06-01-24_JEE-MAIN_Q’P

PHYSICS MAX.MARKS: 100

SECTION – I
(SINGLE CORRECT 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 if not correct.
1. A circular parallel plate capacitor of radius R and distance d between the plate is given.
A capacitor is being charged with a current I flowing through the wire. Neglect fringing
effect.

I I

What is the rate of change of electric flux through plane in middle of capacitor with
d
respect to time (i.e. )–
dt

2I I 4I 6I
A) B) C) D)
0 0 0 0

2. A particle is moving in a circular path. The acceleration and momentum of the particle at
 2 
a certain moment are a  (4î  3 ĵ) m/s and p  (8î  6 ĵ) kg-m/s. The motion of the particle at
that instant is

A) Uniform circular motion

B) accelerated circular motion

C) decelerated circular motion

D) We cannot say anything with 

a and p only

3. A block starts moving up a fixed inclined plane of inclination 60° with a velocity of

20 m/s and stops after 2 sec. The approximate value of coefficient of friction is

(g = 10 m/s2)

A) 3 B) 3.3 C) 0.27 D) 0.33

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4. A ball of mass m approaches a wall of mass M (>> m) with speed 4 m/s along the normal to
the wall. The speed of wall is 1m/s towards the ball. The speed of the ball after an elastic
collision with the wall is -
A) 5 m/s away from the wall B) 9 m/s away from the wall
C) 3 m/s away from the wall D) 6 m/s away from the wall
g
5. The height at which the acceleration due to gravity becomes (where g = the
9
acceleration due to gravity on the surface of the earth) in terms of R, the radius of the
earth, is
R
A) 2R B) C) R / 2 D) 2 R
2

6. An ideal gas is expanded so that amount of heat given is equal to the decrease in internal
energy. The gas undergoes the process TV1/5 = constant. The adiabatic compressibility of
gas when pressure is P, is –
7 5 2 7
A) B) C) D)
5P 7P 5P 3P

7. An ideal gas is held in a container of volume V at pressure P. The average speed of a gas
molecule under these conditions is v. If now the volume and pressure are changed to 2V
and 2P, the average speed of a molecule will be
A) 1/2 v B) v C) 2v D) 4v
8. A wire is 4 m long and has a mass 0.2 kg. The wire is kept horizontally. A transverse
pulse is generated by plucking one end of the taut (tight) wire. The pulse makes four
trips back and forth along the cord in 0.8 sec. The tension in the cord will be -
A) 80 N B) 160 N C) 240 N D) 320 N
9. Two identical simple pendulums A and B are fixed at same point. They are displaced by
very small angles  and  (    ) and released from rest. Find the time after which B
reaches its initial position for the first time. Collisions are elastic and length of strings is
.

 

A B

    2 
A)  B) 2 C) D)
g g  g  g
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10. The figure shows a charge q placed inside a cavity C in an uncharged conductor. Now if
an external electric field is switched on then :

C
q

A) only induced charge on outer surface will redistribute.

B) only induced charge on inner surface will redistribute

C) Both induced charge on outer and inner surface will redistribute.

D) force on charge q placed inside the cavity will change

11. STATEMENT – 1

If potential difference between two points is zero and the resistance between the same
two points is also zero, then current may flow between those two points

STATEMENT – 2

Kirchhoff’s 1st law is conservation of charge.

A) Statement – 1 is True. Statement – 2 is True; Statement – 2 is a correct explanation

for Statement – 1.

B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is not a correct

explanation for Statement – 1.

C) Statement – 1 is True, Statement is False.

D) Statement – 1 is False, Statement is True.

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12. An  particle is moving along a circle of radius R with a constant angular velocity  .
Point A lies in the same plane at a distance 2R from the centre. Point A records magnetic
field produced by  particle. If the minimum time interval between two successive
times at which A records zero magnetic field is 't', then the angular speed  is :
2 2  
A) B) C) D)
t 3t 3t t

13. Curie temperature is the temperature above which -

A) a ferromagnetic material behaves like paramagnetic

B) a paramagnetic material behaves like diamagnetic

C) a ferromagnetic material behaves like diamagnetic

D) a paramagnetic material behaves like ferromagnetic

14. In an LR circuit connected to a battery, the time rate of change of energy stored in the
inductor is plotted against time during the growth of current in the circuit. Which of the
following best represents the resulting curve?
Rate Rate

A) B)

Time Time

Rate Rate

C) D)

Time Time

15. When an object is at distance x and y from a lens, a real image and a virtual image is
formed respectively having same magnification. The focal length of the lens is –
xy
A) B) x – y C) xy D) x + y
2

16. If the zero of the Vernier lies on the right hand side of zero of the main scale and fourth
division coincide with the main scale division when the jaws are in contact, the zero
correction will be __ (assume standard Vernier Calipers)

A) + 0.04 cm B) + 0.06 cm C) –0.04 cm D) –0.06 cm

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17. A zener diode is to be used as a voltage regulator. Identify the correct set up –
Rs Rs
+ +

A) RL B) RL

– –

Rs Rs
+ +

C) RL D) RL

– –

 R 
18. The graph of log  
 versus log A (R = radius of a nucleus and A = mass number) is -
 R0 

A) a circle B) an ellipse C) a parabola D) a straight line

19. Match the entries of Column I with those in Column II

Column I Column II

Mass of products less than mass

(A) (P) Most stable
of reactants

Total internal reflection

(B) Iron nucleus (Q)
take place

(C) Angle of contact is obtuse (R) Nuclear fission

Incidence angle more than critical

(D) (S) Stronger cohesive forces
angle

A) A – R; B – P; C – Q ; D – S B) A – R; B – P; C – S ; D – Q

C) A – Q; B – S; C – R ; D – P D) A – Q; B – S; C – P ; D – R
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20. In a YDSE experiment, I0 is given to be the intensity of the central bright fringe &  is the
fringe width. Then, at a distance y from central bright fringe, the intensity will be -

(y is very small)

 y   y   2y   y 
A) I0 cos    B) I 0 cos 2   C) I 0 cos  D) I 0 cos 2  
         2 

SECTION-II
(NUMERICAL VALUE ANSWER TYPE)
This section contains 10 questions. The answer to each question is a Numerical value. If the Answer in the
decimals , Mark nearest Integer only. Have to Answer any 5 only out of 10 questions and question will be
evaluated according to the following marking scheme:
Marking scheme: +4 for correct answer, -1 in all other cases.
21. A particle of mass 10–2 kg is moving along the positive x-axis under the influence of a
force F(x) = – K/(2x2) where K = 10–2 Nm2. At time t = 0 it is at x = 1.0 m and its
velocity is v = 0. Find its velocity when it reaches x = 0.50 m. (in m/s)

22. A uniform ball of radius R = 10 cm rolls without slipping between two rails such that the
horizontal distance is d = 16 cm between two contact points of the rail to the ball. If the
angular velocity is
5 rad/s, then find the velocity of centre of mass of the ball in cm/s.

23. A disc of radius '5cm' rolls on a horizontal surface with linear velocity v = 1 î m/s and
angular velocity 50 rad/sec. Height of particle from ground on rim of disc which has
velocity in vertical direction is (in cm) -

y

v x

24. A liquid flows out drop by drop from a vessel through a vertical tube with an internal
diameter of 2 mm, then the total number of drops that flows out during 10 grams of the
liquid flow out: [Assume that the diameter of the neck of a drop at the moment it breaks
away is equal to the internal diameter of tube and surface tension is 0.02 N/m,
g = 9.8 m/s2]
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25. A tuning fork is in unison with a sonometer wire vibrating in its fourth overtone. Mass

hanged with wire is 9 kg. When additional mass is hanged wire vibrates in unison with

tuning fork in its 3rd harmonic. Additional mass hanged in kg is.

26. Two conducting concentric spherical shells are present. If the electric potential at the

centre is 2000 V and the electric potential of the outer shell is 1500 V. then the potential

of the inner shell in volts is

27. A series LCR circuit containing a resistance of 120 has angular resonance frequency

4 × 105rads–1. At resonance the voltage across resistance and inductance are 60 V & 40

V respectively. At what frequency the current in the circuit lags the voltage by 45°. Give

answer in -------×105 rad/sec.

28. The energy stored in a 90 cm length of laser beam operating at 10 mW is X × 10–11 J.

Find the value of X.

29. A thin prism P1 with angle 4º and made from glass of refractive index 1.54 is combined
with another thin prism P2 made from glass of refractive index 1.72 to produce
dispersion without deviation. The angle of prism in degrees is

30. An experiment with convex lens gives certain result which is represented by a student in
the shown graph. The power of the lens in diopters is

0.2
1 –1
cm
0.1 v

0.2 0.1
1
cm–1
u
 06-01-24_JEE-MAIN_Q’P
CHEMISTRY MAX.MARKS: 100
SECTION – I
(SINGLE CORRECT 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 if not correct.
31. The element with  Xe  4f 14 5d1 6s 2 belongs to

A) 7th period and Group 4 of the modern periodic table

B) 6th period and Group 3 of the modern periodic table
C) 5th period and Group 3 of the modern periodic table
D) 6th period and Group 5 of the modern periodic table
32. The relative stability of +1 oxidation state of group 13 elements follows the order:
A) A  Ga  T  In B) Ga  A  In  T

C) T  In  Ga  A D) A  Ga  In  T

33. Assertion (A) : SO2 turns acidified K 2Cr2O7 green.

Reason (R) : K 2Cr2O7 is reduced to Cr 3 (green) salt by SO2 .

A) Both A and R are true and R is the correct explanation of A
B) Both A and R are true but R is not the correct explanation of A
C) A is true but R is false
D) A is false but R is true
34. Product (X and Y) of the following reactions (1 and 2) are
1) 2NaOH  C 2  NaC  X  H 2O
( Cold and dilute )

2) 6NaOH  3C 2  NaC  Y  H 2O

( Hot and conc.)

A) X  NaCO3 and Y  NaOC B) X  NaOC and Y  NaCO3

C) X  NaHCO3 and Y  NaOC D) X  NaC O3 and Y  NaHC O3

35. Arrange Ce3 , La 3 , Pm3 , Yb 3 in increasing order of their ionic radii

A) Yb 3  Pm 3  Ce 3  La 3 B) Ce3  Yb 3  Pm 3  La 3

C) Yb 3  Pm 3  La 3  Ce 3 D) Pm3  La 3  Ce 3  Yb 3
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36. Which one of the following does not contain electrons in 5f subshell?
A) Lr  Z  103 B) U  Z  92  C) Cm  Z  96 D) Th  Z  90 

37. Consider octahedral complex of Ca 2 with EDTA 4 .

x = co-ordination number of complex
y = Total number of O  donors . Find (x+y)?
A) 13 B) 14 C) 10 D) 11
38. In which of the following species maximum atoms can lie in same plane?
A) PC 5 B) AsH4 C) XeF4 D) XeO 2 F2
39. Which of following involves maximum C  O bond length:
A) Ni (CO ) 4 B) [ Fe(CO )5 ] C) [ Mn(CO )6 ] D) [V (CO)6 ]
40. Assertion (A) : Coloured cations can be identified by borax – bead test

Reason (R) : Transparent bead  NaBO2  B2O3  forms coloured bead with coloured
cation.

A) Both A and R are true and R is the correct explanation of A

B) Both A and R are true but R is not the correct explanation of A

C) A is true but R is false

D) A is false but R is true

41. Para-coumaric acid, an antioxidant in coffee, is treated with aqueous solution of

bromine. What is the maximum number of bromine atoms that can be incorporated into a
molecule of para-coumaric acid?
O

HO

OH
para  coumaricacid

A) 1 B) 2 C) 3 D) 4
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42. Which of the following reaction produces anisole as a major product?

A) B)

C) D) All
43. The oxidation of toluene with hot KMnO 4 gives

A) Benzoic acid B) Benzaldehyde

C) Benzene D) Benzyl alcohol

44. Primary, secondary and tertiary amines can be separated by

A) Hinsberg’s test B) HNO 2 test

C) Carbylamine test D) Lucas reagent

45. Identify the CORRECT option(s) among the following ?

A) [R] & [S] on treatment with excess of dil NaOH gives same compound
B) Position of double bond in [P] & [Q] in same as per IUPAC
C) [U] & [V] on treatment with excess of dilute NaOH gives different product as [R] &
[S] on same treatment
D) [U] & [R] are constitutional isomers
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46. A 10cm sample of an unknown gaseous hydrocarbon was mixed with 70cm3 of oxygen
3

and the mixture was set on fire by means of an electric spark. When the reaction was
over and water vapours were liquefied, the final volume of gases decreased to 65cm3 .
This mixture then reacted with a potassium hydroxide solution and the volume of gases
decreased to 45cm3 . Find the molar mass of hydrocarbon in gm/mol, if volumes of
gases were measured at standard temperature and pressure (STP) conditions ?
A) 26 B) 36 C) 16 D) 46
47. Radial wave function for 2s orbital of hydrogen – like atoms is given as
3
zr
 z 2  zr 
3
  2 2    2   e 2a 0
 a0   a0 

The radial probability  4r 2  2  for an electron in He  at r  a 0 is

3 3
3
 1 2 3  1 2 1
A)   B) 2 2   C) Zero D)  
 a0   a0   a0 
48. Benzene and toluene are nearly ideal solutions. If at a temperature
0
PBenzene  120mm Hg and PToluene
0
 50mm Hg . Then what will be the mole fraction of
Toluene in vapour phase of this solution if mole fraction of Benzene in liquid phase of
solution is 0.6.
A) 0.70 B) 0.6 C) 0.4 D) 0.22
49. Choose the correct representation of conductometric titration of benzoic acid vs sodium
hydroxide.
Conductance
Conductance

VNaOH  VNaOH 
A) B)
Conductance

Conductance

VNaOH  VNaOH 
C) D)
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50. Consider the reaction 2 H 2  g   2 NO  g   N 2  g   2 H 2O  g  . The rate law for this

reaction is: rate = k  H 2  NO  under what conditions could these steps represents the
2

mechanism?

Step: 1 2NO  N 2O2

Step: 2 N 2O2  H 2  N 2O  H 2O

Step: 3 N 2O  H 2  N 2  H 2O

A) These steps cannot be the mechanism under any circumstances

B) These steps cannot be the mechanism if Step 1 is sloe step

C) These steps cannot be the mechanism if Step 2 is sloe step

D) These steps cannot be the mechanism if Step 3 is sloe step

SECTION-II
(NUMERICAL VALUE ANSWER TYPE)
This section contains 10 questions. The answer to each question is a Numerical value. If the Answer in the
decimals , Mark nearest Integer only. Have to Answer any 5 only out of 10 questions and question will be
evaluated according to the following marking scheme:
Marking scheme: +4 for correct answer, -1 in all other cases.
51. How many of these elements have lower electron affinity than fluorine ?

Cl , S , O, N , P, Br , I , C

52.
Complex Spin-only magnetic moment

 Fe  CO 4  C 2 O 4  
 a

 FeCl4  b


 Fe  CN 6 
3
c

 Cu  NH 3 4 
2
d

Then (a+b+c+d) = _______

(Given : 3  1.7; 2  1.4; 5  2.2; 15  3.8, 35  5.9; 6  2.4 )
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53. The number of correct statements is /are :

(i) In manganate and permanganate ions. The  - bonding takes place by overlap of p-

orbitals of oxygen and d-orbitals of manganese

(ii) Manganate ion is green in colour and permanganate ion in purple in colour

(iii) Manganate and permanganate ions are paramagnetic

(iv) Manganate and permanganate ions are tetrahedral

54. The total number of chiral compound/s from the following is____________

OH

OH
OH OH CH2

HO OH
OH

55. How many of the following reactions will give an alcohol as the major product ?

(i) CH 3CH 2C 

NaOH
aq


(ii) CH3CH  CH 2 

aq Na CO2
 3

(iii) CH3CH 2  C 

CH OH
3

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 i  mCPBA

 ii  LiAlH  4
 iii  H 
(iv)

Br

aq 
AgNO 3

(v)

OH

NH 2 
HC 
NaNO 2

(vi)

OH OH
 i  H SO
CH3  C  C  CH3 
 ii  NaBH 
2 4

4
 ii  H
CH 3 CH 3
(vii)

56. Number of compounds which can evolve H 2 on reaction with Na metal.

OH OH O  CH 2  H

(i) (ii) (iii) (iv)

(v) H 3C  C  C  CH 3 (vi) Ph  SO 3H

CH 3
H  C  OH H 3C  C  H
O CH 3
(vii) (viii) (ix) H 3C  C  CH
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57. Among the following oxides

Cr2O3 , V2 O5 ,FeO,Mn 2O 7 , ZnO,Cu 2 O, MnO 2 ,CrO3 , NiO

If x = number of basic oxides

y = number of acidic oxides

z = number of amphoteric oxides

then the value of

 x  y is
z

58. Find the solubility of Co  NH 3 4 C 2  CO 4 .

 
If the  0m Co  NH3 4 C 2   50 1cm2mol1   0m  CO 4  70  1cm 2 mol1 and the


measured resistance was 33.5 in a cell with cell constant of 0.20cm 1 ?

Represent your answer in milli mol/litre WITH NEARST INTEGER

59. How many of following ethers are difficult to make or cannot be synthesized as a major
product by S N 2 reaction ?

60. At what pH given molecules doesnot migrate towards any electrode when electric field
is applied. (give answer in nearest integer. )

p ka1
 2;p ka  3.90;p ka  10.0 
2 3

NH 3

HOOC  CH 2  CH  COOH
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MATHEMATICS MAX.MARKS: 100
SECTION – I
(SINGLE CORRECT 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 if not correct.
61. Let z1 , z2 , z3 be three distinct complex numbers lying in a circle with a centre at the origin
such that z1  z2 z3 , z2  z3 z1 and z3  z1 z2 are real numbers, then z1 z2 z3 equals to

A) -1 B) 0 C) 1 D) None of these

62. The number of irrational solutions of the equation x 2  x 2  11  x 2  x 2  11  4 is

A) 0 B) 2 C) 4 D) 11

63. If H1 , H 2 ,.....H n are n harmonic means between a and b,  b  a  , then the value
H1  a H n  b
 is equal to
H1  a H n  b

A) n  1 B) n  1 C ) 2n D) 2n  3

1 1 12 12  22 12  2 2  32
64. Let H n  1     then the sum to n terms of the series    .... is
2 n 13 13  23 13  23  33

4 4 1 4 4 2 n
A) Hn 1 B) Hn  C) Hn D) Hn 
3 3 n 3 3 3 n 1

65. If a  log12 18, b  log 24 54 then the value of ab  5  a  b  is

A) 0 B) 4 C) 1 D) 6

66. The number of ordered pairs  m, n  , m, n {1, 2....100} such that 7m  7n is divided by 5 is

A) 1250 B) 2000 C) 2500 D) 5000

The number of irrational terms in the expansion of  51/6  21/8  is

100
67.

A) 96 B) 97 C) 98 D) 99

68. If A, B and C are three sets such that A  B  A  C and A  B  A  C , then

A) A=C B) B=C C) A  B   D) A=B

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69. Two n  n square matrices A and B are said to be similar if there exists a non-singular
matrix Psuch that P 1 AP  B . If A and B are similar and B and C are similar, then

A) AB and BC are similar B) A and C are similar

C) A+C and B are similar D) None of these

17. A letter is taken at random from the letters of the word ‘STATICSTICS’ and another
letter is taken at random from the letters of the word ‘ASSISTANT’. The probability that
they are the same letter is

1 13 19 3
A) B) C) D)
45 90 90 16

71. sin 470  sin 610  sin110  sin 250 is equal to

A) sin 360 B) cos 360 C) sin 70 D) cos 70

72. A1 , A2 , A3 ,... An are n points in a plane whose coordinates are

 x1 , y1  ,  x2 , y2  ,  x3 , y3  ...,  xn , yn  respectively. A1 A2 is bisected at the point G1 , G1 A3 is divided

in the ratio 1:2 at G2 , G2 A4 is divided in the ratio 1:3 at G3 , G3 A5 is divided in the ratio 1:4
at G4 , and so on until all n points are exhausted. The coordinates of the final point G so
obtained are

A)  n  x1  x2  ...  xn  , n  y1  y2  .... yn  

x  x  ...  xn y1  y2  ....  yn 
B)  1 2 , 
 n n 

x  2 x2  3 x3 ...  nxn y1  2 y2  3 y3 ....  nyn 

C)  1 , 
 n n 

D) None of these

73. Let PQ and RS be tangents at the extremities of a diameter PR of a circle of radius r.

Such that PSand RQ intersect at a point X on the circumference of the circle, then
diameter of the circle equals to

PQ  RS 2PQ  RS PQ 2  RS 2
A) PQ.RS B) C) D)
2 PQ  RS 2
 06-01-24_JEE-MAIN_Q’P
a3 x 2 a 2 x
74. The locus of the vertices of the family of parabolas y    2a is
3 2

105 3 35 64
A) xy  B) xy  C) xy  D) xy 
64 4 16 105

75. A triangle ABC is placed so that the mid-point of its sides are on the x, y and z axes
respectively. Lengths of the intercepts made by the plane containing the triangle on these
axes are respectively  ,  ,  , then the coordinates of the centroid of the triangle ABC are

A)   / 3,  / 3,  / 3 B)  / 3,  / 3,  / 3

C)  / 3,  / 3,  / 3 D)  / 3,  / 3,  / 3

e| x|  e  x
76. Let f : R  R be a function defined by f  x  = .Then
e x  e x

A) f is both one-one and onto B) f is one-one but not onto

C) f is onto but not one-one D) f is neither one-one nor onto
cos  sin x   cos x
77. The value of f  0  so that the function f  x   is continuous at each point
x4
in its domain, is equal to
A) 2 B) 1/6 C) 2/3 D) -1/3

f ''  0  f '''  0  f n  0
If f  x   1  x   n  N, x  R  , then the value of f  0   f '  0  
n
78.     is
2! 2! n!

(where f n  0  represents nth derivates of f at 0)

A) n B) 2n C) 2n1 D) None of these

dx
79. x 1/2
 x1/3


A) 2 x  3 x1/3  6 x1/6  log 1  x1/6   C B) 3x1/3  6 x1/6  log 1  x1/6   C

C) x  3x1/3  6 x1/6  log 1  x1/6   C D) 3x1/3  x  6 x1/6  log 1  x1/6   C

80. The solution of the differential equation y ' Ay   By 2 ( A, B  0) is (C is arbitrary constant)

A) y    B / A e Ax  C 
1
B) y   B / A   Ce Ax

C) y  e Ax   B / A e Ax  C  D) y  e Ax   B / A  e Ax  C 
1 1
 06-01-24_JEE-MAIN_Q’P
SECTION-II
(NUMERICAL VALUE ANSWER TYPE)
This section contains 10 questions. The answer to each question is a Numerical value. If the Answer in the
decimals , Mark nearest Integer only. Have to Answer any 5 only out of 10 questions and question will be
evaluated according to the following marking scheme:
Marking scheme: +4 for correct answer, -1 in all other cases.
81. If tan 1  x    tan 1  x    tan 1 then the value of 250 x4 +320 x2 +137 is equal to
2 2 4
 x  x x

82. If O is the origin and the coordinates of A and B are (51, 65) and (75, 81) respectively.
Then OA  OB cos AOB is equal to
k
k k
83. Let 0   r  1  r  1, 2,  and  cos  r = for any k  1 and A   (  r ) r ,then
r 1 2 r 1

lim
498.
1  x  2 1/3
 1  2 x 
1/4

x A
x  x2

7 1
dx  e1   e2  1 is
1
 e  x  1
2 x 1 n
84. A positive integer n  5 such that 0 4 16

85. Let a, b, c are given vectors,  ,  , v are scalars If

d    a  b     b  c   v  c  a  ,  abc   1/ 8 and d   a  b  c   8 then     v is equal to

   
2 3
86. The maximum value of f  x   4  x2  3  4  x 2  1 is

50 n
87. Let U X i  U Yi  T , where each X i contains 10 elements and each Yi contains 5 elements. If
i 1 i 1

each elements of the set Tis an element of exactly 20 of sets X i ' s and exactly 6 of sets
Yi ' s then n is equal to

88. If A  4, 3,5 , B  0,6, 0  , C  8,1, 4  are three consecutive vertices of parallelogram ABCD . If
 is the angle between AC and BD the sum of digits of 3025 sec 2  is

89. Let x1 , x2 ...., x100 be in an arithmetic progression with x1  2 and their mean equal to 200.If
yi  i  xi  i  ,1  i  100. Let y be the mean of y1 , y2 ,.... y100 . Then y  9000.5

90. Let A  1, 2,3, 4 , R be a relation on the set A  A defined by

R   a, b  ,  c, d   :2a  3b  4c  5d  . Then the number of elements in R is

Sec: MODEL-A,B&C Date: 06-01-24
Time: 3 HRS JEE-MAIN Max. Marks: 300

KEY SHEET

PHYSICS
1 B 2 B 3 C 4 D 5 A
6 B 7 C 8 A 9 B 10 A
11 B 12 B 13 A 14 A 15 A
16 C 17 A 18 D 19 B 20 B
21 1 22 30 23 3 24 780 25 16
26 2000 27 8 28 3 29 3 30 20

CHEMISTRY
31 B 32 D 33 A 34 B 35 A
36 D 37 C 38 C 39 D 40 A
41 D 42 B 43 A 44 A 45 A
46 A 47 C 48 D 49 B 50 C
51 7 52 11 53 3 54 2 55 4
56 5 57 2 58 50 59 2 60 3

MATHEMATICS
61 C 62 C 63 C 64 D 65 C
66 C 67 B 68 B 69 B 70 C
71 D 72 B 73 A 74 A 75 D
76 D 77 B 78 B 79 A 80 D
81 1777 82 9090 83 249 84 3 85 64
86 28 87 30 88 31 89 1049 90 6
 06-01-24_JEE-MAIN_KEY&SOL
SOLUTIONS
PHYSICS
1.  = Q A
A 0
Q
=
0
d 1 dQ I
= × =
dt 0 dt 0
 
2. Angle between a and p is :

p 


a

a.p
 = cos–1  
| a || p |
 
32  18
= cos-  

 (16  9) (64  36) 

cos–1  
14
=
 50 
 = 73.73°
Since 0° < 90°, the motion is an acceleration one.
3. Retardation = g (sin  +  cos ) = 5 ( 3  )
Now v = u – at
u
 a= as v = 0
t
 5 ( 3  ) = 10  = 0.27

4 m/s v
4.
1 m/s 1 m/s

Before collision After collision

Let v be the velocity of ball after collision, collision is elastic

 e =1
or
relative velocity of separation = relative velocity of approach
 v – 1 = 4 + 1
or v = 6 m/s (away from the wall)
GH
5.  g=
(R  h ) 2
GM GM
 2
=
9R (R  h ) 2
 3R = R + h
 h = 2R
So option (1) is correct.

6. Dq = – Du
 06-01-24_JEE-MAIN_KEY&SOL
–R R P dV
C = –CV = = +
 –1  –1 n dT
P dV 2R
– 
n dT  – 1
T5V = const.
const.
V=
T5
dV const
=–5 6
dT T
PV = Nrt
P/n = RT/V
RT 5  const  2R
+ T × –5 6  =
const.  T   –1
5 1
=  – 1 = 2/5
2  –1
 = 7/5
adiabatic compressibility
1 5
= =
P 7P
1
7. PV = m o Nv 2rms
3
1
(2P) (2V) = m o Nv 2rms
3
rms = 2vrms = 2v
8. 4 trips means 32 m
v

4m
d d 32
t=  v= = = 40 m/s
v t 0 .8
T
v=

T v2
0 .2 2 16  10
T= × (40)2 =
4 4
T = 80 N

9. Time period of both A and B T = 2
g
After first collision, B acquires amplitude of A and after second collision it acquires its own
amplitude in this process time taken is
T T T T 
= + + + = T = 2
4 4 4 4 g
10. The distribution of charge on the outer surface, depends only on the charges outside, and it distributes
itself such that the net electric field inside the outer surface due to the charge on outer surface and all
the outer charges is zero. Similarly the distribution of charge on the inner surface, depends only on the
charges inside the inner surface, and it distributes itself such that the net, electric field outside the inner
surface due to the charge on inner surface and all the inner charges is zero.
Also the force on charge inside the cavity is due to the charge on the inner surface. Hence answer is
option (A).
11. Conceptual
 06-01-24_JEE-MAIN_KEY&SOL
12. Point A shall record zero magnetic field (due to -particle) is at position P and Q as shown in figure.
The time taken by -particle to go from P to Q is –
P

O 60º
A
60º

12 2
t= or  =
3 3t
13. Curie temperature is temperature above which Ferromagnetic materials obey Curie’s law.
1 2
14. U= LI
2

= LI  
dU dI
Rate =
dt  dt 
At t = 0 , I = 0
 Rate = 0
dI
At t = , I = I0 but = 0, therefore, rate = 0
dt
15. The given lens is a convex lens. Let the magnification be m, then for real image
1 1 1
+ = ... (i)
mx x f
1 1 1
and for virtual image  = ... (ii)
 my y f
From Eq. (i) and Eq. (ii), we get
xy
f=
2
16. e = 4 × LC = 4 × 0.01 cm = 0.04 cm
c = –0.04 cm
17. Zener diode is in parallel to load resistance and
is connected in reverse bias.
R
18. R = R0 A1/3  = A1/3
R0
 R  1
log   = log A
 R0  3
1
or y= x
3
a straight line passing through origin with slope 1/3.
19. Conceptual

20. S1

S2

y 2 2y
x =  = × x =
  
Inet = I + I + 2I cos 
 06-01-24_JEE-MAIN_KEY&SOL
  2y    y 
= 2I 1  cos   = 4I cos2  
      
 y 
= I0 cos2   [ I0 = 4I]
  
k 10 2
21. F=  F = 
2x 2 2x 2
2
F 10 1 dv
a= = a= 2 =v
m (10  2 )2 x 2 2x dx
0.5 v 0.5
dx 1 v2
– =  vdv  –  x  2 dx =
1
2x 2 0
2
1
2
0.5 2
 1   1 v2
  1 =
v 1 1
 =
2  x 1 2 2  0 .5  2
 v = 1 m/s
22.

R
y

d
2
y2 = R2 –  
d
2

vCM/g
× y

vCM/g = × y
= 30 cm/s

23. v = Rcos
v 2
 cos= = cm
R 5
 h = R(1 – cos) = 3 cm.



24. If m is mass of single drop then as it drops

mg = 2 Rt
If number of drops in M = 10 grams is N then,
M Mg Mg
N= = =  779.86  780
m mg 2rT
25. For sonometer wire
F
n

n =
2
 n F = constant
 [are constant for two cases
 of comparison]
 06-01-24_JEE-MAIN_KEY&SOL
n12
 F2 = . F1
n 22
 m2 = 25 kg
 Additional mass = 16 kg
26.

Potential to centre is same as potential at the inner surface of the spherical shell.
27. At resonance reactance = 0
V 60 1
I= = = Amp.
R 120 2
VL = I × XL = I × L
VL
L= ………(i)
I
1
0 =
LC
1
C= ……….(ii)
L 0 2
Calculate L & C from (1) & (2) current will lag
the applied voltage by 45°
1
L –
if tan 45° = c
R
Solve for   = 8 × 105 rad/sec
28. Energy present in 3  108 m is 10 Mj, hence energy in 90 cm is
90  102  10  103
E  3  10 11 J
3 10 8

29.

4º
Aº

deviation = 0
(µ1 – 1) 4 – (µ2 – 1)A = 0
(1.54 – 1) 4 – (1.72 – 1) A = 0
0.54  4
A= = 3º
0.72
1
30. From graph  0.2cm 1
f
 f  5cm
 Power of lens equal to 20 D.
 06-01-24_JEE-MAIN_KEY&SOL
CHEMISTRY
31.  Xe 4f 14 1 2
5d 6s ----------- Lu
32. INERT PAIR effect NCERT (Physical properties of group 13 elements )
33.  K 2SO 4  Cr2  SO 4 3  H 2O
K 2 Cr2O 7  SO 2  H 2SO 4 
34. X : NaC O
Y : NaC O3 .
35. By using Lanthanoid contraction concept we get correct answer.
36. Th  Z  90  : 86  Rn  6d 2 7s 2
37. y  4, x  6
38. PC 5  sp3  TBP  ; AsH 4  sp 3  Tetrahedral  ; XeF4  sp 3d 2 -Square planar;
XeO 2 F2  sp3d  Seesaw 
39. synergic bond concept in complex compounds
40. Conceptual
41. Product is
OH

Br Br

HC  CH  COOH
| |
Br Br
43.

  
44. Hinsberg’s test is used for separation of 1 , 2 and 3 Amines.
46. 2  n  3  2   0 n  4
 y y
C x H y  g   x   O 2  g  
 xCO 2  g   H 2O   
 4 2
10 70 - -
 y 5
- 70  10  x   10x -70  10x  y  10x  65
 4 2
5y  5  2 y  210 x  20 x2
C x H y  C 2 H 2 Molar mass  26g / mol
47. At r  a 0   0  4r 2 2  0
48. In liquid phase
PT  PB0 . B  PT0 .T  120  0.6  50  0.4
= 72 + 20 = 92 mmHg
 06-01-24_JEE-MAIN_KEY&SOL
PToluene 20
In vapour phase  T   0.22
PTotal 92
49. C3H5COOH  NaOH 
 C6 H5COONa  H 2O
 WA  SB  Salt 

D
Conductance

A C

B
VNaOH
50.
51. NCERT data
52. a  3;b  35;c  3;d  3
2
53. Option i) Manganate  MnO 4
Permanganate  MnO 4
O O

Mn Mn
O O O O

O O
hybridisation hybridisation
of Mn  d s 3
of Mn  d 3s

After excitation

2 × 2p   3d 
 06-01-24_JEE-MAIN_KEY&SOL
2 × 2p   3d 
1 × 2p   4p 
2
(ii) MnO 4  green
MnO 4  purple / violet
(iii) Manganate contains I unpaired electron hence it is paramagnetic Where as permanganetic
contains no unpaired electrons hence it is diamagnetic.
3
(iv) Both have d s hybridization hence both have tetrahedral geometry.

54.

55.
 06-01-24_JEE-MAIN_KEY&SOL

56. (i), (ii), (vi), (vii),(ix) can evolve H 2 on reaction with Na metal.
57. X:3
Y:3
Z:3
58.  M  50  70  120 1 cm 2 mol1
C.C 0.2 1 1 K  1000 0.2  1000
K   cm   M  0.4975mol / t
R 33.5 M 120  33.5
M  49.75milli mol / t
60. Average of pK1 and pK 2
pH = 2.95 is isoelectric point

MATHS
61.
 06-01-24_JEE-MAIN_KEY&SOL

62.

63.
 06-01-24_JEE-MAIN_KEY&SOL
64.

65.

66.

67.
 06-01-24_JEE-MAIN_KEY&SOL
69. CONCEPTUAL
70.

71.

72.CONCEPTUAL
73.

74.
 06-01-24_JEE-MAIN_KEY&SOL
75.

76.

77.

78.
 06-01-24_JEE-MAIN_KEY&SOL
79.

80. conceptual
81.

82.

83.
 06-01-24_JEE-MAIN_KEY&SOL
84.

85.