Solved CBSE Class 9 Science Sample paper 2019

SECTION – A
Question 1:
A ball is thrown vertically upwards. What is its momentum at the highest point ?
Answer:
Zero because velocity of ball is zero at the highest point of its motion.

Question 2:
Degree Celsius and Kelvin are two units of measuring temperature. Which of these are SI and non-SI units ?
Answer:
Degree Celsius – Non-SI.
Kelvin – SI.
Question 3:
Why is crystallisation better than evaporation for the separation of mixtures ?
Answer:
In crystallisation, we obtain the components in the form of crystals which are the purest form of a substance and impurities are left in the mother liquor. Crystals are separated from the mother liquor. This is not possible in evaporation. Some solids may decompose on heating to dryness. Such solids cannot be purified by evaporation.
Question 4:
Write names of different kinds of connective tissue :
CBSE Sample Papers for Class 9 Science Solved Set 2 4
Answer:
(a) Blood
(b) Areolar
(c) Bone
(d) Cartilage.
Question 5:
Water hyacinth floats on water. Why ? How is this tissue different from that of succulent plants ?
Answer:
Water hyacinth being aquatic plant floats on water as its parenchyma has lots of air spaces (aerenchyma) which store air and help it in remain afloat.
Whereas in succulent plants it helps in storage of water.
Question 6:
In each of the adjoining diagram, a force F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.
CBSE Sample Papers for Class 9 Science Solved Set 2 6
Answer:
On observing the diagrams carefully we find as follows:
(i) In first (extreme left side) figure, the force is acting in a direction perpendicular to displacement, Hence, work done by the force is zero.
(ii) In second figure, force and displacement both are in same direction. So, the work done by the force is positive.
(iii) In third (extreme right side) figure, the force and displacement are in mutually opposite directions, hence, the work done by the force is negative.
Question 7:
Define power. Derive its SI unit. An electric bulb is rated 10 W. What does it mean? What is the energy consumed in joules if it is used for 5 minutes?
Answer:
The rate of doing work or the rate of transfer of energy is known as the power.
∴ Power =\frac { work }{ time } \frac { w }{ t }
SI unit of power is a watt. Power is said to be 1 watt (1 W) if rate of doing work islJ s-1.
If an electric bulb is rated 10 W, it means that electric bulb supplied 10 J of energy per second.
If the bulb is used for a time t = 5 min = 5 × 60 s = 300 s, then total energy consumed by it E = P.t = 10 × 300 = 3000 J.
Question 8:
(a) What is reverberation of sound?
(b) Draw a diagram depicting low pitched sound and high pitched sound. What is the main difference between the two ?
Answer:
(a) Persistence of sound due to repeated reflection from walls and roof in an auditorium is called reverberation.
(b) The diagrams have been shown below :
CBSE Sample Papers for Class 9 Science Solved Set 2 8
High pitched sound has high frequency but low pitched sound has low frequency.
Question 9:
(a) When an athlete comes running from a distance, he is able to jump longer. Why ?
(b) The velocity-time graph of a ball of mass 20 g moving along a straight line on a long table is given in figure. How much force does the table exert on the ball to bring it to rest ?
Answer:
(a) He gains momentum which helps him in taking a longer jump.
(b) Here mass of ball m = 20 g = 0.02 kg, initial velocity of ball u = 20 m s’1, final velocity v = 0 and time
t = 10 s
CBSE Sample Papers for Class 9 Science Solved Set 2 9
OR
(a) What is the relationship between force applied on a body and the resulting acceleration caused in it?
(b) A force produces an acceleration of 5 m s’2 when applied on a body of mass 2 kg. Find the magnitude of force. How much acceleration will the same force produce when applied to a body of mass 4 kg ?
Answer:
(a) Force (F) = Mass (m) x Acceleration (a).
(b) Here mass m = 2 kg and acceleration a = 5 m s-2
∴ Force F = ma = 2 × 5=10N
Now on applying same force on an object of mass m’ = 4 kg, the acceleration produced is given by
CBSE Sample Papers for Class 9 Science Solved Set 2 9 i
Question 10:
A substance A has high compressibility and can be easily liquefied. It can take the shape of any container. Predict the nature of the substance. Enlist four properties of this state of matter.
Answer:
The substance A is a gas because gases have large empty spaces between the particles. It has high compressibility and can be easily liquefied.
Properties of gases:
(i) Gases have no fixed shape and volume.
(ii) Gases can be compressed on applying pressure to give liquids.
(iii) Particles of a gas are moving in all possible directions with all possible speeds.
(iv) Gases show the property of diffusion.
Question 11:
List four properties of non-metals. Give two examples.
Answer:
(i) They show a variety of colours.
(ii) They are poor conductors of heat and electricity.
(iii) They are not malleable or ductile.
(iv) They are not lustrous or sonorous.
Sulphur and phosphorous are examples of non-metals.
Question 12:
Describe the structure of mitochondria with special emphasis on its membrane coverings. How can we relate the structure of the membrane with the function ?
Answer:
Mitochondria are rod shaped or sausage shaped cell organelles which are known as powerhouse of the cell. They contain enzymes necessary for the oxidation of food during the process of respiration and for release of energy in the form of ATP (Adenosine triphosphate). ATP is also known as the energy currency of the cell. This energy is used by cell to perform various functions such a mechanical work and biosynthesis of new chemical compounds.
Mitochondria to bounded by double membranes. The outer membrane is porous and the inner membrane is deeply folded. These folds are known as cristae and they provide a large surface area for ATP- generating chemical reactions. Mitochondria have their own DNA, ribosomes and enzymes to manufacture their own proteins. So they are also known as semiautonomous bodies.
CBSE Sample Papers for Class 9 Science Solved Set 2 12
Question 13:
How can poultry farming be integrated with crop production ? How improved poultry breeds are developed in poultry farming ?
Answer:
Poultry is rearing of domesticated fowl, ducks, geese, turkeys and pigeons for their meat and eggs.
Poultry feed consists of mashed cereals like wheat, maize, jowar, oil cakes, fish meat etc.
Therefore they should be kept along with such crops in the fields, thus benefitting the farmer.
The cross-breeding programmes between Indian and foreign poultry have developed improved poultry breeds. The new varieties have low maintenance requirements. The size of the egg laying bird is also reduced and it utilises more fibrous cheaper diets consisting of agricultural by products. They produce more eggs and broilers for meat. Hence, it is interesting to note that
poultry is India’s most efficient converter of low fibre food-stuff (which is unfit for human consumption) into highly nutritious animal protein food.
OR
What factors are responsible for storage losses in agricultural produces and how they can be controlled and prevented ?
Answer:
Following factors may be responsible for losses of grains during storage :
(i) Biotic factors : They include insects, rodents, bacteria, fungi and mites.
(ii) Abiotic factors : They include inappropriate moisture and temperature in the place of storage.
These factors cause poor quality, loss in weight, discolouration of produce, poor germinability and marketability of grains.
These factors can be controlled by proper treatment and by systematic management of ware housing.
Preventive measures include strict cleaning of produce before storage, proper drying of the produce first in sunlight and then in shade and fumigation using chemicals to kill pests.
Question 14:
Name the two types of diseases one caused by some external agents and other due to some internal disorder of the body.
Mention any two causative agents.
Answer:
Two types of diseases are :
By external agent – Acquired diseases like Jaundice, Tuberculosis, Polio etc.
By internal disorder – High blood pressure, Diabetes, Cancer etc.
Causative agents – Causes may be malnourishment or pathogen or congenital or stress induced etc.
For e.g., Malaria is caused by Plasmodium. Diabetes is caused due to insufficient insulin or insulin resistance.
Question 15:
Bhola and Rajni, who are studying in Class IX, were travelling in a train. Rajni observed a field with two crops growing simultaneously in a definite pattern. While Bhola was busy in playing with a video game. Rajni noticed that rows of bajra and lobia were grown in alternate rows. She asked her grandfather why bajra and lobia are grown together ?
(i) On what basis are the two crops selected in this pattern ?
(ii) How does this practice benefit the farmer ?
(iii) State any two values in Rajni’s behaviour here that differentiate her from Bhola.
Answer:
(i) This is inter-cropping.
The crops selected for such type of pattern should have different types of nutritional requirements. This will help the crops to utilise maximum nutrients from the soil.
(ii) This pattern prevents pests and diseases to spread in all plants of one crop thereby preventing losses.
(iii) Values that differentiate Rajni from Bhola are :
(a) Awareness
(b) Concern
(c) Inquisitive
(d) Observant etc.
Question 16:
(a) Draw a velocity-time graph for an object in uniformly accelerated motion. Show that the slope of the velocity-time graph gives the acceleration of the object.
(b) An aeroplane starts from rest with an acceleration of 3 m s-2 and takes a run for 35 s before taking off. What is the minimum length of the runway and with what velocity the plane took off ?
Answer:
(a) Velocity-time graph is shown in figure.
CBSE Sample Papers for Class 9 Science Solved Set 2 16
The Slope of the graph is given by \frac { BD }{ AD } . However, from the
graph it is dear that BD = EA = v – u, and AD = OC = t.
∴ slope of velocity -time graph = \frac { BD }{ AD } = \frac { v - u }{ t }
But \frac { v - u }{ t } rate of change of velocity = acceleration ‘a’
So, we conclude that the slope of the velocity-time graph gives the acceleration of the object.
(b) Here u = 0, a = 3 m s’2 and t = 35 s
∴  final velocity v = u + at = 0 + 3 × 35 = 105 m s-1
If minimum length of runway be s m, then
s = ut + \frac { 1 }{ 2 } at2 =0 × 35 +\frac { 1 }{ 2 } × 3 × (35)2 =1837.5
Question 17:
(i) State two factors on which the gravitational force between two objects depends.
(ii) Why is ‘G’ called as universal constant ?
(iii) What happens to the gravitational force between two objects if masses of both the objects are doubled and the distance between them is also doubled ?
(iv) What is the value of ‘G’ on moon ?
(v) What is the value of ‘g’ on moon ?
Answer:
(i) The gravitational force between two objects is (a) directly proportional to the product of their mases, and (b) inversely proportional to the square of the distance between them.
(ii) G is called a universal constant because its value never changes under any condition.
(iii) The force remains unchanged.
CBSE Sample Papers for Class 9 Science Solved Set 2 17
(iv) Value of G on moon is same as on earth having a value 6.673 × 10-11 N m2 kg-2
(v) Value of g on moon is 1/6 th of its value on the earth.
Question 18:
(a) What is meant by molecular formula ? State with example what informations can be derived from a molecular formula ?
(b) Write the names of the compounds represented by the following formulae :
(i) Mg(NO3)2
(ii) K2SO4
(iii) Ca3N2
Answer:
(a) Molecular formula of a compound shows the constituent elements and the number of atoms of each element present in one molecule of the compound.
Take the example of H2SO4. It is the molecular formula of sulphuric acid. It shows that
(i) Sulphuric acid molecule is constituted of hydrogen, sulphur and oxygen.
(ii) One molecule of sulphuric acid contains two atoms of hydrogen, one atom of sulphur and four atoms of oxygen.
CBSE Sample Papers for Class 9 Science Solved Set 2 18
OR
(a) State the basic difference between atoms and molecules.
(b) Why does not atomic mass of an element represent the actual mass of its atoms ?
(c) “The atomic mass of an element is in fraction.” What does this mean ?
Answer:
(a) Atom is the building block of all matter.
A molecule is a general a group of two or more atoms of the same or different elements
that are chemically bonded.
(b) The actual mass of an atom is extremely small. We represent the mass of an atom as
a relative mass compared to 1/12 of the mass of one carbon-12 atom.
(c) Many of the elements have isotopes with different masses. The atomic mass of an element is the average of the masses of these isotopes which comes out to be a fraction.
Question 19:
Give reasons for the following :
(a) Isotopes of an element are chemically similar.
(b) An atom is electrically neutral.
(c) Noble gases show least reactivity.
(d) Ions are more stable than atoms.
(e) Na+ has completely filled K and L shells.
Answer:
(a) Chemical properties of an element depend upon the atomic number. All the isotopes of an element have the same atomic number and therefore are chemically similar.
(b) An atom has the equal number of positively charged protons and negatively charged electrons. Therefore atom is electrically neutral.
(c) Noble gases contain 8 electrons in the outermost shell (2 in case of helium). This is the maximum capacity of the outermost shell. Hence they show least reactivity.
(d) Ions are more stable than atoms because the former have an octet structure.
(e) Electronic configuration of Na is
K         L           M
2          8           1
Na+ is formed from Na by removing the electron from the M shell. Therefore it has the completely filled K and L shells.
Question 20:
Give examples of organisms belonging to Kingdom Protista :
(i) Based on structure for locomotion.
(ii) Based on mode of nutrition,
Answer:
(i) Based on structure for locomotion :
(a) Pseudopodia – e.g., Amoeba.
(b) Flagella – e.g., Euglena.
(c) Cilia – e.g., Paramecium.
(ii) Based on mode of nutrition :
(a) Autotrophic —Euglena.
(b) Heterotrophic – Trypanosoma, Amoeba, etc.
Question 21:
(a) Mention the type of shelters which should be provided to cattles in dairy farming.
(b) Mention the preventive measures taken to control diseases of dairy animals.
(c) What are the food requirements of dairy animals ?
Answer:
(a) The shelter provided for the livestock depends on type of animals to be sheltered, number of animals and location of shelter.
Good animal shelter should be clean, dry, airy and well-ventilated. It should be spacious, have proper disposal of wastes, arrangement of clean drinking water and protection of animals from environment factors, predators and diseases.
(b) Cholera, diarrhoea and tuberculosis are a few examples of bacterial diseases of cattle. Viral disease are pox, rinderpest as well as foot and mouth disease.
Aspergillosis is an example of fungal disease in cattle.
Animal diseases can be prevented if animals are kept in clean, hygienic environment, if they are given regular bath, if their shelter prevent entry of germs and flies; if they are vaccinated at regular intervals.
(c) The cattle feed consists of :
(i) Roughage : It is a coarse and fibrous substance which has low nutrient content e.g., fodder, hay, straw and so on.
(ii) Concentrate : It is rich in nutrients with very little fibrous or cellulose matter. Concentrate is provided by grains, seeds and oil cakes.
SECTION – B
Question 22:
A string is stretched as shown and a ‘pulse’ is created along it. The stop watch is started, from its position A, when the pulse is in position X and is stopped, in its position B, when the ‘pulse’ has travelled back to its position ‘X’. Find the velocity of propagation of the pulse along the string.
CBSE Sample Papers for Class 9 Science Solved Set 2 22
Answer:
Distance covered by pulses – 2m + 2m – 4m Time taken t =45 s – (- 5 s) = 50 s
Speed of pulse v = \frac { s }{ t } =\frac { 4\quad m }{ 50\quad s } =0.08m/s or 8 cm s-1
Question 23:
While determining the density of a copper piece using a spring balance and a measuring cylinder, Seema carried out the following procedure :
(i) Noted the water level in the measuring cylinder without the copper piece.
(ii) Immersed the copper piece in the water.
(iii) Noted the water level in the measuring cylinder with the copper piece inside it.
(iv) Removed the copper piece from the water and immediately weighed it using a spring balance.
Which step in the procedure is wrong and why ? What should have been the correct step ?
Answer:
The step (iv) of the procedure is wrong because the copper piece just removed from the water is wet and on weighing it we shall not get the correct mass.
The copper piece should be weighed only when it is completely dry.
Question 24:
A student while carrying out the separation of a mixture of sand, common salt and camphor arranged the apparatus as shown in the figure but couldn’t collect camphor. What is the defect in his apparatus set up ? Name one more substance that can be purified by this method.
CBSE Sample Papers for Class 9 Science Solved Set 2 24
Answer:
The stem of the funnel is not closed with a cotton plug resulting in the escape of camphor.
Ammonium chloride can also be purified by this method.
Question 25:
On adding zinc to dilute sulphuric acid, zinc sulphate and hydrogen gas are formed.
(a) What is the colour of zinc sulphate solution ?
(b) How do we test hydrogen gas ?
Answer:
(a) Zinc sulphate solution is colourless.
(b) When a burning match stick is brought near the gas, it burns with a pop sound.
Question 26:
Spirogyra, Mosses and Ferns belong to which sub-kingdom ? Give two identifying features.
Answer:
Spirogyra is Thallophyta.
Mosses one Bryophyta.
Ferns are Pteriodophytes.
All these phyla belong to sub-kingdom Cryptogamae.
Features are:
(i) They do not produce external flowers or seeds.
(ii) Common means of reproduction is through spores.
Question 27:
Differentiate dicots from monocots on the basis of roots and leaves.
Answer:
Dicots
Monocots
Roots :
(i) Dicots have tap root system.
(i) Monocots have fibrous roots,
Leaves:
(ii) Leaves are broad with reticulate venation.
(ii) Leaves are thin and blade like with parallel venation.
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