NCERT Most Important Questions For Class-9 Chapter-9 Force and Laws of Motion (Physics)

The start of class 9 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 9 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

1. Give one point of difference between balanced and unbalanced forces.
Answer
When forces acting on a body from all sides are equal, they cancel effect of each other and are known as balanced forces. On the other hand, when forces acting on a body are not equal/do not cancel each other are called unbalanced forces.

2.Mass of a body is doubled. How does its acceleration change under a given force?
Answer
Acceleration becomes half.

3.Mention any two kinds of changes that can be brought about in a body by force.
Answer
Change in speed/change of direction/change of shape.

4.State the SI unit of pressure. Mention the unit which we use to measure pressure exerted by a gas. What do you understand by normal atmospheric pressure?

Answer

Pascal Atmosphere (atm)
Atmospheric pressure at sea level = 1 atm

5.Define SI unit of force. A force of 2N acting on a body changes its velocity uniformly from 2 m/s to 5m/s in 10s. Calculate the mass of the body.

Answer

6.Derive Newton’s first law of motion from the mathematical expression of the second law of motion.

Answer

Newton’s first law states that a body stays at rest if it is at rest and moves with a constant velocity unit if a net force is applied on it. Newton’s second law states that the net force applied on the body is equal to the rate of change in its momentum.
F= ma
F = m(v-u)/t
Ft = mv-mu

8.

What type of force is acting in the cases given above ?

Answer

(i) Accelerating unbalanced force.
(ii) No force.
(iii) Retarding unbalanced force.

15.From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s⁻¹. Calculate the initial recoil velocity of the rifle.

Answer

Mass of the rifle, m₁= 4 kg

Bullet is fired with an initial velocity, v₂= 35m/s
Initially, the rifle is at rest.
Thus, its initial velocity, v= 0
Total initial momentum of the rifle and bullet system= (m₁+m₂)v= 0

= m₁v₁ + m₂v₂ = 0.05 × 35 = 4v₁ + 1.75

According to the law of conservation of momentum:
Total momentum after the firing = Total momentum before the firing  4v₁ + 1.75= 0
v₁ = -1.75/4 = -0.4375 m/s
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.

16.Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms⁻¹ and 1 ms⁻¹, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms⁻¹. Determine the velocity of the second object.

Answer

Mass of one of the objects, m₁ = 100 g = 0.1 kg
Mass of the other object, m₂ = 200 g = 0.2 kg
Velocity of m₁ before collision, v₁= 2 m/s

Velocity of m₁ after collision, v₃= 1.67 m/s

According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision

Hence, the velocity of the second object becomes 1.165 m/s after the collision.

17.When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer

When the carpet is beaten, it is suddenly set into motion. The dust particles tend to remain at rest due to inertia of rest, therefore the dust comes out of it.

18.A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

Answer

Mass of the motor car, m = 1200 kg
Initial velocity of the motor car, u = 90 km/h = 25 m/s
Final velocity of the motor car, v = 18 km/h = 5 m/s
Time taken, t = 4 s
According to the first equation of motion:
v = u + at
⇒ 5 = 25 + a (4)
⇒ a = − 5 m/s²
Negative sign indicates that its a retarding motion i.e. velocity is decreasing.
Change in momentum = mv − mu = m (v−u)
= 1200 (5 − 25) = −24000 kg m s⁻¹
∵ Force = Mass × Acceleration
= 1200 × −5 = −6000 N
Acceleration of the motor car = −5 m/s²
Change in momentum of the motor car = −24000 kg m s⁻¹
Hence, the force required to decrease the velocity is 6000 N.
(Negative sign indicates retardation, decrease in momentum and retarding force)

19.An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s⁻¹ to 8 m
s⁻¹ in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Answer

Initial velocity of the object, u = 5 m/s
Final velocity of the object, v = 8 m/s
Mass of the object, m = 100 kg
Time take by the object to accelerate, t = 6 s
Initial momentum = mu = 100 × 5 = 500 kg ms⁻¹
Final momentum = mv = 100 × 8 = 800 kg ms⁻¹
Force exerted on the object, F = (mv – mu)/ t
= m (v-u)/t
= 800 – 500
= 300/6
= 50 N

Initial momentum of the object is 500 kg ms⁻¹.
Final momentum of the object is 800 kg ms⁻¹.
Force exerted on the object is 50 N.

20. How much momentum will a dumbbell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s⁻².

Answer

Mass of the dumbbell, m = 10 kg
Distance covered by the dumbbell, s = 80 cm = 0.8 m
Acceleration in the downward direction, a = 10 m/s²
Initial velocity of the dumbbell, u = 0
Final velocity of the dumbbell (when it was about to hit the floor) = v
According to the third equation of motion:
v² = u² + 2as
⇒ v² = 0 + 2 (10) 0.8
⇒ v = 4 m/s
Hence, the momentum with which the dumbbell hits the floor is
= mv = 10 × 4 = 40 kgms⁻¹

NCERT Quick revision Notes of Chapter-9 Force & Laws of Motion

NCERT Solution of Chapter-9 Force & Laws of Motion

NCERT MCQs of Chapter-9 Force & Laws of Motion

NCERT Most Important Questions For Class-9 Chapter-8 Motion (Physics)

The start of class 9 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 9 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

1.Give an example of a body which may appear to be moving for one person and stationary for the other.

Answer

The passengers in a moving bus observe that the trees, buildings as well as the people on the roadside appear to be moving backwards. Similarly, a person standing on the roadside observes that the bus (along with its passengers) is moving in forward direction. But, at the same time, each passenger in a moving bus or train observes, his fellow passengers sitting and not moving. Thus, we can tell that motion is relative.

2.How can we describe the location of an object?

Answer

To describe the position of an object we need to specify a reference point called the origin. For example, suppose that a library in a city is 2 km north of the railway station. We have specified the position of the library with respect to the railway station i.e., in this case, the railway station acts as the reference point.

3.What do you mean by average speed? What are its units?

Answer

Average speed is defined as the average distance travelled per unit time and is obtained by dividing the total distance travelled by the total time taken. The unit of average speed is the same as that of the speed, that is, ms^-1.

4.What is the difference between uniform velocity and non-uniform velocity?

Answer

Uniform velocity: An object with uniform velocity covers equal distances in equal intervals of time in a specified direction, e.g., an object moving with speed of 40km h^-1 towards west has uniform velocity.
Non-uniform velocity: When an object covers unequal distances in equal intervals of time in a specified direction, or if the direction of motion changes, it is said to be moving with a non-uniform or variable velocity, e.g., revolving fan at a constant speed has variable velocity.

5.Differentiate between distance and displacement.

Answer

Distance	Displacement
It is the length of the actual path covered by an object, irrespective of its dirction of motion.	Displacement is the shortest distance between the initial and final positions of an object in a given direction.
Distance is a scalar quantity	Displacement is a vector quantity. Displacement may be positive negative or zero.
Distance between two given points may be same or different for different path chosen.	Displacement between two given points is always the same.
Distance covered can never be negative. It is always positive or zero.	Displacement between two given points is always the same.

6.With the help of a graph, derive the relation v = u + at.

Answer

Consider the velocity-time graph of an object that moves under uniform acceleration as shown in the figure (u≠0).


From this graph, we can see that initial velocity of the object (at point A) is u and then it increases to v (at point B) in time t. The velocity changes at uniform rate a. As shown in the figure, the lines BC and BE are drawn from point B on the time and the velocity axes respectively.

The initial velocity is represented by OA.
The final velocity is represented by BC.
The time interval t is represented by OC.

BD = BC – CD, represents the change in velocity in time interval t.

If we draw AD parallel to OC, we observe that BC = BD + DC = BD + OA

Substituting, BC with v and OA with u, we get

v = BD + u
or, BD = v – u — (i)

Thus, from the given velocity-time graph, the acceleration of the object is given by Change in velocity
a = (Change in velocity)/(Time Taken)= BD/AD= BD/OC
Substituting OC with t, we get
a = BD/t ⇒ BD = at — (ii2)

From equations (1) and (2), we have
v-u = at or v =u + at

7.Suppose a ball is thrown vertically upwards from a position P above the ground. It rises to the highest point Q and returns to the same point P. What is the net displacement and distance travelled by the ball?
Answer
Displacement is zero. Distance is twice the distance between position P and Q.

8.A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Answer


Given, Side of the square field= 10m
Therefore, perimeter = 10 m × 4 = 40 m
Farmer moves along the boundary in 40s.
Displacement after 2 m 20 s = 2 × 60 s + 20 s = 140 s =?
Since in 40 s farmer moves 40 m

Therefore, in 1s distance covered by farmer = 40 / 40 m = 1m

Therefore, in 140s distance covered by farmer = 1 × 140 m = 140 m. Now, number of rotation to cover 140 along the boundary= Total Distance / Perimeter
= 140 m /40 m  = 3.5 round
Thus, after 3.5 round farmer will at point C of the field.

Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from intial position.

9.Distinguish between speed and velocity.

Answer

Speed	Velocity
Speed is the distance travelled by an object in a given interval of time.	Velocity is the displacement of an object in a given interval of time.
Speed = distance / time	Velocity = displacement / time
Speed is scalar quantity i.e. it has only magnitude.	Velocity is vector quantity i.e. it has both magnitude as well as direction.

10.What does the odometer of an automobile measure?

Answer

The odometer of an automobile measures the distance covered by an automobile.

11. A bus decreases its speed from 80 km h⁻¹ to 60 km h⁻¹ in 5 s. Find the acceleration of the bus.

Answer

12.What is the nature of the distance – ‘time graphs for uniform and non-uniform motion of an object?

Answer

When the motion is uniform,the distance time graph is a straight line with a slope.

 

When the motion is non uniform, the distance time graph is not a straight line.It can be any curve.

13.What can you say about the motion of an object whose distance – time graph is a straight line parallel to the time axis?

 

Answer

 

If distance time graph is a straight line parallel to the time axis, the body is at rest.

16.An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Answer

Diameter of circular track (D) = 200 m
Radius of circular track (r) = 200 / 2=100 m
Time taken by the athlete for one round (t) = 40 s
Distance covered by athlete in one round (s) = 2π r
= 2 × ( 22 / 7 ) × 100
Speed of the athlete (v) = Distance / Time
= (2 × 2200) / (7 × 40)
= 4400 / 7 × 40
Therefore, Distance covered in 140 s = Speed (s) × Time(t)
= 4400 / (7 × 40) × (2 × 60 + 20)
= 4400 / (7 × 40) × 140
= 4400 × 140 /7 × 40
= 2200 m

Number of round in 40 s =1 round
Number of round in 140 s =140/40
=3 ¹/₂
After taking start from position X,the athlete will be at postion Y after 3 ¹/₂ rounds as shown in figure

 

Hence, Displacement of the athlete with respect to initial position at x= xy

= Diameter of circular track
= 200 m

17.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?

Answer

Total Distance covered from AB = 300 m
Total time taken = 2 × 60 + 30 s
=150 s

Therefore, Average Speed from AB = Total Distance / Total Time
=300 / 150 m s^-1
=2 m s^-1
Therefore, Velocity from AB =Displacement AB / Time = 300 / 150 m s^-1
=2 m s^-1
Total Distance covered from AC =AB + BC
=300 + 200 m

Total time taken from A to C = Time taken for AB + Time taken for BC
= (2 × 60+30)+60 s
= 210 s
Therefore, Average Speed from AC = Total Distance /Total Time
= 400 /210 m s^-1
= 1.904 m s^-1

Displacement (S) from A to C = AB – BC
= 300-100 m
= 200 m

 

Time (t) taken for displacement from AC = 210 s

Therefore, Velocity from AC = Displacement (s) / Time(t)
= 200 / 210 m s^-1
= 0.952 m s^-1

19.A driver of a car travelling at 52 km h⁻¹ applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h⁻¹ in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Answer

As given in the figure below PR and SQ are the Speed-time graph for given two cars with initial speeds 52 kmh⁻¹ and 3 kmh⁻¹ respectively.

Distance Travelled by first car before coming to rest =Area of △ OPR
= (1/2) × OR × OP
= (1/2) × 5s × 52 kmh⁻¹
= (1/2) × 5 × (52 × 1000) / 3600) m
= (1/2) × 5 × (130 / 9) m
= 325 / 9 m
= 36.11 m
Distance Travelled by second car before coming to rest =Area of △ OSQ
= (1/2) × OQ × OS
= (1/2) × 10 s × 3 kmh⁻¹
= (1/2) × 10 × (3 × 1000) / 3600) m
= (1/2) × 10 x (5/6) m
= 5 × (5/6) m
= 25/6 m
= 4.16 m

NCERT Quick Revision Notes OF Chapter-8 Motion

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NCERT Most Important Question For Class-9 Chapter-5 The Fundamental Unit of Life (Biology)

The start of class 9 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 9 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

1.What is plasmolysis? What happens to a plasmolysed cell when it is placed in water?
Answer

Shrinkage of protoplast from the cell wall in presence of hypertonic solution due to exosmosis is know as plasmolysis. When a plasmolysed cell is placed in water, the concentration of water in the outside medium is more than the concentration in the cell. Hence, water moves inside the cell leading to its swelling.

2.What is plasma membrane ? What are its functions?

Answers

Plasma membrane also called as cell membrane, is the outer covering of a cell that separates its contents from the surrounding medium. It is made up of lipids and proteins, and provides a mechanical barrier to protect the inner contents of the cell. It encloses the nucleus and cytoplasm of the cell.

3.What do you mean by a nucleoid?

Answer

In prokaryotes and lower organisms like bacteria, the nuclear region of the cell may be poorly defined because of the absence of a nuclear membrane. Such an undefined and incipient nucleic region containing only naked nucleic acids without any membrane covering them is called a nucleoid.

4.(a) Why is the cell called the structural and functional unit of life?
(b) Why is the plasma membrane called a selectively permeable membrane ?
(c) Name the factor which decides the movement of water across the plasma membrane.

Answer

(a) Because shape and size of cells are related to specific functions they perform. Cells constitute various components of plants and animals.

(b) Because it permits exit and entry of some selected materials in and out of the cells.

(c) Amount of substance dissolved in water or solute concentration.

5. (i) Where are chromosomes present in the cell? What is their chemical composition?
(ii) How many pairs of chromosomes are present in humans?

Answer

(i) Chromosomes are present in the nucleus of a cell. Their chemical composition is of DNA, RNA and proteins.

Prokaryotic cell	Eukaryotic cell
Most prokaryotes are unicellular.	Most eukaryotes are multicellular.
Size of the cell is generally small (0.5- 5 µm).	Size of the cell is generally large (50- 100 µm).
Nuclear region is poorly defined due to the absence of a nuclear membrane or the cell lacks true nucleus.	Nuclear region is well-defined and is surrounded by a nuclear membrane, or true nucleus bound by a nuclear membrane is present in the cell.
It contains a single chromosome.	It contains more than one chromosome.
Nucleolus is absent.	Nucleolus is present.
Membrane-bound cell organelles such as plastids, mitochondria, endoplasmic reticulum, Golgi apparatus, etc. are absent.	Cell organelles such as mitochondria, plastids, endoplasmic reticulum, Golgi apparatus, lysosomes, etc. are present.
Cell division occurs through binary fission	Cell division occurs by mitosis.
Prokaryotic cells are found in bacteria and blue-green algae.	Eukaryotic cells are found in fungi, plants, and animal cells.

 

 

 

 

 

 

 

11. Why are lysosomes known as suicide bags?

Answer

Lysosomes are called suicide bags because in case of disturbance of their cellular metabolism they digest their own cell by releasing own enzymes.

12.Where are proteins synthesized inside the cell?

Answer

The proteins are synthesized in the Ribosome inside the cell.

13.Make a comparison and write down ways in which plant cells are different from animal cells.

Animal cell	Plant cell
They do not have a cell wall.	They have a cell wall made up of cellulose.
They do not have chloroplast.	They contain chloroplast.
They have centrosome.	They do not have centrosome.
Vacuoles are smaller in size.	Vacuoles are larger in size.
Lysosomes are larger in number.	Lysosomes are absent or very few in number
Prominent Golgi bodies are present.	Subunits of Golgi bodies known as dictyosomes are present.

14.Which organelle is known as the powerhouse of the cell? Why?

Answer

Mitochondria are known as the powerhouse of cells because energy required for various chemical activities needed for life is released by mitochondria in the form of ATP (Adenosine triphosphate) molecules.

15.Where do the lipids and proteins constituting the cell membrane get synthesized?

Answer

Lipids are synthesized in Smooth endoplasmic reticulum (SER) and the proteins are synthesized in rough endoplasmic reticulum (RER).

16.How does an Amoeba obtain its food?

Answer

Amoeba takes in food using temporary finger-like extensions of the cell surface which fuse over the food particle forming a food-vacuole as shown in figure. Inside the food vacuole, complex substances are broken down into simpler ones which then diffuse into the cytoplasm. The remaining undigested material is moved to the surface of the cell and thrown out.

Animal Cell

Plant Cell

NCERT Quick Revision Notes of Chapter-5 The Fundamental Unit of Life

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NCERT Most Important question & Solutions for class-9 Chapter-2 Is Matter Around us Pure (Chemistry)

The start of class 9 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 9 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

1.Describe any three properties of colloid.

Answer

(i) It is a heterogenous mixture.
(ii) Size of particles is too small to be seen by naked eye.
(iii) They scatter light passing through them making its path visible.
(iv) They do not settle down when left undisturbed.
(v) They cannot be separated by the process of filtration.

2.State the principle of separating two immiscible liquids by separating funnel. Describe an activity with diagram to separate a mixture of water and kerosene oil.

Answer

Immiscible layers separate out in layers depending on their densities in the separating funnel.


Activity to separate kerosene oil from water using a separating funnel:

• Pour the mixture of kerosene oil and water in the separating funnel as shown in the figure.

3.Differentiate between a true solution and a colloid.

Answer

True solution	Colloid
A true solution is a homogeneous mixture of two or more substances.	A colloidal solution is a heterogeneous mixture of two substances.
The size of the particles is less than one nanometer.	The range of particle size is between one nanometer to 1000 nanometer.
It is always transparent.	It is translucent.
The particles cannot be seen even with microscope.	The particles of a colloidal solution can be seen with microscope.
It does not show Tyndall effect.	It shows Tyndall effect.

4.Distinguish between physical change and chemical change.

Answer

Physical change	Chemical change
In a physical change, only physical properties such as colour, physical state, density, volume etc. change, chemical properties remain unchanged.	In a chemical change, the chemical composition and chemical properties undergo a change.
No new substance is formed in a physical change.	A new substance is formed in a chemical change.
Very little or no energy in the form of heat, light or sound is usually absorbed or given out in physical change.	A chemical change is always accompanied by a absorption or evolution of energy.
A physical change is a temporary change.	A chemical change is a permanent change.
The original form of substance can be regained by simple physical methods.	Original substance cannot be obtained by simple physical methods.
A physical change is reversible.	A chemical change is irreversible.

5. ‘Sea water can be classified as homogeneous as well as heterogeneous mixture.’ Comment.

Answer

Sea water is a mixture of salts and water which cannot be separated except by evaporation. Therefore, sea water is considered homogeneous. Sea water also contains mud, decayed plant, etc., other than salts and water, so it is heterogeneous also.

6.Based on which factor a solution is said to be diluted, concentrated or saturated?
Answer
A solution is said to be diluted, concentrated or saturated on the basis of the amount of solute dissolved in the solution.

7.List the points of differences between homogeneous and heterogeneous mixtures.

Answer

Homogeneous mixtures	Heterogeneous mixtures
Homogeneous mixtures have uniform composition.	Heterogeneous mixtures have non uniform composition.
It has no visible boundaries of separation between its constituents.	It has visible boundaries of separation between its constituents.

8.How are sol, solution and suspension different from each other?

Answer

Sol	Solution	Suspension
They are heterogeneous in nature.	They are homogeneous in nature.	They are heterogeneous in nature.
They scatter a beam of light and hence show Tyndall effect.	They do not scatter a beam of light and hence do not show Tyndall effect	They scatter a beam of light and hence show Tyndall effect.
They are quite stable.	Examples of solution are: salt in water, sugar in water.	Examples of suspension are: sand in water, dusty air

9.To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.

Answer

Mass of solute (sodium chloride) = 36 g (Given)
Mass of solvent (water) = 100 g (Given)
Then, mass of solution = Mass of solute + Mass of solvent
= (36 + 100) g
= 136 g

 

10.How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than 25°C), which are miscible with each other?

Answer

Kerosene and petrol are miscible liquids also the difference between their boiling point is more than 25 ºC so they can be separated by the method of distillation.

 

In this method, the mixture of kerosene and petrol is taken in a distillation flask with a thermometer fitted in it. We also need a beaker, a water condenser, and a Bunsen burner. The apparatus is arranged as shown in the above figure. Then, the mixture is heated slowly. The thermometer should be watched simultaneously. Kerosene will vaporize and condense in the water condenser. The condensed kerosene is collected from the condenser outlet, whereas petrol is left behind in the distillation flask.

NCERT Quick Revision Notes For Ch-2 Is Matter Around As Pure

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NCERT Most Important question & Solutions for class-9 Chapter-1 Matter in Our Surrounding (Chemistry)

The start of class 9 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 9 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

1. Arrange the three states of matter in the increasing order of: (i) rate of diffusion(ii) particle motion.
Answer
(i) Rate of diffusion: solid < liquid < gas.

(ii) Particle motion: solid < liquid < gas

2.Explain why heat energy is needed to melt a solid. Define latent heat of fusion.
Answer
Heat energy is needed to melt a solid because there exist forces of attraction between the particles of solid. The heat energy helps in overcoming the forces of attraction between the particles and thus helps in melting a solid.Latent heat of fusion is the amount of heat energy required to change 1 kg of solid into liquid at atmospheric pressure at its melting point.

3.Why does the temperature remain constant during the change of state of matter? Explain it on the basis of change of solid state into liquid state.

Answer

The temperature remains constant as the heat gets used up in changing the state by overcoming the forces of attraction between the particles.For example, a solid melts on heating. Its temperature does not rise until the entire solid is converted into liquid. This heat energy gets hidden into the contents and is known as the latent heat.

4.List any four factors on which evaporation depends. Explain in short any three factors.

11. Convert the following temperature to Celsius scale:

(a) 300 K
► 300 K = (300 – 273)°C
= 27°C
(b) 573 K► 573 K = (573 – 273)°C
= 300°C

12.Why does our palm feel cold when we put some acetone or petrol or perfume on it?

ANSWERS

Acetone, petrol, and perfume evaporate at low temperatures. When some acetone, petrol, or perfume is dropped on the palm, it takes heat from the palm and evaporates, thereby making the palm cooler.

13.Give reason for the following observations.

(a) Naphthalene balls disappear with time without leaving any solid.
(b) We can get the smell of perfume sitting several metres away.

Answer

(a) Naphthalene balls disappear with time without leaving any solid because of they undergoes sublimation easily i.e., the change of state of naphthalene from solid to gas takes place easily.

(b) Perfumes have high degree of vaporisation and its vapour diffuses into the air easily. Therefore, we can get the smell of perfume sitting several metres away.

14.Name A, B, C, D, E and F in the following diagram showing change in its state.

Answer

NCERT Quick Revision Notes For Chapter-1 Matter in Our Surrounding

NCERT Solution For Chapter-1 Matter in Our Surrounding

NCERT MCQs For Chapter-1 Matter in Our Surrounding

NCERT MCQs For Class-9 Chapter-15 Probability

1) The probability of each event, when a coin is tossed for 1000 times with frequencies: Head:455 & Tail: 545 is:

a. 0.455 & 0.545

b. 0.5 & 0.5

c. 0.45 & 0.55

d. 455 & 545

Answer: a

Explanation: Let E and F are the event of the occurrence of Head and Tail, respectively.

Probability of Occurrence of Head P(E) = No. of heads/total number of trials

P(E) = 455/1000 = 0.455

Similarly,

P(F) = No. of tails/total number of trials

P(F) = 545/1000 = 0.545

2) The sum of all probabilities equal to:

a. 4

b. 1

c. 3

d. 2

Answer: b

3) The probability of each event lies between:

a. 1 & 2

b. 1 & 10

c. 0 & 1

d. 0 & 5

Answer: c

4) If P(E) = 0.44, then P(not E) will be:

a. 0.44

b. 0.55

c. 0.50

d. 0.56

Answer: d

Explanation: We know;

P(E) + P(not E) = 1

0.44 + P(not E) = 1

P(not E) = 1 – 0.44 = 0.56

5) If P(E) = 0.38, then probability of event E, not occurring is:

a. 0.62

b. 0.38

c. 0.48

d. 1

Answer: a

Explanation: P(not E) = 1 – P(E) = 1-0.38 = 0.62

6) The probability of drawing an ace card from a deck of cards is:

a. 1/52

b. 1/26

c. 4/13

d. 1/13

Answer: d

Explanation: There are 4 aces in a deck of card.

Hence, the probability of taking one ace out of 52 card = 4/52 = 1/13

7) If the probability of an event to happen is 0.3 and the probability of the event not happening is:

a. 0.7

b. 0.6

c. 0.5

d. None of the above

Answer: a

Explanation: Probability of an event not happening = 1 – P(E)

P(not E) = 1 – 0.3 = 0.7

8) A dice is thrown. The probability of getting 1 and 5 is:

a. ⅙

b. ⅔

c. ⅓

d. ½

Answer: c

Explanation: The probability of getting 1 and 5 = 2/6 = ⅓

9) A batsman hits boundaries for 6 times out of 30 balls. Find the probability that he did not hit the boundaries.

a. ⅕

b. ⅖

c. ⅗

d. ⅘

Answer: d

Explanation: No. of boundaries = 6

No. of balls = 30

No. of balls without boundaries = 30 – 6 =24

Probability of no boundary = 24/30 = ⅘

10. Three coins were tossed 200 times. The number of times 2 heads came up is 72. Then the probability of 2 heads coming up is:

a. 1/25

b. 2/25

c. 7/25

d. 9/25

Answer: d

Explanation: Probability = 72/200 = 9/25

11.Which of the following cannot be the empirical probability of an event?
(a) 2/3
(b) 3/2
(c) 0
(d) 1

Answers: (b)

12.The sum of the probabilities of all events of a trial is
(a) 1
(b) greater than 1
(c) less than 1
(d) between 0 and 1

Answers: (A)

13.Two coins are tossed 1000 times and the outcomes are recorded as below:

The probability of getting at the most one head is:
(a) 1/5
(b) 1/4
(c) 4/5
(d) 3/4

Answers: (b)

14.A card is selected at random from a deck of 52 cards. The probability of its being a red face card is
(a) 3/13
(b) 1/2
(c) 2/13
(d) 3/26

Answers: (D)

15.In a series of 6 cricket matches, the number of runs scored by the captain of a team are 54, 32, 48, 55, 29, 35. So in the next match, the probability that he will cross the half century is
(a) 0.33
(b) 0.24
(c) 0.35
(d) 0.48

Answers: (a)

NCERT Quick revision Notes For Chapter-15 Probability

NCERT Solutions For Chapter-15 Probability

NCERT Most Important Questions For Chapter-15 Probability

NCERT MCQs For Class-9 Chapter-14 Statistics

1) The ratio of the sum of observations and the total number of observations is called:

a. Mean

b. Median

c. Mode

d. Central tendency

Answer: a

2) The mean of x+2, x+3, x+4 and x-2 is:

a. (x+7)/4

b. (2x+7)/4

c. (3x+7)/4

d. (4x+7)/4

Answer: d

Explanation: Mean = (x+2+x+3+x+4+x-2)/4 = (4x+7)/4

3) The median of the data: 4, 6, 8, 9, 11 is

a. 6

b. 8

c. 9

d. 11

Answer: b

4) The median of the data: 155 160 145 149 150 147 152 144 148 is

a. 149

b. 150

c. 147

d. 144

Answer: a

Explanation: First arrange the data in ascending order.

144 145 147 148 149 150 152 155 160

Since, the number of observations here is odd, therefore,

Median = (n+1)/2 th = (9+1)/2 = 10/2 = 5th number = 149

5) The median of the data: 17, 2, 7, 27, 15, 5, 14, 8, 10, 24, 48, 10, 8, 7, 18, 28 is:

a. 10

b. 24

c. 12

d. 8

Answer: c

Explanation: Arrange the given data in ascending order:

2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48

Since, the number of observations givere here is even, hence,

Median will be average of two middle terms.

n/2th = 16/2 = 8th term

(n/2 +1)th = (16/2 + 1)th = 9th term

Therefore,

Median = (10+14)/2 = 12

6) The mode of the given data: 4, 6, 5, 9, 3, 2, 7, 7, 6, 5, 4, 9, 10, 10, 3, 4, 7, 6, 9, 9 is;

a. 7

b. 9

c. 10

d. 6

Answer: b

Explanation: First arrange the data in order:

2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 9, 9, 9, 9, 10, 10

Hence, mode = 9

7) The value which appears very frequently in a data is called:

a. Mean

b. Median

c. Mode

d. Central tendency

Answer: c

8) The collection of information, collected for a purpose is called:

a. Mean

b. Median

c. Mode

d. Data

Answer: d

9) The mean of the data 2, 3, 4, 5, 0, 1, 3, 3, 4, 3 is

a. 2

b. 2.2

c. 2.4

d. 2.8

Answer: d

Explanation: Mean = (2+3+4+5+0+1+3+3+4+3)/10 = 28/10 = 2.8

10) Which of the following is not a measure of central tendency?

a. Standard deviation

b. Mean

c. Median

d. Mode

Answer: a

11)The class mark of the class 90-130 is:
(a) 90
(b) 105
(c) 115
(d) 110

Answers: (d)

12.The range of the data:
25, 81, 20, 22, 16, 6, 17,15,12, 30, 32, 10, 91, 8, 11, 20 is
(a) 10
(b) 75
(c) 85
(d) 26

Answers: (c)

13.In the class intervals 10-20, 20-30, the number 20 is included in:
(a) 10-20
(b) 20-30
(c) both the intervals
(d) none of these intervals

Answers: (b)

14.A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class interval is constructed for the following data:
268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304,402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236.
The frequency of the class 370-390 is:
(a) 0
(b) 1
(c) 3
(d) 5

Answers: (a)

15.If x1¯, x2¯, x3¯, …….., xn¯ are the means of n groups with n₁, n₂, ……. n_n number of observations respectively, then the mean x of all the groups taken together is given by:

Answer: (c)

NCERT Quick revision Notes Chapter-14 statistics

NCERT Solutions Chapter-14 statistics

NCERT Most Important Questions Chapter-14 statistics

NCERT MCQs For Class-9 Chapter-3 Coordinate Geometry

1. The points (–4,–8) lies in:

a) First quadrant

b) Second quadrant

c) Third quadrant

d) Fourth quadrant

Answer: (d)

2. The point (0, –5) lies:

a) On the x-axis

b) On the y-axis

c) In the first quadrant

d) None of the above

Answer: (b)

3. Ordinate of all the points in the x-axis is:

a) 0

b) 1

c) –1

d) Any natural number

Answer: (a)

4. Points (1, -2), (1, -3), (-4, 5), (0, 0), (3, -3)

a) Lie in III quadrant

b) Lie in II quadrant

c) Lie in IV quadrant

d) Do not lie in the same quadrant

Answer: (d)

5. If the x-coordinate of a point is zero, then this point lies:

a) In II quadrant

b) In I quadrant

c) On x-axis

d) On y-axis

Answer: (d)

6. If the perpendicular distance of a point P from the x-axis is 7 units and the foot of the perpendicular lies on the negative direction of x-axis, then the point P has:

a) y-coordinate = 7 or –7 only

b) y-coordinate = 7 only

c) y-coordinate = –7 only

d) x-coordinate = –7

Answer: (a)

7. On plotting P (–3, 8), Q (7, –5), R (–3, –8) and T (–7, 9) are plotted on the graph paper, then point(s) in the third quadrant are:

a) P and T

b) Q and R

c) Only R

d) P and R

Answer: (c)

8. If the coordinates of the two points are P (–7, 5) and Q (–6, 9), then (abscissa of P) – (abscissa of Q) is

a) –3

b) 1

c) –2

d) –1

Answer: (d)

9. Abscissa of a point is positive in:

a) I and II quadrants

b) I and IV quadrants

c) I quadrant only

d) II quadrant only

Answer: (b)

10. The point whose ordinate is 8 and lies on y-axis:

a) (0, 8)

b) (8, 0)

c) (5, 8)

d) (8, 5)

Answer: (a)

11. The coordinates of any point on the y-axis are of the form (0, k), where |k| is the distance of the point from the:

a) y-axis

b) x-axis

c) (0, 1)

d) (1, 0)

Answer: (b)

12. The mirror of a point (3, 4) on y-axis is:

a) (3, 4)

b) (–3, 4)

c) (3, –4)

d) (–3, –4)

Answer: (b)

13. The distance of the points (5, 0) and (–3, 0) from x-axis is:

a) –3

b) 5

c) 0

d) 2

Answer: (c)

14. The perpendicular distance of a point P (5, 8) from the y-axis is:

a) 5

b) 8

c) 3

d) 13

Answer: (a)

15. A point (x + 2, x + 4) lies in the first quadrant, the mirror image for which for x-axis is (5, –7). What is the value of x?

a) (–5, –7)

b) (–5, 7)

c) (5, –7)

d) (5, 7)

Answer: (d)

NCERT Quick revision Notes Of Chapter-3 Coordinate Geometry

NCERT Solutions Of Chapter-3 Coordinate Geometry

NCERT Most Important Questions Of Chapter-3 Coordinate Geometry

NCERT MCQs For Class-9 Chapter-4 Linear Equation in Two Variables

1. The linear equation 4x – 10y = 14 has:

a) A unique solution

b) Two solutions

c) Infinitely many solutions

d) No solutions

Answer: (c)

2. Find the number of solutions of the following pair of linear equations. x + 2y – 8 = 0 and 2x + 4y = 16:

a) 0

b) 1

c) 2

d) Infinite

Answer: (d)

3. If (2, 0) is a solution of the linear equation 2x +3y = k, then the value of k is:

a) 4

b) 6

c) 5

d) 2

Answer: (a)

4. The graph of the linear equation 2x +3y = 6 cuts the y-axis at the point:

a) (2, 0)

b) (0, 3)

c) (3, 0)

d) (0, 2)

Answer: (d)

5. The equation y = 5, in two variables, can be written as:

a) 1 .x + 1 .y = 5

b) 0 .x + 0 .y = 5

c) 1 .x + 0 .y = 5

d)0 .x + 1 .y = 5

Answer: (d)

6. Any point on the line y = x is of the form:

a) (a, –a)

b) (0, a)

c) (a, 0)

d) (a, a)

Answer: (d)

7. The graph of x = 5 is a line:

a) Parallel to x-axis at a distance 5 units from the origin

b) Parallel to y-axis at a distance 5 units from the origin

c) Making an intercept 5 on the x-axis

d) Making an intercept 5 on the y-axis

Answer: (b)

8. x = 9, y = 4 is a solution of the linear equation:

a) 2x + y = 17

b) x + y = 17

c) x + 2y = 17

d) 3x – 2y = 17

Answer: (c)

9. Any point on the x-axis is of the form:

a) (0, y)

b) (x, 0)

c) (x, x)

d) (x, y)

Answer: (b)

10. If a linear equation has solutions (–3, 3), (0, 0) and (3, –3), then it is of the form:

a)y – x = 0

b)x + y = 0

c) –2x + y = 0

d) –x + 2y = 0

Answer: (b)

11. The positive solutions of the equation ax + by + c = 0 always lie in the:

a) 1st quadrant

b) 2nd quadrant

c) 3rd quadrant

d) 4th quadrant

Answer: (a)

12. The graph of the linear equation 5x + 3y = 10 is a line which meets the x-axis at the point:

a) (0, 3)

b) (3, 0)

c) (2, 0)

d) (0, 2)

Answer: (c)

13. The point of the form (a, –a) always lies on the line:

a) x = a

b) y = –a

c) y = x

d) x + y = 0

Answer: (d)

14. The graph of x = 9 is a straight line:

a) Intersecting both the axes

b) parallel to y-axis

c) parallel to x-axis

d) Passing through the origin

Answer: (b)

15. Equation of the line parallel to x-axis and 6 units above the origin is:

a) x = 6

b) x = –6

c)y = 6

d)y = –6

Answer: (c)

16.The value of y at x = -1 in the equation 5y = 2 is
(a) 5/2
(b) 2/5
(c) 10
(d) 0

Answers: (2/5)

17. To which equation does the graph represent?

(a) 3x – 7y = 10
(b) y – 2x = 3
(c) 8y – 6x = 4
(d) 5x +35/2 y = 25

Answers: (C)

18. Cost of book (x) exceeds twice the cost of pen (y) by Rs 10. This statement can be expressed as linear equation.
(a) x – 2y – 10 = 0
(b) 2x – y – 10 = 0
(c) 2x + y – 10 = 0
(d) x – 2y + 10 = 0

Answers: (a)

19.Equation of a line passing through origin is
(a) x + y = 1
(b) x = 2y – 4
(c) x + y = 0
(d) y = x – 1

Answers: (c)

20.The condition that the equation ax + by + c = 0 represents a linear equation in two variables is
(a) a ≠ 0, b = 0
(b) b ≠ 0, a = 0
(c) a = 0, b = 0
(d) a ≠ 0, b ≠ 0

Answers: (d)

NCERT Quick revision Notes Of Chapter-4 Linear Equation In two Variable

NCERT Solutions Of Chapter-4 Linear Equation In two Variable

NCERT Most Important Questions Of Chapter-4 Linear Equation In two Variable

NCERT MCQs For Class-9 Chapter-2 Polynomial

1.  Which one is not a polynomial     

 (a)   4x² + 2x – 1 (b)    (c)   x³ – 1 (d)   y² + 5y + 1

Answers:-(b)

2.  The polynomial px² + qx + rx⁴ + 5 is of tyep     

 (a)   linear (b)   quadratic (c)   cubic (d)   Biquadratic

Answers:-(d)

3.  Identify the polynomial      

(a)   x^–2 + x^–1 + 5 (b)    c)    (d)   3x² + 7

Answers:-(d)

4.  The zero of the polynomial p(x) = 2x + 5 is  

    (a)   2 (b)   5 (c)    (d)   

Answers:-(d)

5.  The number of zeros of x² + 4x + 2     

 (a)   1 (b)   2 (c)   3 (d)   none of these

Answers:-(b)

6.  The polynomial of type ax² + bx + c, a = 0 is of type     

 (a)   linear (b)   quadratic (c)   cubic (d)   Biquadratic

Answers:- (a)

7.  The value of k, if (x – 1) is a factor of 4x³ + 3x² – 4x + k, is 

    (a)   1 (b)   2 (c)   –3 (d)   3

Answers:-(c)

8.  The degree of polynomial  is  

    (a)   0 (b)   2 (c)   1 (d)   3

Answers:-(c)

9.  If 3 + 5 – 8 = 0, then the value of (3)³ + (5)³ – (8)³ is    

  (a)   260 (b)   –360 (c)   –160 (d)   160

Answers:-(b)

10.  If value of 104 × 96 is    

  (a)   9984 (b)   9469 (c)   10234 (d)   11324

Answers:-(a)

11.  The value of 5.63 × 5.63 + 11.26 × 2.37 + 2.37 × 2.37 is    

  (a)   237 (b)   126 (c)   56 (d)   64

Answers:-(d)

12.  The value of      

 (a)   300 (b)   500 (c)   400 (d)   600

Answers:-(b)

13.  If x + y = 3, x² + y² = 5 then xy is    

  (a)   1 (b)   3 (c)   2 (d)   5

Answers:-(c)

14.  If x + 2 is a factor of x³ – 2ax² + 16, then value of a is  

    (a)   3 (b)   1 (c)   4 (d)   2

Answers:-(b)

15.  If one of the factor of x² + x – 20 is (x + 5). Find the other   

   (a)   x – 4 (b)   x + 2 (c)   x + 4 (d)   x – 5

Answers:-(A)

NCERT Quick revision Notes Of Chapter-2 Polynomials

NCERT Solutions Of Chapter-2 Polynomials

NCERT Most Important Questions Of Chapter-2 Polynomials