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Class IX – Science – 6 – MS

Practice Paper – TERM II (2021 – 2022)

Class – IX

Science (086)

Time: 2 hours                                                                                                  Maximum Marks: 40

General Instructions:

i) All questions are compulsory.

ii) The question paper has three sections and 15 questions. All questions are compulsory.

iii) Section–A has 7 questions of 2 marks each;

iv) Section–B has 6 questions of 3 marks each; and

v) Section–C has 2 case based questions of 4 marks each.

vi) Internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.

SECTION – A

1. Atomic numbers of two elements A and B are given below. Atomic number of element A: 18, Atomic

number of element B: 17

(a) Which element has zero valency?

Element A has zero valency because electronic configuration of element A is 2, 8, 8 i.e., complete octet.

(b) What is the valency of other element?

The atomic number of element B = 17, i.e., it has 17 electrons. Hence its electronic configuration will be 2, 8, 7.

Since it has 7 electrons in its valence shell, so, its valency = 8 – 7 = 1

2. Two students of class IX are discussing a law that states the mass of reactants is always equal to that

of the products.

(a) Identify the law.

Law of conservation of mass: The law states that matter can neither be created nor destroyed or mass of reactants is always equal to that of product.

(b) Based on the above law find out if 12 g of carbon is burnt in the presence of 32 g of oxygen, how much carbon dioxide will be formed?

C + O2 –> CO2

Carbon + Oxygen –> Carbon dioxide

Mass of reactants = 12 + 32 = 44 g

Mass of products (CO2) = 44 g

Thus, one mole of carbon (C) reacts with one mole of Oxygen (O2) to form one mole of CO2

Mass of CO2 formed = 44g

3. Ernest Rutherford was interested in knowing how the electrons are arranged within an atom. Rutherford designed an experiment for this. In this experiment, fast moving alpha (α)-particles were made to fall on a thin gold foil. State the observations in the experiment, which led Rutherford to make the following conclusions:

(a) Most of the space in an atom is empty.

Most of the alpha-particles passed through the gold foil without getting deflected.

(b) Whole mass of an atom is concentrated in its centre.

Very few particles were deflected from their path by 1800, indicating that whole mass of the atom is present at its centre.

4. A person is suffering from an incurable disease. His reports say that he is infected with HIV.

(a) Identify the disease the person is suffering from.

The person is suffering from the disease called AIDS (Acquired immunodeficiency syndrome)

(b) State any two ways by which this disease spreads from an infected person to a healthy person

(i) Sexual contact with infected person

(ii) From a pregnant mother to her foetus.

(iii) Blood contact with infected person

(iv) Using needle or syringe of infected person.

5. Although Archana has been suffering from cold and cough she decided to appear for her class test.

Classmates seated close to her had an exposure to the infection being carried by Archana. However,

only one of them actually suffered from cold and cough.

(a) Which system is responsible for fighting with infection carrying microbes?

Immune system of an individual fight from infection carrying microbes.

(b) What prevented the rest of those classmates from catching cold and cough in spite of their exposure to the infection.

Immune system of those who does not suffer with cold and cough successfully fought against the microbes to which they were exposed.

OR

“Making antiviral medicines is harder than making antibacterial medicines”. Justify this statement.

Viruses have very few biochemical mechanisms of their own. They depend on their host to complete all the biochemical reaction. Thus, it is very difficult to find the targets where these drugs get interfered by not making any harm to the host.

6. A box of about 10 kg is placed at a point A on a horizontal surface. It is moved to a point B which is at a

distance of 2 m from A by the action of force, 4N applied on the box at an angle of 60o.

(a) Find the work done by the gravitational force on the box.

Work done is zero because force due to gravity is perpendicular to the displacement that occurred due to applied external force.

W = Fgd cos 900 = 0 J

[⸫ cos 900 = 0, Fg= Force due to gravity]

(b) Calculate the work done by the externally applied force. Justify your answer. (Given, g = 10 m/s2)

Given, m = 10 kg

Work done = F.d cos 600

                  = 4 x 2 x ½

                  = 4 J

OR

You might have seen that when a blacksmith hits a sheet of iron by fast-moving hammer, the

iron sheet heats up.

(a) Which type of energy does the fast-moving hammer possess?

Kinetic energy

(b) State the type of energy conversion that takes place in the above process.

The kinetic energy get converted into heat energy during hitting a metal by a fast moving hammer due to which metal gets heated.

7. An astronaut carried a pot containing soil weighing 60 N from the earth to the surface of moon. He

kept it there and just before the return journey from moon to earth he weighed the soil on the surface

of moon and found that it was only 10 N. Why did its weight decrease and how much was the loss in

mass of the soil? (g earth = 10 ms–2) (g moon= g earth / 6)

Given, the weight of the soil on earth is 60 N

gearth = 10 m s-2

Mass on the earth m1 = 60/10 = 6 kg

Weight of the soil on the moon = 10 N

gmoon = gearth/6 = 10/6 ms -2

Mass on the moon m2 = (10 x 6)/10 = 6 kg

Because m1 = m2, hence there has been no loss in mass of the soil on the surface of moon and decrease in weight was due to difference in the gravity.

OR

A stone and the Earth attract each other with an equal and opposite force. Then, why we see only the

stone falling towards the earth but not the earth rising towards the stone?

From Newtons second law of motion,

F = ma => a = F/m

The mass of a stone is very small due to which the gravitational force produces a large acceleration in it. Due to very large mass of earth, the same gravitational force produces very small acceleration in the earth.

SECTION – B

8. As a part of practical examination paper, Nisha has to classify the following elements on the basis of

their valencies. But she could not recollect what is valency and what is cation. Help her to answer the

below questions.

(a) Define valency.

The combining power (or capacity) of an element to displace or combine with number of hydrogen atoms is known as its valency.

(b) Define cation.

Cation is an ion with positive electric charge.

(c) Classify the following cations on the basis of their valencies: NH4+, Al+++

NH4+ = Monovalent

Al+++ = Trivalent

9. Write the chemical formulae of the following:

(a) (i) Aluminium nitrate (ii) Magnesium hydrogen carbonate.

(b) Give the names of the elements present in the following compounds: (i) Quick lime (ii) Hydrogen

Bromide

(i) Calcium and oxygen (CaO)

(ii) Hydrogen and bromide (HBr)

OR

During an experiment performed in a laboratory, 14 g of sodium bicarbonate was allowed to react with 10 gm of acetic acid. After the reaction was completed, it was found that only 16.67 g of the solution was left because a gas escaped from the container.

(a) What was the mass of the gas that escaped into the atmosphere?

Mass of the gas that escaped into the atmosphere = 24 – 16.67 = 7.33 gm

(b) Name and state the law applied to find the answer.

The law applied was the law of conservation of mass. The law states that matter can either be created nor destroyed or mass of reactants is always equal to that of product.

10. Given that the percentage abundance of the isotope 10Ne20 is 90% and that of isotope 10Ne22 is 10%.

Calculate the average atomic mass of Neon.

Average atomic mass is given by:

20 x 90/100 + 22 x 10/100

(18 + 2.2) u = 20.2 u

11.(a) From the symbol 16S32 state:

(i) Atomic number of sulphur

16

(ii) Mass number of sulphur

32

(iii) Electronic configuration of sulphur

2, 8, 6

(b)Which of the two elements given below would be chemically more reactive, ‘X’ of atomic number 18

or element ‘Z’ of atomic number 16 and why?

Z is more reactive than X

This is because, electronic configuration of Z = 2, 8, 6 which may gain 2 electrons and thus is more reactive whereas X = 2, 8, 8 has complete octet and is inert.

12. Find the momentum of a body of mass 100 g having kinetic energy 500 J

Using the formula for K.E. we get,

K.E = ½ mv2

M = 100 g = 0.1 kg

K.E = 500 J

v2 = (2 x 500)/ 0.1 = 10000

v = 100 m/s

Momentum = Mass x Velocity

                   = 0.1 x 100 = 10 kg m/s

OR

A boy is pulling a cart by supplying a constant force of 8 N on a straight path of 20 m. On a round about

of 10 m diameter he forgets the path and takes 112turns and then continues on the straight path for

another 20 m. Find the net work done by the boy on the cart.

Work done, F = 8N

W = F x s

W1 = 8 x 20 = 160 J

D = 10 m

So, radius r = D/2 = 5 m

Circumference of a circle = 2πr

                                    = 2 x 22/7 x 5 m

                                    = 31.43 m

Distance in ½ circle = πr

                                    = 22/7 x 5 = 15.71

Total distance covered in 1.5 circle = 31.43 + 15.71 = 47.14 m

Therefore, work done will be,

W2 = F x s = 8 x 47.14 = 377.12 J

Again travelling for 20 m, work done will be,

W3 = 20 x 8 = 160 J

Total work done = 160 + 377.12 + 160

                        = 697.12 J

13. Weight is the force by which an object is attracted towards the earth. A man’s weight when taken at

the poles are 600 N.

(a) Will his weight remain the same when measured at the equator?

No, his weight will not remain same as that at the poles.

(b) Will there be an increase or decrease in his weight? Explain.

There will be a decrease in his weight at the equator. As the radius of the earth increases from the poles to the equator, the value of ‘g’ becomes greater at poles decreasing towards equator. Also, the force of gravity decreases from poles to the equator.

SECTION – C

14. Read the passage and answer the following questions.

Human beings live in societies. Our social environment is an important factor in our individual health.

We live in villages, towns or cities. In such places, even our physical environment is decided by our social

environment. At any place, our body may feel sickness. There are many tissues in the body which make up physiological systems to carry out functions. When there is a disease some changes give

rise to symptoms and signs of disease.

(a) What is a disease?

Any condition which impairs the health, or interferes with the normal functioning of the body is called disease.

(b) What are called the symptoms of a disease?

A symptom is a phenomenon that is experienced by the individual affected by the disease.

(c) What can be various causes for a person getting diseased?

Causes:

(i) Extrinsic factor: Water, food

(ii) Intrinsic factor: Poor eating habits, poor nourishment.

(iii) Genetic constitution: Weak immune system.

(iv) Social reason: Poor public services.

OR

Based on duration what are the two main types of diseases? Explain giving one example for each.

Types of diseases:

(i) Acute diseases: Last for short period of time. Ex. Common cold.

(ii) Chronic diseases: Lasts for long time even for lifetime. Eg. diabetes

15. An experiment based on simple pendulum was conducted in a class. The following diagram shows

the set up. It shows a simple pendulum consisting of a bob of mass 100 g. Initially the bob of the pendulum is at rest at ‘O’. It is then displaced to one side at A. The height of ‘A’ above ‘O’ is 5 cm. (Take g = 10 m/s2)

(a) What is the value of potential energy of bob at ‘A’?

Given, m = 100 g = 0.1 kg

g = 10m/s2

The value of potential energy of the bob at ‘A’ = m x g x h

                                                                        = 0.1 x 10 x 0.05 = 0.05 J

(b) What is the value of kinetic energy of the bob at mean position ‘O’?

At mean position, total P.E. change to K.E. So, kinetic energy of the bob at mean position = 0.05 J

(c) What is the value of kinetic energy and potential energy of the bob at the position ‘P’ whose height above mean position is 2 cm?

P.E at 2 m height = m x g x h

                        = 0.1 x 10 x 0.02

                        = 0.02 J

K.E at 2 m height = Total energy – P.E at 2 m height

                        = 0.05 – 0.02 = 0.03 J

OR

(c) The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

The law of conservation of energy states that energy can neither be created nor be destroyed, but can be converted from one form to another.

The given condition does not violate the law. This is because potential energy is converted into kinetic energy.