CBSE Class 9 Science Sample paper 2019

SECTION – A

Question 1:
If on a round trip you travel 6 km and then arrive back home, calculate your displacement after completing the trip.
Answer:
Displacement is zero.
Question 2:
The atomic number of three elements A, B and C are 9,10 and 13 respectively. Which of them will form a cation ?
Answer:
The element with atomic number 13 will form a cation. It has the electronic configuration 2, 8, 3. It loses three electrons to attain stable octet and forms the cation.

Question 3:
(i) What is the similarity in the electronic structure of the following set of atoms :
(a) Lithium
(b)Sodium
(c) Potassium
(ii) Which of the above elements is most reactive and why ?
Answer:
(i) Electronic configuration of the elements are given as under:

Question 4:
What are Phanerogams ? Point out differences between its two groups. Ans. Phanerogams are flowering plants. Its two groups are Gymnosperms and Angiosperms:
Answer:

Gymnosperms	Angiosperms
(i) They produce naked seeds. (ii) Considered more primitive in evolutionaiy term.	(i)   They produce well protected seeds, inside a fruit. (ii)   Considered most highly evolved in evolutionary terms.

Question 5:
Why are antibiotics effective against bacteria but not against viruses ?
Answer:
Antibiotics act by preventing one of the bio-chemical events towards formation of a new pathogen cell. Thus, most antibiotics prevent cell wall formation, which happens in bacterial cell only, thereby preventing bacterial diseases.
Question 6:
(a) Write four phenomena which were successfully explained using universal law of gravitation.
(b) The gravitational force between two objects is 100 N. How should the distance between the objects be changed so that force between them becomes 50 N ?
Answer:
(a) The universal law of gravitation successfully explained several phenomena. Some such phenomena are:
(i) the force which binds us to the earth
(ii) motion of moon around the earth
(iii) motion of planets around the sun
(iv) occurrence of sea tides due to attraction of moon and of sun.
(b) Let distance between the two objects be changed from r to r’ so that gravitational force between them changes from F = 100 N to F’ = 50 N.
Then as per inverse square law, we have

Question 7:
Give reason for the following :
(a) The sound of a thunder is heard a little later than the flash of light is seen,
(b) The ceilings of the concert hall are curved.
(c) Some animals get disturbed before the earthquakes.
Answer:
(a) In lightning process, flash and thunder are produced simultaneously. Flash is seen almost immediately because speed of light is extraordinarily large (c = 3 × 10⁸ m s^-1). But thunder is heard a few seconds later because speed of sound is much less (about 346 m s^-1 at 25 °C) and requires time to cover up the distance from the site of thunder in sky to us.
(b) The ceilings of concert halls and conference halls are generally curved as shown in figure. It is done to ensure that the sound after reflection from the ceilings reaches all corners of the hall.
(c) Before earthquakes generally infrasonic shock waves are produced. Some animals, who are sensitive to infrasonic waves, get disturbed due to them.

Question 8:
The velocity-time graph of an object is shown in the figure.
(a) State the kind of motion that object has from AtoB and from B to C.
(b) Identify the part of graph where the object has zero acceleration. Give reason for your answer.
(c) Identify the part of graph where the object has negative acceleration. Give reason for your answer.
Answer:
(a) Region AB of graph shows uniform motion but the region BC shows non-uniform motion.
(b) In region AB of the graph the object has zero acceleration because velocity in this
region is constant at 40 m s^-1.
(c) In region BC the object has negative acceleration because the velocity is falling from 40 m s^-1 at point B (time = 25 s) to zero at point C (time = 45 s).
Question 9:
(a) Suggest two ways to decrease pressure on a surface.
(b) Density of an object is 1.8 g/cm³. Express it in kg/m³.
Answer:
(a) Two suggested ways so as to decrease pressure on a surface are : decrease the force acting on the surface, and increase the surface area of given surface.
Density of an object in CGS system is d – 1.8 g/cm³ 1 g/cm³ = 10³ kg/m³, hence density in SI system will be d = 1.8 x 10³ kg/m³

Question 10:
(a) Define atomicity.
(b) State the atomicity of the following molecules :
(i) Oxygen
(ii) Phosphorous
(iii)Sulphur
(iv) Argon

Answer:
(a) The number of atoms in one molecule of the substance is called atomicity. Atomicity of the substances is given as under:

OR
State three points of difference between anion and cation.
Answer:

Anion	Cation.
It is negatively charged. Anion is bigger than the atom from which it is formed. It is obtained by adding electrons to the atom. Non-metals generally form the anions.	It is positively charged. It is smaller than the atom from which it is formed It is obtained by removing electrons from the atom. Metals generally form the cations

Question 11:
An element X forms the following compounds with hydrogen, carbon and phosphorous : P₂X₃, P₂X₅, H₂X₂, H₂X, CX₂, CX. Find the valencies of X and other element present in the compound.
Answer:

Question 12:
(a) How are CFCs harmful for the environment and living beings ?
(b) Mention any two human activities which would lead to an increase in the carbon dioxide content of air.
Answer:
(a) CFCs result in damaging the ozone layer. Depletion of ozone layer in upper atmosphere of earth results in more UV radiations reaching the earth’s atmosphere. This results in global warming.
More UV radiations indicate more high energy radiations that can mutate our DNA. This has resulted in increased number of skin cancer.
(b) CO₂ content can increase due to :
(i) burning of fossil fuel.
(ii) burning of dead leaves and other plant material.

Question 13:
(i) Write the definition of health as given by WHO.
(ii) List two factors that affect health.
Answer:
(i) According to WHO, health is a state of complete physical, mental and social well being, and not merely an absence of disease or infirmity.
(ii) Factors that affect health are –
(a) Clean physical environment – as it would result in pathogen-free environment.
(b) Malnourishment leads to inability to fight diseases or results in deficiency diseases.

OR
Explain the basis of Principle of Treatment for any disease.

Answer:
There are two ways to treat an infectious disease. They are as follows :
(i) Reduce the effect of disease : The principle involved is that the effect of disease is lessened without killing the infectious agent which is done either by taking medicines and antibiotics as well as taking appropriate rest so that the body heals.
(ii) Kill the microorganisms of infectious agents : The infectious agents like bacteria, viruses, fungi, helminths and protozoan’s have some essential biochemical life processes which are peculiar to this group. These processes may be pathways for respiration or synthesis of new substances. Drugs are available which block these processes and kill the infectious agents. Antibiotics are chemicals produced by microbes which kill or prevent the growth of other microbe by blocking life processes without harming human cells.

Question 14:
Differentiate between xylem and phloem tissues.
Answer:

Xylem	Phloem
It conducts water and inorganic solutes in vascular plants. Conduction usually unidirectional i.e.,  from roots upwards. Conducting channels arc tracheids and vessels, Tracheids, vessels, fibres and parenchyma are its components of which only parenchyma is living. It also provides mechanical strength.	It conducts organic solutes in vascular plants. Conduction in all directions Le.t from leaves to other plant parts. Conducting channels are sieve tubes. Sieve tubes, companion cells, fibres and parenchyma are its components of which only fibres are dead, It does not provide any mechanical strength.

Question 15:
Ravi is a poor labourer. For last few days he used to feel pain in his lower abdomen as well as in his back. When he went to a doctor, the doctor prescribed some clinical tests as well as ultrasonography of lower abdomen. When Rita, who lived nearby came to know of this, she decided to give financial help to Ravi. After tests and ultrasonography it was found that there was a stone in the left kidney of Ravi. Rita asked her neighbours to contribute for the operation. All the neighbours contributed for the operation as well as subsequent medical expenses. Now Ravi is absolutely fit.
(i) What is ultrasound ? Explain its working principle.
(ii) What values are shown by Rita and her neighbours ?
Answer:
(i) Ultrasound is a diagnostic technique, used to get images of internal organs. It is based on the principle called piezoelectric (pressure electricity) effect. The transducer probe is the main part of the ultrasound machine which makes sound waves and receives reflected echoes from the patient’s tissues.
(ii) Values shown by Rita and her neighbours are :
(a) Comradaric
(b) Concern
(c) Generosity
(d) Compassion.

Question 16:
(a) How would you arrive at a mathematical formula to measure force using second law of motion ? Define the unit of force using this formula.
(b) A bullet of mass 10 g travelling horizontally with a velocity of 150 ms^-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer:
(a) Newton’s second law of motion states that the rate of change of momentum of an object is directly proportional to the external unbalanced force applied on it and takes place in the direction of applied force.
Let an object of mass m was initially moving with a velocity u. It means that its initial
momentum
p1 = mu.
If on applying a constant force F for time t the velocity of the object changes to v then final momentum of given object p₂ = mv.
Change in momentum = P₂ – P₁ = mv – mu = m(v – u)
Rate of change of momentum = ,
where a = = acceleration of the given object
According to statement of second law of motion, the rate of change of momentum is directly proportional to the applied force. Hence, we have
F ∝ or F ∝ ma
F = kma, where k is a constant of proportionality.
Generally the units of force are so selected that if m = 1 and a = 1, then force F = 1. In such a case value of constant k becomes 1 and the relation becomes :
F = ma
The SI unit of force is a newton (1 N), which is the force which produces an acceleration of 1 m s^-2 when applied on an object of mass 1 kg.

and the force exerted by the wooden block on the bullet F = ma = (0.01) × (- 5000) = – 50 N
The -ve sign of force shows that the force is a retarding force or it is opposing the motion.

Question 17:
(a) Define work. Give SI unit of work done. Write an expression for positive work done.
(b) Calculate the work done in pushing a cart 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 ?
Answer:
(a) Work done on an object is defined as the magnitude of the force acting on the object multiplied by the distance moved by the object in the direction of the applied force.
∴ Work (W) = Constant force applied (F) × Displacement along the direction of force (s).
SI unit of work is called joule (J or N m). Work is said to be 1 joule if under the influence of a force of 1 N the object moves through a distance of 1 m along the direction of applied force.
(b) Force applied in forward direction so as to overcome force of friction = F = 250 N and displacement in the direction of force s = 50 m
∴ Work done W = Fs = 250 N × 50 m = 12500 J
The work done is positive.
(c) If displacement is perpendicular to the direction of force the work done is zero.
(d) Work done on an object is zero when it moves on a circular path because displacement is perpendicular to the direction of force.
OR
(a) Define potential energy of an object. Give an expression for gravitational potential energy.
(b) A car of mass 2000 kg is lifted up a distance of 30 m by a crane in 2 minutes. What is the power supplied by the crane ? (g = 9.8 m s^-2)
Answer:
(a) The potential energy of an object is the energy present in it by virtue of its position or configuration or change thereoff.
If an object of mass m is situated at a height h from a reference ground level, then its gravitational potential energy is
Ep = mgh
(b) Here mass of car m = 2000 kg, vertical distance covered by car s = h = 30 m and time t = 2
minutes = 2 x 60 s = 120 s
∴ Total work done by crane W = mgh
∴ Power supplied by the crane t = = = = 4900 W or 4.9 kW

Question 18:
Wet clothes dry up similarly when we spill water on the floor it dries up after sometime. In both the cdses change of state from liquid to vapour takes place without reaching the boiling point.
(i) What is this phenomenon called ?
(ii) Explain how the change occurs at temperatures lower than the boiling point.
(iii) Mention three factors which determine the rate at which the change of state from water to vapours occurs at room temperature.
Answer:
(i) This phenomenon is called evaporation.
(ii) Molecules of water which are moving at speed higher than the average speed leave the surface and go into the atmosphere. This is how evaporation takes place.
(iii) Factors that determine the rate of evaporation :
(a) Surface Area – Greater the surface area, greater will be the rate of evaporation.
(b) Temperature – Higher the temperature, greater will be the rate of evaporation.

Question 19:
Rahul’s mother mixed oil and water in kitchen by mistake. Rahul told her that he can separate the mixture. Name the technique used by Rahul and explain how he will do. Draw the diagram and write the principle of this technique.
Answer:
The technique used by Rahul to separate the mixture of oil and water was to use separating funnel as shown in figure.
Kerosene oil does not mix with water and forms a separate layer. Oil being lighter than water, forms the upper layer. Steps involved in the separation of kerosene oil from water are:
(i) Pour the mixture of kerosene oil and water in the separating funnel.
Let it stand undisturbed for sometime so that separate
layers of oil and water are formed.
(ii) Open the stopcock of the separating funnel and pour out the lower layer of water carefully.
(iii) Close the stopcock of the separating funnel as the oil reaches the stopcock.
(iv) Transfer oil from the neck of the separating funnel to a separate vessel.

Question 20:
The immune system of Hari is damaged by the attack of pathogen on his body.
(i) Name the disease he is suffering from.
(ii) Name the pathogen that has attacked his body.
(iii) Mention any three modes of transmission of this disease.
Answer:
(i) Hari is suffering from AIDS – Acquired Immuno – Deficiency Syndrome.
(ii) The pathogen that has caused AIDS is HIV – Human Immuno Deficiency Virus.
(iii) This virus is transmitted through following ways :
(a) Sexual contact with an infected person carrying AIDS virus.
(b) Transfusion of blood infected with HIV.
(c) Use of unsterilized needles, blades or razors.
(d) Transplacental transmission i.e., AIDS infected mother to the foetus developing in her womb.
Question 21:
Answer the following –
(i) Draw a labelled diagram of smooth muscle.
(ii) Differentiate between parenchyma and sclerenchyma.
(iii) Name the epithelium which has hair-like projection on outer surface of the cells.
(iv) Name a tissue that stores fat in the body.
Answer:
(i)

(ii) Distinction between parenchyma and sclerenchyma on the basis of their cells wall:

Parenchyma	Sclerenchyma
(a) A simple permanent plant tissue with thin walls. Cells are loosely packed with large intercellular spaces and form the packing tissue of all plant organs. (b) It is primarily involved in storage of food material. It also facilitates movement of gases. Parenchyma cells remain turgid and provide rigidity to softer parts.	(а) Sclerenchyma is the chief mechanical tissue of plants which comprises of highly thick walled cells with narrow lumen. The walls are thick on all sides due to presence of lignin. (b) It provides strength to the plants and enables them to bear various stresses. It also forms a protective covering around seeds and nuts.

SECTION – B

Question 22:
While determining the density of the material of a metallic sphere using a spring balance and measuring cylinder, a student noted the following readings :
(1) Mass of the sphere = 81 g.
(2) (i) Reading of water level in the cylinder without sphere = 54 mL.
(ii) Reading of water level in the cylinder with sphere = 63 mL.
On the basis of these observations calculate the density of the material of the sphere.
Answer:
Here mass of sphere M = 81 g and volume of sphere V = (63 – 54) mL = 9 mL
∴Density of the material of the sphere
Question 23:
What happens in your experiment of verification of the laws of reflection of sound if pipe through which sound is heard is lifted vertically through a small distance ?
Answer:
If the pipe, through which sound is heard in our experiment, is slightly lifted vertically then the reflected sound beam does not remain in same plane as of incident sound beam and hence either a weakened sound of table clock is heard or no sound is heard at all.
Question 24:
Ramesh prepared the solution of chalk powder in water and Mahesh prepared the solution of milk in water. What type of solutions were prepared by them and whether the solutions are homogeneous or heterogeneous ?
Answer:
Solution prepared by Ramesh is suspension.
Solution prepared by Mahesh is colloidal solution.
Both the solutions are heterogeneous.
Question 25:
A horse magnet was moved over a mixture of iron filings and sulphur as shown in the figure.
(a) Tell whether the mixture of the two substances is homogeneous or heterogeneous.
(b) Which substance will be left on the watch glass after the experiment ?

Answer:
(a) The mixture of iron filings and sulphur is heterogeneous.
(b) Sulphur will be left on the watch glass after the experiment.
Question 26:
State any two specific features of Earthworm.
Answer:
Earthworm belongs to Phylum Annelida.
Its specific features are :
(i) Body is segmented i.e., shows metamerism.
(ii) Presence of clitellum.
Question 27:
Name two features which you would examine to categorise a plant into dicot or monocot.
Answer:
(i) Leaf : Are broad, with reticulate venation in dicots, and blade-like with parallel venation in monocots.
(ii) Flower: Flowers are pentamerous in dicots and trimerous in monocots