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FIITJEE
Solutions to JEE(Main) -2023
Test Date: 1st February 2023 (First Shift)

PHYSICS, CHEMISTRY & MATHEMATICS
Paper - 1
Time Allotted: 3 Hours


Maximum Marks: 300

Please read the instructions carefully
...


Important Instructions:
1
...


2
...
Each subject (PCM) has 30 questions
...


3
...
Part-A is Physics, Part-B is Chemistry and
Part-C is Mathematics
...


4
...


5
...


6
...
Each question carries +4 marks for correct answer and –1 mark for wrong
answer
...


Section-B (1 – 10) contains 10 Numerical based questions
...
Each question carries +4 marks for correct
answer and –1 mark for wrong answer
...
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...
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...
Each question has four choices (A), (B), (C) and
(D), out of which ONLY ONE option is correct
...


The average kinetic energy of a molecule of the gas is
(A) Proportional to absolute temperature
(B) dependent on the nature of the gas
(C) proportional to pressure
(D) proportional to volume

Q2
...
Then the
electric fields in the three different region E1, EII and EIII
are
...


Surface Charge
Density 

I

II





III

d

a 

 P  2   V  b   RT represents the equation of state of some gases
...
The physical quantity, which has
b2
dimensional formula as that of
, will be :
a
(A) Compressibility
(B) Bulk modulus
(C) Energy density
(D) Modulus of rigidity

e
x ms 1 will be the
3
minimum velocity required by a rocket to pull out of gravitational force of P, where e is escape
velocity on earth
...


If earth has a mass nine times and radius twice to that of a planet P
...


An object moves with speed 1,  2 and 3 along a line A
B
segment AB, BC and CD respectively as shown in
figure
...
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Match List I with List II :
List-I
List-II
A
...

Fermi-level near the valence band
B
...

Fermi-level in the middle of valence and conduction band
C
...
Fermi-level near the conduction band
D
...
Fermi-level inside the conduction band
Choose the correct answer from the options given below :
(A) A – II, B – II, C – III, D – IV
(B) A – I, B – II, C – III, D – IV
(C) A – III, B – I, C – II, D – IV
(D) A – II, B – III, C – I, D – IV

Q7
...

An unpolarized light of intensity I is incident into this arrangement
...
The value of n will be :
64
(A) 6
(B) 5
(C) 4
(D) 3

Q8
...
Neglecting the air resistance, the speed with which the stone hits the
ground will be ______ms-1
(given, g = 10 ms-2)
...


A sample of gas at temperature T is adiabatically expanded to double its volume
...


A steel wire with mass per unit length 7
...
The speed of
transverse waves in the wire will be :
(A) 10 m/s
(B) 100 m/s
(C) 50 m/s
(D) 200  m / s

Q11
...
Microwaves
I
...
Gamma rays
II
...
Radio waves
III
...
X-rays
IV
...


A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of 
...
The ratio of
kinetic energy of proton and that of alpha particle is :
(A) 1 : 2
(B) 1 : 4
(C) 2 : 1
(D) 4 : 1

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JEE-MAIN-2023 (1st February-First Shift)-PCM-4

Q13
...


Which of the following frequencies does not belong to FM broadcast
...


The mass of proton, neutron and helium nucleus are respectively 1
...
0087u and 4
...

The binding of helium nucleus is :
(A) 28
...
8 MeV
(C) 7
...
2 MeV

Q16
...
Now, if a force of 30N is applied
in the direction parallel to surface of the table, the block slides through a distance of 50m in an
interval of time 10s
...
60
(B) 0
...
25
(D) 0
...


Match List I with List II :
List I
List II
A
...
Presence of both L and C
B
...
Electromagnetic Induction
C
...
Quality factor
D
...
Mutual Induction
Choose the correct answer from the options given below :
(A) A – IV, B – II, C – I, D – III
(B) A – IV, B – III, C – I, D – II
(C) A – II, B – IV, C – I, D – III
(D) A – II, B – I, C – III, D – IV

Q18
...
Surface tension of mercury
is 0
...
The gain in surface energy is :
(A) 5  105 J
(B) 28  105 J
5
(C) 17
...
26  105 J

Q19
...

Statement II : Acceleration due to gravity increases as we go down below the earth’s surface
...
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Find the magnetic field at the point P in figure
...

 i
2
(A) 0  1  
2r 

 i
1
(B) 0  1  
2r 

0i  1 1 
(C)
  
2r  2  
 i 1 1 
(D) 0  

2r  2 2 

i
p
r

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JEE-MAIN-2023 (1st February-First Shift)-PCM-6

SECTION - B
(Numerical Answer Type)
This section contains 10 Numerical based questions
...

Q1
...
If the
cylinder rolls without slipping, its speed upon reaching
the bottom of the inclined plane is _______ ms 1
...


Two equal positive point charges are separated by a distance 2a
...
The value of x is ________
...


A light of energy 12
...
The atom absorbs
the radiation and reaches to one of its excited states
...
The value of x is _______(use h = 4
...



Q4
...
The value of work done will be _________J
...


A charge particle of 2 C accelerated by a potential difference of 100V enters a region of uniform
magnetic of magnitude 4mT at right angle to the direction of field
...
The mass of the charge particle is_________
1018 kg
...


A series LCR circuit is connected to an ac source of 220V, 50Hz
...
6
...


Q7
...
The rod is placed in a such a way that mid point of the rod is at 40cm
x
from the pole of mirror
...
The value of x
3
is ______
...


The amplitude of a particle executing SHM is 3cm
...


Q9
...
Water gets
compressed to 0
...
03%
...
The value of x is ________
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In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf
1
...
If this cell is replaced by another cell of emf E, the length-of null point
x
increases by 40cm
...
The value of x is _______
...
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Each question has four choices (A), (B), (C) and
(D), out of which ONLY ONE option is correct
...


How can photochemical smog be controlled?
(A) By using catalyst
...

(C) By using catalytic convertors in the automobiles / industry
...


Q2
...


O

NHCOOEt

NH2
(C)

C

(D)

CH2COOEt

CH2OH

But-2-yne is reacted separately with one mole of Hydrogen as shown below:
B

Na
CH3 C
liq
...
A is more soluble than B
...
The boiling point & melting point of A are higher and lower than B respectively
...
A is more polar than B because dipole moment of A is zero
...
Br2 adds easily to B than A
...


Given below are two statements: one is labelled as Assertion A and the other is labelled as
Reason R
Assertion A: Hydrogen is an environment friendly fuel
...

In the light of the above statements, choose the correct answer from the options given below
(A) A is true but R is false
(B) Both A and R are true but R is NOT the correct explanation of A
(C) Both A and R are true and R is the correct explanation of A
(D) A is false but R is true

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JEE-MAIN-2023 (1st February-First Shift)-PCM-9

Q5
...
FeSO4
...
6H2O
B
...
4NH3
...
K2SO4
...
24H2O
D
...
4KCN
Choose the correct answer
(A) A and C only
(B) A and B only
(C) B and D only
(D) A, B and D only

Q6
...
Beryllium oxide is purely acidic in nature
...
Beryllium carbonate is kept in the atmosphere of CO2
C
...

D
...

Choose the correct answer from the options given below:
(A) B, C and D only
(B) A, B and C only
(C) A only
(D) A and B only

Q7
...


Match List I with List II
List-I
List-II
A
...

NaOH
B
...
Ca(OH)2
C
...
Na2CO3,10H2O
D
...
CaSO4
Choose the correct answer from the options given below:
(A) A- III, B-II, C-IV, D- I
(B) A-III, B-IV, C-II, D-I
(C) A-II, B-IV, C-I, D-III
(D) A-I, B-IV, C-II, D-III

Q9
...

3
Reason R: The critical volume Vc(cm mol1) and critical pressure PC (atm) is highest for Krypton
but the compressibility factor at critical point ZC is lowest for Krypton
...


Match List I with List II
List-I
List-II
A
...

Anti blood clotting
B
...
Salvarsan
C
...
Antidepressant drug
D
...
soframicine
Choose the correct answer from the options given below:
(A) A- IV, B-II, C-I, D-III
(B) A-III, B-I, C-II, D-IV
(C) A-II, B-IV, C-I, D-III
(D) A-II, B-I, C-III, D-IV

to

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the

JEE-MAIN-2023 (1st February-First Shift)-PCM-10

Q11
...

(B) CO32 has a single structure i
...
resonance hybrid of the above three structures
...

(D) All these structure are in dynamic equilibrium with each other
...


Which of the following complex will show largest splitting of d- orbtials?
(A) Fe  CN6 

3

(C) Fe  C2O4 3 
Q13
...


Given below are two statements: one is labelled as Assertion A and the other is labelled as
Reason R
Assertion A: In an Ellingham diagram, the oxidation of carbon to carbon monoxide shows a
negative slope with respect to temperature
...

In the light of the above statements, choose the correct answer from the options given below
(A) Both A and R are correct and R is the correct explanation of A
(B) Both A and R are correct but R is NOT the correct explanation of A
(C) A is correct but R is not correct
(D) A is not correct but R is correct

Q15
...
The correct statements about Mn2O7 are
(A) Mn is tetrahedrally surrounded by oxygen atoms
...

(C) Contains Mn-O-Mn bridge
(D) Contains Mn-Mn bond
...
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Match List I with List II
List-I
List-II(Functional group / Class of
(Test)
compound
A
...

Peptide
B
...
Carbohydrate
C
...
Primary amine
D
...
Aldehyde
Choose the correct answer from the options given below:
(A) A- III, B-IV, C-I, D-II
(B) A-III, B-IV, C-II, D-I
(C) A-II, B-I, C-III, D-IV
(D) A-I, B-II, C-III, D-IV

Q17
...

Statement II: Chemical reactivity of an element can be determined by its reaction with oxygen
and halogens
...


Which of the following represents the lattice structure of A0
...


B
...


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JEE-MAIN-2023 (1st February-First Shift)-PCM-12

(A) B only
(C) A and B only
Q19
...


(D)

O
OH
H
H HO
H
OH
OH H

Decreasing order of dehydration of the following alcohols is
OH
a
(A) a > d > b > c
(C) d > b> c > a

OH
b

OH

OH
c

d
(B) b > d > c > a
(D) b > a > d > c

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JEE-MAIN-2023 (1st February-First Shift)-PCM-13

SECTION - B
(Numerical Answer Type)
This section contains 10 Numerical based questions
...

Q1
...


Q2
...


25 mL of an aqueous solution of KCl was found a require 20 mL of 1M AgNO3 solution when
titrated using K2CrO4 as an indicator
...
0 K kg mol1)
Assume 1) 100% ionization and 2) density of the aqueous solution as 1 g mL1

Q4
...
1M) | Mn2 (0
...
282V?
_________
...
303RT
o
Given EMnO
 0
...
54V,
4 |Mn
F

Q5
...
The value of x is
 p2 
___________(Nearest integer)
Q6
...
(iii) do not give orange precipitate with 2,4-DNP (iv) on hydrogenation
give identical compound with molecular formula C9H12O is_______
...


Electrons in a cathode ray tube have been emitted with a velocity of 1000 ms
...

31
34
Given: h  6  10 Js , me  9  10 kg
(A) The de-Broglie wavelength of the electron emitted is 666
...

(B) The characteristic of electrons emitted depend upon the material of the electrodes of
the cathode ray tube
...

(D) The nature of the emitted electron depends on the nature of the gas present in
cathode ray tube
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The total number of chiral compound/s from the following is__________
...


The density of 3 M solution of NaCl is 1
...
Molality of the solution is_______
10-2m
...
5 g mol-1 respectively
...


A and B are two substance undergoing radioactive decay in a container
...
if the initial concentration of B is 4 times that of A and they both start
decaying at the same time, how much time will it take for the concentration of both of them to be
same?____________ min
...
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Each question has four choices (A), (B), (C) and
(D), out of which ONLY ONE option is correct
...


The area enclosed by the closed curve C given by the differential equation
dy x  a

 0, y 1  0 is 4
...
If normals at P and and Q
on the curve C intersect x-axis at points R and S respectively, then the length of the line segment
RS is
(A) 2 3
(B) 2
(C)

4 3
3

(D)

2 3
3

Q2
...
If 3 observations are 1, 3, 5,
then the sum of cubes of the remaining two observations is
(A) 1456
(B) 1792
(C) 1072
(D) 1216

Q3
...
x   , 
...


 3
(A)   sin1 
 4 



(B)

(C) 0

 3
(D)   2sin1 
 4 



Let R be a relation on  , given by R 

2
3

a,b  : 3a  3b 



7 is an irrational number
...


Q6
...
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For a triangle ABC, the value of cos 2A  cos 2B  cos 2C is least
...


If y  y  x  is the solution curve of the differential equation
dy


 y tan x  x sec x, 0  x  , y  0   1 , then y   is equal to
dx
3
6

Q9
...






(B) 9
(D) 12

If the orthocentre of the triangle, whose vertices are 1, 2  ,  2,3  and

 3,1 is  , ,

then the

quadratic equation whose roots are   4 and 4   , is
(A) x 2  19x  90  0
(C) x 2  20x  99  0
Q11
...
Then

(A) there exists xˆ  0,3  such that f '  xˆ   g'  xˆ 
(B) min f '  x   1  max g'  x 
(C) there exist 0  x1  x 2  3 such that f  x   g  x  , x   x1, x 2 
(D) max f  x   max g  x 
Q12
...



Let S   x : x  R and

(A) 4
(C) 6



3 2



x2  4





3 2



x2  4


 10 
...
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Q15
...
is
2
4
2
4
1 1  1 1 2  2
1  32  3 4
56
(A)
111
59
(C)
111

55
111
58
(D)
111

(B)

The combined equation of the two lines ax  by  c  0 and a ' x  b ' y  c '  0 can be written as

 ax  by  c  a' x  b' y  c '   0
...


(A) x 2  y 2  10xy  0

(B) 3x 2  5xy  2y2  0

(C) 3x 2  xy  2y 2  0

(D) x 2  y 2  10xy  0

1  sin2 x
cos2 x
2
Let f  x   sin x
1  cos 2 x
sin2 x

cos 2 x

sin2x
 
sin2x , x   , 
...


The value of
250
51!
251
(C)
51!

(A)

Q18
...


is :
1!50! 3! 48! 5! 46!
49!2! 51!1!
251
(B)
50!
250
(D)
50!

Let the image of the point P  2,  1, 3  in the plane x  2y  z  0 be Q
...


(B) 3 14
(D) 2 14

The shortest distance between the lines
x5 y2 z4
x  3 y  5 z 1


and


is
1
2
3
1
4
5
(A) 5 3
(B) 6 3
(C) 7 3

(D) 4 3

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JEE-MAIN-2023 (1st February-First Shift)-PCM-18

Q20
...

is equal to
2n 
1  n 2  n 3  n

n

(A) 0

2
(B) loge  
3

(C) loge 2

3
(D) loge  
2

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JEE-MAIN-2023 (1st February-First Shift)-PCM-19

SECTION - B
(Numerical Answer Type)
This section contains 10 Numerical based questions
...

Q1
...






ˆ 
Let v   ˆi  2ˆj  3k,
 2 ˆi  ˆj  kˆ and u be a vector such that u    0
...


Q3
...


A  2,6,2  , B  4,0,  , C  2,3, 1 and D  4,5,0 ,   5 are the vertices of a quadrilateral ABCD
...

2

Let f : R  R be a differentiable function such that f '  x   f  x    f  t  dt
...


0

2f  0   f  2  is equal to………
1

Q6
...


Q7
...
Then 12 A is equal to………

Q8
...
,an be an A
...
If the sum of its first four terms is 50 and the sum of its last four
terms is 170, then the product of its middle two terms is……
...


The number of words, with or without meaning, that can be formed using all the letters of the
word ASSASSINATION so that the vowels occur together, is……
...


If f  x   x 2  g' 1 x  g"  2  and g  x   f 1 x 2  xf '  x   f "  x  , then the value of f  4   g  4  is
equal to……
...
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A

2
...


A

4
...


D

6
...


A

8
...


A

10
...


C

12
...


D

14
...


A

16
...


C

18
...


A

20
...


2

2
...


828

4
...


144

6
...


32

8
...


1

10
...


C

2
...


DROP

4
...


A

6
...


A

8
...


B

10
...


B

12
...


A

14
...


B

16
...


B

18

B

19
...


B

SECTION - B
1
...


499

3
...


3

5
...


2

7
...


2

9
...


15

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JEE-MAIN-2023 (1st February-First Shift)-PCM-21

PART – C (MATHEMATICS)
SECTION - A
1
...


C

3
...


A

5
...


A

7
...


D

9
...


C

11
...


A

13
...


B

15
...


D

17
...


B

20
...


514

2
...


29

4
...


1

6
...


62

8
...


50400

10
...
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...
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...
The average kinetic energy of a molecule of the gas = kT
2
The average kinetic energy of a molecule of the gas  T ( absolute temperature)

Sol2
...
Dimension of b =  V   M 0 L3T 0 
0 6 0
 MLT 2  6
b2  M L T 
5 2


PV 2  a  

L

ML
T


  M 1 LT 2 

2


5 2
L
a


ML
T




1 dV 1
Compressibility 
   M 1 LT 2 
V dP P 
Sol4
...


2GM
r

9
3

2
2

ve 2 ve

3
3

x x2
V1

AB  BC  CD  x and S  ut
x
x
x
t AB 
, tBC 
and tCD 
v1
v2
v3

 total time 

A

V2

B

V3
C

D

3x
 t AB  tBC  t CD
v avg

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...


JEE-MAIN-2023 (1st February-First Shift)-PCM-23



x
x
x


v1 v 2 v 3



v v  v1  v 2  v1v 2
3
 2 3
v avg
v1v 2 v 3

 v avg 

3v1v 2 v 3
v 2 v 3  v1v 2  v1v 3

Sol6
...


1st Plate 

I0
2
I
2nd plate  0 cos2 
2
I
th
 n plate  0 cos2 
2
 Total no of plates = 6



Sol8
...


PV   cons tan t
TV 1  cons tan t
1

TV 2  cons tan t
1/2

T1V11/2  T2  2V1 

T2 

T1
2

nR
 T1  T2 
r 1


T 

T 
RT 
 RT 

2
2
Work done = 
 
 RT 2  2
3
1
1
2
2





Sol10
...
0  103 kg / m
T  70
v 

T



70
7  10

3

 104  100

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JEE-MAIN-2023 (1st February-First Shift)-PCM-24

Sol11
...
v 
10

Sol12
...
Balanced Wheat stone Bridge
R

R

3R

3R
A

2R

9R

9R
A

B
6R

2R

6R

8
1
1 2 1

  Re q 
Re q 4  2 8
3

Sol14
...
 Helium  2 proton and 2 neutron

 m  4
...
0087  1
...
0016   2  2
...
0016  4
...
0305 amu  mass defect
 BE  mc 2  0
...
2745  1016
N

Sol16
...
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...
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...
D
...
Resonance phenomenon  I Presence of L & C
...
Transformer  IV Mutual induction
...
AC generator  EMI
Sol18
...
45  4  10 8  180  10 8
UFinal  0
...
14  106  28
...
Statement 1 is true
...


d


g'  g  2r cos 

 i 1

d

gd  ge  1  
 R
O

 i
Sol20
...


ICylinder 

mR2
2

1
1
3
21
1 mR2 v 2
mv 2  I2 
mv 2  

 mv 2
2
2
2
22
2
2
4
R
3
30
 Conversation of energy  mv 2  mg 
4
100

 Total KE at bottom



 v2  4
v2

Sol2
...
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...


1 1 
E  12
...
6   
1 n22 
13
...
75eV  13
...
6
 0
...
6  100
n2 
 16
0
...
14  1015  1017
x  828

Sol4
...
S  F
...


mv
3
mv


 8  3  1011  mv  24  10 11  mv
qB
100 2  4  109
qV  KE

r

2m  qV  m2 v 2  24  24  1022
m

24  24  10 22  1018
100  2  2  10

Sol6
...


4

 144  1018

At resonance, any rate of energy supplied is max
1
1
1
XL 
C

 C  40F
C
  L 1000   70
...


32
, Hence x  32
3

U = (constant)  sin2 t
K = (constant)  cos2 t
We know total energy = (constant) = K + U
1
...
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B

5 2
2
x  4A 2  4x2  5x 2  4A 2  9x 2  x  A
4
3

1 dp
V dV


1  dV   
 V   0
...
01 1
  dP L  V L  
3 3
Hence   x  1
1 x
B
 W 
BL

Sol10
...


Photochemical smog is controlled by using catalytic converters in the automobile industry
...


NH2

O
EtO

C

OEt

CH2OH

O
O
NH

C

CH2

OH

NH

OEt

C
O

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JEE-MAIN-2023 (1st February-First Shift)-PCM-28

Sol3
...
NH3

H
CH3
(B)
Correct options are not given because Trans is more stable than
Cis is polar and trans is non-polar
B
...

M
...

Sol4
...


Sol5
...
Double salts are
FeSO4 (NH4)2SO4,6H2O and K2SO4
...
2HH2O

Sol6
...

(C) Beryllium sulphate is readily soluble in water due to high degree of hydration
...


Sol7
...


Salaked lime  Ca(OH)2
Dead burnt plaster  CaSO4
Caustic soda  NaOH
...
10H2O

Sol9
...

3
ZC  for all real gases
...
Tranquilizers
Aspirin
Antibiotic
Antiseptic






Anti depressant drug
Anti Blood clotting
Salvarsan
Soframincine
...
Resonating structure are hypothetical structure and resonance hybrid is real structure which is
the actual representative
...
Splitting of d-orbitals takes place in pressure of strong field ligand and CN is the ligand in which
maximum splitting occurs
...

Br
+ KOH (alc)

No reaction

Elimination reaction takes place with alcohol KOH and -hydrogen should be present
...
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...
2C(s)  O2 (g)  2CO(g)
S0r is +ve, G0r  Hr0  T
...


Gr0 become more –ve thus it has lower tendency to get

Sol15
...
Test
(A) Molisch’s Test
(B) Biuret test
(C) Carbylamine Test
(D) Schiff’s Test

Functional group / class of compound
...


Sol17
...


tend to

Sol18
...

Sol19
...
Dehydration of alcohol is directly proportional to stability of carbocation
...


HBrO3 ( Bromic acid)
O
...
S  7
Sum of the O
...


2H2 O(g)  2H2 (g)  O2 (g)   242  2  kJ / mol

H2 (g)  O2 (g) 
 2OH  78kJ / mol
H2 (g) 
 2H  430kJ / mol

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JEE-MAIN-2023 (1st February-First Shift)-PCM-30

2H2 O 
 2H  2OH  998kJ / mol
H2 O  H  OH  998 

Sol3
...
mol of KCl = m mol of AgNO3 = 20 m
...
5   23
...
85
23
...
085  3
...


MnO 4  8H  5e  Mn2   4H2O
Mn2  
0
...
059
103
1
...
54 
log
8
5
101  H 
0
...
059
H 
21
...
98  3
Sol5
...
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As per information given in the question the compound should not be alcohol, aldehyde, ketone
and it is aromatic as well
 Ph CH CH
OCH3
Cis and trans form exist only
...


(A) Ve  1000m / s, h  6  1034 Js
Me  9  10 31Kg
h
6  1034

 666
...
67nm
(B) The characteristics of electrons emitted is independent of the material of the
electrodes of the cathode ray tube
...

(D) The nature of the emitted electrons is independent of the nature of the gas present in
the cathode ray tube
...


OH
O

O

Cl

No POS & COS (Chiral)
OH

COOH
OH

HO

H
C

HO
OH
(Achiral)
Sol9
...
wt
...
64
1000  1   3  58
...
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For A
Ct  Co
...
e kt  x
...
x
...
e

e

 n2 

t
 15 

 n2 


 5 

 n2 

t
 15 

t

 4x
...


dy x  a

0
dx y  2

  y  2  dy   x  a  dx  0
2

2

  y  2   x  a  c
y 1  0  c  4  1  a 
2

2

2

c :  y  2    x  a   4  1  a 

2

Area enclosed by c is 4  1  a  0
 a  1
2

2

c :  x  1   y  2   4

Q
C(1,2)

R

P
N

S

2

x  0   y  2  3

y  2 3



P 0, 2  3

 & Q 0, 2  3 

Equation QS
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JEE-MAIN-2023 (1st February-First Shift)-PCM-33

y  2   3  x  1



2
,0 
Cuts x axis at S  1 
3 

2
NS 
3
4
 RS 
3

Sol2
...
(i)
1
1  9  25  x 24  x 52  25  8
5
 x 24  x52  33  5  35  130 ………
...


cos 1  2x   2cos 1 1  x 2  

 cos1 2x    2cos 1 1  x 2
as 0  cos1 y  
 only possibility
cos 1 2x  
 2x  cos   1
1
x
2
&
2cos1 1  x 2  0
 cos1 1  x 2  0  1  x 2  1
x0
 no such value of x satisfy simultaneously
...




R :  a,b  : 3a  3b  7 is an irrational number



3  a  b   7  Irrational number
 3  a  a   7  Irrational
 R is reflexive
...

R is not transitive too
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Sol6
...


cos 2A  cos 2B  cos 2C  1  4 cos A cosB cos C which

is minimum for A  B  C 
3
r=3

AM  r cosec A / 2  3 cos  6 as ABC is equilateral
6
 M is centroid too
...
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dy
 y tan x  x sec x
dx

I
...
= e 
 en sec x  sec x
solution y sec x   x sec 2 x dx
tan x dx

y sec x  x tan x  n sec x  c
1 c
y sec x  x tan x  nsec x  1


y  
6
y

Sol9
...
Equation AD y  2 

5 7
Solving (i) & (ii) H  ,     , 
3 3
  4  11 & 4    9
 the required quadratic equation is
x 2  20x  99  0

D
H

A(1, 2)

E

B(2, 3)

Sol11
...
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S   x : x  R


Let y 
y







3 2

3 2



x2  4





x2  4



1

y





3 2

3 2





x2  4


 10 


x2  4

1
 10  y  5  2 6
y

3 2



x2  4

 52 6 



3 2





3 2



2

x2  6  x   6



3 2



x2  4

52 6 

2

 x 2  4  2

x 2
n 5   4
 1 1
Sol13
...

   2





2

     42  6
S

Sol14
...
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2x 2  xy  3y 2  0 ……
...
f  x   sin2 x
1  cos2 x
sin2 x

cos2 x

sin 2x
sin 2x
1  sin2x

C1  C2  C3
1 cos2 x
f  x    2  sin2x  1 1  cos2 x
1

cos2 x

sin 2x
sin 2x
1  sin2x

R3  R1, R2  R1
1 cos2 x sin 2x
f  x    2  sin2x  0
1
0
0

0

1

f  x   2  sin2x
  2  1  3 for x 



...


1
1
1


...



51!  1! 50! 3! 48!
51! 1! 

1 51
 C1  51C3 
...
Image of P  2,  1, 3  in x  2y  z  0 be

x  2 y  1 z  3 2  2  2  3 



1
1
2
1
1 4  1

x  3, y  1, z  2
 Q  3, 1, 2 

 distance of 3x  2y  z  29  0 from Q 

9  2  2  29
9  4 1



42
14

 3 14

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JEE-MAIN-2023 (1st February-First Shift)-PCM-38

Sol19
...
D
...
lim 



...


Let E(i) denote number of 3 digit numbers divisible by i
...


ˆi
 
vw  

ˆj

kˆ

2 3  ˆi  5ˆj  3kˆ

2 1 1

 
  
  
u v w   u  v  w  u v  w cos    34 2  1 cos 
which is minimum for cos   1





  34 2  1   3401
 34 2  1  3401
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JEE-MAIN-2023 (1st February-First Shift)-PCM-39

  2  100    10  0

 u   ˆi  5ˆj  3kˆ

 u   3401





 2
2
100
m
 u  i  2 


3401 3401 n
 m  n  3501

Sol3
...
 200 C198  21  2200 


201
 All terms except 2 is divisible by 49
...
 67 C67 767

divisible by 49

Now to check only
1  67  7  470
470  49  9  29
 Remainder = 29
Sol4
...


f '  x   f  x    f  t  dt
0

dx
I
...
= e   ex
 solution e x f  x    a ex dx

2

where a   f  x  dx
0

 ex f  x   a ex  c
f  0   e 2  e 2  a  c

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JEE-MAIN-2023 (1st February-First Shift)-PCM-40

 c  e2  a
e x f  x   ae x  e 2  a
 x 2
 f  x   a  e    ae x ……
...


21

 x14  x 7

 2x

14

 3x 7  6

1/7



dx

0

1



 2x

  x 20  x13  x 6

21

 3x14  6x 7

1/7



dx

0

Let 2x 21  3x14  6x 7  t 7





 42 x 20  x13  x6 dx  7t 6 dt
1

6

1
117


0

11

1

m/n

t 8 7 118/7 11
t  t dt 


48 0
48

6

  48,m  8, n  7
  m  n  63

Sol7
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website: www
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com
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a1  8, a2 ,a3 ,
...
P
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ASSASSINATION
A3 T1
S4 O1
I2
N2 vowels AAAIIO
SS AAAIIO SSNTN
Required no
...
f  x   x 2  ax  b
we have a  g' 1 & b  g"  2 
f '  x   2x  a & f "  x   2
g  x   1  a  b  x 2  x  2x  a   2
  a  b  3  x 2  ax  2
g'  x   2  a  b  3  x  a
g"  x   2  a  b  3 

Now, a  g' 1  2  a  b  3   a
 a  b  3  0 …………
...
, FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
website: www
...
com

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