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Previous Year Question Papers Class 9 Science With Solutions Set 5

Solved CBSE Sample Papers for Class 9 Science Set 5

Time Allowed: 3 hours

(GENERAL INSTRUCTIONS)

(i) The question paper comprises of two sections. A and 13. You are to attempt both the sections.
(ii) All questions are compulsory. However, internal choice has been provided in two questions of three marks each and one. question office marks. Only one option in. such questions is to be attempted.
(iii) All questions of section A and all questions of section B are to be attempted separately.
(iv) Question numbers 1 and 2 in section A are one mark questions. These are to be answered in one word or in one sentence.
(v) Question numbers 3 to 5 in section A are two marks questions. These are to be answered in about 30 words each.,
(vi) Question numbers 6 to 15 in section A are three marks questions. These are to be answered in about 50 words each..
(vii) Question numbers 16 to 21 in section A are five marks questions. These arc to be answered in about 70 words each.
(viii) Question numbers 22 to 27 in Section B are two marks questions based on practical skills. These are to he answered in brief.

SECTION – A

Question 1:
List the combining elements
Answer:
CBSE Sample Papers for Class 9 Science Solved Set 5 1

Question 2:
State the universal law of gravitation.
Answer:
Everybody in this universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of distance between their centres.

Question 3:
23 g of salt is dissolved in 100 g of water at 293 K. What is the concentration of solution at this temperature ?
Answer:
CBSE Sample Papers for Class 9 Science Solved Set 5 3

Question 4:
(a) When observed under microscope, small pores here and there in the epidermis of the leaf are seen. What are these pores called ? Write two functions of these pores.
(b) Name the simple permanent tissue which :
(i) forms the basic packing tissue.
(ii) provides flexibility in plants.
Answer:
(a) • These pores are called stomata.
• These pores help in exchange of gases and in transpiration.
(b) (i) Parenchyma,
(ii) Collenchyma.

Question 5:
Give two applications based on the concept of pressure.
Answer:
(i) Tyres of heavy vehicles are wide.
(ii) The strap of bags are wide.

Question 6:
The discovery of sub-atomic particles led to a revolution in the study of matter. Name the scientists who discovered these sub-atomic particles.

OR

Define the following terms :
(a) Electronic configuration
(b) Valence shell
(c) Valency
Answer:

S. No.

Sub-atomic particles of matter

Discoverer

1.

Electron

JJ. Thomson

2.

Proton

E. Goldstein

3.

Neutron

j.Chadwick

OR

(a) The distribution of electrons into different orbits (shells) of an atom is known as its electronic configuration.
(b) The outermost shell of an atom is known as valence shell.
(c) Combining capacity of an atom is called its valency.

Question 7:
From where does the term Echinoderma originate ? Mention two animals belonging to Echinodermata.

OR

(a) Why was the need to write scientific names for organisms ? Who introduced the system of scientific naming ?
(b) What are the two Latin forms in a scientific name of an organism ?
Answer:
In Greek, echinos means hedgehog, and derma means skin. Thus, echinodermates are spiny
skinned organisms.
e.g., Antedon (feather star), Asterias (star fish), Holothuria (sea cucumber).

OR

(a) Because it would be difficult for people speaking or writing in different languages to know when they are talking about the same organism. This problem was resolved by agreeing upon a ’scientific’ name for organisms which is unique and can be used to identify an organism anywhere in the world. The system of scientific naming or nomenclature was introduced by Carious Linnaeus in the eighteenth century.
(b) A generic name and a specific name are the two Latin forms in a scientific name of an organism.

Question 8:
(a) Mention four characteristic features of the cells of meristematic tissue.
(b) Which elements of xylem :
(i) help in transport of water and minerals,
(ii) store food, and
(iii) provide mechanical support ?
Answer:
(a)
(i) Meristematic tissue consists of repeatedly dividing cells.
(ii) The cell wall of these cells are thin.
(iii) Vacuoles are not found in these cells.
(iv) The protoplasm of these cells are very dense.
(b)
(i) Tracheids and vessels,
(ii) Xylem parenchyma,
(iii) Xylem fibres.

Question 9:
(a) A body thrown vertically upwards reaches a maximum height fa. It then returns to ground.
Calculate the distance travelled and its displacement.
(b) In a long distance race, the athletes were expected to take four rounds of the track such that the line of finish was same as the line of start. Suppose the length of the track was 200 m.
(i) What is the total distance to be covered by the athletes ?
(ii) What is the displacement of the athletes when they touch the finish line ?
(iii) Is the motion of the athletes’ uniform or non-uniform ?
Answer:
(a) Here, Distance travelled = h + h = 2 h Displacement = 0
(b) Length of the track – 200 m.
(i) Total distance to be covered by the athletes in four rounds = 200 x 4 = 800 m.
(ii) Displacement of the athletes when they touch the finish line = 0.
(iii) Non-uniform.

Question 10:
What is velocity-time graph ? State how it can be used to find:
(i) the acceleration of a body,
(ii) the displacement of the body,
(iii) the distance travelled in a given time.
Answer:
In the velocity-time graph, time is taken on the x-axis velocity is taken along the y – axis.
(i) Since acceleration = change in velocity/time, therefore, the slope of the velocity¬time graph gives the acceleration. If the slope is positive, it is accelerated motion and if the slope is negative, the motion is retarded.
(ii) Since velocity x time = displacement. The area enclosed above the time axis represents positive displacement and the area enclosed below the time axis represents negative displacement. The total displacement is obtained by adding them numerically with proper sign.
(iii) The total distance travelled by the body is their arithmetic sum (without sign).

Question 11:
Calculate the ratio of momentum, when :
(i) velocity of an object is doubled.
(ii) mass of an object is halved.
(iii) both mass and velocity are increased by three times.
Answer:
Let a body of mass m is moving with velocity.
Then its momentum, pi – mv
CBSE Sample Papers for Class 9 Science Solved Set 5 11

Question 12:
Explain the meanings of the following desirable factors for which crop variety improvement is done:
(i) Biotic and abiotic resistances.
(ii) Wider adaptability.
(iii) Desirable agronomic traits.
Answer:
(i) Biotic and abiotic resistances : Varieties resistant to biotic and abiotic stresses can improve crop production.
Biotic factors — Diseases, insects and nematodes.
Abiotic factors — Drought, salinity, water logging, heat, cold, etc.
(ii) Wider adaptability : Developing varieties for wider adaptability help in stabilising the crop production under different environmental conditions.
(iii) Desirable agronomic traits : Tallness and profuse branching are desirable traits for fodder
crops. Dwarfness is desired in cereals.

Question 13:
Waves of frequency 150 Hz are produced in a string. Find the :
CBSE Sample Papers for Class 9 Science Solved Set 5 13
(a)wavelength,
(b) amplitude and
(c) velocity of the waves in SI units. Ans.
Answer:
(a) Wavelength = 25 cm
(b) Amplitude = 6 cm
(c) v = v X -150 x 0.25
= 37.5 m s-1

Question 14:
When a rubber band is stretched it acquires some energy while stretching.
(a) Name the type of energy gained by the stretched rubber band.
(b) How it gained this energy ? Explain.
Answer:
(a) Elastic potential energy.
(b) When the band is stretched work is done in stretching it. This work is stored in the rubber band as elastic potential energy.

Question 15:
Vidushi was watching a programme based on space on television. She saw, the astronauts who landed on moon were not able to talk directly to each other. She asked her science teacher about it.
(a) Why astronauts on moon were not able to talk directly to each other ?
(b) What value of Vidushi’s nature is exhibited in this incident ?
Answer:
(a) Sound requires a medium to propagate. As there is no air on the moon sound cannot propagate.
(b) Inquisite nature, zeal to learn.

Question 16:
(a) Differentiate between bone and cartilage with respect to structure, function and location,
(b) (i) Why is connective tissue called so ?
(ii) What is the function of the areolar connective tissue?
(iii) Which substance is present in the adipocytes ? How does it help ?
Answer:

Point of Difference

Bone

Cartilage

Structure

It is strong and non-flexible tissue whose cells are embedded in a hard matrix which is composed of calcium and phosphorus compounds.

It is soft and flexible tissue whose solid matrix is composed of proteins and sugars. Also, it has widely spaced cells.

Function

It forms the framework that supports the body and anchors the muscles that support the main organs of the body.

It smoothens bone surfaces at joints.

Location

It is present in skeletal system.

It is present in nose, ear, trachea and larynx.

(b) (i) These tissues serve the functions of binding and joining one tissue to another, provide support and help packing together different organs of the body.
(ii) Areolar tissue fills the space inside the organs, supports internal organs and helps in repair of tissues.
(iii)
• Adipocytes are fat storing tissues.
• These tissues are found below the skin and between internal organs.
• The cells of this tissue are filled with fat globules.
• They act as insulator because they store fats.

Question 17:
Compare in tabular form the properties of solids, liquids and gases with respect to :
(i) Volume
(ii) Diffusion
(iii) Fluidity or Rigidity
(iv) Shape
(v) Kinetic energy of particles at a given temperature
Answer:

Properties

SolidsLiquids

Gases

(i) Volume

Definite volume, as intermolecular forces between the consti­tuent particles are very strong.

Definite volume, as intermolecular forces between the consti­tuent particles are strong.

No definite volume, as intermolecular forces between the consti­tuent particles are weak.

(ii) Diffusion

Can diffuse into liquids.

Diffusion is higher than solids.

Highly diffusible as particles move rand­omly at high speed.

(iii) Fluidity or rigidity

Very rigid and cannot flow.

Less rigid and can flow easily.

No rigidity and can flow most easily.

(iv) Shape

They have a definite shape.

They do not have a definite shape.

They do not have a definite shape.

(v) Kinetic energy of particles at a given tempe­rature.

Least energy.

Higher than solids.

Maximum energy.

Question 18:
Define the following terms :
(a) Atom
(b) Molecule
(c) Avogadro’s number
(d) Mole s
(e) Molar mass
Answer:
(a) An atom is the smallest particle of an element that can exist independently and retain all its chemical properties.
(b) A molecule is the smallest particle of an element or a compound capable of independent existence under ordinary conditions. It shows all the properties of the substance.
(c) The number of particles (atoms, molecules or ions) present in 1 mole of any substance is fixed, with a value of 6.022 x io23. This number is called the Avogadro number or Avogadro constant (represented by No). This is an experimentally obtained value.
(d) The mole is the amount of substance that contains the same number of particles (atoms/ions/ molecules/formula units, etc.), as there are atoms in exactly 12 g of carbon-12.
(e) Mass of 1 mole of a substance is called its molar mass, e.g., 12 g of carbon is its molar mass.

Question 19:
Give the appropriate term for each of the following:
(a) complex sugar that makes the fungal cell wall.
(b) naked seeds.
(c) blue-green algae.
(d) basic unit of classification.
(e) group of unicellular eukaryotic organisms.
Answer:
(a) Chitin
(b) Gymnosperms i
(c) Cyanobacteria
(d) Species
(e) Protista

Question 20:
What happens to the gravitational force between two objects if:
(i) the mass of one object is doubled ?
(ii) the distance between the objects is doubled ?
(iii) the masses of both the objects are doubled ?
(iv) the distance between them is halved ?
(v) mass of one of the objects is halved ?
Give reason in each case.
Answer:
(i) Force is directly proportional to the mass of each object. The force becomes twice of the original force.
(ii) Force is inversely proportional to the square of distance between objects. The force becomes l/4th of the original force.
(iii) Force is directly proportional to the product of masses of the two objects. The force becomes four times of the original force.
(iv) If the distance between the two objects is reduced to half, then the force is four times of the original force, i.e., it increases.
(v) The gravitational force of attraction becomes half.

Question 21:
Define density and relative density. Give their mathematical formulae and SI units. Relative density of gold is 19.5. The density of water is 1000 kgmr3. What is the density of gold in SI unit and in g cc-1.

OR

Calculate the electricity bill amount for a month of 30 days, if the following devices are used as specified:
(a) 3 bulbs of 40 W for 6 hours.
(b) 4 tubelights of 50 W for 6 hours.
(c) A refrigerator of 320 W for 24 hours.
Given the rate of electricity is Rs 2.50 per unit.
Answer:
CBSE Sample Papers for Class 9 Science Solved Set 5 21

SECTION – B

Question 22:
Mention the position of bulb of thermometer in the following experiments :
(i) In an experiment to determine the melting point of ice.
(ii) In an experiment to determine the boiling point of water.
Answer:
(i) The bulb of the thermometer should remain inside the crushed ice.
(ii) The bulb of the thermometer should just be held above the surface of water.

Question 23:
Write the procedure you need to follow for preparing a solution of starch in water. State the precaution which is most important and one must follow essentially. Answer:
(i) Procedure:
(1) Taken water in a beaker and mixed some starch in it.
(2) Stirred the mixture gradually, using glass rod.
(3) Left the mixture undisturbed for sometime.
(ii) Precaution:
For making starch solution, always distilled water should be used.

Question 24:
Mention the four sequential steps of procedure required to calculate the percentage of water absorbed by raisins in the laboratory.
Answer:
(i) Find the initial weight of raisins.
(ii) Keep the raisins in a petri dish for 2-3 hours so that they absorb water and swell up.
(iii) Wipe water on the outer surface of raisins and weigh them.
(iv) Find the percentage of water absorbed by raisins by using formula:
CBSE Sample Papers for Class 9 Science Solved Set 5 24

Question 25:
A student observed that a block of mass 100 g displaced 50 mL of water when dipped in measuring cylinder. Calculate the density of the block.
Answer:
Mass of block = 100 g
Volume of block = 50 mL = 50 cm³
CBSE Sample Papers for Class 9 Science Solved Set 5 25

Question 26:
A student uses a spring balance of least count 10 g and range 500 g. He records the weight of a small lead sphere in air, in tap water and in a concentrated solution of salt water respectively. If the readings are in order Wu W2 and W3. Write the relation between them with reason.
Answer:
W1 > W2 > W3

Question 27:
(a) Which feature of the body design helps a fish to swim in water ?
(b) How many gills are present in a bony fish ?
Answer:
(a) Streamlined body.
(b) The vast majority of bony fish have 5 pairs, and a few have 6 or 7 pairs.

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