class: sz1 jee-main model date: 10-05-2020 time ......sz1_jee-main_wtm-18_qp_exam.dt.10-05-2020...
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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
Sec – I(Q.N : 26 – 45) Questions with Single Answer Type 4 -1 20 80
Sec – II(Q.N : 46 – 50) Questions with Numerical Answer Type
(+/ - Decimal Numbers) 4 0 5 20
Total 25 100
MATHEMATICS Section Question Type
+Ve Marks
- Ve Marks
No.of Qs
Total marks
Sec – I(Q.N : 51 – 70) Questions with Single Answer Type 4 -1 20 80
Sec – II(Q.N : 71 – 75) Questions with Numerical Answer Type
(+/ - Decimal Numbers) 4 0 5 20
Total 25 100
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
Narayana CO Schools
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–: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.
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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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
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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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
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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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
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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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
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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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 +
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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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
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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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 __________
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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SECTION-I (26 TO 45)
(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.
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
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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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
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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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)
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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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
SECTION-II (46 TO 50)
(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.
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
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
Narayana CO Schools
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SECTION-I (51 TO 70)
(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.
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
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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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− +
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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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
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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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+ = −
SECTION-II (71 TO 75)
(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.
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
SZ1_JEE-MAIN_WTM-18_QP_Exam.Dt.10-05-2020
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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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