Class: SZ1 JEE-MAIN MODEL Date: 10-05-2020 Time ......SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020 Narayana CO Schools 4 04. Flux through the cone when a charge q is placed h distance

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Class: SZ1 JEE-MAIN MODEL Date: 10-05-2020

Time: 3hrs WTM-18 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

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SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020

Narayana CO Schools

2

–:SZ1 Jee-Main Exam Syllabus (10-05-20):–

TOPICS

PHYSICS

PRESENT WEEK 50%:

Gauss's law, Calculation of electric field using Gauss's law for Charged

sphere,non conducting unifromly charged sphere, Calculation of electric

field using Gauss's law for Long charged wire, Long uniformly charged

conducting and non conducting cylinder,, Calculation of electric field

using Gauss's law for Non conducting uniformly charged sheet, Charged

conducting sheet), Applications of Gauss's law in the region of varying

electric field Applications, Electric field inside a cavity of conducting and

non-conducting charged body, Applications.

PREVIOUS WEEK 50%:

ELECTRO STATICS : Introdution to electric flux, Flux in non uniform elelctric

field, Flux through a (Circular disc, Lateral surface of a cylinder due to

point charge), Flux produced by a point charge, Applications, Concept

of solid angle, Relation between half angle of cone and solid angle at

vertex, Flux calculation using solid angle, Applications.

TOPICS

CHEMISTRY

PRESENT WEEK 50%:

Complex formation, Interstitial compounds, Alloy formation, Preparation

and properties and uses of K2Cr2O7 and KMnO4.

PREVIOUS WEEK 50%:

d- BLOCK ELEMENTS: d- Block elements General characteristics,

Oxidation states, Colour, Magnetic properties,

TOPICS

MATHS

PRESENT WEEK 50%:

DEFINITE INTEGRATION: Introduction, Fundamental Theorem of Integral

Calculus - I and II, Fundamental Theorem of Integral Calculus - I and II,

Properties of Definite Integrals (Improper Integrals are to be strictly

avoided), Related Problems.

PREVIOUS WEEK 50%:

COMPLEX NUMBERS: Section Formula, Area of a Triangle. Locus

Problems ,Rotation and Coni's Formula, Equation of a Line and a Circle

in Complex form, Geometrical Applications, Related Problems.


Narayana CO Schools

3

SECTION-I (1 TO 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. The field potential in a certain region of space depends only on the

x coordinate as 3ax b = − + , a and b are constants, the distribution

of the space charge ( )x is

1) 0

6 ax 2) 0

2 ax 3) 0

4 ax 4) 0

8 ax

02. A uniform field E is parallel to axis of a hollow hemisphere of radius

r.

Electric flux through the hemisphere surface is

1) 24 r E 2)

33

8r E 3)

2r E 4) 22 r E

03. A linear charge having linear charge density , penetrates a cube

diagonally and then it penetrates a sphere diametrically as

If ratio of flux through cube and through sphere is r then value of

2

3r is

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


Narayana CO Schools

4

04. Flux through the cone when a charge q is placed h distance from

the base of centre cone is 0

2

3

q =

.

If charge is placed h distance above the base on the centre line, flux

through the cone is

1) 0

2

3

q 2)

0

3

2

q 3)

0

3

q 4)

0

2

q

05. Consider an electric field 0ˆE E x= , where 0E is a constant. The flux

through the shaded area (as shown in the figure below) due to this

field is

1) 2

02E a 2)

2

02E a 3)

2

0E a 4)

2

0

2

E a


Narayana CO Schools

5

06. In given situation, radius of disc is 3band distance of point charge

from the disc is b. Ratio of electric flux not going through the disc

and electric flux of charge through the disc is k. The value of k is

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

07. A sphere of radius R surrounds a point charge Q, located at its

centre. Net outward flux through of semi-vertical angle and of

slant length R with vertex at centre is

1) ( )0

2 1 cos

Q

− 2)

0

cos

4

kQ

3) ( )0

1 cosQ

− 4) ( )0

1 cos2

Q

−

08. A small charged sphere with charge q mounted over one of the

vertices of the cube as shown. Then flux through shaded face of

cube is ____________

1) 0

q

2)

024

q

3)

06

q

4)

08

q


Narayana CO Schools

6

09. Electric charges are distributed in a small volume. The flux of the

electric field through a spherical surface of radius 10 cm

surrounding the total charge is 25 vm. The flux over a concentric

sphere of radius 20 cm will be __________

1) 25 vm 2) 50 vm 3) 100 vm 4) 200 vm

10. A charge q is placed at the centre of the open end of a cylindrical

vessel shown in figure. The flux of the electric field through the

surface of the vessel is

1) Zero 2) 0

q

3)

02

q

4)

0

2q

11. Figure shows a charge q placed at the centre of a hemisphere. A

second charge Q is placed at one of the positions A, B, C and D. In

which position(s) of this second charge, the flux of the electric field

through the hemisphere remains unchanged?

I) A II) B III) C IV) D

1) I,III 2) I,II 3) II,III 4) III,IV


Narayana CO Schools

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12. If the flux of the electric field through a closed surface is zero, then

the charge inside the surface must be

1) 2) 0 3) 1 4) 0

q

13. Mark the correct option.

1) The flux of the electric field through a closed surface due to all

the charges is equal to the flux due to the charges enclosed by

the surface

2) Gauss’ law is valid only for charges placed in vacuum

3) Gauss’ law is valid only for symmetrical charge distributions

4) Electric flux is a vector quantity

14. A charge ( )q is placed at the centre of a sphere. Taking outward

normal as positive, the flux of the electric field through the surface

of the sphere due to the enclosed charge is _______

1) 02

q

2)

0

q

3)

03

q

4) 0

15. A uniform electric field exists in space. The flux of this field through

a cylindrical surface with the axis parallel to the field is ___________

1) 2) 0

3) 1 4) All of these

16. If the electric flux entering and leaving an enclosed surface

respectively is 1 and 2 , then electric charge inside the surface will

be

1) ( )2 1 0 − 2) ( )1 2

0

+

3) ( )2 1

0

−

4) ( )1 2 0 +


Narayana CO Schools

8

17. A charged particle q is placed at the centre O of cube of

L(ABCDEFGH). Another same charge q is placed at a distance L

from O. Then, the electric flux through ABCD is

1) 04

q

L 2)

02

q

L

3) 08

q

L 4) None of these

18. A uniform electric field of magnitude 100 /E N C= exists in the space

in X-direction. The flux of this field through a plane square area of

edge 20cm placed in the yz plane is __________ 2 / .Nm C ( Take the

normal along the positive X-axis to be positive)

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

19. The electric field in a region radially outward with magnitude

.E Aa= The charge contained in a sphere of radius ( )a centred at

the origin is _____________(nearly)

(Take 2100 / , 20A v m a cm= = )

1) 118.89 10 C− 2) 129 10 C− 3) 119.69 10 C− 4) 0

20. A charge(Q) is placed at the centre of a cube. The flux of the electric

field through the six surfaces of the cube is __________

1) 0

Q

2)

06

Q

3)

03

Q

4) None


Narayana CO Schools

9

SECTION-II (21 TO 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 charge(Q) is situated at the centre of a cube. The electric flux

through one of the faces of the cube is 0

.Q

N Then the value of N is

22. A cube of side ( )l is placed in a uniform electric filed ( ) ,E where

,E Ei= the net electric flux through the cube is ____________

23. A point charge ( )q+ is placed at the centre of a cube of side(L). The

electric flux emerging from the cube is 0

.Nq

p Then the value of N p−

is

24. The net flux emerging from given enclosed surface is

12 2__________ 10 / .Nm C

25. The electric field in a region of space given by 5 2 / .E i j N C= + The

electric flux due to this field through an area 24m lying in the yz

plane, in S.I units is __________


Narayana CO Schools

10

SECTION-I (26 TO 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. The general electronic configuration of d- block elements is

1) 1 10 1 2(n-1)d ns− − 2) 1 2 1 10(n-1)d ns− −

3) ( )1 10 1 2nd 1n s− −− 4) ( )1 2 1 10nd 1n s− −−

27. Manganese has the highest oxidation state in

1) 3 4Mn O 2) 2MnO

3) 2 4K MnO 4) 4KMnO

28. Which of the following transition elements shows the highest

oxidation state?

1) Fe 2) Mn 3) Cr 4) V

29. How many unpaired electrons are present in 2Ni + ?

1) 0 2) 2 3) 4 4) 8

30. Which of the following are trapped inside the crystal lattices of

metals to form interstitial compounds?

1) Hydrogen 2) Carbon

3) Nitrogen 4) All of these

31. The magnetic nature of elements depends on the presence of

unpaired electrons. Identify the configuration of transition element,

which shows highest magnetic moment.

1) 73d 2) 53d 3) 83d 4) 23d

32. Transition elements form complexes easily because of

1) Large ionic charge 2) Small cation size

3) Availability of d-orbitals 4) All of the above


Narayana CO Schools

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33. Brass is an alloy of

1) copper -zinc 2) copper -tin

3) Zinc-tin 4) Iron-carbon

34. 4KMnO acts as an oxidising agent in alkaline medium. When

alkaline 4KMnO is treated with KI, iodide ion is oxidised to

1) 2I 2) IO− 3) 3IO − 4) 4IO −

35. When acidified 2 2 7K Cr O solution is added to 2Sn + salts then 2Sn +

changes to

1) Sn 2) 3Sn + 3) 4Sn + 4) Sn+

36. Highest oxidation state of manganese in fluoride is +4 ( )4MnF but

highest oxidation state in oxides is +7 ( )2 7Mn O because

____________.

1) fluorine is more electronegative than oxygen.

2) fluorine does not possess d-orbitals.

3) fluorine stabilises lower oxidation state.

4) in covalent compounds fluorine can form single bond only while

oxygen forms double bond.

37. Which of the following are the characteristics of interstitial

compounds

1) They are chemically inert

2) They retain metallic conductivity

3) They are very hard

4) All of the above


Narayana CO Schools

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38. Why is HCl not used to make the medium acidic in oxidation

reactions of 4KMnO in acidic medium?

1) Both HCl and 4KMnO act as oxidising agents.

2) 4KMnO oxidises HCl into 2Cl which is also an oxidising agent.

3) 4KMnO is a weaker oxidising agent than HCl .

4) 4KMnO acts as a reducing agent in the presence of HCl .

39. Which of the following metal does not show variable oxidation state

1) Zn 2) Fe 3) Cu 4) Cr

40. Potassium dichromate is prepared by

1) Treating the solution of sodium dichromate with potassium

chloride

2) Heating powdered chromite ore with quicklime in the presence

of air

3) Heating K2CrO4 with NaOH

4) All of these

41. In which of the following does manganese have an oxidation state

of 4?

1) Mn2O3 2) MnO2 3) MnO3 4) MnO

42. Which of the following statements are true?

1) Most compounds formed by transition elements are coloured.

2) Transition metals does not show magnetic behaviour.

3) Compounds of transition metals show diamagnetic behaviour.

4) Most compounds formed by transition elements are colourless.

43. Transition elements

1) do not form alloys 2) form coloured salts

3) show variable oxidation states 4) Both (2) & (3)


Narayana CO Schools

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44. When 4KMnO acts as an oxidising agent and ultimately forms

2 2

4 2 MnO , MnO ,Mn− + then the number of electrons transferred in

each case respectively is

1) 4, 3, 5 2) 1, 5, 7 3) 1, 3, 5 4) 3, 5, 1

45. Potassium permanganate is prepared by

1) Fusion of MnO2 with an alkali metal hydroxide followed by

oxidation with air

2) Manganese (II) ion salt is oxidised by peroxodisulphate

3) Electrolytic oxidation of potassium manganate in alkaline

solution

4) All of these



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.

46. The number of unpaired electrons in 2Fe + is

47. Electronic configuration of a transition element X in +3 oxidation

state is 53Ar d . What is its atomic number?

48. In the conversion of 4KMnO into 2 3Mn O the number of electrons

transferred is

49. The oxidation state of Cr in K2Cr2O7 is

50. The maximum oxidation state of osmium is


Narayana CO Schools

14

SECTION-I (51 TO 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. The value of ( )

1

2 1

2 20

sin

1 1

xdx

x x

−

− − is equal to

1) 1

ln 24 2

− 2)

1ln 2

2 2

−

3) 1

ln 24 2

+ 4)

1ln 2

2 2

+

52. The value of ( )

441

2

0

1

1

x xdx

x

−

+ is equal to

1) 0 2) 22

7− 3)

22

7− 4)

22

7+

53. If ( )

2 2

2 3

cos 2 cos2

sec sin ,

1 2 tan

x xx e x

f x x x x x

x x

= +

+

then ( ) ( ) ( )2

2

2

1 ''x f x f x dx

−

+ + =

1) 2) 2

3)

3

2

− 4) 0

54. The value of 1

2

0

1

2 cos 1dx

x x=

+ +

1) 2sin

2) sin

2

3) .sin 4)

2sin

55. The value of ( )4

4

ln sin cosx x dx

−

+ is equal to

1) ln 22

− 2) ln 2

2

3) ln 2

4

− 4) ln 2

4


Narayana CO Schools

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56. If 22

0

sin,

sinn

nxI dx

x

= then 2 1 3 2 4 3, ,I I I I I I− − − are in

1) A.G.P 2) G.P 3) H.P 4) A.P

57. ( )( ) _________

b

a

x a b x dx− − =

1) 2) ( )2

8b a

− 3) ( )

2b a

− 4) ( )

2a b

+

58. 3

2

0

.sin_______

1 cos

x xdx

x

=+

1) ( )22

+ 2)

2 4

− 3) ( )2

2

− 4)

2

2

59. ( )

22

5

0

sec

sec tan

xdx

x x

=+

1) 3

8 2)

4

15 3)

2

3 4)

5

24

60. 4

0

sin cos

9 16sin 2

x xdx

x

+=

+

1) 1

ln 310

2) 1

ln 320

3) 1

ln 340

4) 1

ln 325

61. Let 1 2 3 4 5, , , ,z z z z z and 6z be the vertices of a regular hexagon in anti-

clockwise sense. Then 3z is equal to

1) ( ) ( )1 2

1 11 3 3 3

2 2i z i z− + + + 2) ( ) ( )1 2

1 13 3 1 3

2 2i z i z− − −

3) ( )1 23 1 3i z i z− + + 4) ( ) 1 21 3 3i z i z− +


Narayana CO Schools

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62. The mirror image of the curve given by arg1 4

z i

z

+ =

− in the line y x=

is

1) arg3 6

z i

z

+ =

− 2) arg

1 4

z i

z

− =

+

3) 1

arg6 4

z

z

+ =

− 4) arg

1 4

z i

z

+ =

+

63. The complex number 3 4z i= − is rotated through an angle 0180

about the origin in anti-clockwise direction and its magnitude is

increased by two and half times. In new position z is equal to

1) 15

102

i+ 2) 15

102

i−

+ 3) 15 10i− + 4) 15

102

i− −

64. ABC is a triangle inscribed in the circle .z r= The internal bisector

of angle A meets the circle at D. If , , ,A B C D are represented by

1 2 3 4, , ,z z z z , then

1) 1 4 2 3z z z z= 2) 1 3 2 4z z z z= 3) 1 2 3 4z z z z= 4) 2

2 3 4z z z=

65. Given that the two curves ( )arg6

z

= and 2 3z i r− = intersect in two

distinct points, then ([.] denotes G.I.F)

1) 3r 2) 2r = 3) 0 3r 4) 2r

66. Let 'a be the reflection of a point a in the line 0bz bz c+ + = where

, ',a a b are non-zero complex numbers and ,c R then 'a is equal to

1) ba c

b

+ 2)

ba c

b

+−

3) ab c

b

+ 4)

ab c

b

−

67. Let complex numbers and 1

lie on the circle 0z z r− = and

0 2z z r− = respectively, where 0 0 0.z x iy= + If 2 2

02 2z r= + then is

equal to

1) 1

7 2)

1

7 3)

2

7 4) 7


Narayana CO Schools

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68. If ( )arg , 02

z

= and 3 3,z i− = then 6

cotz

− is equal to

1) 1 2) 1− 3) i 4) i−

69. Let A,B and C represent the complex numbers 1 2,z z and 3z in the

argand plane. If circumcentre of the triangle ABC is at the origin,

then the complex number corresponding to orthocentre is

1) ( )1 2 3

1

4z z z+ + 2) ( )1 2 3

1

3z z z+ +

3) ( )1 2 3

1

2z z z+ + 4) ( )1 2 3z z z+ +

70. Equation of the line making equal intercepts on the axes and

touching 2 1z = is

1) 1x y+ = 2) 9x y− =

3) 4x y− = 4) 10x y+ = −



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.

71. The minimum distance between the circles 1z = and

1

2

arg ,4

z z

z z

− =

− 1 29 5 , 3 5z i z i= + = + is ' 'd then d is ([.] denote G.I.F)

72. Let 1z and 2z vary over the curves ( )3 1 2 2 2 7z i− + = and

( ) ( )3 1 2 3 9 18z i z i− − − = − + respectively. If the minimum value of

2

1 2 ,z z d− = then the greatest integer value of d is equal to


Narayana CO Schools

18

73. If ( )( ) ( )102

0

1 1 11 2 .... 100 ..... ! !

1 2 100;x x x dx p q

x x x

− − − + + + = − − − −

, ,p q N then p q− is equal to

74. If ( )( )

1

2 2 4

0

1

5 2 2 1 xdx

x x e −=

+ − +

1 1ln

2 1

b

a b

+ −

where , ,a b N then a b+

is equal to

75. 2

0

cos9 cos6

2cos5 1

x xdx

x

+

− is equal to

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Documents

Class: SZ1 JEE-MAIN MODEL Date: 10-05-2020 Time ......SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020 Narayana CO Schools 4 04. Flux through the cone when a charge q is placed h distance