Class 9 Science Sample Papers 2022-23

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Although CBSE does not issue model papers for class 9 yet these are always similar to class 10. So we have prepared Class 9 Science Sample Papers 2022-23 as per CBSE guidelines. You can download them as PDFs from the myCBSEguide app or our student dashboard.

Class 9 Science Sample Paper 2022-23

Sample Paper of class 9 Science – in PDF

CBSE has already released a new marking scheme and blueprint for Class 9. Based on that, we are providing Class 9 Science Sample Question Papers for the academic session 2022-23. These model question papers are available for free download on the myCBSEguide app and website in PDF format. This helps students to find the answer to the most repeatedly asked questions in CBSE board exams.

However, the new competency-based design is very dynamic students have to go deep into concepts to answer such questions. Class 9 Science involves practical questions too. So, you can’t mug up the theoretical part without understanding it’s practical implications.

A simple suggestion to students is that you must practice more and more full syllabus model papers to achieve your goals. Just install the myCBSEguide app and download the latest sample papers for class 9 Science today.

Sample Papers of Class 9 Science 2023 with solution

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Class 9 Science Sample Papers 2022-23

We at myCBSEguide provide Class 9 Science Sample Papers for the session 2022-23 with solutions. These model question papers follow the course content given in NCERT textbooks and the CBSE curriculum. In class IX Science, schools conduct their annual examinations. So, it’s purely home exams in classes 9th and 11th. But still, the schools have to follow guidelines issued by CBSE and so as the format of the question paper.

Here is the annual exam question paper format of CBSE class 9 Science. It is exactly the same as CBSE provided for class 10 board exams.

Class 09 – Science Sample Paper – 01 (2022-23)

Maximum Marks: 80
Time Allowed: : 3 hours

General Instructions:

This question paper consists of 39 questions in 5 sections.
All questions are compulsory. However, an internal choice is provided in some questions. A student is expected to attempt only one of these questions.
Section A consists of 20 objective type questions carrying 1 mark each.
Section B consists of 6 Very Short questions carrying 02 marks each. Answers to these questions should in the range of 30 to 50 words.
Section C consists of 7 Short Answer type questions carrying 03 marks each. Answers to these questions should in the range of 50 to 80 words.
Section D consists of 3 Long Answer type questions carrying 05 marks each. Answer to these questions should be in the range of 80 to 120 words.
Section E consists of 3 source-based/case-based units of assessment of 04 marks each with sub-parts.

Class 9 Science Sample Paper Section A

Match the following with the correct response:-

(1) Sublimation	(A) Heat required to change the state of 1Kg of solid into liquid
(2) Evaporation	(B) Change of liquid state into vapours without coming to its boiling point
(3) Latent heat of vaporisation	(C) Change of solid directly into vapours
(4) Latent heat of fusion	(D) Heat required to change the state of 1Kg liquid into vapours

a) 1-C, 2-B, 3-D, 4-A

b) 1-A, 2-C, 3-B, 4-D

c) 1-B, 2-D, 3-A, 4-C

d) 1-D, 2-A, 3-C, 4-B

The primary function of smooth endoplasmic reticulum in liver cells is:
a) detoxification
b) protein synthesis
c) carbohydrate metabolism
d) catabolism of proteins
Match the following with the correct response:
(a) Velocity (i) m/s²
(b) Displacement (ii) m/s
(c) Acceleration (iii) s
(d) Time (iv) m
a) (a) – (i), (b) – (iii),(c) – (ii), (d) – (iv)
b) (a) – (ii), (b) – (iv),(c) – (i), (d) – (iii)
c) (a) – (iii), (b) – (ii),(c) – (iv), (d) – (i)
d) (a) – (iv), (b) – (i),(c) – (iii), (d) – (ii)
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Preventive and control measures adopted for the storage of grains include
a) strict cleaning
b) fumigation
c) all of these
d) proper drying
Latex of which of the following plants yield chewing gum?
a) Achras sapota
b) Bougainvillea
c) Cucurbita
d) Castor
Cell arises from the pre-existing cell was stated by
a) Virchow
b) Purkinje
c) Robert Hook
d) Robert Brown
Match the following with the correct response:
(1) 1 mole of {tex}{O_2}{/tex} (A) {tex}6.022 \times {10^{23}}{/tex}molecules
(2) 1 mole of N (B) 12 g
(3) 1 mole of C (C) {tex}6.022 \times {10^{23}}{/tex}atoms
(4) 1 mole of {tex}{H_2}O{/tex} (D) 18 g
a) 1-C, 2-B, 3-D, 4-A
b) 1-D, 2-A, 3-C, 4-B
c) 1-A, 2-C, 3-B, 4-D
d) 1-B, 2-D, 3-A, 4-C
Haversian canals are present in _____.
a) Ligament
b) Bone
c) Cartilage
d) Tendon
A heavier and a lighter body have equal momentum, then
a) heavier will have more K.E.
b) lighter will have more K.E.
c) they will have equal K.E.
d) K.E. will be independent of momentum

Match the following with the correct response:

(a) I equation of motion	(i) v = u + at
(b) II equation of motion	(ii) v² = u² + 2as
(c) III equation of motion	(iii) S = ut +{tex}\frac{1}{2}a{t^2}{/tex}
(d) Retardation	(iv) {tex} – \left( {\frac{{v – u}}{t}} \right){/tex}

a) (a) – (i), (b) – (iii),(c) – (ii), (d) – (iv)

b) (a) – (iii), (b) – (ii),(c) – (iv), (d) – (i)

c) (a) – (iv), (b) – (i),(c) – (iii), (d) – (ii)

d) (a) – (ii), (b) – (iv),(c) – (i), (d) – (iii)

Calculate the number of moles in 17 g of hydrogen peroxide (H₂O₂).
a) 2 mole
b) 1 mole
c) {tex}1\over2 {/tex} mole
d) 3 mole
Match the following with the correct response
(a) Aerenchyma (i) Permanent tissue
(b) Parenchyma (ii) Thin-walled packing cells
(c) Chlorenchyma (iii) Buoyancy
(d) Cork (iv) Photosynthesis
a) (a) – (i), (b) – (iii), (c) – (ii), (d) – (iv)
b) (a) – (iii), (b) – (ii), (c) – (iv), (d) – (i)
c) (a) – (ii), (b) – (iv), (c) – (i), (d) – (iii)
d) (a) – (iv), (b) – (i), (c) – (iii), (d) – (ii)
Organelle without a cell membrane is
a) Ribosome
b) Golgi apparatus
c) Chloroplast
d) Nucleus
Nitrogen is:
a) Diatomic
b) Monatomic
c) Triatomic
d) Tetratomic
Which of the following settles down when allowed to stand undisturbed doe sometimes?
a) Copper sulphate solution
b) Blood
c) Muddy water
d) Solution of egg albumin in water
‘Operation Flood’ in India refers to:-
a) Controlling floods during the rainy season.
b) Increase in milk production by the development of co-operative dairy on a large scale.
c) Increase in fish production during floods.
d) The large increase in egg production.
Assertion (A): A tiger can accelerate from rest at the rate of 4 m/s².
Reason (R): The velocity attained by it in 10s is 40 m/s.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is not the correct explanation of A.
c) A is true but R is false.
d) A is false but R is true.
Assertion (A): During evaporation of liquids, the temperature remains unaffected.
Reason (R): Kinetic energy of the molecules is inversely proportional to absolute temperature.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is not the correct explanation of A.
c) A is true but R is false.
d) A is false but R is true.
Assertion (A): 1-day Vinita went to see his brother’s competition and he saw his brother in pain and not able to run comfortably.
Reason (R): Two bones can be connected to each other by another type of connective tissue called the ligament. This tissue is very elastic. It has considerable strength.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is not the correct explanation of A.
c) A is true but R is false.
d) A is false but R is true.
Assertion (A): Isotopes are electrically neutral.
Reason (R): Isotopes are species with same mass number but different atomic numbers.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is not the correct explanation of A.
c) A is true but R is false.
d) A is false but R is true.
Class 9 Science Sample Paper Section B
A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
OR
A horse pulls a cart with a force of 300 N, such that the system of horse and cart moves with uniform speed of 18 kmh^-1 on a level road. Calculate the power developed by the horse in watt and also find its equivalent horse power. (Take 1 hp = 746 W)
What are the factors which are responsible for bringing a change in the physical state of a substance?
Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child. Given velocity of sound in air and aluminium are 346 ms^-1 and 6420 ms^-1 respectively.
Why are we able to sip hot tea or milk faster from a saucer rather than a cup?
Explain why the sole of a shoe is worn out.
OR
Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
If bromine atom is available in the form of, say, two isotopes {tex}_{35}^{79}Br(49.7\% ){/tex} and {tex}_{35}^{81}Br\left( {50.3\% } \right){/tex}. Calculate the average atomic mass of bromine atom.
Class 9 Science Sample Paper Section C
Observe the following graphical diagram and answer the following questions:
1. What is represented by the graphical diagram shown above?
2. Which wave characteristic determine the pitch of sound?
3. What is the relationship between pitch and frequency?
1. Identify the ion from the given figure.
2. Write the electronic configuration of the ion and atom mentioned in the figure.
3. How do we get the number of protons as 12?
A body is dropped from a height of 320 m. The acceleration due to the gravity is 10 m/s².
(a) How long does it take to reach the ground?
(b) What is the velocity with which it will strike the ground?
OR
Two cars A and B are moving along a straight line. Car A is moving at a speed of 80 Km/h while car B is moving at a speed 50 Km/h in the same direction. Find the magnitude and direction of:
(a) the relative velocity of car A with respect to B
(b) The relative velocity of car B with respect to A.
Due to heavy rains and speedy winds a tree broke and fell on the road. Two persons from local shops came and begin to remove that broken part of the tree, but in vain as the log was too heavy. A passerby came to their help with a few solid heavy sticks. Consequently, the heavy log was put aside and the commutation on the road began as usual.
(a) Comment upon the value(s) displayed by the passerby?
(b) Name the energy stored when the heavy log of the tree was lifted to some height?
Two bodies as shown in the figure collide with each other and join thereafter, with what velocity will they move after combining together?
What is the difference between plasma membrane and cell wall ? Give the functions of each one.
OR
Differentiate between active and passive transport.
Observe the given below diagram and answer the following questions:
1. What does A represent in the given diagram? How does cell ‘A’ of root hairs cells help in water absorption?
2. How does B in the given diagram help the plants?
3. Out of A, B, and C cells in the above diagram, which cell helps in the closing and opening of the stomata? Write the name of the cell.
Class 9 Science Sample Paper Section D
1. At some moment, two giant planets Jupiter and Saturn of the solar system are in the same line as seen from the earth. Find the total gravitational force due to them on a person of mass 50 kg on the earth. Could the force due to the planets be important?
  Mass of the Jupiter = 2 {tex}\times{/tex} 10²⁷ kg
  Mass of the Saturn = 6 {tex}\times{/tex} 10²⁶ kg
  The distance of Jupiter from the earth = 6.3 {tex}\times{/tex} 10¹¹m
  The distance of Saturn from the earth = 1.28 {tex}\times{/tex} 10¹²m
2. A bag of sugar weighs ‘w’ at a certain place on the equator. If this bag is taken to Antarctica, then will it weigh the same or more or less. Give a reason for your answer.
OR
1. A cube of side 5 cm is immersed in water and then in saturated salt solution. In which case, will it experience a greater buoyant force? If each side of the cube is reduced to 4 cm and then immersed in water, what will be the effect on the buoyant force experienced by the cube as compared to the first case for water. Give the reason for each case.
2. A ball weight 4 kg of density 4000 kg m^-3is completely immersed in water of density 10³kg m^-3.Find the force of buoyancy on it. (Given g = 10 ms^-2.)
Given below is shown the structure of an animal cell.

Using the above diagram, answer the following questions:
1. In which cell organelle cellular respiration takes place?
2. Why the nucleus is called the brain of the cell?
3. Why lysosomes are called suicide bags?
4. Which is the fluid content that contains specialised cell organelles inside the plasma membrane?
5. What is the composition of the cell membrane?
OR
The labelled diagram of the nucleus is shown below:

Using the above diagram, answer the following questions:
1. Which nuclear structures are embedded in the nucleoplasm?
2. What is the function of the nucleolus in the nucleus?
3. What is the composition of chromosomes in the nucleus?
4. When are the chromosomes seen in the nucleus?
5. Which structure transmits the heredity information from parents to offsprings?
1. Distinguish among the true solution, suspension and colloid in a tabular form under the following heads:
  1. Stability
  2. Filterability
  3. Type of mixture
2. Give the expression for the concentration of a solution. How will you prepare a 10% solution of glucose by mass in the water?
Class 9 Science Sample Paper Section E
Read the text carefully and answer the questions:
Animal tissues are of many types such as epithelial tissue, connective tissue, muscular tissue and nervous tissue. Blood is a type of connective tissue, and muscle forms muscular tissue. The nature of the matrix differs in concordance with the function of the particular connective tissue. Blood has a fluid (liquid) matrix called plasma, in which red blood corpuscles, white blood corpuscles and platelets are suspended. Blood flows and transports gases, digested food, hormones and waste materials to different parts of the body. Bone is another example of connective tissue. It forms the framework that supports the body. It also anchors the muscles and supports the main organs of the body. Another type of connective tissue, cartilage, has widely spaced cells.
1. Identify the following tissue.
2. Are the cells of connective tissues loosely spaced?
3. OR
  What are the components of the matrix of bone?
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Read the text carefully and answer the questions:
Poultry is the rearing of domesticated fowl (chicken), ducks, geese, turkey and some varieties of pigeon for their meat and eggs. Poultry birds are of two types that is broilers and layers. One is specialized meat-producing poultry birds while other is egg-laying poultry birds. The tremendous rise in the availability of poultry products is called Silver Revolution.
1. What is the meaning of layers regarding poultry?
2. There are different breeds of hens, so give some information about broiler.
3. We know that different types of revolution regarding animal husbandry. So, what is the silver revolution explain?
4. OR
  There are different breeds of poultry birds, mention two examples of indigenous and exotic breeds of poultry birds.
Read the text carefully and answer the questions:
A mole is defined as the mass of the substance which consists of an equal quantity of basic units. Example atoms in 12 grams are the same as 12C. The basic units can be molecules, atoms or formula units based on the substance. A mole fraction indicates the number of chemical elements. One mole of any substance is equal to the value of 6.023 {tex}\times{/tex} 10²³. It can be used to measure the products obtained from the chemical reaction.
1. Calculate the number of moles of sodium in a sample containing 10²⁰ atoms of sodium.
2. How do we express number of particles in terms of moles?
3. How do we express mass of a substance in terms of moles?
4. OR
  Calculate the number of moles in 56 g of He.

Class 09 – Science Sample Paper – 01 (Solution)

Solution

Class 9 Science Sample Paper Section A (Solved)
(a) 1-C, 2-B, 3-D, 4-A
Explanation:
1. A change of state directly from solid to gaseous without changing into liquid or vice versa is called Sublimation. E.g. Naphthalene, camphor and ammonium chloride change directly from solid state to gaseous state, and vice-versa, without changing into the liquid state.
2. Evaporation is a change of liquid state into vapours below the boiling point of the liquid. It is a surface phenomenon, and produces the effect of cooling.
3. Latent heat of vaporisation is the heat energy required to change 1 kg of a liquid to gas at atmospheric pressure at its boiling point.
4. Latent heat of fusion is the amount of heat energy required to change 1 kg of solid into liquid.
(a) detoxification
Explanation: Smooth Endoplasmic Reticulum (SER) is responsible for the synthesis and repair of membranes. It also has a detoxification function.
(b) (a) – (ii), (b) – (iv),(c) – (i), (d) – (iii)
Explanation:
1. The SI unit of velocity is meter per second (m/s).
2. The SI unit of displacement is the meter (m).
3. The SI unit of acceleration is the metre per second squared (m/s²).
4. The SI unit of time is the second (s).
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(c) all of these
Explanation: Preventive and control measures adopted for the storage of grains include strict cleaning of the produce before storage, proper drying of the product first in sunlight and then in a shade, and fumigation using chemicals that can kill pests.
(a) Achras sapota
Explanation: Latex of Achras sapota plants yields chewing gum. Latex is obtained by making small cuts in the stem to secrete juice like latex.
(a) Virchow
Explanation: Rudolf Virchow presented the idea in 1855 that every cell arises from another (pre-existing) cell. This was an addition to the cell theory that was proposed earlier by Matthias Jakob Schleiden and Theodor Schwann.
(c) 1-A, 2-C, 3-B, 4-D
Explanation:
(1) 1 mole of {tex}{O_2}{/tex} (A) {tex}6.022 \times {10^{23}}{/tex}molecules
(2) 1 mole of N (C) {tex}6.022 \times {10^{23}}{/tex}atoms
(3) 1 mole of C (B) 12 g
(4) 1 mole of {tex}{H_2}O{/tex} (D) 18 g
(b) Bone
Explanation: Haversian canals are found in the bone matrix of a long bone, like humerus of mammals. Haversian canals contain artery and veins, osteoblasts in areolar tissue, nerves, and lymph. It is also called an osteon.
(b) lighter will have more K.E.
Explanation: The lighter body will have more K.E because kinetic energy is inversely proportional to the mass of the body.
(a) (a) – (i), (b) – (iii),(c) – (ii), (d) – (iv)
Explanation: Three equation of motion and their formula is given.
Negative acceleration is known as retardation. Therefore, the formula of retardation is the same as acceleration with a negative sign.
(c) {tex}1\over2 {/tex} mole
Explanation: The molecular mass of H₂O₂is 34 u. So, 34 g of hydrogen peroxide will contain 1 mole. Therefore, 17 g of hydrogen peroxide will contain {tex}17\over34{/tex} moles or {tex}\frac{1}{2}{/tex}mole.
(b) (a) – (iii), (b) – (ii), (c) – (iv), (d) – (i)
Explanation:
- Aerenchyma provides buoyancy.
- Parenchyma cells are thin-walled.
- Chlorenchyma is responsible for photosynthesis in green plants.
- Cork is a permanent tissue.
(a) Ribosome
Explanation: Ribosomes are non membrane bound organelles that are found freely occurring in the cytoplasm.
(a) Diatomic
Explanation: Nitrogen (N₂) is diatomic. A nitrogen atom has 5 electrons in its outer shell. It forms three covalent bonds with another nitrogen atom and makes a nitrogen molecule.
(c) Muddy water
Explanation: Muddy water will settle down because particles are heavy and settle due to gravity. Setting down of coarse particles under the influence of gravity is called sedimentation. During sedimentation, heavier particles settle down faster than finer particles.
(b) Increase in milk production by the development of co-operative dairy on a large scale.
Explanation: Operation Flood was a rural development program started by India’s National Dairy Development Board (NDDB) in 1970. It resulted in making India one of the largest producers of milk and milk products and hence is also called the White Revolution of India.
(a) Both A and R are true and R is the correct explanation of A.
Explanation: Initial velocity (u) = 0 , acceleration (a) = 4 m/s²
v = u + at
v = 0 + 4 × 10
v = 40 m/s
(d) A is false but R is true.
Explanation: During the evaporation of liquids, the temperature decreases due to the escape of molecules with high energy. The kinetic energy of the molecules is directly proportional to absolute temperature.
(b) Both A and R are true but R is not the correct explanation of A.
Explanation: Two bones can be connected to each other by connective tissue called the ligament. Vinita’s brother might be suffering from a ligament tear due to which he has pain in the knee while running.
(c) A is true but R is false.
Explanation: Isotopes are species with same atomic number but different mass numbers.
Class 9 Science Sample Paper Section B (Solved)
Energy consumed by a certain household = 250 kWh
since 1 kWh {tex}= {\rm{ }}3.6 \times {10^6}J{/tex}
therefore 250 kWh {tex}= {\rm{ }}250 \times 3.6 \times {10^6}{/tex} {tex}= {\rm{ }}9 \times {10^8}J\;{/tex}
OR
Given : Force acting on the horse-cart system = 300 N, Initial velocity
= 18 kmh^-1= {tex}\frac{5}{18}{/tex}× 18 = 5ms^-1
Therefore power developed by the horse = Force × velocity = 300 ×5 = 1500 W
Therefore equivalent horse power = {tex}\frac{1500}{746}{/tex} = 2.01 hp
Temperature and pressure are the two factors which can bring about the change in physical state. But they have opposing effects. Increase in temperature pulls the constituents of a substance apart. Increase in pressure tends to bring them closer.
{tex}\frac{Time\;taken\;by\;sound\;travel\;aluminium\;}{Time\;taken\;by\;sound\;in\;air}=\frac{Speed\;of\;sound\;in\;air\;}{Speed\;of\;sound\;in\;aluminium}{/tex}
Speed of sound in air Speed of sound in aluminium
{tex}\Rightarrow\frac{T_{alu}\;}{T_{air}}=\frac{346\;}{6420}{/tex}
⇒ T_Alu: T_Air = 346:6420 = 1: 18.55
Saucer has a bigger surface area as compared to cup. Since evaporation is a surface phenomenon, by using a saucer instead of cup we are increasing the surface area for evaporation to occur. Faster evaporation of particles of tea or milk allows cooling and taking a sip becomes easier.
When we walk on the ground the force of friction between the sole of our shoe and the ground acts opposite to the direction of our motion. It is this force of friction which gradually wears out the sole of our shoe.
OR
We can explain the above observation on the basis of law of conservation of momentum. When a system of the hose and water is not ejecting any water its momentum is zero. When the water issues out from the hose with a high velocity, it has momentum in the forward direction. Therefore, in order to conserve momentum the hose tends to move in the backward direction and hence is difficult to hold.
The average atomic mass of bromine:
= 0.01 [(Atomic mass of isotope-1) (Occurence of isotope-1) + (Atomic mass of isotope-2) (Occurence of isotope-2)] = 0.01 [(Atomic mass of bromine-79) (Percentage occurence of bromine-79) + (Atomic mass of bromine-81) (Percentage occurence of bromine-81)] = {tex}\frac{{\left( {79\times49.7} \right) + \left( {81\times50.3}\right)}}{{100}}{/tex}
= {tex}\frac{{3926.3 + 4074.3}}{{100}}{/tex} = {tex}\frac{{8000.6}}{{100}}{/tex} = 80 u
The average atomic mass of bromine atom is 80 u.
Class 9 Science Sample Paper Section C (Solved)
1. The two graphical figure represents the same amplitude but different frequency.
2. The pitch of the sound is determined by the frequency wave.
3. Lower pitch indicates low frequency and higher pitch indicates the higher frequency that is{tex}Pitch \propto Frequency{/tex}.
1. Mg²⁺ ion is mentioned in the given figure.
2. The electronic configuraton of Mg²⁺ion = 2, 8 and that of ₁₂Mg atom = 2, 8, 2
3. Number of protons in Mg atom = 2 + 8 + 2 = 12
Height = h
Distance = s = 320 m
Acceleration due to gravity = g = 10 m/s²
Initial velocity = u =0
(i) from s = ut + {tex}\frac{1}{2}a{t^2}{/tex}
{tex}h = ut \times \frac{1}{2}g{t^2}{/tex}
{tex}320 = 0 \times t + \frac{1}{2} \times 10 \times {t^2}{/tex}
{tex}\frac{{320 \times 2}}{{10}} = {t^2}{/tex}
{tex}64 = {t^2}{/tex}
t = {tex}\sqrt{64}{/tex}
t = 8 sec
(ii) from v = u + at
{tex}v = 0 + 10 \times 8{/tex}
v = 80 m/s
OR
On taking the formula when both bodies are moving along same direction, we have
(a) Relative velocity of A with respect to B is –
V_AB = V_A – V_A = 80 – 50 =30 km/h
Hence, the velocity of car A with respect to car B is 30 km/h in same direction.
(b) Relative velocity of B with respect to A is –
V_BA = V_B – V_A = 50- 80 = -30 km/h
Hence, the velocity of car B with respect to car A is 30 km/h in opposite direction.
(a) Social value, conceptual value, presence of mind.
(b) Potential energy.
Given, m₁ = 65 kg, m₂ = 60 kg
u₁ = +5 m/s positive direction
u₂ = -6 m/s negative direction, v = ?
Total momentum of two bodies before collision = m₁u₁ + m₂u₂
= 65 {tex}\times{/tex} 5 + 60 {tex}\times{/tex} (-6) = -35 kg-m/s
If v is the velocity of two combined bodies.
Agter collision, the total momentum will be = m₁v₁ + m₂v₂ [{tex}\because{/tex} v₁ = v₂ = v] =(65 + 60) kg {tex}\times{/tex} v m/s = 125 v kg-m/s
{tex}\therefore{/tex} According to the law of conservation of momentum,
m₁u₁ + m₂u₂= m₁v₁ + m₂v₂ {tex}\Rightarrow{/tex} -35 = 125v
{tex}\therefore{/tex} {tex}v = \frac {-35}{125} = -0.28 \ m/s{/tex}
So, two bodies will move with a velocity of 0.26 m/s in the direction of the second body.
1. Plasma Membrane is the phospholipid layer, found in all types of cells; it helps in protecting the protoplasm and checks the passage of molecules inside the cell, Though cell wall is found in the plant cell, fungi, bacteria only and protects the cell from external shocks, and provide rigidity and shape to the cell.
2. The cell wall is the outermost boundary of the cell (if present), and plasma membrane is present in the inner lining of the cell. The plasma membrane is delicate thin layer while cell wall is the thick and rigid layer. Plasma Membrane is selectively permeable membrane allowing small molecules entry only; their layer is made up of lipids and proteins and few carbohydrates, while Cell wall constituents may vary from chitin, peptidoglycon, and cellulose.
3. Plasma membrane is the living membrane made up of lipids and proteins, whereas cell wall is non-living made up of cellulose.

Function of Plasma membrane: It acts as semi permeable membrane which allows only selective substances to pass through it.
Function of Cell Wall: It provides rigidity and protection to cell.

Active transport	Passive transport
1. It involves movement of molecules against the concentration gradient. 2. It requires energy in the form of ATP molecules. 3. It is a rapid movement. 4. Movement of large molecules occur by active transport.	1. It involves movement of molecules along the concentration gradient. 2. No energy is required 3. It is a slow movement. 4. Small molecules or water molecules only are transported passively.

In the given diagram of the epidermis, A represents the epidermal cells of the roots bear long hair-like parts called root hairs. With the help of these cells, root hairs greatly increase the total absorptive surface area and help in water absorption.
B represents the stomata. Stomata are the pores present in the epidermis of the leaves. Stomata help in the exchange of gases with the atmosphere during photosynthesis and respiration. Also, the process of transpiration (loss of water in the form of water vapour) takes place through stomata.
C cell that represents the guard cells. These cells are kidney-shaped that enclose the stomata and thus help in the opening and closing of stomata.

Class 9 Science Sample Paper Section D (Solved)
1. 1. Gravitational force acting on the 50 kg,
    F = mg = 50 {tex}\times{/tex} 9.8 = 490 N
  2. Gravitational force acting on the 50 kg mass due to jupiter,
    F_Jupiter = {tex}\frac{G \times M_{\text { jupiter }} \times M_{\text { person }}}{(\text { distance of jupiter from the earth })^{2}}{/tex}
    F_Jupiter = {tex}\frac{6.67 \times 10^{-11} \times 2 \times 10^{27} \times 50}{6.3 \times 10^{11} \times 6.3 \times 10^{11}}{/tex}
    F_Jupiter = 1.68 {tex}\times{/tex} 10^-5 N
  3. Gravitational force acting on the 50 kg mass due to saturn
    F_saturn = {tex}\frac{G \times M_{\text { saturn }} \times M_{\text { person }}}{\left(\text { distance of saturn from the earth) }^{2}\right.}{/tex}
    F_saturn = {tex}\frac{6.67 \times 10^{-11} \times 6 \times 10^{26} \times 50}{1.28 \times 10^{12} \times 1.28 \times 10^{12}}{/tex}
    F_saturn = 1.12 {tex}\times{/tex} 10^-6 N
    {tex}\therefore{/tex} Total gravitational force due to the Jupiter and the Saturn = (1.68 {tex}\times{/tex} 10^-5 + 1.12 {tex}\times{/tex} 10^-6) = 1.8 {tex}\times{/tex} 10^-5 N
    Thus, the combined force due to the planets Jupiter and Saturn (1.8 {tex}\times{/tex} 10^-5) N is negligible as compared to the gravitational force i.e. 490 N due to the earth.
2. We know that g at the equator is less than g at poles (Antarctica). Thus, weight at the equator is less than weight at the pole (Antarctica). A bag of sugar weighs ‘w’ at a certain place on the equator. If this bag is taken to Antarctica, then it will weigh more due to the greater value of g.
OR
1. The cube will experience a greater buoyant force in saturated salt solution than in water because density of saturated salt solution is more than the density of water. If each side of the cube is reduced to 4 cm, it will result in reduction in volume of the cube. Hence, the buoyant force experienced by it will reduce in water.
2. Buoyant force = weight of displaced water
  = density of water {tex}\times{/tex} volume of displaced water {tex}\times{/tex} g
  = 1000 {tex}\times{/tex} {tex}\frac {4}{4000}{/tex} {tex}\times{/tex} 10 [{tex}\because{/tex} volume = {tex}\frac {weight}{density}{/tex}] = 10 N
1. Cellular respiration takes place in mitochondria resulting in the release of energy.
2. The nucleus is called the brain of the cell because it controls the various activities of the cell.
3. As lysosomes are organelles which on bursting can kill cells possessing them, they are called suicide bags.
4. The cytoplasm is the fluid content inside the plasma membrane. It also contains many specialised cell organelles which perform a specific function for the cell.
5. The cell membrane or plasma membrane is flexible and is made up of organic molecules called lipids and proteins.

Within nucleoplasm two types of nuclear structures are embedded – the nucleolus and chromatin material.
The nucleolus is rich in protein and RNA molecules and acts as the site for ribosome formation.
Chromosomes are composed of DNA and protein.
In a non-dividing cell, the DNA is present as part of chromatin material which is visible as an entangled mass of thread-like structures. Whenever the cell is about to divide, that is, at the time of cell division, the chromatin material gets organized into chromosomes.
Genes are the functional units of chromosomes which transmit the heredity information from parents to offsprings.

Distinctions between true solution, suspension and colloid are:

Property	Solution	Suspension	Colloid
Stability	It is stable. Constituting particles do not settle down on keeping undisturbed.	It is unstable. Constituting particles settle down on keeping undisturbed.	It is quite stable. Constituting particles do not settle down on keeping undisturbed.
Filterability	Particles cannot be separated by filtration. Means passes through filter paper.	Particles are large, so they can be easily separated by ordinary filtration. Means do not pass through filter paper.	It cannot be separated by ordinary filter paper but can be separated by ultrafiltration. Means passes through filter paper.
Type of mixture	Homogeneous	Heterogeneous	Heterogeneous but appears to be homogeneous.

Concentration is defined as the number of moles (amount of substance) per unit volume (often liters/ dm³ )
The methods by which the concentration of a solution can be expressed are:
a) Mass by mass% of solution = {tex}\frac{Mass \ of \ solute}{Mass \ of \ solution} \times 100{/tex}
b) Mass by volume % of solution = {tex}\frac{Mass \ of \ solute}{Mass \ of \ solution} \times 100{/tex}
Thus, a 10 percent solution of glucose can be prepared by dissolving 10 g of glucose in 90 g of water.

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Class 9 Science Sample Paper Section E (Solved)
1. Adipose tissue.
2. The cells of the connective tissue are loosely spaced and invaded in and intracellular matrix.
3. OR
  Calcium and phosphorus compounds.
1. Egg-laying poultry birds are called layers.
2. The specialized meat-producing poultry birds are called broilers. Broilers are quick growing birds which are raised for 6-8 weeks. Their food is rich in vitamin A and K.
3. The tremendous rise in the availability of poultry products is called Silver Revolution.
4. OR
  Following are the example of poultry birds
  Indigenous breed: Assel and Kadaknath.
  Exotic breed: Rhode island red and Light Sussex.
1. Number of moles = {tex}\frac{Number \ of \ atoms \ of \ sodium}{Avogadro’s \ number \ of \ particles}{/tex}
  = {tex}\frac{1.0 \times 10^{20} \text { atoms }}{6.022 \times 10^{23} \text { atoms }}{/tex} = 1.66 {tex}\times{/tex} 10^-4 mol
2. Number of particle is expressed by Number of moles = {tex}\frac{Given \ number \ of \ particles}{Avogadro’s \ number}{/tex}
3. Mass of a substance in terms of moles expressed by Number of moles = {tex}\frac{Given \ mass}{Gram \ atomic \ mass}{/tex}
4. OR
  Number of moles in 56g of He = {tex}\frac{Mass \ of \ He}{Gram \ atomic \ mass \ of \ He}{/tex} = {tex}\frac{56}{4}{/tex} = 14 mol.

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Class 9 Science Sample Papers 2022-23

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Class 9 Science Sample Papers 2022-23

Class 09 – Science Sample Paper – 01 (2022-23)

Class 9 Science Sample Paper Section A

Class 9 Science Sample Paper Section B

Class 9 Science Sample Paper Section C

Class 9 Science Sample Paper Section D

Class 9 Science Sample Paper Section E

Class 09 – Science Sample Paper – 01 (Solution)

Class 9 Science Sample Paper Section A (Solved)

Class 9 Science Sample Paper Section B (Solved)

Class 9 Science Sample Paper Section C (Solved)

Class 9 Science Sample Paper Section D (Solved)

Class 9 Science Sample Paper Section E (Solved)

Sample Papers for Class 9 2023

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(1) 1 mole of {tex}{O_2}{/tex}	(A) {tex}6.022 \times {10^{23}}{/tex}molecules
(2) 1 mole of N	(B) 12 g
(3) 1 mole of C	(C) {tex}6.022 \times {10^{23}}{/tex}atoms
(4) 1 mole of {tex}{H_2}O{/tex}	(D) 18 g

(a) Aerenchyma	(i) Permanent tissue
(b) Parenchyma	(ii) Thin-walled packing cells
(c) Chlorenchyma	(iii) Buoyancy
(d) Cork	(iv) Photosynthesis

(1) 1 mole of {tex}{O_2}{/tex}	(A) {tex}6.022 \times {10^{23}}{/tex}molecules
(2) 1 mole of N	(C) {tex}6.022 \times {10^{23}}{/tex}atoms
(3) 1 mole of C	(B) 12 g
(4) 1 mole of {tex}{H_2}O{/tex}	(D) 18 g