Class 9 Science Unit test -2 for the preparation of Term-2 CBSE Exam (2021-22) - Future Study Point

Class 9 Science Unit test -2 for the preparation of Term-2 CBSE Exam (2021-22)

class 9 unit test-2 for term 2 science preparation

Class 9 Science Unit test -2 for the preparation of Term-2 CBSE Exam (2021-22)

class 9 unit test-2 for term 2 science preparation

Class 9 Science Unit test -2 for the preparation of Term-2 CBSE Exam (2021-22) is the solution of a question paper taken from a public school. In this test question paper there are 20 questions from three chapters atoms and molecules,structure of atom and gravitation.The Class 9 Science Unit test -2 for the preparation of Term-2 CBSE Exam (2021-22) is helpfull for the preparation of term-2 CBSE Board exam 2021-22. All questions of Class 9 Science Unit test -2 for the preparation of Term-2 are created by a reputed public school G.D Lancer Public School,Mohan Garden New Delhi and solved by a science expert of Future Study Point.

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Class 9 CBSE maths and science sample papers and last years question papers

Class 9 Science Unit test -2 for the preparation of Term-2 CBSE Exam (2021-22)

Q1.The molecular mass of ammonium hydroxide (NH4OH) is :

(a) 26 u      (b) 45 u      (c) 35 u       (d) 14 u

Ans.     (c) 35 u

The number of atoms in (NH4OH)

N -1, H-4, O-1 and of H-1

Atomic mass of atoms ,N-14 u, O-16u, H-1 u

The molecular mass of (NH4OH)= 14 +4×1+16 +1=14+4 +17 =35 u

Q2.The chemical formula of Ammonium sulphate is:

(a) (NH4)2SO4        (b) NH4SO4   (c) (NH4)3SO4    (d) NHSO4

Ammonium sulphate contains two ions which are following with their valancy

NH4 → +1

SO→ -2

Cross multiplying their valancy,we get the required chemical formula

(NH4 )2 SO4

Q3. If the distance between two objects is tripled, the force of gravitation becomes:

(a) 1/4        (b) 1/6          (c) 1/3      (d) 1/9 

Ans. (d) 1/9 

The force of gravitation between two objects is represented by

F = G(M.m/d²)………(i), where G-universl constant of gravitation force,M and m are the masses of the objects

ATQ, distance between two objects is tripled (i.e d→3d),then F’ is

F’ = G(M.m/(3d)²)= GM.m/9d²……..(ii)

F/F’ = 9/1

9F’ = F

F’ = (1/9) F’

Q4.How does the value of acceleration due to gravity varies at poles and at equator ?

(a) Increase at poles and decrease at equator

(b)Decrease at poles and increase at equator

(c) Remains the same

(d) None of the above

Ans.(b)Decrease at poles and increase at equator

The formula for evaluating g is

g = G M/r²

The radius at the poles is smaller than the radius at equator because the earth is flattened at the poles

Here g is inversely proportional to the square of the radius of the earth,therefore the value of g is more in equator and lesser at the poles.

Q5.The mass of 1 mole of nitrogen atoms is:

(a) 12      (b) 13          (c) 14      (d) 15

Ans.(c) 14

The atomic mass of nitrogen is =14 u

∴ The molar mass of nitrogen is =14 g

No.of moles = given mass/molar mass

1 = m/14

m = 14 g

Q6.Two objects of different masses falling freely near the surface of the moon would.

(a) Have same velocities at any instant

(b) have different accelerations

(c) Experience forces of same magnitude

(d) Undergo a change in their intertie 

Ans.(a) Have same velocities at any instant

In the codition of free fall both objects are under the impact of gravity of the moon ,so their motion don’t  depend on their mass as an example, let the velocity of one object is v(t) after the time t then it is reprsented by

v(t) = u + gt =0+gt =gt

Therefore if both objects are allowed to fall from a same height then they same velocities at any instant.

Q7.Suppose a planet exists whose mass and radius are half that of the earth. The value of accelerarion due to gravity on this planet is:

(a) 10.9      (a) 19.6         (a) 18.9        (a) 19.9 

Ans.Let let the mass of the earth is M and radius of the earth is R then accelerarion due to gravity on the earth is given by

g = G.M/R²……..(i)

The mass of the given planet is M/2 and radius  is R/2, then accelerarion due to gravity on the planet is given by

g’ = G.M/2/(R/2)²= 4G.M/2R²=2G.M/R²……(ii)

Putting the value of g from equation (i)

g’ = 2×g

g’ = 2×9.8 =19.6 m/s²

Q8. The atomicity of H2SO4 is :

(a) 4     (b) 5        (c)6       (d) 7

Ans. (a) 7

The atomicity of a molecule is the number of atoms in it

In H2SO4 is there are

No. of H atoms =2

No. of S atoms =1

No. of O atoms =4

Total atoms =2+1 +4 =7

∴ Atomicity of H2SO4 is 7

Q9.According to law of conservation of mass:

(a) Mass of a body is always the same

(b) Mass is neither created nor destroyed.

(c) Mass is always constant.

(d) All of these

Ans.(b) Mass is neither created nor destroyed.

According to law of conservation of mass ,during chemical reaction the mass of reactants is always equal to the mass of reactants,hence mass is neither created nor destroyed.

Q10. An element X forms a compound X2O3 . The valency of X is :

(a) 3     (b) 4        (c)6       (d) 8

Ans.(a) 3

The formula X2O3 . shows that X holds 3 O atoms ,therefore valency of X is 3

Q11. An ion M3+ contains 10 electrons and 14 neutrons .The atomic number and mass number of the element M respectively will be:

(a) 10, 24        (a) 13, 24      (a) 13, 27       (a) 10, 27 

Ans. (a) 13, 27

The number electrons given in M3+ ion are 10 and nuetrons are 14

Since M3+ ion is formed after donating 3 electrons,therefore no. of electrons in M atom is 10 +3 =13

Atomic number= no. of electrons =no of protons=13

Mass number of the element M = No of protons +No. of neutrons=13+14 = 27

Q12.An element M forms the oxide M2O3 .The formula of its phosphate will be:

(a) M2PO4         (b) M3PO4      (c) MPO4       (d) M2(PO4 )3

Ans.(c) MPO4

The oxide of M, M2O3 .shows that M has the valency of 3

Phosphate has the valency of 3

∴ Chemical formula of phosphate of M = MPO4

Q13.Mass of silver nitrate that will react with 5.85 g of sodium chloride to produce 14.35 g of silver chloride and 8.5 g of sodium nitrate will be:

(a) 22.85 g      (b) 17 g      (c) 14.35 g     (d)85g

Ans. (b) 17 g 

Let the mass of silver nitrate react with 5.85 g of sodium chloride

AgNO3(m)+ NaCl( 5.85 g) → AgCl (14.35 g) +NaNO3(8.5 g)

According to conservation of the mass

m +5.85 g =14.35 g +8.5 g=22.85g

m = 22.85g -5.85 g =17.00 g

Q14. Which of the following has maximum number of atoms ?

(a) 18 gH2O       (b) 18 g CO2      (c) 18 g O2     (d) 18 g CH4

Ans.(d) 18 g CH4

The number of atoms depends on the number of moles in a substance since avagadro no. is a constant

Hence,we have to find out no. of moles in the given substance

No. of moles of H2O =given mass /Molar mass =18/18 = 1 mole

No. of moles of CO2  =given mass /Molar mass =18/44 =.40 mole

No. of moles of O2  =given mass /Molar mass =18/32 =.56 mole

No. of moles of CH4=given mass /Molar mass =18/16 =1.13mole

Q15. The number of electrons, protons, and neutrons in Narespectively will be:

(a) 11,11,12        (b) 10,12,13      (c) 11,11,11      (d) 10,11,12

Ans.(d) 10,11,12

The number of electrons in Na atom are = 11

The number of electrons in Na+ are = 11 -1 =10

The number of protons in Na+ ion and Na atom remains the same i.e  11

The number of protons in Na+ are = 11

The number of neutrons in Na+ ion and Na atom remains the same i.e  12

Q16. An apple falls from a tree because of gravitation between earth and apple. If F1 is the magnitude of the force exerted by the earth on the apple and F2 is the force exerted by the apple and F2 is the force exerted by the apple on the earth.

(a) F1 is very much greater than F2 .      

(b) F2 is very much greater than F1.     

(c) F1 is only a little greater than F2.

(d) F1 and F2 are equal.

Ans. (d) F1 and F2 are equal.

According to Newton’s universal law of gravitational force, two objects exert an equal amount of force to each other. Apple falls down to the earth because its mass is very less as compared to the earth, so it is accelerated towards the earth by following the second law of motion.

Q17. The law of gravitation  gives the gravitational force between :

(a) The earth and a point mass only.

(b) The earth and the sun only.

(c) Any two bodies having some mass.

(d) Two charged bodies only.

Ans. (c) Any two bodies having some mass.

The law of gravitation  gives the gravitational force(F) between two bodies in the universe having some mass(m and M)

F = GmM/d²,,where d is the distance between the objects

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Q18. The mass of a body is 50 kg . The weight of the object on the earth and moon is

(a) 48 N,9N          (a) 49 N,8.16N      (c) 84 N,8.16N    (d) 94 N,4N  

Ans. (a) 49 N,8.16N

The given mass of a body is 50 kg

The acceleration due to gravity on the earth is =9.8 m/s²

The weight of the object on the earth is = mg =50×9.8=49 N

The weight of the object on the moon is = mg/6 =50×9.8/6 =49/6 =8.16 N

Q19. The value of G is found by:

(a) Isaac Newton           (b) Blaise Pascal       (c) Henry Cavendish      (d) Galileo Galilei

Ans.  (c) Henry Cavendish 

Henry Cavendish was the first physicist who discovered the value of G in 1798 using a torsion balance also called the Cavendish apparatus,he computed it 6.754 ×10-11 Nm²/kg² as compared to todays value of G =6.67259 ×10-11 Nm²/kg²

Q20.The value of G on -Jupitor will be:

(a) Same as that of the earth       

(b) About one -fifth that of the earth

(c) About one sixth that of the earth

(d)  About one -fourth that of the earth

Ans. (a) Same as that of the earth

The value of G is same everywhere, it is called the universal constant of gravitational force,it is small g which depends on mass and size of the heavenly body.

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NCERT Solutions of  Science and Maths for Class 9,10,11 and 12

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Chapter 1- Number SystemChapter 9-Areas of parallelogram and triangles
Chapter 2-PolynomialChapter 10-Circles
Chapter 3- Coordinate GeometryChapter 11-Construction
Chapter 4- Linear equations in two variablesChapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
Chapter 6-Lines and AnglesChapter 14-Statistics
Chapter 7-TrianglesChapter 15-Probability
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Chapter 1-Matter in our surroundingsChapter 9- Force and laws of motion
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Chapter3- Atoms and MoleculesChapter 11- Work and Energy
Chapter 4-Structure of the AtomChapter 12- Sound
Chapter 5-Fundamental unit of lifeChapter 13-Why do we fall ill ?
Chapter 6- TissuesChapter 14- Natural Resources
Chapter 7- Diversity in living organismChapter 15-Improvement in food resources
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Chapter 1-Real numberChapter 9-Some application of Trigonometry
Chapter 2-PolynomialChapter 10-Circles
Chapter 3-Linear equationsChapter 11- Construction
Chapter 4- Quadratic equationsChapter 12-Area related to circle
Chapter 5-Arithmetic ProgressionChapter 13-Surface areas and Volume
Chapter 6-TriangleChapter 14-Statistics
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Chapter 1- Chemical reactions and equationsChapter 9- Heredity and Evolution
Chapter 2- Acid, Base and SaltChapter 10- Light reflection and refraction
Chapter 3- Metals and Non-MetalsChapter 11- Human eye and colorful world
Chapter 4- Carbon and its CompoundsChapter 12- Electricity
Chapter 5-Periodic classification of elementsChapter 13-Magnetic effect of electric current
Chapter 6- Life ProcessChapter 14-Sources of Energy
Chapter 7-Control and CoordinationChapter 15-Environment
Chapter 8- How do organisms reproduce?Chapter 16-Management of Natural Resources

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Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
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Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 15- Statistics
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Chapter 1-Relations and FunctionsChapter 9-Differential Equations
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