CBSE Class 9 Science Exam 2020: Important Long Answer Type Questions with Solutions

Important long answer type questions are provided here to prepare for CBSE Class 9 Science Exam 2020. Practicing these thoroughly solved questions will help you fine-tune your preparation for the final exam.

Nov 19, 2019 15:39 IST
CBSE Class 9 Science Exam 2018

In CBSE Class 9 Science Exam, around 40 percent marks will be kept for long answer type questions. Section-C of Class 9 Science question paper will have six long answer type questions with each carrying 5 marks. Students will have to answer these questions in about 80-90 words each. So, if you want to score good marks in your class 9 Science exam, then it’s important to get acquainted with the topics from which long answers type questions are frequently asked.

We are providing here a set of important long answer type questions along with their detailed answers. The questions are available free of cost which students can practice to enhance their chances of securing high marks in CBSE Annual Exam 2020.

Given below are some very important long answer type questions for CBSE class 9 Science exam:

Question 1.  (a) Which method will you use to separate : (i) components of air (ii) acetone and water? Explain each.

(b) What is the difference in the apparatus used for distillation and fractional distillation?

(a) (i) Separation of constituents of air by fractional distillation:

Air consists mainly of nitrogen and oxygen. Carbon dioxide and other gases are also present in trace amounts. To separate these different gases first we need to liquify the air. When the air is compressed and cooled down, at 0°C, water present in air freezes to ice. On further cooling, carbon dioxide separated as dry ice at about -78°C. on further cooling under pressure, air gets liquefied at -200°C. After this, through fractional distillation each gas can be separated on the basis of their boiling points. The gas having lower boiling point will be obtained first and the gas having highest boiling points will be obtained at last.

Nitrogen has a boiling point of -196°C while oxygen has -183°C. The nitrogen gas will start to escape through the outlet and it is collected. The liquid oxygen will be collected in the fractionating column.

(ii) Separation of acetone and water:

It can be done by simple distillation method. In this method, the components of mixture are separated on the basis of difference in boiling points of constituent liquids. First, the lower boiling point liquid will start boiling and its vapours can be collected and condensed to obtain the liquid. In this case acetone will come out first and left over will be water only.

(b) Fractional distillation uses a complex apparatus with a fractionating column which is long tube packed with metal wire, metal ribbon or glass beads that give extra condensing surfaces, allowing the liquid to vaporize and condense at every minute change reduce in temperature.

Simple Distillation includes a simple apparatus with a flask to contain the mixture, a condenser and a flask to collect purified components.

Question 2. (a) List any four properties of a colloid and mention any two properties in which colloids differ from suspension.

(b) State what is Tyndall effect ? Which of the following solutions will show Tyndall effect ? Starch solution, sodium chloride solution, Tincture iodine, air

(a) Properties of a colloid (any four)

(i) A colloid is a heterogeneous mixture.

(ii) The size of particles of a colloid is too small to be individually seen by naked eyes.

(iii) Colloids are big enough to scatter a beam of light passing through it and make its path visible.

(iv) They do not settle down when left undisturbed, that is, a colloid is quite stable.

(v) They cannot be separated from the mixture by the process of filtration. But, a special technique of separation known as centrifugation can be used to separate the colloidal particles.

Two properties in which colloids differ from suspension are (ii) and (iii) as the particles of a suspension are large enough to be observed with a naked eye and these particles settle down well when the mixture is left undisturbed.

(b) Tyndall effect is the phenomenon of scattering of a beam of light by the particles of a colloid.

Starch solution and tincture iodine are colloid and thus will show Tyndall effect.

NCERT Solutions for Class 9 Science: All Chapters

Question 3. Give reasons for the following:

(a) It is difficult to balance our body, when we accidentally step on a peel of a banana.

(b) Pieces of bursting cracker fall in all possible directions.

(c) A glass pane of a window is shattered, when a flying pebble hits it.

(d) It is easier to stop a tennis ball than a cricket ball moving at the same speed.

(e) Javelin thrower is marked foul, if an athlete crosses over the line marked for the throw. Athletes often fail to stop themselves before the line.

(a) The reason is that one cannot exert action force effectively on the slippery peel of banana in the backward direction. Hence, in response, the ground does not exert sufficient reaction force in forward direction and hence we lose our balance.

(b) This can be explained by the law of conservation of momentum. When due to explosion some pieces move in the same particular direction, then in order to conserve momentum, the remaining pieces move in the opposite direction.

(c) The reason is that, the glass pane of the window is a hard solid. The flying pebble suffers change in momentum in a very short time, so the force exerted by the glass window on the pebble will be large. Consequently the glass pane of the window will shatter.

(d) The reason is that mass of the cricket ball is more than that of a tennis ball. Thus, momentum is more in case of the cricket ball due to larger mass as compared to the tennis ball. So, less force has to be applied in the case of the tennis ball to stop.

(e) The reason is that when athletes are running for the throw then due to inertia of motion they often fail to stop themselves before the line.

Question 4. (a) What is transformation of energy? Explain with any two suitable examples.

(b) What must be the velocity of a moving body of mass 2kg so that its KE is 25J?

(c) Represent graphically a constant force acting on a body producing a displacement along the direction of motion on a force-displacement graph. What is the significance of force-displacement graph?

(a) The conversion of energy from one form to another is called transformation of energy.

For example:

(i) When we throw a ball, muscular energy which is stored in our body, gets converted into kinetic energy of ball.

(ii) The wound spring in the toy car possesses potential energy. As the spring is released, its potential energy changes into kinetic energy due to which, toy car moves.

(b) Given, mass of the body, m = 2 kg

Kinetic energy, KE = 25 J

We know, KE = &frac12; mv2

⟹ &frac12; (2)(v)2 = 25

⟹ v = 5 m/s

(c) Graphical representation of force( f) applied along the direction of displacement (s), is shown below:

From the graph,

Area of rectangle = f.S = Work done

Significance of force-displacement graph is that, area under F–S curve gives the work done.

Question 5. (a) What is immunisation ?

(b) Define immunity and vaccination.

(c) Define vaccine.

(d) What type of diseases can be prevented through vaccination?

(a) Immunisation: It is a process of inoculation (injecting) of substance (vaccine) into a healthy person in order to develop immunity against the disease.

(b) Immunity: Immunity is the ability of a body to recognise, destroy and eliminate external disease-causing agents.

Vaccination: Vaccination is the administration of antigenic material (a vaccine) to stimulate an individual's immune system to develop immunity to a pathogen.

(c) Vaccine: A vaccine is a solution containing the disease-causing organisms in a diluted or weakened form. It may have organisms in living or even dead form.This does not actually cause the disease but this would prevent any subsequent exposure to the infecting microbe from turning into actual disease.

(d) Protection against diseases like smallpox, rabies, polio, diphtheria, chickenpox and hepatitis is provided through vaccination. It has been possible to eradicate smallpox from all regions of the world through a massive vaccination programme.

Question 6. (a) Define

(i) Vector

(ii) Carrier

(b) What are the modes of transmission of AIDS?

(a)  (i) Vector: It is an organism that does not cause disease itself but spreads infection by conveying pathogens from one infected person to the other acting as host. For example; mosquito with malarial parasite.

(ii) A carrier is an organism who has no symptoms of a disease but harbors the pathogen in its body and passes it on to someone else. For Example: human with HIV virus but not AIDS; or human with pathogen Salmonella typhi but not Typhoid.

(b) Modes of transmission of virus of AIDS disease are given below.

(i) By sexual intercourse.

(ii) By use of infected syringes.

(iii) Through blood

(iv) From mother to child through birth or by breast feeding

CBSE Class 9 Science: Chapterwise important topics and questions

Question 7. (a) (i) Why denitrification of nitrogenous compounds is necessary?

(ii) What is the special structure found in legumes and what is its function

(b) What is biomass?

(c) What is humification? why it is important?

(a) (i) Denitrification of nitrogenous compounds is necessary so thatnitrogen cycle does not stop and free nitrogen remainsavailable to living organisms.

(ii) Root nodules are found in the roots of legumes that containnitrogen fixing bacteria which help in fixing nitrogen from thesurroundings

(b) The amount of living matter in each trophic level is called biomass.

(c) Addition of incomplete decomposed organic matter, the humus into the soil is known as Humification.

It increases the fertility of soil as it enriches soil with organic matter.

Question 8. (a) Oxygen has three isotopes of atomic masses 16, 17 and 18 respectively. Explain the following:

(i) They have same chemical properties.

(ii) They are all electrically neutral.

(b) Name the isotopes ot hydrogen.

(c) Give one point of similarity and one point of difference between isotopes 146C and 126C?

(a) (i) Mass numbersof isotopes of oxygen are different, i.e.,16, 17 and 18 but they have same atomic number 8.Thus, the number of electrons is same.Therefore, they have same the electronic configuration 2, 6 and same number of valence electrons.So,their chemical properties are same.

(ii) They are electrically neutral because number of negatively charged electrons is same to the number of positively charged protons.

(b) There are three isotopes of hydrogen-(i) Protium (A=1) (ii) Deuterium (A=2) (iii) Tritium( A=3)

(c) One point of similarity between isotopes 146C and 126C: They have the same number of protons.

One point of difference between isotopes 146C and 126C: They have different number of neutrons.

Question 9. (a) Write the rules followed for filling the electrons in various energy shells of any atom, as proposed by Bohr and Bury.

(b) Write the electronic configuration of an atom of sulphur. Also draw a schematic diagram of its atom showing the distribution of electrons in its shells.

(c) State the law of constant Proportion ith the help of an example.

(a) The rules followed for filling the various electrons in various energy shells of any atom, as proposed by Bohr and Bury are :

(i) The maximum number of electrons present in a shell is given by the formula 2n2, where ‘ri is the orbit number or energy level index.

(ii) The maximum number of electrons that can be accommodated in the outermost orbit is 8.

(iii) Electrons are not accommodated in a given shell, unless the inner shells are filled. That is, the shells are filled in a stepwise manner.

(b) Atomic number of sulphur(S) is = 16

Therefore, electronic configuration of sulphur = 2,8,6

Following diagram shows the distribution of electrons in various shells of sulphur:

(c) According to the law of constant proportion , a chemical substance always contains the same elements in a fixed proportion by mass, irrespective of its source.

For example: Pure water obtained from any source will always contain two hydrogen atoms and one oxygen atom. Hydrogen and oxygen respectively combine together in the ratio of 1:8 by mass to form water. The ratio by the number of atoms for water will always be H : O = 2 : 1. Thus, 18 g of water contains 2 g of hydrogen and 16 g of oxygen.

Question 10. (a) Why are angiosperms so called ? In which structures do the seeds develop ? How are angiosperms different from gymnosperms ?

(b) Give an appropriate term for each of the following:

(i) Complex sugar that makes the fungal cell wall.

(ii) Plants which bear naked seeds.

(iii) Basic unit of classification.

(iv) Group of unicellular eukaryotic organism.

(a) Angiosperms are so called because these plants have covered seeds.
Seeds develop within ovary which later modify into fruit.

Angiosperms can be differentiated from gymnosperms as follows:

 S. No. Angiosperms Gymnosperms 1. They are flowering plants. They are non-flowering plants. 2. The plants of this group contains seeds which develop inside an organ which is further modified to become a fruit . The plants of this group bear naked seeds. 3. The plants of this group may be annual, evergreen and woody. The plants of this group ane usually perennial, woody or non-woody.

(b)

(i) Complex sugar that makes the fungal cell wall - Chitin

(ii) Plants which bear naked seeds - Gymnosperms

(iii) Basic unit of classification - Cell structure

(iv) Group of unicellular eukaryotic organism - Protista.

CBSE Class 9 Science Practice Papers for Annual Exam 2020

Question 11. (a) Write any four characteristics exhibited by a pure substance?

(b) What happen to sugar when it is dissolved in water? Where does the sugar go? What information do you get about the nature of matter from the this solution of sugar in water.

(a) Four characteristics exhibited by a pure substance are:

(i) A pure substance contains only one kind of atoms or molecules.

(ii) It is perfectly homogenous

(iii) It has definite composition which does not vary with time.

(iv) It has definite melting point, boiling point, density etc.

(b) When sugar is dissolved in water, the particles of sugar gets settled in between the spaces of the particles of water and fills that space. From this process we conclude that the particles of water have space between them and that is why the sugar particles gets into those spaces resulting in no change in the water level.

Question 12. 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.

(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 coconut 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 13. (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?

(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 14. (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?

(ii) A swimmer push water backward with his hands, in order to swim in forward direction?

(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.

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.

CBSE Class 9 Science Syllabus 2019-2020

Question 15. (a) What are the conditions required by an object to float on water?

(b) Give reason for the following:

(i) A ship made of iron floats while a piece or nail sinks.

(ii) Solid fom of water (ice) floats on water.

(a) For an object to float on water, mass of the object must be equal to the mass of liquid displaced by it. Or the overall density of the object must be less than or equal to the density of liquid in which it has to float.

(b) (i) A ship though made of iron has  a lot of empty space. Due to which the overall density of the given volume of  the boat remains lower than the density of the same volume of water. Now, since an object with density lower than that of water floats on water therefore a boat also floats on water.
on the other hand, there are no such air spaces in a nail. It is a complete soild. Therefore, its average density is greater than that of water, hence it sinks down when immersed in water.

Question 16. (a) Defne work. Give SI unit of work. Write an expression for positive work done.

(b) Calculate the work done in pushing a box through a distance of 50 m against the force of friction equal to 250 N. Also state the type of work done.

(c) What will be the work done, if displacement of the object is perpendicular to the direction of force?

(d) When an object moves on a circular path, what will be the work done?

(a) Work is the energy required to move an object against a force. The SI unit for work is joule. Positive work is equal to the force times the distance the object moves, i.e.,

Work = Force × Displacement

(b) Here, Distance moved by box = 50 m

Force acting on box = 250 N

Thus work done on box = force × displacement

= 250×50= 12500 J

Since the box moves against the force of friction, thus work done on box is negative.

(c) When the force and displacement are perpendicular to each other, then work done is equal to zero. As work done = Force x cos θ x Displacement

Where θ is the angle between force and displacement

Since here displacement of the object is perpendicular to the direction of force.

&there4; θ = 90

⟹ cos θ = cos 90 = 0

⟹ Work = f x 0 x s = 0

(d) The work done on a body moving in a circular path is zero. This is because, when a body moves in a circular path, then the centripetal force acts along the radius of the circle, and it is at right angles to the motion of the body.

Question 17. (a) What happens to sugar when it is dissolved in water? What information do you get about the nature of matter from the dissolution of sugar in water?

(b) Why does diffusion occur more quickly in a gas than a liquid?

(c) What is Brownian motion? Write the characteristics of  matter is demonstrated by Brownian motion?

(a) When we dissolve sugar into water, the particles of sugar occupy the spaces between the molecules of water and eventually all the sugar disappears.

We can conclude from this behavior of sugar in water that, a sugar crystal is comprised of millions of tiny particles which keep on dividing themselves into smaller particles. We can also draw a conclusion here that liquids are made of particles which can move freely and have space between them.

(b) The speed of particles in gas is higher as compared to that of liquids. Also the space between the particles in gases is also large as compared with that of liquids, as a consequence the particles in gases have higher kinetic energy and larger area to occupy and hence the diffusion in gases is faster as compared to that of liquids.

(c) The random motion of particles in liquid or gases is known as Brownian motion. Brownian motion demonstrates that particles of any have kinetic  energy which is higher in gases as compared to liquids and due to larger space between the particle of gases, there is higher rate of random motion occurring in gases as compared to liquids. Due to this random motion, the particles keep colliding amongst each other. This Brownian motion aids in mixing of two substances as well.

Question 18. (a) Explain the following statements giving appropriate reasons:

(i) Fertile soil has a lot of humus.

(ii) Lichens help in the formation of soil.

(iii) During summer, walking around a lake or a river, will give you relief from heat.

(b) How will you prove the dust to be a major pollutant?

(a) (i) Humus is the organic matter contained in soil, formed by the decomposition of leaves and other plant material by soil microorganisms. It provides nutrients to the plants and increases the ability of soil to retain water. Now, fertile soil contains lots of microorganisms, which decompose the dead animal and plant material and thus add humus to the fertile soil.

(ii) Lichens are a symbiosis of  fungi and green algae living and growing on rocks where they help in the breakdown of the rocks’ surface by releasing some chemicals. These chemicals produce pores and crevices and thus help in weathering of rocks and formation of soil.

(iii) During summer days, the land near the lake or river gets heated up quickly but the water in lake or river remains cool.  The wind above the land heats up, and hot air, being lighter, rises up, and the cool air above the lake/river flows to take its place causing a cool breeze to blow from lake/ river to land hence reduces the heat in the adjoining land area.

(b) Dust is a common air pollutant generated by many different sources and activities like natural erosion of soil, Pollen, microscopic organisms, concrete crushing, cement batching and road stone plants. Smaller dust particles are light enough to be carried by air which when inhaled penetrate deeply into the lungs and even the ultra fine particles can be absorbed directly into the blood stream causing the respiratory problems like coughing, sneezing, asthma attacks, etc.

Question19. (a) What do the following symbols/formulae stand for:

(i) 2O (ii) O2 (iii) O3 (iv) H2O

(b) Give the chemical formulae of the following compounds:

(i) Calcium carbonate
(ii) Calcium chloride.

(c) Calculate the formula unit mass of Al2(SO4)3.

(Given atomic mass of Al = 27 u, S = 32 u, O = 16 u)

(a) (i) 2O = 2 atoms of the element Oxygen

(ii) O2 =  One molecule of oxygen. This is gaseous form.

(iii) O3 One molecule of ozone.

(iv) H2O= One molecule of water.

(b) (i) ‎CaCO3

(ii) CaCl2

(c) Unit mass of Al2(SO4)3 = 2 × Al + 3 (S + 4 × O)

= 2 × 27 + 3 (32 + 4 × 16)

= 54 + 3 × 96

= 54 + 288

Question 20. (a) What were the observations concluded by Rutherford in his alpha particle scattering    experiment?

(b) This experiment led to the discovery of which sub atomic particle.

(a) Rutherford concluded from the α-particle scattering experiment that:

(i) Most of the space inside the atom is empty because most of the α-particles passed through the gold foil without getting deflected.

(ii) Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space.

(iii) A very small fraction of α-particles were deflected by 180o, indicating that the entire positive charge and mass of the gold atom were concentrated in a very small volume within the atom.

(b) Rutherford, using the conclusions of his α-particle scattering experiment discovered a sub atomic particle called ‘nucleus.’

(c) The important points of information given by Rutherford about nucleus are:

(i) Nucleus is a positively charged centre in an atom and almost all the mass of an atom resides in the nucleus.

(ii) Nucleus of an atom is very dense and hard.

(iii) The size of the nucleus is very small as compared to the size of the atom as a whole.

21. (a) Define the terms and give one example of each:

(i) Bilateral symmetry

(ii) Coelom

(iii) Diploblastic

(b) Identify the group of animals with

(i) Spiny body and radial body symmetry

(ii) Four pairs of jointed legs and no wings

(a) (i) Bilateral symmetry: The type of body symmetry in which the two sides of the body are irror images of one another is called bilateral symmetry. Example: Earthworm

(ii) Coelom: Body cavity lined with an epithelium derived from the mesoderm is called coelom. Example: Spider

(iii) Diploblastic: Animals which have two germ layers: outer ectoderm and inner endoderm, in the embryo are said to be diploblastic.Example: Hydra

(b) (i) Echinodermata

(ii) Arachnida

22. (a) How does cork act as a protective tissue?

(b) Observe the figure carefully and label the parts marked A and B.

(a)

• They are arranged very closely so that there is no inter-cellular space.
• There is deposition of suberin on the walls which make them impervious to water and gases. As a result, the cork cells prevent desiccation, infection and mechanical injury to the plant body.

(b) A– Apical meristem

B– Intercalary meristem

CBSE Class 9 Science Solved Practice Paper 2019-2020

23. (a) What are the characteristics of the particles of matter?

(b) What happens when sugar is dissolved in water? Where does the sugar go? What information do you get about the nature of matter from the dissolution of sugar in water?

(a) The characteristics of particles of matter are:

• They have intermolecular space between them.
• They are continuously moving.
• They attract each other.

(b) When sugar is dissolved in water, its crystals break down into tiny particles. The sugar particles go into the spaces between the particles of water and mix with them to form sugar solution. Sugar particles occupy the space between water particles. From the dissolution of sugar in water, we can infer

(i) Matter (consisting of sugar and water) is made of small particles.

(ii) Particles of matter have paces between them.

24. A man weighs 300 N on the surface of the Earth. If he were taken to the Moon, his weight would be 50 N. Calculate the mass of this man on the Moon (g = 10 m/s2).

(b) A man hears an echo of thunder 2 seconds after lightning strikes. Calculate the distance of lightning from the man (Speed of sound in air = 330 m/s).

(a) Given, weight of man on the Earth (WEarth) = 300 N

Acceleration due to gravity (gEarth) = 10 m/s2

Now,   WEarth = massEarth × gEarth

But mass of the object is constant. It will be the same on moon.

Therefore, weight of man on moon = 30kg

(b) We know that

So, the distance between the man and the point of lightning is 660 m

(a) Why solid carbon dioxide is called dry ice? Give its uses.

(b) Why is dry ice more effective for cooling purposes than ordinary ice?

(c) Why is dry ice stored under high pressure?

(a) Solid carbon dioxide directly changes to carbon dioxide and does not melt to produce a liquid (like ordinary ice. Therefore, it is called dry ice.

Dry ice is used to deep freeze food and to keep ice cream cold.

(b) Dry ice can produce much lower temperatures as its freezing point is (–80 degree Celsius) than ordinary ice which freezes at 0 degree Celsius. So, it is more effective for cooling purposes than ordinary ice.

(c) Dry ice is stored under high pressure because if the pressure is released it would sublime back to its original form, i.e., carbon dioxide gas.

Keep practicing with the questions provided here to strengthen all important concepts so that you are able to write accurate answers in exam and obtain desired marks.

Also Check:

CBSE Class 9 Science: Important Short Answer Type Questions

CBSE Class 9 Science: Important Practical Based Questions and Answers