:
$PE=mgh$
After striking the ground, $40\%$ of energy is lost. Hence, it will have only $60\%$ of energy.
With the help of this $60\%$ energy, the object will attain maximum height.
Therefore, if height attained is H, then we can say:
$60\%(mg(10))=mgh$
$\Rightarrow \frac{60}{100}(mg(10))=mgh$
$\Rightarrow h=6m$
The height attained is 6m.
Question:20
Answer:
The electrical energy consumed is calculated in units.We know that one unit is equal to the energy of 1 kilowatt-hour.
Power of heater = 1200 watt =1.2 kilowatt
Per day usage = 30 minutes = 0.5 hour
Energy consumed per day = (1.2 kilowatt) (0.5 hour) = 0.6 units.
In April, the total number of days = 30
Therefore, total consumption will be: Energy = (30) (0.6) = 18 units
It can be converted into joules:
$1unit=1000\times 3600J$
$\Rightarrow 18unit=(18\times 1000\times 3600)J=64800000J$
NCERT Exemplar Class 9 Science Solutions Chapter 11-Long Answer
The long-answer type questions of Chapter 11: Work, Energy, and Power assist students in grasping the concepts in a more elaborate and detailed way. The NCERT Exemplar Class 9 Science Solutions Chapter 11 are also presented in a step-by-step format, thus simplifying the process of solving questions based on applications and numerical problems. These questions are practised to enhance conceptual clarity and improve performance at the exams.
Question:21
Answer:
For any object of mass m, the kinetic energy and momentum are defined as:
$KE=\frac{1}{2}mv^{2}$
$P=mv$
Here v is the speed of the object.
We try to eliminate speed and find out the relation between kinetic energy and momentum.
$KE=\frac{1}{2}m\left ( \frac{p}{m} \right )^{2}$
$\Rightarrow KE=\frac{1}{2}\frac{p^{2}}{m}$
If two bodies have the same momentum, then kinetic energy will be inversely proportional to the mass of the body.
Therefore, the lighter body will have more kinetic energy than the heavier body.
Question:22
Answer:
Mass of the car = 1000 kg
Speed of car A = 36 km/h = 10 m/s
Frictional force =100 N
We know that power is defined as:
Power = Fv
Hence, the power of the engine will be:
Power = (100N) (10m/s) =1000 watt.
In the collision, no external force is involved. Therefore, the total momentum of both cars will be conserved.
$m_{A}u_{A}+m_{B}u_{B}=m_{A}v_{A}+m_{B}v_{B}$
Where $m_{A}=1000kg;m_{B}=1000kg;u_{A}=10m/s;u_{B}=0m/s;v_{A}=0m/s;$
By putting the values, we will get:
$v_{B}=10m/s;$
Question:23
A girl having mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of $4 ms^{-1}$ by applying a force.
The trolley comes to rest after traversing a distance of 16 m.
(a) How much work is done on the trolley?
(b) How much work is done by the girl?
Answer:
Mass of the girl = 35 kg
Mass of the trolley = 5 kg
Initial velocity of trolley = 4m/s
Total distance travelled by trolley = 16 m
We know that
$\\\;v^2-u^2 = 2as \\\; a=\frac{v^2-u^2}{2s} \\\; = \frac{-16}{2\times 16}= -0.5$
$Force=ma=40\times (-0.5) = -20N$
$\\\;work\; done\; on\; the\; trolley\;=20N\times 16m=320 J.\\\; work \;done \;by \;the\; girl =0 J.$
Question:24
Four men lift a 250 kg box to a height of 1 m and hold it without raising or lowering it. (a) How much work is done by the men in lifting the box? (b) How much work do they do in just holding it? (c) Why do they get tired while holding it? $(g = 10 ms^{-2})$
Answer:
Mass of the box = 250 kg
Height attained = 1 m
(a) Work done in lifting the block will be stored as potential energy
$W=PE=mgh$
After calculating, the work done will be 2500 joules.
(b) When they hold it in the air without any movement, the displacement of the box will be zero.
We know work is defined as:
$W=Fd\cos \theta$
Hence, work done will be zero.
(c) When they hold it in the air without any movement, their muscles are stretched.
These stretched muscles require energy and cause fatigue and tiredness.
Question:25
Answer:
The power of anything is defined as the rate of work done by that thing or energy supplied per unit time.
Mathematically, power is defined as the product of force and velocity.
Power = Fv
Power is defined as work done per unit time, and its unit will be watts or kilowatts.
If we multiply power by time, we will get energy or work done.
Therefore, kilowatt-hour will be the unit of energy.
[Only kilowatt is the unit of power, and that is not a unit of energy.]
In one minute, all the potential energy of water can be utilised.
Energy utilised = Potential energy = mgh
$PE=(2000\times 10^{3}kg)\times 10ms^{-2}\times 20m$
$=4\times 10 ^{8}Joule$
Now time t = 1 minute = 60 seconds
Hence power will be
$power=\frac{energy}{time}=\frac{4\times 10^{8}Joule}{60seconds}=6.67\times 10^{6}watt$
Question:26
How is the power related to the speed at which a body can be lifted? How many kilograms will a man working at the power of 100 W, be able to lift at constant speed of $1 m s^{-1}$ vertically? $(g = 10 m s^{-2})$
Answer:
The power of anything is defined as the rate of work done by that thing or energy supplied per unit time.
Mathematically, power is defined as the product of force and velocity.
Power = Fv
Power is defined as work done per unit time, and its unit will be watts or kilowatts.
If we multiply power by time, we will get energy or work done.
If a man is lifting a body with constant speed, he has to apply a force equal to his weight.
Force applied: $F = mg = 10m N$
Speed attained: $V = 1 m s^{-1}$
Power used: P = 100W
By using the formula of power:
$100 = (10m) (1)$
$m = 10kg.$
Question:27
Define watt. Express kilowatt in terms of joule per second. A 150 kg car engine develops 500 W for each kg. What force does it exert in moving the car at a speed of $20 m s^{-1}$?
Answer:
The power of anything is defined as the rate of work done by that thing or energy supplied per unit time.
Mathematically, power is defined as the product of force and velocity.
Power = Fv
Power is defined as work done per unit time, and its unit will be watts (in SI units) or kilowatts.
If we multiply power by time, we will get energy or work done.
$power=\frac{energy}{time}$
$unit of power=\frac{joule}{second}=watt$
$1 kilowatt = 1000 watt = 1000 joule/second.$
For the car:
Mass of the car: $m = 150 kg$
Speed attained: $V = 20 m s^{-1}$
Power used: $P = 500W per kg$ hence,
Total power will be:
$Power=500\times 150=75000watt$
By using the formula of power:
$75000 = (Force) (20)$
$Force = 3750 N.$
Question:28
Compare the power at which each of the following is moving upwards against the force of gravity? $(given: g = 10 m s^{-2})$
(i) a butterfly of mass $1.0 g$ that flies upward at a rate of $0.5 m s^{-1}$.
(ii) a 250 g squirrel climbing up a tree at a rate of $0.5 m s^{-1}$
Answer:
The power of anything is defined as the rate of work done by that thing or energy supplied per unit time.
Mathematically, power is defined as the product of force and velocity.
Power = Fv
In lifting a body with constant speed, it has to apply a force equal to its weight.
(i)For the Butterfly:
Mass of the butterfly: $m = 1 gram = 0.001 kg$
Speed attained: $V= 0.5 m s^{-1}$
Force required: $F = mg = 0.01 N$
By using the formula of power:
$Power=0.01\times 0.5=5\times 10^{-3}watt$
(ii)For the Squirrel:
Mass of the squirrel: m =250 gram = 0.250 kg
Speed attained: V = 0.5 m s–1
Force required: F = mg = 2.5 N
By using the formula of power:
$Power=2.5\times 0.5=1.25watt$
NCERT Exemplar Solutions Class 9 Science Chapter 11: Important Concepts and Formulas
Important Concepts and Formulas of Class 9 Science Chapter 11 - Work, Energy and Power guide students to learn how the processes of physics require energy and work to be transferred. Mastering the following important concepts will ensure that questions are easier to solve and will assist in exams. Such formulas are commonly applied in numericals, and a regular revision increases the accuracy and confidence in solving problems.
1. Work Done (W)
-
Work is done when a force causes displacement.
-
Formula:
$
W=F \times d
$(Where: $F=$ Force,$d=$ displacement)
- Unit: Joule (J)
2. Work Done at an Angle
W=Fdcosθ
(Where θ is the angle between force and displacement)
3. Positive, Negative and Zero Work
- Positive Work: Force and displacement in the same direction
- Negative Work: Force and displacement in opposite directions
- Zero Work: No displacement or force perpendicular to displacement
4. Energy
- The capacity to do work.
- Unit: Joule (J)
-
Kinetic Energy (KE): Energy of an object due to motion
$
K E=\frac{1}{2} m v^2
$
(Where $m=$ mass, $v=$ velocity)
-
Potential Energy (PE): Energy due to position or height.
PE=mgh
(Where g = acceleration due to gravity, h = height)
-
Mechanical Energy: Total of kinetic and potential energy.
ME = KE + PE
5. Law of Conservation of Energy
- Energy cannot be created or destroyed; it only changes from one form to another.
6. Power (P)
- Rate of doing work.
- P = W/t
- Unit of Power: Watt (W) = 1 Watt = 1 Joule/second
-
Commercial Unit of Energy
Kilowatt-hour (kWh) = 1kWh = 3.6 x 106 J
Advantages of NCERT Exemplar Class 9 Science Chapter 11 Work, Energy and Power solutions
The NCERT Exemplar Class 9 Science Solutions Chapter 11 are very useful in developing a sound platform in chapters such as work, energy, and power. These NCERT Exemplar Class 9 Science Chapter 11 Solutions offer exam-oriented explanations which are detailed enough and enhance the level of understanding and problem-solving. Another thing is that they teach students the proper method to use in numerical problems, which is highly essential during exams. The scoring of good marks becomes easier, and confidence is gained through practice.
- Enhances understanding of concepts in work, energy and power with clear answers.
- Enables students to know the variety of numerical and solutions in steps.
- Increases the speed and the accuracy of problem-solving with practice.
- Strengthens exam preparation by exposing students to commonly asked questions.
- It will help make a quick revision before exams and end-of-course exams.
- Develops analysis and reasoning abilities that will be essential in the higher classes.
- Answers are made based on the most current CBSE syllabus, and hence are accurate and relevant.
- Builds confidence in the students by having them answer high-level questions in the same style as the exams.
NCERT Class 9 Exemplar Solutions for Other Subjects
NCERT Class 9 Science Exemplar Solutions for Other Chapters
NCERT Class 9 Science solutions of other chapters provide well-organised solutions to high-level problems that enhance conceptualisation. These solutions enable the students to revise on major topics, practice exam-type questions, and gain confidence to score well in CBSE exams.
|
Chapter-wise solutions |
Features of NCERT Exemplar Class 9 Science Solutions Chapter 11
Understanding Work, Energy, and Power becomes much easier with the help of NCERT Exemplar Solutions. These solutions are created to strengthen conceptual clarity and help students solve different types of exam-based questions confidently. The solutions are structured in a step-by-step manner so that students can learn effectively and revise quickly.
-
Concept Clarity - As in work, energy, power, and mechanical energy, these concepts are deeply explained in very simple words.
-
Step-by-Step Solutions - Each practical and theoretical question is solved in a specific manner so that strong problem-solving abilities can be developed.
-
Based on CBSE Syllabus - Completely integrated with the most recent Class 9 CBSE Science syllabus.
-
Covers All Question Formats - Contains MCQs, short answer questions, long answer questions, and questions that require application of concepts.
-
Focus on Real-Life Applications - Concept of energy conversion, machines, and power is explained with the help of relevant examples.
-
Clear Mathematical Derivations - Important derivations such as Work-Energy Theorem and power and energy formulas are included.
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 | Motion |
| Chapter 8 | Force and Laws of Motion |
| Chapter 9 | Gravitation |
| Chapter 10 | Work and Energy |
| Chapter 11 | Sound |
| Chapter 12 | Improvement in Food Resources |
Also, Read NCERT Solution Subject Wise
Check NCERT Notes Subject Wise
Also Check NCERT Books and NCERT Syllabus here
Frequently Asked Questions (FAQs)
Q: What is the scientific definition of work?
A:
Work is said to be done when a force is applied on an object and the object moves in the direction of the force.
Q: How is average power different from instantaneous power?
A:
Average power is total work done divided by total time, while instantaneous power is power at a specific moment in time.
Q: Why is energy important in daily life?
A:
Energy is required for every activity—from walking and cooking to running machines and lighting homes.
Q: Can energy be destroyed?
A:
No, according to the law of conservation of energy, it can only be transformed from one form to another.
Q: Why do we use kilowatt-hours instead of joules in electricity bills?
A:
Because joules are very small units; kilowatt-hours are more practical for measuring large-scale energy consumption.
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