class: sz2-a jee-main model date: 14-06-2020 time: 3hrs ... · previous week 50%: transformations:...

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Class: SZ2-A JEE-MAIN MODEL Date: 14-06-2020

Time: 3hrs WTM-02 Max. Marks: 300

IMPORTANT INSTRUCTIONS

PHYSICS

Section Question Type +Ve

Marks - Ve

Marks No.of

Qs Total marks

Sec – I(Q.N : 1 – 20) Questions with Single Answer Type 4 -1 20 80

Sec – II(Q.N : 21 – 25) Questions with Numerical Answer Type

(+/ - Decimal Numbers) 4 0 5 20

Total 25 100

CHEMISTRY

Section Question Type +Ve

Marks - Ve

Marks No.of

Qs Total marks




Total 25 100

MATHEMATICS Section Question Type

+Ve Marks

- Ve Marks

No.of Qs

Total marks




Total 25 100

mailto:[email protected]

http://www.narayanagroup.com/

SZ2-A_JEE-MAIN_WTM-02_QP_Dt.14-06-2020

Narayana CO Schools

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–:SZ2-A Jee-Main Exam Syllabus (14-06-20):–

TOPICS

PHYSICS

PRESENT WEEK 50%:

Newtons Laws of Motion: Newtons Laws,Concept of Force, Frame of

Reference (inertial and Non-inertial), introduction of Linear

Momentum and Linear Impulse, Free Body Diagrams, Types of Forces

(Normal, Weight, …….),Connected Bodies (Tension in Strings,

Contact Force),Atwood Machines, Applications.

PREVIOUS WEEK 50%:

Projectile motion on an inclined plane, Applications, Relative motion

between projectiles, condition for collision between projectiles,

Applications.

TOPICS

CHEMISTRY

PRESENT WEEK 50%:

HYBRIDISATION, VSEPR THEORY, BOND PARAMETERS

PREVIOUS WEEK 50%:

Resonance, characteristics of resonance, Properties of covalent

compound,s Co - ordinate covalent bond (dative bond) Properties

of co - ordinate covalent bond, VBT

TOPICS

MATHS

PRESENT WEEK 50%:

2D Co-Ordinate System: Co-Ordinate Geometry, 2d, Co Ordinate

System, Location Of Point In A Plane, Distance Formula, Section

Formula, Area Of A Triangle, Quadrilateral, Polygonrelated

Problems.

PREVIOUS WEEK 50%:

Transformations: Standard Formulae of Trigonometric

Transformations, Conditional Identities, Related Problems.

SZ2-A _JEE-MAIN_WTM-02_QP_Dt.14-06-2020

Narayana CO Schools

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SECTION-I(1-20)

(Single Answer Type)

This section contains 20 multiple choice questions. Each question has 4 options

(1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and -1 in all other cases.

01. An inclined plane makes an angle 0

0 30 = with the horizontal. A

particle is projected from this plane with a speed of15ms− at an angle

of elevation with the horizontal as shown in figure. Find the

range of the particle for 0120 = . ( )210 /g m s=

1) 1

3m 2)

2

3m 3)

4

3m 4)

5

3m

02. A body has maximum range 1R when projected up the plane. The

same body when projected down the inclined plane, it has

maximum range 2R . Assume 10 3 /u m s= in each case. Find 1 2

1 2

R R

R R+

( )210 /g m s=

1) 5m 2) 10m 3) 15m 4) 20m

03. A particle is projected on an inclined plane with a speed 10 2 /u m s=

as shown in figure. If it strikes the inclined plane normally then its

time of flight is ( )210 /g m s=

1) 1sec 2) 2sec 3) 3sec 4) 4sec


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04. A particle is projected up an inclined plane of inclination at an

elevation to the horizontal. Find the ratio between tan tanand

, if the particle strikes the plane horizontally.

1) 1 2) 2 3) 3 4) 4

05. A particle is thrown at time t=0 with a velocity of 110ms− at an angle

060 with the horizontal from a point on an inclined plane, making

an angle of 030 with the horizontal. The time when the velocity of the

projectile becomes parallel to the incline is

1) 2

3s 2)

1

3s 3) 3s 4)

1

2 3s

06. In the given figure, the angle of inclination of the inclined plane is

030 . A particle is projected with horizontal velocity 0 4 /V m s= from a

height H=4m. Find its velocity when the particle hits the inclined

plane perpendicularly. ( )210 /g m s=

1) 8m/s 2) 6 m/s 3) 2 m/s 4) 9m/s


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07. A particle P is projected from a point on the surface of smooth

inclined plane. Simultaneously another particle Q is released on the

smooth inclined plane from the same position. P and Q collide on

the inclined plane after t=4 second. The speed of projection of P is

(Take g=10 2/m s )

1) 5 m/s 2) 10 m/s 3) 15 m/s 4) 20 m/s

08. Two bodies A and B are projected from the same place in same

vertical plane with velocities 1 2v and v from a long inclined plane as

shown, find the ratio of their times of flight

1) 1

2

sinv

v

2) 1

2

2 sinv

v

3) 1

2

sin

2

v

v

4) 1

2

cosv

v

09. A ball is projected horizontally on the inclined plane as shown in

the figure. If 0 10 /V m s= then find the time in which its velocity is

parallel to the plane ( )( )2 010 / sin37 3/ 5g m s= =

1) 0.25 sec 2) 0.5 sec 3)0.75 sec 4)1 sec


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10. Angle of an inclined plane is with the horizontal. From the foot of

this plane a body is projected up the plane at an angle with the

plane. For maximum range =

1) 2 2

− 2)

4 2

− 3)

2 2

+ 4)

4 2

+

11. A body is projected down inclined plane of angle at an angle

with plane with a velocity u. Its maximum range is

1) ( )

2

1 sin

u

g + 2)

( )

2

1 sin

u

g − 3)

( )

2

1 cos

u

g + 4)

( )

2

1 cos

u

g −

12. In the direction shown in figure two bodies with velocities 20 /Av m s=

and 10 /Bv m s= are projected. They collide in air after 1

2s . Find the

distance x.

1) 2 3m 2) 3 3m 3) 4 3m 4) 5 3m

13. Inertia of a body has direct dependence on

1) velocity 2) mass 3) area 4) volume

14. A body is acted upon by a constant force then it will have a uniform

1) speed 2) momentum

3) velocity 4) acceleration

15. When a horse pulls a cart, the force which helps the horse to move

forward is the force exerted by

1) cart on the ground 2) ground on the cart

3) horse on the ground 4) ground on the horse

16. When we jump out of a boat standing in water, it moves

1) backward 2) forward

3) sideways 4) All the above


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17. A man is at rest in the middle of a pond on perfectly smooth ice. He

can get himself to shore by use of Newton’s

1) first law 2) second law 3) third law 4) all of above

18. A cricketer drags his hands backward to catch the ball, because

1) by increasing the time in which he stops the ball, reduces force

on his hands

2) by increasing distance, he allows ball to travel a greater

distance for more runs

3) it is an art to catch the ball

4) it is rule in cricket.

19. A force of 200N acts in the opposite direction on a body moving with

uniform velocity and brings it to rest in 0.25 sec. what is the initial

momentum of the body (in kg-m/s) (only magnitude)

1) 50 2) 75 3) 100 4) 125

20. A constant force F is applied on a stationary particle of mass m. The

velocity attained by the particle in a certain displacement will be

proportional to

1) m 2) 1

m 3) m 4)

1

m

SECTION-II(21-25)

(Numerical Value Answer Type)

This section contains 5 questions. The answer to each question is a Numerical values comprising of positive or negative decimal numbers (place value

ranging from Thousands Place to Hundredths Place). Eg: 1234.56, 123.45, -123.45, -1234.56, -0.12, 0.12 etc. Marking scheme: +4 for correct answer, 0 in all other cases.

21. A force-time (F-t) graph for a linear motion is shown in adjoining

figure. The segments shown are circular. The linear momentum

gained between 0 and 8 seconds is _______N – S.


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22. At time t second, a particle of mass 3kg has position vector r metre,

where ˆ ˆ3 4cosr ti tj= − . The impulse of the force during the time

interval 02

t

is ___________ ˆ .j Ns

23. A constant force 2 /2F m g= is applied on the block of mass 1m as

shown. The string and the pulley are light and the surface of the

table is smooth. Then the acceleration of 1m is

( )+

2

1 2

.m g

P m mThen

the value of P is

24. The force excreted by string on the pulley is ______ 2 .N ( )210 g ms−=

25. If 1 10m kg= , 2 4m kg= and 3 2m kg= the tension in the string

connecting 1m and 2m is .Ag

B Then the value of A B− is


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SECTION-I(26-45)


This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be

correct. Marking scheme: +4 for correct answer, 0 if not attempted and -1 in all other cases.

26. Covalent compounds are generally soluble in

1) Polar solvents 2) Non-polar solvents

3) Concentrated acids 4) All solvents

27. Most important concept of valence bond theory is

1) Overlap of atomic orbitals results in the bond

2) Sharing of odd number of electrons for bonding

3) Sharing of electrons follow the octet rule

4) Transfer of electron follow the octet rule

28. 4CCl is insoluble in water because

1) 2H O is non-polar

2) 4CCl is non-polar

3) They do not from inter molecular H-bonding

4) They do not form intra molecular H-bonding

29. The type of bonds present in ammonium chloride are

1) Only ionic and dative

2) Only covalent and electrovalent

3) Only covalent and coordinate

4) Ionic, covalent and coordinate

30. 3PH and 3BF form an adduct readily because they form

1) A coordinate bond 2) A covalent bond

3) An ionic bond 4) A hydrogen bond

31. Dative bond is present in the molecule of

1) 3NH 2) 2CO 3) CO 4) 5PCl


Narayana CO Schools

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32. According to valence bond theory, water molecule has

1) Two dative bonds and bond angle 090

2) Two covalent bonds and bond angle 090

3) Two dative bonds and bond angle 0104.5

4) Two covalent bonds and bond angle 0104.5

33. According to V.B theory, the bonds in methane are formed due to

the overlapping

1) 1 ,3s s s p − − 2) 1 ,3s p s s − −

3) 2 , 2s s s p − − 4) 34 sp s −

34. Which of the following does not contain coordinate covalent bond?

1) 4NH + 2) 3H O+ 3) 3CH − 4) All of these

35. Molecules with 2sp hybridisation will have the following shape

1) Linear 2) Triangle planar

3) Tetrahedral 4) Pyramidal

36. The type of hybrid orbitals employed in the formation of 6SF

molecules is

1) 3sp d 2) 3sp 3) 3 2sp d 4) 2 3d sp

37. The % of ' 'p character in hybrid orbital of the central atom of water

molecule

1) 25% 2) 50% 3) 75% 4) 33.3%

38. Which hybridisation is found in 3NH and 2 ?H O

1) 3sp 2) 2dsp 3) sp 4) 2sp

39. In the molecule of 4 ,XeF hybridisation of ' 'Xe atom is

1) 3sp 2) 3sp d 3) 3 2sp d 4) 2 3d sp

40. Which one of the following is a linear molecule?

1) 2NO 2) 2SO 3) 2CO 4) 2H S

41. In 3BCl molecule, the Cl B Cl− − bond angle is

1) 090 2) 0120 3) 0109 28' 4) 0180


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42. Bond angle ( )H O H− − in 2H O is

1) 090 2) 0104 30 '

3) 0107 18 ' 4) 0109 28'

43. In 5 ,PCl bond angle in plane is

1) 090 2) 0120 3) 0180 4) 0109 .28 '

44. The shape of 3AB E is

1) Pyramidal 2) Tetrahedral

3) Angular 4) Linear

45. In the formation of 6SF molecules, the sulphur atom is in

1) first excited state 2) second excited state

3) third excited state 4) fourth excited state

SECTION-II(46-50)




46. In 4 2.5CuSO H O total number of 2H O molecule which form

coordinate bond with metal ion.

47. Total number of bonds in 5PCl which are at 090 to each other, is

48. The total number of electrons that take part in forming bonds in 2O

molecule according to V.B.T

49. The number of possible resonance structures for 2

3CO− is

50. The number of possible resonance structures for 2CO is


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SECTION-I (51-70)


This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be

correct. Marking scheme: +4 for correct answer, 0 if not attempted and -1 in all other cases.

51. If ( ) ( )2cos 3cos ,+ = − then tan .tan is

1) 1

5 2) 2 3) 5− 4)

1

5−

52. 2 0 2 0 0 0sin 14 cos 16 sin14 cos16+ − =

1) 1

4 2)

3

4 3)

1

2 4)

3

2

53. 0 0 0 0cos 20 sin 50 sin 5 cos65− =

1) 5 1

4

+ 2)

5 1

4

− 3)

3 1

4

− 4)

3 1

4

+

54. 0 0 0

2 0 01 1 12sin 8 4cos16 sin 7 sin 8 cos32

2 2 2+ + =

1) 3 1

2 2

− 2)

3 1

2 2

+ 3)

3 1

4 2

− 4)

3 1

2

+

55. If 4

cos cos ,5

x y+ = 2

cos cos ,7

x y− = then 14 tan 5cot2 2

x y x y− + + =

1) 0 2) 1

4 3)

5

4 4)

3

4

56. If 3

cos cos2

+ = and 1

sin sin ,2

+ = is the A.M of and , then

sin2 cos2+ is equal to

1) 5

2)

4

5 3)

7

5 4)

8

5


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57. ( ) ( ) ( )

( ) ( )

sin 1 2sin sin 1

cos 1 cos 1

n A nA n A

n A n A

+ + + −=

− − +

1) tan2

A 2) 0tan 90

2

A −

3) tan A 4) cot A

58. If 0180 ,A B C+ + = then 2 2 2cos cos cosA B C+ + =

1) 1 2cos cos cosA B C+ 2) 1 2sin sin sinA B C+

3) 1 2cos cos cosA B C− 4) 1 2sin sin sinA B C−

59. In sin sin sin

,sin sin sin

A B CABC

A B C

+ +

+ − is equal to

1) tan .cot2 2

A B 2) cot . tan

2 2

A B

3) cot .cot2 2

A B 4) tan . tan

2 2

A B

60. If 0270 ,A B C+ + = then cos2 cos2 cos2 4sin sin sinA B C A B C+ + + =

1) 0 2) 1 3) 2 4) 3

61. The coordinates of a point which divides externally the line joining

( )1, 3− and ( )3,9− in the ratio 1:3 is

1) ( )9, 3− 2) ( )9,3− 3) ( )3, 9− 4) ( )3,9

62. If the distance between the points ( ), 2a and ( )3,4 is 8, then ' 'a is

equal to

1) 2 3 15+ 2) 2 3 15− 3) 5 2 15 4) 3 2 15

63. The three points ( ) ( ) ( )2,2 , 8, 2 4, 3and− − − − are the vertices of

1) an isosceles triangle 2) an equilateral triangle

3) a right angled triangle 4) Scalene triangle


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64. The points ( ) ( ) ( ) ( )4, 1 , 2, 4 , 4,0 , 2,3− − − − taken in order are the vertices

of a

1) Parallelogram 2) Rhombus

3) Rectangle 4) Square

65. If ( ) ( ) ( )2, 3 , 6,5 , 2,1− − are three consecutive vertices of a rhombus,

then its area is (in sq.units)

1) 24 2) 36 3) 18 4) 48

66. If ( ) ( )2,1 , 2,5− are two opposite vertices of a square, then its area is

(in sq.units)

1) 4 2) 12 3) 16 4) 36

67. The points ( ) ( )1,2 , ( 3,4), 7, 1A B C− − are collinear. The ratio in which A

divides BC is

1) 2:3 2) 3:2 3) – 2:3 4) –3:2

68. The coordinates of three consecutive vertices of a parallelogram are

( ) ( )1,3 , 1,2− and ( )2,5 , then the coordinates of fourth vertex are

1) ( )6, 4 2) ( )4, 6− 3) ( )4, 6− − 4) ( )4,6

69. The points with the coordinates ( ) ( )2 ,3 , 3 ,2a a b b and ( ),c c are

collinear

1) for no values of , ,a b c 2) for all values of , ,a b c

3) if , ,5

ca b are in H.P 4) if

2, ,

5

ca b are in H.P


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70. Let ( ) ( )1, , 1,1A k B and ( )2,1C be the vertices of a right angled triangle

with AC as its hypotenuse. If the area of the ABC is 1 sq unit. Then

set of values which k can take is given by

1) 1,3 2) 0, 2 3) 1,3− 4) 3, 2− −

SECTION-II(71-75)




71. The value of the expression 0 0

0

1 4sin10 sin 7019

2sin10

−

is equal to

72. If 0 0 0cos55 , cos65 , cos175 ,x y z= = = then ( )

264 xy yz zx+ + is equal to

73. The area of the pentagon whose vertices are ( ) ( ) ( ) ( )4,1 , 3,6 , 5,1 , 3, 3− − −

and ( )3,0− is ( in sq.units)

74. The area of the quadrilateral formed by the points ( ) ( ) ( )2, 1 , 4,3 , 1,2− −

and ( )3, 2− − is ( in sq.units)

75. If ( ) ( ) ( ) ( )3,4 , 1, 2 , 5,6 , , 4A B C D x= − = − − = = − are vertices of a

quadrilateral such that area of triangle 2ABD = are of triangle ACD,

then x is equal to

" When you can't control what is happening, challenge yourself to control

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class: sz2-a jee-main model date: 14-06-2020 time: 3hrs ... · previous week 50%: transformations:...

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