Cbse Class 9 Science Solved Practice Paper Set i

CBSE Class 9th Science Practice Paper

Solved Practice Paper Class – IX Subject – Science Set - I Time Allowed: 3 Hours Max. Marks: 90 General Instructions: (i) The question paper comprises two sections, A and B. You are to attempt both sections. (ii) All questions are compulsory. (iii) All questions of Section A and all questions of Section B are to be attempted separately. (iv) Question numbers 1 to 3 A are one mark questions. (v) Question numbers 4 to 6 are two marks questions, to be answered in about 30 words each. (vi) Question numbers 7 to 18 are three marks questions, to be answered in about 50 words each. (vii) Question numbers 19 to 24 are five marks questions, to be answered in about 70 words each. (viii) Question numbers 25 to 36 in Section B are practical based questions. (ix) Question numbers 25 to 33 are of one mark and 34 to 36 are of two marks each. Section A Question 1. Define the commercial unit of electrical energy. Answer. Kilowatt - hour is the commercial unit of electrical energy which is defined as the amount of electrical energy consumed when an electrical appliance having a power rating of 1 kelowatt is used for 1 hour. Question 2. Name the characteristics of sound wave on which the following properties depend? (a) Loudness (b) Pitch Answer. (a) Loudness depends on amplitude (b) Pitch depends on frequency. Question 3. Find the relative density of copper block of mass 216g having volume of 80cm3 (Density of Water = 1g/ cm3). Answer. Get SOLVED & UNSOLVED question papers, updated Syllabus, Sample papers and study material and much more…


Density of block = Mass/volume = 216/80 = 2.7 g/cm3 Relative density = Density of a substance / Density of water = 2.7/1 = 2.7 Question 4. Differentiate between transverse and longitudinal waves and give one example of each. Answer. Transverse waves: Transverse waves are the waves in which particles of the medium vibrate at right angles to the direction of propagation of wave. For example: Water waves set up in a rope whose one end is fixed and the other is jerked, are an example of transverse waves. Longitudinal waves: Longitudinal waves are those waves in which particles of the medium vibrate in the same direction as that of the wave. For example: Waves set in a compressed spring, are an example of longitudinal waves. Question 5. What is the causal organism for Swine flu? What protective measures should be taken to prevent the spread of such viral diseases? Answer. Virus (H1N1) is the casual organism for Swine flu. Following are the measures that must be taken to prevent the spread of viral diseases:  

Spraying pesticides. Cleaning of garbage dumps, drains and sewers.

Question 6. (a) Identify two features possessed by all Chordates. (b) In which class would you place any organism which has: (i) a scaly exoskeleton and a bony endoskeleton (ii) a scaly exoskeleton and lay eggs outside water. Answer. (a) Some of the important features possessed by all Chrodates are:  

They have notochord / dorsal nerve chord / paired gill pouches They possess bilaterally symmetrical body

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 They are triploblastic and are coelomate. (any two) (b) (i) Animals exhibiting a scaly exoskeleton and a bony endoskeleton form Class pisces. (ii) Animals exhibiting a scaly exoskeleton and lay eggs outside water form Class Reptilia. Question 7. A flask contains 4.4g of CO2 gas. Calculate (a) How many moles of CO2 gas does it contain? (b) How many molecules ot CO2 gas are present in the sample. (c) How many atoms of oxygen are present in the given sample. [Atomic mass of C is12u and that of O is16u] Answer. 1 moleof CO2 =12 +16 × 2 = 44g (a) Number of moles of CO2 =

4.4  0.1mol 44g mol 1

(b) Number of molecules of CO2 = 0.1 × 6.022 × 1023 = 6.022 × 1022 molecules (c) Number of atoms of oxygen = 2 × 0.1 × 6.022 × 1023 =2 × 6.022 × 1022 = 1.204 × 1023 atoms Question 8. (a) Define Power of a body. (b) Two children A and B both weighing 32 kg start climbing up a rope separately reach a height of 8m, ‘A’ takes 15s and ‘B’ takes 20s to reach that level. Calculate the amount of work done by A and B. Which of the two has more power. Show by calculation. [g = 10 m/s2] Answer. (a) Power: Power is defined as the rate of doing work or work done per unit time. W i.e., P = t (b) ForA: W=m×g×h =32kg × 10 m/s2 × 8m =2560 J Both have same mass and climb the same height.

 Work done by B = 2560 J Now power of A , PA =

2560J  170.7W 15s



And, power of B , PB =

2560J  128W 20s

Question 9. A cyclist goes around a circular track once every 2 minutes. If the radius of the circular track is 105 metre. Calculate his speed ? (Given π = 22/ 7) Answer. Given, radius of circular track, r = 105m Distance travelled by the cyclist in going around the circular track, D = 2πr = 2 ×

22 × 105 = 660 m 7

Time taken by cyclist to complete 1 round, t = 2 minutes = 2 × 60 = 120 seconds Now, Speed =

Distance 660 =  5.5m/s Time 120

Therefore, speed of cyclist = 5.5 m/s. Question 10. Distribution of electrons, protons and neutrons in atoms of four elements A,B,C and D is given in the following table: Element Protons Neutrons Electrons A

19

21

19

B

17

18

17

C

17

20

17

D 18 22 Observe the table and answer the following questions: (i) Describe the electronic distribution in atom of element B. (ii) Is elements B a metal or a non-metal? Why? (iii) Which two elements form a pair of ISOTOPES? (iv) Which two elements form a pair of ISOBARS?

18

Answer. (i) Number of electrons in element B = 17 Get SOLVED & UNSOLVED question papers, updated Syllabus, Sample papers and study material and much more…


Therefore, distribution of electrons is represented as: B (17) = 2 8 7 (ii) Element B is a non-metal as it is short of one electron to complete the octet. (iii) As isotopes are the elements having same atomic numbers. Therefore, here Brand C are the isotopes with atomic number 17. (iv) Isobars are the elements with same mass number, i.e., sum of protons and neutrons. Thus, here AandD are the neutrons with same mass number, 40. As, mass number of A = 19 + 21 = 40 And, mass number of B = 18 + 22 = 40 Question 11. (a) List four main processes involved in the water cycle. (b) Give a diagrammatic representation of Carbon Cycle in nature. Answer. (a) Four main processes involved in water cycle are: 1. Evaporation: Water from oceans, lakes, rivers and other water bodies get warmed up due to sun’s heat and converted into water vapour (gas). Water vapour droplets join together to make clouds. 2. Condensation: In this step vapour gets cooled down and turns back into liquid water. 3. Precipitation: Here water (in the form of rain, snow, hail or sleet) falls from clouds in the sky. 4. Runoff: The water from clouds falls down the hills, mountains, or other inclines to join oceans, rivers, lakes, streams, etc. Most will infiltrate (soak into) the ground and will collect as underground water. (b) Diagrammatic representation of Carbon Cycle in nature:



Question 12. When we open the cap of a cola drink (or any carbonated beverage), why does excess of bubbles come out? Answer. A cola (or carbonated) beverage is produced by dissolving carbon dioxide in the drink solution under pressure. Increase in pressure increases the solubility of gases. Therefore, more carbon dioxide dissolves at the higher pressure. When this pressure is suddenly released, e.g. by removing the cap of the bottle, carbon dioxide is less soluble, and it comes out in as bubbles from the solution. Question 13. Calculate the mass of glucouse and mass of water required to make 200g of 25% solution of glucouse. Answer. Given mass of solution(M) = 200g Concentration of solution = 25% Since, Mass by Mass percentage of solution =

Mass of solute  m  ×100 Mass of solution  M 



 

m 100 200 m  50 g 25 

∴ Mass of solute (glucose) = 50g Mass of solvent (water) = M - m = 200g - 50g = 150g Question 14. Why air is a mixture not a compound? Give reasons. Answer. Air is a mixture for of the following reasons: 1. Air does not have a fixed composition. Its composition varies from place to place. 2. The components of air can be separated by a physical method such as fractional distillation of liquid air. 3. Liquid air does not have a definite boiling point. 4. We can prepare air artificially by mixing the various components of air in the same proportions in which they occur at a place. This is a physical process because no energy changes occur. Question 15. What is supersaturated solution? Answer. Any solution containing more solute than required to prepare a saturated solution at any fixed temperature is called supersaturated solution. Supersaturated solution is a meta-stable state i.e. it will remain in supersaturated state so long it is left undisturbed. A slight disturbance (e.g. slight rise or fall of temperature) will turn it into a saturated solution. Question 16. Give three examples of the range of variations that you see in life forms around you. Answer. The following range of variations can be seen in different life forms living on Earth:



1. Variation in size: Some microscopic bacteria are few micrometre in size that can not be even observed with naked eye whereas another creatrure like blue whale is about 30 metres length and is one among the largest animal on earth. 2. Variation in lifespan : Some pine trees live for thousands of years while insects like mosquitoes die within a few days. Question 17. 10 bulbs of 50 W each, are being used for 6 hours a day along with 5 bulbs of 100 W each, for 2 hours a day. How many `units' (kWh) of electrical energy are consumed in one day by all bulbs together. Answer. We know that, Energy consumed by electric bulb = Power of the bulb × Time For 50 W bulbs: Energy consumed by 1 electric bulb in a day = 50 W × 6 h = 300 watt-hours Energy consumed by 10 electric bulbs in a day = 300 × 10 = 3000 watt-hour For 100 W bulbs: Energy consumed by 1 electric bulb in a day = 100 W × 2 h = 200 watt-hour Energy consumed by 5 electric bulb in a day = 200 × 5 = 1000 watt-hour Thus, total energy consumed by all 50 W bulbs and 100 W bulbs, in a day = 3000 + 1000 = 4000 watt-hour = 4 kWh Question 18. When does an object float or sink when placed in water? A sheet of tin weighing 1 kg sinks in water but if the same sheet is converted into a boat, it floats. Why? Answer. When an object is immersed in water, the water exerts an upward force on the object. This upward force equal to the weight of water displaced by object is called the buoyant force. If on completely immersing the object, the buoyant force is more than the weight of object. it will float on water. In other words, if density of object is less than that of water, then the object will float on water. On the other hand if buoyant force, on completely immersing the object in water, is less than the weight of the object, the object will sink in water. In other words, if the density of object is greater than water, then the object will sink in water.



The density of tin is higher than that of water, so the sheet of tin sinks in water. But when the same sheet of tin is converted into a boat, a lot of air is trapped in the boat causing the average density of boat to becomes lower than that of water. Therefore, it floats on water. Question 19. Give reasons for the following statements: (a) Meristematic cells have a prominent nucleus and dense cytoplasm but they lack vacuole. (b) Intercellular spaces are absent in sclerenchymatous tissues. (c) We get a crunchy and granular feeling, when we chew pear fruit. (d) Branches of a tree move and bend freely in high wind velocity. (e) It Is difficult to pull out the husk of a coconut tree. Answer. (a) Vacuoles have a function of storing food and other nutrients that a cell might need to survive. But, as Meristematic cells have an ability to divide and form new cell so there is no point in storing food and other nutrients when the cell has to divide. So they lack vacuole. (b) There are no intercellular spaces in the sclerenchyma cells as these cells are lignified to provide strength to the plants. (c) This is due to the presence of cells known as sclereids or stone cells. The sclereids give a crunchy feeling to the pear fruit because it provides support and hardens the tissue. (d) ) Branches of a tree move and bend freely because of the presence of a simple permanent tissue called collenchyma (e) The husk of a cocunut tree is made up of scelerenchyma tissues which gives rigidity and stiffness to the plant cells due to which we find it tough to pull the husk out. Question 20. (a) Calculate the number of ethane (C2H6) molecules contained in a gas which is having the same mass as that of 1.5 × 1022 molecules of methane (CH4). (b) Calculate the number of atoms of gold present in 1g of its sample which contains 90% gold and the rest being copper. Answer. (a) An Avogadro's number = 6.02 × 1023 Now, 6.02 × 1023 molecules of methane = 1 mole 1 ∴ 1.5 × 1022 molecules of methane = 1.5 1022  2.49 102 moles 23 6.02 10 Molar mass of Methane (CH4) = 12 + 1 × 4 = 16 g Mass of methane = Molar mass × Number of moles = 16 x 2.49 × 10−2 = 3.984 x 10−1 g Get SOLVED & UNSOLVED question papers, updated Syllabus, Sample papers and study material and much more…


According to question, Mass as Ethane = Mass of methane = 3.984 x 10−1 g Now, molar mass of ethane (C2H6) = 12 × 2 + 1 × 6 = 30 ⟹ Number of molecules in 30 g of Ethane = 6.02 × 1023 ∴ Number of molecules in 3.984 × 10−1g of Ethane =

6.02 1023  3.984 101 30

 0.799 1022  8 1021 molecules of ethane

(b) Atomic Mass of Gold, Au = 197 Thus, 197 g of Gold contains 6.02 × 1023 atoms of Gold Given, mass of gold in the sample = 90% × 1 g = 0.9 g As, 197 g of gold = 6.02 × 1023 atoms ∴ 0.9 g of gold =

6.02 1023  0.9 197

 0.0275 1023  2.75 1021 atoms Question 21. (a) We can’t hear the sound produced by a vibrating pendulum but the sound produced by the humming bees can be heard clearly. Why? (b) What makes the voices of different people distinguishable? (c) Name the three tiny bones present in the middle of the ear. What function do they perform in working of ear? Answer. (a) The frequency of sound produced by humming bear is greater than 20 Hz, while the frequency of vibrations produced by pendulum is less than 20 Hz. Since the audible range for human beings is 20 Hz to 20000 Hz, so the sound produced by the humming bees can be heard, while the sound produced by the vibrating pendulum cannot be heard. (b) It is the pitch of the sound that makes the voices of different people distinguishable. (c) Three tiny bones present in the middle of the ear are: Hammer, Anvil and Stirrup Function:



(i) Hammer: This is the largest bone in the middle ear. It's attached to the inner side of the eardrum. When the eardrum vibrates due to the incoming sound waves, it causes hammer to vibrate which inturn passes these vibrations to the next bone, Anvil. (ii) Anvil: Anvil which is located in between Anvil and Stirrup, passes the sound vibrations from Anvil to Stirrup. (iii) Stirrup: The vibrating stirrup strikes on the membrane of the oval window and passes its vibrations to the liquid in cochlea which then produces electric impulses to be interpreted by the brain. Question 22. (a) How is cell wall different from plasma membrane? (b) What do you mean by selectively permeable membrane? Which of the two cell wall or plasma membrane have this feature? (c) State the functions of plasma membrane. Answer. (a) Difference between cell wall and plasma membrane: S. No.

Cell Wall

1.

It occurs in plant cells only.

2.

It is non-living and quite thick.

Plasma membrane It is found in both plant and animal cells. It is a living part of the cell and is quite thin. It is flexible. It is formed of lipids and proteins.

It is rigid. It is formed of cellulose, hemicelluloses and pectin. It provides protection and strength to It holds cellular contents and controls 5. the cell. the passage of materials in and out of the cell. (b) A selectively permeable membrane is a membrane which allows the entry and exit of some selected materials only. Since plasma membrane allows only a few or selected materials to move in and out through the cell so it is terned as selectively permeable membrane. Whereas, cell wall does not facilitate the restricted flow of materials so, it is simply permeable. 3. 4.

(c) Functions of plasma membrane:  

It holds the various contents lying inside the cell. It controls the movement of selected substances in and out of the cell.



 

It helps to maintains the shape of the cell. It provides protection to cell from microbes and foreign substance.

Question 23. (a) What is Rutherford’s Nuclear Model of atom? (b) Why did Rutherford select the gold foil for his α -particle scattering experiment? (c) Define the term scintillations? Answer. (a) The main points of Rutherford’s Nuclear Model are as follows: 

An atom is made up of two parts, nucleus and extra nuclear part. Nucleus is a small, heavy, positively charged portion of the atom and located at the centre of the atom. whereas the extra nuclear part is the space around the nucleus in which the electrons are located.



The whole mass of the atom is located in the nucleus. Since the electrons have negligible mass, the mass of the atom is mainly due to the protons present inside the nucleus.



The electrons revolve around the nucleus in fixed orbits.



An atom is electrically neutral due to the presence of same number of protons and electrons.



Most of the atom is empty space.

(b) Gold is a highly malleable metal which can be hammered and converted into very thin sheets or foil. Thus, it is easier for the α -particle to pass through the gold foil rarely deviated by nucleus. As the thickness of the foil decrease, the possibility of correctness of experiment increases. (c) Scintillations are bright flashes produced through the α -particles in Rutherford model of an atom. When these particles passed through a slit and strikes against the gold foil, they get scattered and produce bright flashes known as scintillations. Question 24. (a) State the Law of Conservation of Momentum. Deduce it from Newton’s second law of motion. (b) Give reason for the following: (i) An air filled balloon rise up slightly when punctured from below? Get SOLVED & UNSOLVED question papers, updated Syllabus, Sample papers and study material and much more…


(ii) A swimmer push water backward with his hands, in order to swim in forward direction? Answer. (a) The law of conservation of momentum states that for two objects colliding in an isolated system (external forces are absent), the total momentum before and after the collision is equal. This is because the momentum lost by one object is equal to the momentum gained by the other. Derivation of Law of Conservation of Momentum from Newton’s second law of motion: Let p1 and p2 represent the sum of momenta of a group of objects before and after the collision, respectively. Let t be the time elapsed during the collision. p  p1 According to Newton’s secon F  2 d law of motion, t External force = Rate of change of momentum p  p1 Or F 2 t If there is no external force, F=0 p2  p1 ⟹ 0 t Hence, in the absence of an external force, the total momentum of a group of objects remains unchanged or conserved during collision. This is the Law of Conservation of Momentum. (b) (i) When the air at the point of puncture moves out with a certain momentum in the downward direction, in order to conserve momentum the balloon moves with the same momentum in opposite direction. Thus, balloon rise up slightly, before falling down. (ii) When the fuel burns in the ignition chamber of a rocket, the hot gases pass out from its exhaust with a certain momentum in backward direction. Thus, in order to conserve momentum the rocket moves with the same momentum in forward direction. Section B Question 25. In which of the following situations, waves propagate well? (a) Loaded slinky (b) unloaded slinky (c) equally in (a) and (b) Get SOLVED & UNSOLVED question papers, updated Syllabus, Sample papers and study material and much more…


(d) None of these. Answer. (a) Loaded slinky Question 26. A pulse was created in a string/slinky of length 4m by a group of 4 students. They observed that it returned, after reslection, at the point of creation 6 times in 10 seconds and calculated the speed as follows: Student Speed in m/s

A 0.4

B 2.4

C 4.8

D 9.6

The correct speed was calculated by the student (a) A (b) B (c) C (d) D Answer. (c) We know that speed is given as

Distance Time

Here, distance = Length of string = 4m And Time = Time taken by the pulse to travel one to and fro movement. Here time taken for 6 to and fro movements = 10 s ⟹ Time taken for one one to and fro movement =  Speed =

10 s 6

4 24   2.4 m/s 10 / 6 10

Question 27. Observe the following figure of a pendulum:



At which position is the kinetic energy of pendulum the maximum? (a) At position O (b) At position A (c) At position B (d) Both at A and B Answer. (a) At position O The bob of a pendulum has maximum kinetic energy at the mean position. The total energy at any point remains constant. At the lowest point, Potential energy is zero. So the entire energy is in form of K.E. Hence K.E. is maximum at the lowest or mean point. Question 28. In an experiment to find the focal length of a concave mirror, a graph is drawn between v and u. Choose the correct graph from the graphs given below?



(a) A (b) B (c) C (d) D Answer. (C) Graph between u and v is obtained as a rectangular hyperbola. Question 29. In which of the following conditions, the distance between the molecules of hydrogen gas would increase? (i) Increasing pressure on hydrogen contained in a closed container (ii) Some hydrogen gas leaking out of the container (iii) Increasing the volume of the container of hydrogen gas (iv) Adding more hydrogen gas to the container without increasing the volume of the container (a) (i) and (iii) Get SOLVED & UNSOLVED question papers, updated Syllabus, Sample papers and study material and much more…


(b) (i) and (iv) (c) (ii) and (iii) (d) (ii) and (iv) Answer. (c) (ii) and (iii) In both the cases due to the availability of larger space molecules of hydrogen gas get scattered all around to create larger intermolecular spaces. Question 30. Making anti-viral drugs is more difficult than making anti-bacterial medicines because (a) virus make use of host-machinery (b) viruses are on the border line of living and non-living (c) viruses have very few biochemical mechanisms of their own (d) viruses have a protein coat Answer. (c) viruses have very few biochemical mechanisms of their own. Question 31. Find out the false statement (a) Golgi apparatus is involved with the formation of lysosomes (b) Nucleus, mitochondria and plastid have DNA, hence they are able to make their own structural proteins (c) Mitochondria is said to be the power house of the cell as ATP is generated in them (d) Cytoplasm is called as protoplasm Answer. (d) Cytoplasm is called as protoplasm Cytoplasm is the jelly like substance that occupies the space between the cell membrane and the nucleus. Whereas, the cytoplasm surrounded by the cell membrane and enclosing the nucleus together constitute the protoplasm. Question 32. Given below is a diagram showing the structure of a neuron tissue.



Choose the correct labeling for the parts A, B, C, D and E. (a) A – Nucleus; B – Cell body; C – Dendrite; D – Axon; E – Nerve ending. (b) A – Nucleus; B – Dendrite; C – Cell body; D – Nerve ending; E – Axon. (c) A – Nucleus; B – Axon; C – Cell body; D – Dendrite; E –Nerve ending. (d) A – Nucleus; B – Dendrite; C – Cell body; D – Axon; E – Nerve ending Answer. (d) A – Nucleus; B – Dendrite; C – Cell body; D – Axon; E – Nerve ending Question 33. An object is put in three liquids having different densities, one by one. The object floats with 1/9, 2/11 and 3/7 parts of its volume outside the surface of liquids of densities d1, d2 and d3 respectively. Which of the following is the correct order of the densities of three liquids? a) d1 > d2 > d3

b) d2 > d3 > d1

c) d1 < d2 < d3

d) d3 > d2 > d1

Answer. Here, volumes of object outside the liquids of densities d1, d2, d3 respectively, are given as: 1 2 3  0.11;  0.18;  0.43 9 11 7 This means buoyant face is maximum in liquid with density d3 and minimum in liquid with density d2. As buoyant force exerted by a liquid is directly proportional to its density, Therefore, correct order of the densities of three liquids is: d1 < d2


Answer. Thallophyte has a simple plant body structure with no distinct root system, stem and leaves. Such simple body is called thallus. Thallophytes have unicellular gametes. After fertilisation, the zygote does not form an embryo. Therefore, these plants are called non-embryonic plants. Question 35. A student recorded the mass of dry raisins as 2.0 g and mass of raisins after soaking as 3.5 g. Calculate the percentage of water absorbed by raisins. Write one precaution for this experiment. Answer. Mass of dry raisins = 2.0 g Mass of raisins after soaking = 3.5 g Mass of water absorbed = 3.5 – 2.0 = 1.5 g Percentage of water absorbed by raisins =

1.5 100  75% 2.0

Precaution: Raisins should be well immersed in water instead of lying not on surface. Question 36. Write any two charateristics of the animals belonging to phylum chrodata. Answer. Two charateristics of the animals belonging to phylum chrodata are: (i) They are bilaterally symmetrical,triploblastic and have body cavity. (ii)they exhibit notochord, a flexible rod between the nerve cord and the digestive track, at some stages during development.


Cbse Class 9 Science Solved Practice Paper Set i

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