05-01-24_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. The acceleration of the block B shown in the figure will be (Assuming the surfaces and
the light pulleys P1 and P2 all are smooth)

F F F 3F
A) B) C) D)
4m 6m 2m 17m
2. The distance between two parallel plates of a capacitor is a. A conductor of thickness b
(b < a) is inserted between the plates as shown in the figure. The variation of effective
capacitance between the plates of the capacitors as a function of the distance (x) is best
represented by

A)
B)

D)
C)
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3. One mole of an ideal monoatomic gas is taken through a cyclic process ABCA as shown
in the P-V diagram. The efficiency (in percentage) of the cyclic process is
A) 8.33 B) 12.33 C)16.33 D) 20.33

4. Two infinitely long conductors carrying equal currents are shaped as shown . The short
sections are all of equal lengths. The point P is located symmetrically with respect to the
two conductors. The magnetic field at point P due to any one conductor is B. The total
magnetic field at point P is

A) zero B) B C) 2B D) 2B
5. At the initial moment three point A,B and C are on a horizontal plane along a straight
line such that AB = BC . Point A begins to move upward with a constant velocity ‘v’ and
point C downward without any initial velocity at a constant acceleration ‘a’. If the point
begin to move simultaneously, then the initial velocity and acceleration of point B for all
the three particles to be constantly on same straight line must be:
v a v a
A) upwards, downwards B) upwards, upwards
2 2 2 2
v a v a
C) downwards, downwards D) downwards, upwards
2 2 2 2
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6. Assertion (A): The magnetic field at the center of the current carrying circular coil
shown in the fig. is zero.

Reason (R): The magnitudes of magnetic fields are equal and the directions of magnetic
fields due to both the semicircles are opposite.
A) Both A and R are true but 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 also false
7. A particle projected at an angle  grazes the inclined surface BC at point P as shown.
Find the time required to reach P from O.

2u sin  u sin  u sin      2u sin     

A) B) C) D)
g cos  g cos  g cos  g cos 
8. Objects A and B that are initially separated from each other and well isolated from their
surroundings are then brought into thermal contact . Initially temperature of A and B are
00 C and 100 0 C respectively. The specific heat of A is less than the specific heat of B.
After some time, the system comes to an equilibrium state. The final temperatures are

A) TA  TB  500 C B) TA  TB  500 C

C) TA  TB  500 C D) TB  TA  500 C
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9. A large plane mirror with its bottom on the floor is tilted at an angle 30° to the vertical.
A boy whose eyes are at height √3 m above the floor is standing in front of the mirror.
At what maximum distance (in m) from mirror should the boy be to see his full image in
mirror?

A) 1m B) 2m C) 2.3m D) 1.5m

10. A heavy nucleus X having mass number 200 gets disintegrated into two small fragments
Y and Z of mass numbers 80 and 120 respectively. If binding energy per nucleon for the
parent atom X is 6.5 MeV and for daughter nuclei Y and Z are 7 MeV and 8 MeV
respectively. Energy released in the decay will be

A) 200 MeV B) 240 MeV C) 220 MeV D) 180 MeV

11. Which of the following circuits will provide a full wave rectification of an AC input?

A) B)

C) D)
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12. A vessel of volume of V0 is evacuated by means of a piston air pump. One piston stroke
captures the volume V  0.2V0 . If process is assumed to be isothermal then find the

 1 
minimum number of strokes after which pressure in the vessel becomes    Pinitial  .
 1.728 

A) 2 B) 3 C) 5 D) 7

13. A balloon of volume V, contains a gas whose density is  and the density of the air at
the earth’s surface is 15 . If the envelope of the balloon be of weight w but of negligible
volume. Find the acceleration with which it will begin to ascend.

 7Vg  w   2Vg  w 
A)  g B)  g
 Vg  w   Vg  w 

14Vg  w   7Vg  w 
C)  g D)  g
 Vg  w   Vg  w 

14. Assertion (A): A small body suspended by a light spring, perform SHM. When the entire
system is immersed in a non-viscous liquid, the period of oscillation does not change.

Reason (R): The angular frequency of oscillation of the particle does not change.

A) Both A and R are true but 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 also false

15. A satellite is revolving around the earth in an orbit such that it time period of revolution
as same as that of earth and it revolve in same sense as of earth. To make it escape from
gravitational field of earth, its velocity must be increased by

A) 100% B) 41.4% C) 50% D) 59.6%

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16. Match the List-I with the List-II.

List-I List-II


Phase difference between current and ; current leads
A. 1. 2
voltage in a purely resistive AC circuit
voltage

Phase difference between current and

B. 2. Zero
voltage in a pure inductive AC circuit


Phase difference between current and ; current lags
C. 3. 2
voltage in a pure capacitive AC circuit
voltage

Phase difference between current and  X  XL 

D. 4. tan 1  C 
voltage in an LCR series circuit  R 

Choose the most appropriate answer from the options given below:

A) A-2, B-3, C-1, D-4 B) A-1, B-3, C-4, D-2

C) A-2, B-3, C-4, D-1 D) A-2, B-4, C-3, D-1
17. In the given figure two concentric cylindrical region in which time varying magnetic
field is present as shown. From the centre to radius R magnetic field is perpendicular
dB
into the plane varying as  2k 0 and in a region from R to 2R magnetic field is
dt
dB
perpendicular out of the plane varying as  4k 0 . Find the induced emf across an arc
dt
AB of radius 3R.

A) 6R 2 k 0 B) 5R 2 k 0 C) 7R 2 k 0 D) None of these
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18. If a string of length  fixed at both ends vibrates with a standing wave
 2 
y  Asin  x  sin 2t in resonance. Then the minimum time (from t=0) after which
  
energy is maximum at mid-point of string will be –

1 1 1 1
A) sec B) sec C) sec D) sec
4 5 8 6

19. Figure shows the graph of stopping potential versus the frequency of a photosensitive
metal. The plank’s constant and work function of the metal are (V and  0 are two
different constant.)

(3V)e (2V)e
A) Wc  (2V)e; h  B) Wc  (2V)e; h 
0 0

(3V)e (2V)e
C) Wc  (3V)e;h  D) Wc  (3V)e; h 
0 0

20. There are two bulbs B1 (P, V), B2 (P,2V) their rated power and voltages are mentioned
W1
with them. Calculate the ratio of consumed power ?
W2

25 4 10 4
A) B) C) D)
4 25 4 10
 05-01-24_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.
21. A system of two planks and a sphere of radius R is in motion as shown in figure. Radius
of the sphere is R and there is no slipping anywhere. It is given that R   2a 0 where 
is angular acceleration of sphere and acceleration of upper block is a 1  ka 0 where k is a
+ve constant then the value of k will be

22. In the system shown, the mass m  2 kg oscillates in a circular arc of amplitude 600 .
The minimum value of coefficient of friction between mass = 8 kg and surface of table
to avoid slipping is  . Then find 10 .

23. Consider a circuit with an alternating source and contains inductor and capacitor. Given
reading of A 1 and A 2 as 3 ampere and 5 ampere respectively. Find the magnitude of
reading of A in ampere.

24. The electric field associated with e.m. waves in vacuum is given by

 
E  ˆi40cos kz  6  108 t , where E, z and t are in volt/m, meter and seconds

respectively. The value of wave vector k is _______per m

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25. A point like sound source emits sound in all direction uniformly. An observer who is at a
distance of 50 m from the source detects sound of intensity 10-2 watt/m². If the bulk
modulus of air is 1.6 x 105 N/m² and velocity of sound is 340 m/s. Find pressure
amplitude (in N/m²) at the position of observer in nearest integer.
26. An x-ray beam of monochromatic photon are incident on a metallic surface having a
negligible work function  take   0  . It is seen that the wavelength of most energetic
photoelectrons is equal to the wavelength of x-ray photons. Find the wavelength (in pm).
Round off to nearest integer.

27. Figure shows a right angle solid prism ( O  ) and mass of prism is m. Moment of
2
ma 2
inertia of prism about axis perpendicular to plane passing through O is , find value
k
of k.

28. Calculate time constant (in milli-sec) of the circuit.

29. A vertical capillary tube with inside radius 0.25 mm is submerged into water so that the
length of its part protruding over the water surface is equal to 25 mm. Surface tension of
water is 73 x 10-3 N/m and angle of contact is zero degree for glass and water,
acceleration due to gravity is 9.8 m/s². Then value of 10R approximately (in mm) is
(where R is radius of meniscus and h is height of water in capillary tube)
30. In the figure, if a parallel beam of white light is incident on the plane of the slits
kd
S1 & S2 then the distance of the central maxima on the screen from O is . Find the
8
value of k.  Assume D  d,d    .
 05-01-24_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. There are four elements 'p', 'q', 'r' and 's' having atomic numbers Z-1, Z, Z+1 and Z+2
respectively. If the element 'q' is an inert gas, select the correct answers from the
following statements.

(i) 'p' has most negative electron gain enthalpy in the respective period.

(ii) 'r' is an alkali metal.

(iii) 's' exists in +2 oxidation state.

A) (i) and (ii) only B) (ii) and (iii) only

C) (i) and (iii) only D) (i), (ii) and (iii)

32. The following are some statements related to VA group hydrides. INCORRECT
statement is:

A) Reducing property increases from NH 3 to BiH3 .

B) Tendency to donate lone pair decreases from NH 3 to BiH3 .

C) Thermal stability of hydrides decreases from NH 3 to BiH 3 .

D) Bond angle of hydrides increases from NH 3 to BiH 3 .

33. The incorrect statement is

A) The first ionization enthalpy of K is less than that of Na and Li

B) Xe does not have the lowest first ionization enthalpy in its group

C) The first ionization enthalpy of element with atomic number 37 is lower than that of
the element with atomic number 38.

D) The first ionization enthalpy of Ga is higher than that of the d-block element with
atomic number 30.
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34. Which one of the following statements is incorrect?

A) Highest oxidation state of manganese in fluorides is +4 (MnF4) but highest oxidation

state in oxides is 7 (Mn2O7).

B) VO2+< Cr2O7 2  < MnO4 is the correct order of oxidizing power.

C) When I  is oxidized by MnO4 in alkaline medium, I  converts into I2.

D) Silver nitrate on heating decomposes to give two types of paramagnetic gases along
with a residue.

35. Consider the following co-ordination compounds.

(i) Ni(CO)4 (ii) [Co(CO)4]- (iii) [Fe(CO)4]2-
The M-C bond strength follows the order :

(A) (ii)>(iii)>(i) (B) (iii)>(ii)>(i) (C) (i)>(ii)>(iii) (D) (i)>(iii)>(ii)

3 4
36. Assertion (A):  Fe  CN 6  is more stable than  Fe  CN 6 

Reason (R): Complexes where the iron is in the (III) oxidation state are generally more
stable than those in (II) oxidation state.

A) Assertion is True, Reason is True; Reason is correct explanation for Assertion

B) Assertion is True, Reason is True; Reason is NOT a correct explanation for Assertion
C) Assertion is True, Reason is False
D) Assertion is False, Reason is True
37. Which of the following statements is CORRECT?
A) In the formation of dioxygen from oxygen atoms, 10 molecular orbitals will be
formed.
B) All the molecular orbitals in the dioxygen will be completely filled.
C) Total number of bonding molecular orbitals will not be same as total number of anti-
bonding orbitals in dioxygen.
D) Number of filled bonding orbitals will be same as number of filled anti bonding
orbitals.
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38. Yellow coloured solution of metal ion [which belongs to group-III of basic radicals] +
SCN   Blood red colour solution.

The intensity of red colour will not change on addition of:

A) Hg(NO3)2 B) F- ions C) C2O42- ions D) Br- ions

39. The IUPAC name of the following compound is

CN

OHC COOH

A) 5-cyano-3-formylcyclohex-3-en-1-carboxylic acid

B) 3-cyano-5-formylcyclohex-4-ene-1-carboxylic acid

C) 5-cyano-3-oxocyclohex-3-ene-1-carboxylic acid

D) 5-carboxy-3-formylcyclohex-2-ene-1-carbonitrile
Br2 , NaHCO3
40    A , Major organic product

The CORRECT structure of A is

OH Br

OH OH O
O O
Br
O O
A) Br B)Br C) D) Br

41. Compound X(C7H13Br) is a tertiary bromide. On treatment with potassium ethoxide in

ethanol, X is converted into Y(C7H12). Ozonolysis of Y gives Z as the only product. Z
can also be obtained by the oxidation of heptane-2,6-diol.

Select the INCORRECT statement from the following:

A) Z can also be obtained by oxidation of Y with conc. hot KMnO4 solution.

B) Z gives negative 2,4-DNP test.

C) Z gives positive iodoform test.

D) Z can be reduced to a diol with NaBH4.

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42. Find out correct product of reaction:

HBr
(Excess)
O

Br

O OH Br
A) B)

Br Br Br OH
C) D)
43. Analyse the following reaction sequence:
O

H3C i) Me3COK/THF i) dil. KOH

X Y
ii) CH 3COCl +
COOC 2H5 ii) H 2O/H , 

The compound Y is the above sequence of reaction is

O O
O O O
COOC 2H5 COOH
A) B) C) D) COOH
H3C CH3
COCH 3 COcH 3

44. The nitrogen atom in each of the following tertiary amines may be removed as trimethyl
amine by repeated Hoffmann eliminations (exhaustive methylation followed by heating
with moist Ag2O). Which of the amines requires the greater number of such Hoffmann
sequences to accomplish complete removal of nitrogen.

A) B) C) D)

45. Which of the following statement is correct?

A) α-D-glucose and β -D-glucose are enantiomers.

B) Glycine is optically active.

C) Cellulose on hydrolysis yields only -D-glucose.

D) Arginine is most basic amino acid

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46. One litre of a mixture of CO & CO2, is burnt in excess of oxygen. The reduction in
volume after combustion and cooling was found to be 100 ml at same temperature and
pressure. Calculate the volume ratio of CO and CO2 in initial mixture.
A) 1 : 9 B) 9 : 1 C) 1 : 4 D) 4 : 1
47. The entropy change can be calculated by using the expression S  qrev / T . When water
freezes in a glass beaker, choose the correct statement amongst the following.
A) ∆S (system) decreases but ∆S (surroundings) remains the same.
B) ∆S (system) increases but ∆S (surroundings) decreases.
C) ∆S (system) decreases but ∆S (surroundings) increases.
D) ∆S (system) decreases and ∆S (surroundings) also decreases.
48. A liquid mixture contain 10 moles of A(PA = 200 mmHg) and 10 moles of
B(PB = 100 mmHg). The vapour pressure over liquid mixture is 160 mm Hg. Which is
correct statement?
A) ∆Gmix = +ve B) ∆Vmix= -ve
C) ∆Ssurrounding= +ve D) ∆Ssurrounding = -ve

49. For the reaction N2 O4  g  

 2NO2  g  , the value of K is 50 at 400 K and 1700 at 500
K. Which of the following options is INCORRECT?
A) The reaction is endothermic.
B) The reaction is exothermic.
C) If NO2 (g) and N2O4 (g) are mixed at 400 K at partial pressures 20 bar and 2 bar
respectively, more N2O4 (g) will be formed.
D) The entropy of the system increases in the forward direction.
50. Match the items of Column I and Column II.
Column I Column II
(i) Leclanche cell (a) cell reaction 2H2 + O2→ 2H2O

(ii) Ni–Cd cell (b) does not involve any ion in solution and is used in hearing aids.

(iii) Fuel cell (c) rechargeable

(iv) Mercury cell (d) reaction at anode, Zn → Zn2+ + 2e–

(e) converts energy of combustion into electrical energy
 05-01-24_JEE-MAIN_Q’P
A) (i) →(d); (ii) →(c); (iii) →(a), (e); (iv) →(b)

B) (i) →(c); (ii) →(d); (iii) →(e); (iv) →(b)

C) (i) → (c); (ii) →(e); (iii) →(a), (c); (iv) →(b)

D) (i) → (d); (ii) →(d); (iii) →(a); (iv) →(c)

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. Total number of geometrical isomers for the square planar complex [RhCl(CO)
(PPh3)(NH3)] is /are _______ .

52. The total number of compounds having pp bonding among the molecules given
below are ________ .

SO 2 , SO3 , CO2 , C3O2 , N 2 O5 , Cl2 O7 and Cl2 O6

53. Find the number of resonating structures of the given carbanion where negative charge is
on 2 carbon.
N
–
CH2

54. From which position does NO2+ replace a hydrogen from the following compound
predominantly?
H
N
1 O
2

3 4
5 8

6 7

55.

The degree of unsaturation in major product (B) will be:

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56. How many of the following diazonium salts are less reactive than (I) towards diazo
coupling reaction?

+  + + +
N 2 Cl N2 Cl  N2 Cl
 N Cl 
2
NO2 O 2N NO2
I. II. III. IV.

NO2 NO2 NO2

+
N  +N Cl 
2 Cl 2

V. VI.
CH3
CH3 CH3

57. Analyse and complete the following reactions

(i) CH2CH=CH2 + H2 –(Ni) →
(ii) CH3CH2CH2Cl+ H2 –(Zn/H+) →

(iii) CH2CH2CH2COONa + NaOH –(CuO/∆) →

Find the sum of molar masses (in g) of byproducts (if any) formed in above reactions

[Take atomic masses – H = 1, C = 12, O = 16, Na = 23, Cl = 35, Cu = 63, Zn =65

58. A light of wavelength 200nm falls upon a surface and two different wavelength photons
λ = 800nm and λ =400nm are emitted from the surface. 80% of the energy absorbed is
re-emitted in the form of photon. Number of photons emitted as λ = 800nm is 3 times
that of number of photons emitted as λ = 400nm. If the ratio of total absorbed photon to
total emitted photon is x. then find the numerical value of (12.8x)

59. When 1 mole of an ideal gas at 20 atm and 15 L volume expands such that the final
pressure becomes 10 atm and volume is 60 L. The entropy change of the process in

J/K/mole is (Round off your answer to nearest double digit integer)

Given that: Cp,m = 30.96 J/mole/K, R = 8.314 J/mol/K, ln 2 = 0.69

60. pH of 0.08 moldm–3HOCl solution is 2.85. If its ionization constant is A × 10–B

Find the value of B. [antilog (0.15) = 1.413] (Given : 1  A  10, log 2  0.3 )
 05-01-24_JEE-MAIN_Q’P
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 C be the set of all complex numbers and

 z z 
R   z1 , z2   C  C : 1 2 is real 
 z1  z2 

Then, on C, R is a

A) reflexive relation B) symmetric relation

C) transitive relation D) equivalence relation

62. Let A be a 3  3 real matrix such that A  I  2 BBT , where BT is transpose of column

matrix B, whose sum of the squares of elements is unity. Given the statements

I) Tr  A   1( Tr  A  is sum of the elements in the principal diagonal of matrix A)

II) A is a symmetric matrix

III) A is an orthogonal matrix

Then the number of the correct statements among the above is

A) 0 B) 1 C) 2 D) 3

63. If  is real and   2    2  x 2     2  x  1 for all real x, then  belongs to the interval

A)  2,1 B)  2, 2 / 5 C)  2 / 5,1 D) 1, 2 

64. Let f  x   x 2  3 x  3 and x1 , x2 , x3 and x4 be solutions of the equation f  f  x    x . Then

the number of arrangements of x1 , x2 , x3 and x4 taken all at a time, is

A) 24 B) 4 C) 6 D) 1
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65. Let S  {1, 2,3,4,5,6,7} .A subset of A is selected from ‘S’ and now keeping back
elements in S, again a subset of S,B is selected. Let E1  event that
A  B  {1,2,3,4,5}, E2  event that A  B  {1,2}, if

E  a 1 b 
P  2    a, b  N   G.C.D of  a, b   1 ,then equals___
 E1  b 10  a 

([.] represents G.I.F)

A) 2 B) 3 C) 4 D) 5

x2 y2
66. Let H n :   1 , n  N . Let k be the smallest even value of n such that the
1 n 3  n
eccentricity of H k is a rational number. If l is the length of the latus rectum of H k , then
21l is equal to

A) 101 B) 204 C) 102 D) 306

67. Two adjacent sides of parallelogram ABCD are given by AB  2iˆ  10ˆj  11kˆ and
AD  ˆi  2ˆj  2kˆ . The side AD is rotated by an acute angle  in the plane of the

parallelogram so that AD becomes AD ' . If AD ' makes a right angle with the side AB,
then the cosine of the angle  is given by

8 17 1 4 5
A) B) C) D)
9 9 9 9

68. Consider a complex number z on the argand plane satisfying

i 2
  
arg  z      arg  z       e 3  . If minimum value of z  2  2i z  2  2i is
2 2 2

2  
a b ab
 a, b  N  ,then the value of 
2 52
A) 2 B) 5 C) 10 D) 20
tan x cot x
t dt
69.  dt   
1/ e t 1  t 
2
1/ e
1 t 2

A) 2  tan e  1 B) 2 tan e C) 1 D) tan e  cot e

 05-01-24_JEE-MAIN_Q’P
70. The area enclosed between the curves y 2  4 x and x 2  4 y inside the square formed by
the lines x  1, y  1, x  4, y  4 is

8 16 13 11
A) B) C) D)
3 3 3 3

71. The curve passing through the point  0,  / 4  satisfying the differential equation

dy
 x sin 2 y  x 3 cos 2 y is
dx

1 2 3 2 1 2 3 2
A) tan y 
2
 x  1  e x / 2
2
B) tan y 
2
 x  1  e  x / 2
2

1 2 3 2 1 2 3 2
C) tan y 
2
 x  1  e x
2
D) tan y 
2
 x  1  e  x
2

72. Statement – 1: If a,b,c are non zero real numbers such that

3  a 2  b 2  c 2  1  2  a  b  c  ab  bc  ca  , then a,b,c are in A.P. as well as in

G.P.

Statement – 2: A series is in A.P. as well as in G.P. if all the terms in the series are equal
and non zero.

A) Both Statement - 1 and Statement - 2 are true

B) Both Statement - 1 and Statement - 2 are false

C) Statement - 1 is true, Statement - 2 is false

D) Statement - 1 is false, Statement - 2 is true

73. If a1 , a2 , a3 and a4 are the coefficients of any four consecutive terms in the expansion of
a1 a2 a3
1  x  , then are in
n
, ,
a1  a2 a2  a3 a3  a4

A) AP B) GP C) HP D) AGP

74. x1 , x2 ,...........x10 are ten observations such that x i  50 and x x i j  1100 1  i  j  10 , then

standard deviation of x1 , x2 ........x10 is equal to

A) 5 B) 10 C) 5 D) 10
 05-01-24_JEE-MAIN_Q’P
75. Match the following

Column I Column II
If tan  is the G.M. between sin  and cos
A) P) 1
then 2  4 sin 2   3sin 4   sin 6  can be

B) 3 cot 200  4 cos 200  Q) 0

C) cot160 cot 44 0  cot 440 cot 760  cot 760 cot160  R) 3

9
 r 
D)  sin
r 1
2
 
 18 
S) 5

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

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

76. Let

23  1 33  1 n3  1
Pn  . ........... ; n  2,3, 4......
23  1 33  1 n3  1

Then lim
n 
Pn is equal to

1 7 3 2
A) B) C) D)
2 11 4 3

77. Let f  x   a0 x  a1 x  a2 x  a3 . Then

3 2

A) f is differentiable at x = 0 if a2  0

B) f is not differentiable at x=0, whatever be a0 , a1 , a2 , a3

C) f is differentiable at x  0 if and only if a2  0

D) If f is differentiable at x  0 , then a0  0 and a2  0

78. Let A, B and C be finite sets such that A  B  C   and each one of the sets AB, BC
and C A has 100 elements. The number of elements in A  B  C is

A) 250 B) 200 C) 150 D) 300

 05-01-24_JEE-MAIN_Q’P
79. Let Z be the set of all integers,

A  x, y   Z  Z :  x  2  2

 y2  4

B   x, y   Z  Z : x 2  y 2  4 and

C  x, y   Z  Z :  x  2 2
  y  2  4
2

If the total number of relations from A  B to A  C is 2p , then the value of p is

A) 16 B) 49 C) 25 D) 9

80. Let A,B and C be three events such that the probability that exactly one of A and B
occurs is 1  k  , the probability that exactly one of B and C occurs is 1  2k  , the
probability that exactly one of C and A occurs is 1  k  and the probability of all A, B
and C occur simultaneously is k 2 , where 0  k  1 . Then the probability that at least one
of A, B and C occur is

1 1
A) Greater than B) Exactly equal to
2 2

1 1 1 1
C) Greater than but less than D) Greater than but less than
8 4 4 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.
81. P is a point satisfying arg z=  / 4 , such that sum of its distances from two given points
k
 0,1 and  0, 2  is minimum, then P must be 1  i  . Then numerical value of k is _____
3

82. The number of ways can 14 identical toys distributed among three boys so that each one
n
gets atleast one toy and no two boys get equal number of toys is n, then is equal
10
to____
 05-01-24_JEE-MAIN_Q’P

83. Let P be an interior point of a triangle ABC such that PA  2PB  3PC  0 . Then the ratio

Area of triangle ABC

is equal to ______
Area of triangle APC

84. The number of values of x at which the function f  x    x  1 x 2/3 has extremum values is

π π
If the complete solution set of the inequality  cosec1 x   2  cosec 1 x   cosec 1 x 
2
85. is
6 3
(, a]  [b, ) , then (a  b) is equal to

86. A straight line L with negative slope passes through the point (8, 2) and cuts the positive
coordinate axes at points P and Q. Then the absolute minimum value of OP+OQ as L
varies (Where O is the origin) is

87. The sum of first 100 terms of the series is

 1   1   1   1 
tan 1  2 
 tan 1  2 
 tan 1  2 
 tan 1  2 
 .... (if x>0) is
 1  x  x   3  3x  x   7  5x  x   13  7x  x 

 100 
tan 1  2  , then a  b  ____
 1  ax  bx 

88. Let the curve C be the mirror image of the parabola y 2  4x with respect to the line
x  y  4  0 . If A and B are the points of intersection of C with the line y  5 , then the

distance between A and B is

5 x8  7 x 6 1
89. If f  x    dx,  x  0  f  0   0 and f 1  , then the value of K is ______
x 2
 1  2x 
7 2 K

90. Suppose x1 and x2 are the point of maximum and the point of minimum respectively of
the function f  x   2 x 3  9ax 2  12a 2 x  1 respectively,  a  0  then for the equality x12  x2
to be true the value of ‘a’ must be
Sec: MODEL-A,B&C Date: 05-01-24
Time: 3 HRS
JEE-MAIN Max. Marks: 300

KEY SHEET

PHYSICS
1 D 2 C 3 A 4 A 5 A
6 A 7 C 8 A 9 A 10 C
11 D 12 B 13 C 14 A 15 B
16 A 17 B 18 A 19 A 20 A
21 5 22 5 23 2 24 2 25 3
26 1 27 3 28 120 29 6 30 3

CHEMISTRY
31 D 32 D 33 D 34 C 35 B
36 A 37 A 38 D 39 A 40 D
41 B 42 B 43 C 44 A 45 D
46 C 47 C 48 D 49 B 50 A
51 3 52 5 53 4 54 3 55 4
56 3 57 4 58 5 59 27 60 5

MATHEMATICS
61 B 62 D 63 B 64 B 65 B
66 D 67 B 68 B 69 C 70 D
71 C 72 A 73 A 74 C 75 C
76 D 77 C 78 C 79 C 80 A
81 2 82 6 83 3 84 2 85 1
86 18 87 101 88 4 89 4 90 2
 05-01-24_JEE-MAIN_KEY&SOL
SOLUTIONS
PHYSICS
1.

F  2T  2ma 1
3T  4ma 2
2a 1  3a 2
solving we get
3F
a2 
17m
A 0 A0

cc x (a  b  x)
2. c 1 2 
c1  c2 A0  A 0
x (a  b  x)
A0
c independent of x
(a  b)
1 PV
3. Wcyclic  V0P0  0 0
2 2
For theprocess AB,P  KV  PV 1  constant
Molar heat capacity of the gas in the process AB,
R 3R R
C  Cv     2R
(1  x) 2 2
Q AB  nCT  n2R(4T0  T0 )  6nRT0  6P0 V0
Q BC  0 and Q CA  0
Wcyclic P0 V0
The efficiency of the cyclic process,    100   100
Qsup plied 2  6P0 V0
25
  8.33%
3 
4. By the symmetry, B total  0
5. Let us observe the motion of A and B relative to C.
AA BB

2d d
 05-01-24_JEE-MAIN_KEY&SOL
1 2
AA vt  at
BB   2
2 2
Vt 1 2
BB'   at
2 4
V
is initially speed of B w.r.t C as well as ground.
2
a
is acceleration of B w.r.t C
2
  
aB/ g  aB/C  aC/ g
a 1
 a  a
2 2
6. Conceptual
7. v cos   u cos 
u cos 
v
cos 
v sin   u sin   gt
u cos  sin 
 u sin    gt
cos 
u cos  sin   u sin  cos 
  gt
cos 
u sin    
t
g cos 
8. m1s1T1  m 2s 2 T2
Since m1s1  m 2s 2
T1  T2
9.

3
tan 2 
D
10. Q  (80  7)  (120  8)  (200  6.5) MeV  220 MeV
11. By the property of full wave rectifier
n
 V0 
12. Pfinal  Pinitial  
 V0  V 
w 
13. 15Vg  Vg  w    V  a
g 
 14Vg  w 
a  g
 w  Vg 
 05-01-24_JEE-MAIN_KEY&SOL
14. Time period of a spring mass system will remain constant when fluid is non-viscous.
15. v c  2v 0  1.414 v 0 % increase in orbital
v0  v0
velocity   100  41.4%
v0
16.

A) phase difference between current and voltage in a purely resistive AC circuit is zero.

B) phase difference between current and voltage in a pure inductive AC circuit is ; current lags
2
voltage.

C) phase difference between current and voltage in a pure capacitive AC circuit is ; current leads
2
voltage.
 X  XL 
D) phase difference between current and voltage in an LCR series circuit is tan 1  C .
 R 
1 1
17.    R 2B1  B2 3R 2
2 2
d R  dB1 3R 2 dB2
2
  
dt 2 dt 2 dt
R 2 3R 2
 .2K 0   4K 0
2 2
 emf  5R 2 K 0
18. At node, energy is maximum when all particle reach to there extreme position.
19. Vs  tan .   C
4V  tan . 2 0  C
10V  tan . 4 0  C
3V
6V  20 tan   tan  
0
Wc
4V  6V 
e
Wc  (2V)e
(3V)e
h
0
 05-01-24_JEE-MAIN_KEY&SOL
20.

v2 V 2
w1   P
R v2
P
2
 V  4V 2 .P 4P
w2    .4R  
 5R  25V 2 25
21.

a 0  a cm  R........1
a1  a cm  R....... 2 
Solivng equation 1 &  2 
a 1  a 0  2R   4a 0
a 1  5a 0
K  5.00
22. Tension of rope is maximum at lowest point
mv 2
Tmax  mg  ........(1)

By energy conservation,
 1
mg  mv2  v  g
2 2
From (1),
m(g)
Tmax  mg   Tmax  2mg

For 8 kg block, Tmax  f L
2mg   (8g)
4g   (8g)
  0.5
23.

Phasor diagram
 05-01-24_JEE-MAIN_KEY&SOL

I8  I C  I L  5  3  2A
24. E  E0 cos  kz   t  i
 6  108
k  2
Vwave 3  108
P02 V
25. I
2B
P02  340
102 
2  1.6  105
320 160
P0  
34 17
P0  3
p 2 hc hc h2
26.     
2m   2m 2
h 6.60  1034 11
  31
  1011 m  1.2pm
2mc 2  9.1  10  3  10 91
8

27. Consider prism of mass 4 m by joining 4 prism given in question.Total MOI of this system will be

 4m   
2
2a ma 2
I   4I prism    I prism 
6 3
28.   R Th. C, R Th is the Thevenin’s resistance at the capacitor terminals.
R Th  8  (20 (9  (70 30))  20 k
  0.12 s
2T cos 
29. h  59.6mm
 rg
Here h is greater than protruding part of tube hence water will rise to maximum length of tube such
hr
that radius of meniscus is given by R 
l
30. White spot on screen would be central maxima
Where
d d 3d
x  0 y   
2 8 8
 05-01-24_JEE-MAIN_KEY&SOL
CHEMISTRY
31. As 'q' is noble gas, p, r and s having atomic number Z-1, Z+1 and Z+2 should belong to halogen,
alkali metal and alkaline earth metal respectively. As halogen has one electron less than stable noble
gas configuration it has greater tendency to accept an additional electron forming anion. Alkaline
earth metal having valence shell configuration ns2 exists in +2 oxidation state.
32. VA group hydrides: (a) reducing properties increases down the group.
(b) Basic nature decreases down the group.
(c) Thermal stability decreases down the group.
(d) Bond angles decreases down the group.
34. Mn2O7: multiple bonds

AgNO 3   Ag   NO 2   O 2 
 5  6   7 
V O  Cr2 O  Mn O 4

2
2
7
When I is oxidized by MnO4 in alkaline medium, I converts into IO3
37. Conceptual
38. Yellow coloured solution of metal ion belonging to basic radicals of group-III → Fe³ is metalion
Fe+3 + SCN– → [Fe(SCN)2]+ (Red solution) + other species
Fe+3 + C2O42– → [Fe(C2O4)3] (stable complex)
Fe3+ + F– → [FeF6]3- (stable complex)
F- and C2O42– are stronger ligand than SCN–.
Hg2+ forms stable complex with SCN–
39. Conceptual
41. X = 1-Bromo 1,2-dimethylcyclopentane, Y= 1,2 -dimethylcyclopentene,
Z = Heptane -2,6 –dione
43. Decarboxylation of betaketo acid after ester hydrolysis
44.
CH 3
N
3 2
N

N (CH 3 )2 CH 3
N

1 2

46. Initial volume of gas  1000  VO 2

x
Final volume of gas  1000  VO2 
2
 vol of CO  20ml
47. Freezing is exothermic process. The heat released increases the entropy of surrounding.
48. Psolution  160
Solution have positive deviation from Raoult’s law.
Ptotal  200  0.5  100  0.5  150
G m ix  0 Vmix  0 H mix  0
Ssurr  0
49. Justification: K increases with increase in temperature.
Q > K, Therefore, reaction proceeds in the backward direction.
∆n > 0, Therefore, ∆S > 0.
50. Conceptual
 05-01-24_JEE-MAIN_KEY&SOL
51. Two cis, one trans
52. Conceptual
53. Conceptual
55.

56. III, V & VI are less reactive

 B  3k1  3   at any time
57.  
C 8k1 8
C  8k1  8   at any time
 
 D 7.5k 2 7.5
   0.4
n1 3n 2 n
58.  0.8   2
2000 8000 4000
n  0.8 5n 2
 1 
2000 8000
n 5
 1 
n 2 4  0.8
n 5 5
 1  
4n 2 16  0.8 12.8
P1V1 P2 V2 T 6
59.   2 
T1 T2 T1 3
 T P
S  2.303  n CP log 2  R log 1 
 T1 P2 
S  27.22 J / K / mole
60. pH of HOCl  2.85
But,  pH  log  H  
2.85  log  H  
3.15  log  H  
For weak mono basic acid  H +   K a  C

1.413  10 3 
2 2
 H  
Ka  
C 0.08
 24.957  10  2.4957  105
6
 05-01-24_JEE-MAIN_KEY&SOL
MATHS
61. Since  0, 0   R, R is not reflexive, we have
z1  z2
 z1 , z2   R  is real
z1  z2
z2  z1
 is real   z2 , z1   R
z1  z2
Therefore R is symmetric,
Since  0, z   R and  z , 0   R , but  0, 0   R there fore R is not transitive. Hence R is not an
equivalence relation.
63 Suppose that   2    2  x 2     2  x  1  0
For all real x,
 2    2  0 and    2   4   2    2   0
2

   2    1  0 and 5 2  8  4  0
2    1 and    2  5  2   0
2
2    1 and 2   
5
These inequalities imply
 2
   2, 
 5
64. Note that every solution of f  x   x is also a
solution of f  f  x    x
f  x   x  x 2  4 x  3  0  x  3 or 1
Therefore, 3 and 1 are roots of f  x   x , also
f  f  x    x   x 2  3 x  3  3  x 2  3 x  3  3  x
2

 x 4  6 x3  12 x 2  10 x  3  0
Since 3 and 1 are roots of f  x   x , then are roots of
f  f  x    x also and therefore.
f  f  x    x   x  3 x  1  x 2  2 x  1   x  3 x  1
3

Therefore 3, 1, 1, 1 are solutions of f  f  x    x . Hence

The number of arrangements of the solutions is
4!
4
3!
3 n 2n  4
66. e  1 
1 n n 1
Put n  48
10
e is a rational number
7
x2 y 2
 1
49 51
2b 2 102
l 
a 7
 05-01-24_JEE-MAIN_KEY&SOL
69. Let
tan x cot x
t dt
F  x   dt   t 1  t 
1/ e
1 t2 1/ e
2

Then
 tan x  2 1
F ' x    sec x 
 1  tan x 
2
cot x 1  cot x 
2   cos ec 2 x 

 tan x  1 / cot x  =0
Therefore F is a constant function. Now
 
1 1
t 1
F   dt   dt
 4  1/ e 1  t 1/ e t 1  t 
2 2

1
t2 1
1/ e t 1  t 2  dt  log e t 1/e
1


= 0   0  log e e  =1
Hence F  x   1
70. The two curves intersect at  4, 4  which is a vertex of the given square. Therefore
4
Required area (Shaded portion ) =  2 x
1
4 2
x 2 1
dx   2  1  1 = 2   x 3/ 2    x3   1
4 4

1
4 3 1 12 2

Y
(1,4)
(4,4)
y2  4x

x2  4 y
(1,1)
(2,1)
O 1 2 4 X
4 1 28 56 112  56  12
  8  1   64  8   1 =   1 
3 12 3 12 12
44 11
 
12 3
71. The given equation is
dy
sec 2 y  x  2 tan y   x3
dx
Put tan y  z , Therefore
dz
  2 x  z  x3 (Linear in z)
dx
The integrating factor is
I.F = e 
2 xdx 2
 ex
Therefore
ze x   x 3e x dx  c
2 2
 05-01-24_JEE-MAIN_KEY&SOL
1 2 x2
  x 2 e x xdx  c  x e  2 x  dx  c
2
2

1 1
  tet dt  c where t  x 2  et  t  1  c
2 2
1 x2 2
 e  x  1  c
2
So
1
tan y   x 2  1  ce x
2

2
The curve passes through  0,  / 4  . This implies
1 3
1  c  c 
2 2
1 2 3  x2
Therefore tan y   x  1  e
2 2
72. 3  a 2  b 2  c 2  1  2  a  b  c  ab  bc  ca   0

 a  1   b  1   c  1   a  b    b  c    c  a   0
2 2 2 2 2 2

 a=b=c=1
73. Let a1 , a2 , a3 , a 4 be the coefficients of rth,  r  1 th ,  r  2  th and  r  3 th terms, respectively.
Then
a1  n Cr 1 , a2 n Cr , a3  n Cr 1 , a4 n Cr  2
We know that
n
CK n  K 1
n

CK 1 K
Therefore
a2 n  r  1 a n 1
  1 2 
a1 r a1 r
a3 n  r a n 1
  1 3 
a2 r  1 a2 r  1
a4 n  r  1 a n 1
  1 4 
a3 r 1 a3 r  2
And hence
a1 a3 r r2  r 1   a2 
    2   2 
a1  a2 a3  a4 n  1 n  1  n 1   a2  a3 

74.  x   x 
2 2
 2 x1 x j  300;
x 2
1
 30
1 i
10
2
x12  x1 
   ;   30  25  5
10  10 
75. (A) tan 2   sin  cos   sin   cos 3 
 (1  sin 2  )  (1  3sin 2  )  3sin 4   sin 6 
 cos 2   (1  sin 2  )3  cos 2   cos 6   cos 2   sin 2   1
(B) sin 400  sin(600  200 )
3 1
2sin 200 cos 200  cos 200  sin 200
2 2
 05-01-24_JEE-MAIN_KEY&SOL
4cos 20  3 cot 20  1
0 0

3  cot 760 cot160 3sin 760 sin160  cos 760 cos160

(C) 
cot 760  cot160 sin(760  160 )
2sin 760 sin160  cos(760  160 ) cos 600  cos 920  cos 600 1  cos 920
    tan 460  cot 440
sin(76  16 )
0 0
sin 92 0
sin 92 0

   2  2  
(D) sin 2    sin 2    ....  sin   = 5
 18   18  2
76. We have
k 3  1  k  1  k  k  1  k  1   
2
k 2  k 1
     
k 3  1  k  1  k 2  k  1  k  1    k  12   k  1  1 
For k  2,3,.............n Therefore
 2 1 3 1 4 1 n  2 n 1   7 13 21 n2  n  1 
Pn   , , ..... .    . . ......... 
 2 1 3 1 4 1 n n 1  3 7 13  n  1
2
  n  1  1 
 
1 2 3 n  2 n  1   7 13 21 n  n 1
2 
=  . . ..... .   . . ..... 
n n  1   3 7 13  n  1   n  1  1 
2
3 4 5
 2   n2  n  1  2  1 
      1  
 n  n  1   3  3  n  n  1 
2 2
Therefore lim Pn  1  0  
n  3 3
3
77. We have seen that x is not differentiable at x  0 , Whereas x is differentiable at x  0 . Also
2
x  x 2 is differentiable for all real x. If a2  0 , then
f  x   a0 x  a1 x  a3
3 2

Is differentiable at x  0 . Conversely, if f  x  is differentiable at x  0 , then

a2 x  f  x   a0 x  a1 x  a3
3 2

Is differentiable at x  0 which is possible when a2  0

78. Let n  X  denote the number of elements in X
Then,
n  A  B  C   n  A  n  B   n  C   n  A  B 
n  B  C   n  C  A  n  A  B  C 
 n  A   n  A  B 
(since A  B  C   )
Now
AB   A  B    B  A    A  B    A  B 
Therefore
n  AB   n  A  B   n  A  B 
n  A   n  B   2n  A  B 
And
300  n  AB     n  A   n  B   2n  A  B    2 n  A   n  A  B  
Therefore n  A  B  C   n  A   n  A  B   300 / 2  150
 05-01-24_JEE-MAIN_KEY&SOL
79.

80.

81.
B  0, 2
A  0,1
yx

A ' 1, 0
PA  PB will be minimum
Where A and A ' are mirror image
A ' , P, B are collinear equation of line A ' B : 2x+y=2 Solve A ' B with y =x
2 2
x  , y 
3 3
2
 P  1  i 
3
k  2
 05-01-24_JEE-MAIN_KEY&SOL
82. Number of ways to distribute at least one toy to each 141 C31 13 C2  78
If toys are distributed in the following way then two will get equal number of toys
No.of ways
3!
1 1 12  3ways
2!
2 2 10 3 ways
338 3 ways
446 3 ways
554 3 ways
662 3 ways
 Required number of ways = 78-18= 60
83.

84. Differentiating the given function we have

2
f '  x   x 2/3   x  1 x 1/3
3
3x  2  x  1 5 x  2
 
3 x1/3 3 x1/3
Now f '  x   0  x  2 / 5 . Also f  x  is defined and continuous at x  0 , f is not differentiable
at x  0 . Thus, zero is a critical point. Therefore, 0 and 2/5 are critical points of f  x 
(i) x  0  f '  x   0 and x  0  f '  x   0 . Therefore at x  0 , f is maximum and the
maximum value = f  0   0
ii) x  2 / 5  f '  x   0 and x  2 / 5  f '  x   0 . Thus f is minimum at x  2 / 5 and the
minimum value
1/3
2 3 4 
f    
5 5  25 
86. Let the equation of the line L, by hypothesis, be y  2  m  x  8 
Where m  0 . Therefore
 2 
p   8  , 0  and Q   0, 2  8m 
 m 
Now,
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 2
OP  OQ   8     2  8m  m  0 
 m
2  2
10    8m   10  2   8m 
m  m
 10  8  AM  GM 
And equality occurs if and only if
2 1
  8m or m  
m 2
Hence, the absolute minimum of OP  OQ is 12 + 6 = 18
87.

88.
 05-01-24_JEE-MAIN_KEY&SOL
5x  7
8 6
5 x  7 x 8
6
89.  14  1 1  dx    1 1 2 dx
x  5  7  2 2 5  7 
x x   x x 
1 1
Put 2  5  7  t
x x
 5 x  7 x  dx  dt
6 8

 dt 1
  c
t2 t
x7
f  x  7 c
2 x  x2  1
1
f  0   0  C  0  f 1  k 4
4
90. 6  x 2  3ax  2a 2 
= 6  x  a  x  2a  ; a  0
x = a is point of maxima
x = 2a is point of minima
 a 2  2a
 a  0 or a  2
But a  0  a  2