Sri Chaitanya IIT Academy.,India.
 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
JEE-ADVANCE-2022_P2
Sec: Sr. Elite(C-120, C-IPL, IPL-IC)
RPTA-08
Time: 02.30PM to 05.30PM
Date: 08-10-2023
Max. Marks: 180
KEY SHEET
MATHEMATICS
1
9
2
6
3
8
4
8
5
2
6
1
7
0
8
4
9
ABD
10
AD
11
ABCD
12
CD
13
ABD
14
BCD
15
A
16
B
17
A
18
C
PHYSICS
19
2
20
8
21
2
22
5
23
6
24
8
25
4
26
8
27
AB
28
AC
29
ABCD
30
ABC
31
AD
32
AD
33
A
34
B
35
C
36
D
3
CHEMISTRY
37
9
38
5
39
9
41
5
42
5
43
2
44
3
45 ABCD 46
AB
47
BC
48
ACD
49
CD
50
CD
B
D
53
A
54
C
51
40
52
Sri Chaitanya IIT Academy


08-10-23_Sr.Elite(C-120, C-IPL, IPL-IC)_Jee Adv(2022-P2)_RPTA-08_Paper 2_Key & Sol's
SOLUTIONS
MATHEMATICS
1
1
1



 9  4  36 .
 det ad  B 1 A1 
1 1 2
1
1
B A
j

2
2
B
A
01.
det  ad j  B 1 A1 

02.
 3 4   3 4   1 0 
BC  



 2 3  2 3   0 1 
1


ABC=A
 A  BC 2 
 ABC 
Tr  A  Tr 

  Tr 

4
 2 




 1 
 A
 A
= Tr  A  Tr    Tr          Tr  A 
 2Tr  A  2  3  6
1
2
4
1



 2
03.
A2  2 I  0; 2C  A2  32
A2  A3  B   2C  A3  B   0
A
3
 B  A2  2C   0
As A2  2C  0; A3  B  0
[2C  A2 is non singular]
B   A3   A  A2    A  2 I   2 A
2
2
B  A  2A  A  A
2
A2  2 I
A2   2  I  8
3
2
 B  A  A2  8
 Absolute value=8.
04.
 adj  A 
T
 adj A 
T
1 1
 adj  AT   AT  A
2

2
1
1
2

 A
2
1
adj A
A1
Sec : Sr.Elite(C-120, C-IPL, IPL-IC)
Page 2
Sri Chaitanya IIT Academy
1  b  x 1  c  x
1  b x 1  c  x
1  b  x 1  c x
1  a2 x
05.
08-10-23_Sr.Elite(C-120, C-IPL, IPL-IC)_Jee Adv(2022-P2)_RPTA-08_Paper 2_Key & Sol's
f  x   1  a 2  x
1  a  x
2
2
2
2
2
2
2
R3  R3  R1
1  b  x 1  c  x
1  a2 x
2
R2  R2  R1  x  1
1 x
0
x 1
1  a2 x
2
0
1 x
  x  1
2
1
1
1  b  x 1  c  x
2
2
1
0
0
1
2
  x  1 1  a 2 x  1  0   1  b2  x  1  0   1  c 2  x  0  1 
2
2
  x  1 1  a 2 x  x  b2 x  x  c 2 x    x  1 1  2 x   a 2  b 2  c 2  x 
But a 2  b 2  c 2  2 .   x  1 1  2 x  2 x    x  1
2
2
Degree 2.
06.
l1
m1
n1 l1
m1
n1
  l2
l3
m2
m3
n2 l2
n3 l3
m2
m3
n2
n3
2
l12  m12  n12
 l1l2  m1m2  n1n2
l1l3  m1m3  n1n3
l1l2  m1m2  n1n2
l m n
2
2
2
2
2
2
l2l3  m2 m3  n2 n3
l1l3  m1m3  n1n3
1 0 0
l2l3  m2 m3  n2 n3  0 1 0  1   2  1    1
0 0 1
l32  m32  n32
   1.
07.
Put n=3
But A2  A1
3
B  A2  A8
C  A2
3 1
A3  I
 A2
 B  A6 . A2  I . A2  A2
Hence det  B  C   A2  A2  0  0 .
08.
A is 3 x 3 matrix and A  2
Now  adjA1  
1
09.
1
1
2

 A 4
2
1
adjA
A 1
A2  4 A  5I 3  0
A A  4 I 3   5I 3
A1 
1
 A  4 I3  .
5
Sec : Sr.Elite(C-120, C-IPL, IPL-IC)
Page 3
Sri Chaitanya IIT Academy
10.
9
2 A9 B 1  22 A .
08-10-23_Sr.Elite(C-120, C-IPL, IPL-IC)_Jee Adv(2022-P2)_RPTA-08_Paper 2_Key & Sol's
1
 2
B
9
A  1
3  4  1    1 .
11.
 2   2   2  2  2   2  3    1     1    1  0        1
2
x
12.
 b
a
a
2
2
b
x a
b x
a  b  x 
1
a
b
a
b
c1  c1  c2  c3  a  b  x
x
ab x
b
a  a  b  x 1 x a
1 b x
x
R2  R2  R1 , R3  R3  R1
1
a
b
 a  b  x  0  x  a a  b
0 b  a x  b
2
  a  b  x    x  a  x  b    a  b     a  b  x   x 2   a  b  x  a 2  b 2  ab 


13.
a,b,c are roots of x 2  x 2  4  0
a  b  c  1 ab+bc+ca=0, abc=4
a b
c
A  c a b  a 3  b3  c 3  3abc   a  b  c   a 2  b 2  c 2  ab  bc  ca 
b c a
2
2
  1   a  b  c   3  ab  bc  ca     1   1  0  1  0




AAT  I  A1  AT .
14.
A

2
3
2
 2  6 .
(i) Unique solution if A  0  2  6  0
(ii) Infinitely many solutions if A  0    3
Equations becames 3x  2 y   and 3x  2 y  
3 x  2 y  
        0 .
Sec : Sr.Elite(C-120, C-IPL, IPL-IC)
Page 4
Sri Chaitanya IIT Academy
15.
08-10-23_Sr.Elite(C-120, C-IPL, IPL-IC)_Jee Adv(2022-P2)_RPTA-08_Paper 2_Key & Sol's
By definition A  I
2
A2 
A
 2 A  I
2
 A
   1  2 A.
2
16.


i
1  j x        1 i
Lt 


x 0
x
j
 a11 a12
 A  a21 a22

 a31 a32
a13   1 1 / 2 1 / 3 
a23    2 1
2 / 3
 

a33   3 3 / 2
1 
A2  3 A .
17.
f(x)=
1
x
x 1
2x
3x  x  1
x  x  1
x  x  1 x  2 
 x  1 x
x  x  1 x  1
x
2
x
1
2
x
1
x
2
x
x 1
 x  1 2
x 1 x 1
3 x  2 x 1
1
 1 2 x  1 1
3 x2 1
C2  C2  C1  x
2
x
1 x 1
2
 1 2 x 1  0
3 x 1
 f  x   0  f  500  0 .
18.
4 as non zero solution  A  0
p 1 1
1
q 1 0
1
1 r
Put p=2, q=3
2 1 1
1 3 1 0
1 1 r
2  3r  1  1  r  1  1 1  3  0
6r  2  r  1  2  0
5r  3
r  3/5
1
1
1
1
1
1
1 5






 1 
 =-1+2=1
1 p 1 q 1 r 1 2 1 3 1 3 / 5
2 2
Sec : Sr.Elite(C-120, C-IPL, IPL-IC)
Page 5
Sri Chaitanya IIT Academy
19.
08-10-23_Sr.Elite(C-120, C-IPL, IPL-IC)_Jee Adv(2022-P2)_RPTA-08_Paper 2_Key & Sol's
PHYSICS
Shearing stress, F / A  10 N / m
6
2
Length of side, L=10cm=0.1m
Shearing displacement, x  0.05cm  0.005 cm
The modulus of rigidity is

F
FL

A Ax
x  106  0.1

 2  108 N / m 2



tan




L 
0.005

20.
Y  FL / Ae
21.
consider an element as shown in the figure.
Stress in the element =
Force xA g

 x g
Area
A
Now, electric potential energy stored in the wire is
dx
1
dU  (stress) (strain) (volume)
2
x
1  stress 
 .
 volume 
2
Y
2
1  x g 
1  2g2 A 2
dU  .
Adx  .
x dx
2
Y
2
Y
2
1  2g2 A 2
 2 g 2 AL3 12  106
x
dx


 2  106 J
2
Y 0
6Y
6
Total electric potential energy= .
22.
L
Here, r1  30cm and r2  4cm .
Since a saop bubble has two free surfaces, total surface area of the coalesced
bubble= 2  4 r 2  8 r 2
Potential energy of the coalesced bubble, i.e. W  8 r 2T
Similarly if W1 and W2 are the potential energies of the two given bubbles,
W1  8 r12T and W2   r22T
Since the total energy of the system is conserved.
8 r 2T  8 r12T  8 r22T
or r 2  r12  r22   3   4   25
2
2
or r  5cm .
Sec : Sr.Elite(C-120, C-IPL, IPL-IC)
Page 6
Sri Chaitanya IIT Academy
23.
08-10-23_Sr.Elite(C-120, C-IPL, IPL-IC)_Jee Adv(2022-P2)_RPTA-08_Paper 2_Key & Sol's
4
4
 R 3   r 3  K (Conservation of volume)
3
3
R=radius of bigger drop
r=radius of smaller drop
R3  r3  K
U i  S  4 R 2 
U f  KS  4 r 2 
KS  4 r 2   S  4 R 2  103
4 S  Kr 2  R 2   10 3
K 1/3 R 2  R 2  102
10a /3  1  100
10a /3  101
a 6
24.
4
4
 R 3  n  r 3 or R 3  nr 3 or R  n1/3r
3
3
Or R  2n1/3mm
v0 r 2 , v01 R 2
Now,
v01 R 2 4n 2/3
32
 2 
or
 n 2/3 or n 2/3  4
v0 r
4
8
or n  43/2  64 or n=8
25.
Mass of air entering in 1s is m   b2v
Change in momentum in 1s  mv  0   b2  v 2
Thus, F   b 2  v 2
Pressure due to this force is p 
F  b2  v 2

  v2
A
 b2
When this excess pressure equals to
  v2 
4T
, the bubble gets separated
R
4T
4T
R 2
R
v
So,   4 .
Sec : Sr.Elite(C-120, C-IPL, IPL-IC)
Page 7
Sri Chaitanya IIT Academy
08-10-23_Sr.Elite(C-120, C-IPL, IPL-IC)_Jee Adv(2022-P2)_RPTA-08_Paper 2_Key & Sol's
26.
 r 4 P1   2 r  P2
P

 1 8
8 I
8  2l 
P2
27.
Conceptual
28.
Conceptual
29.
Conceptual
30.
Conceptual
31.
Conceptual
32.
Conceptual
33.
Taking g constant, work done in both cases is –mg (1000) since the max velocity of the
4
journey is attained at a height of 1km, and the second half journey is travelled by max
velocity the time taken for second half journey will be less.
34.
F  F ' A
dv
 mg sin 
dz
Velocity gradient
dv v

dz h
F
mg sin 
v
h
F
300
35.
m  10  2  3  102 
10
10 

or m  6  104 kg  6  104  103 g  0.6 g
36.
A longer pipe offers a greater resistance to the flow of a viscous fluid than a shorter pipe
does. The volume flow rate depends inversely on the length of the pipe. For a given pipe
radius and pressure difference between the ends of the pipe, the volume flow rate is less
in longer pipes. In this case the longer pipe is twice as long, so its volume flow rate QB is
one-half that of QA .
Sec : Sr.Elite(C-120, C-IPL, IPL-IC)
Page 8
Sri Chaitanya IIT Academy
08-10-23_Sr.Elite(C-120, C-IPL, IPL-IC)_Jee Adv(2022-P2)_RPTA-08_Paper 2_Key & Sol's
CHEMISTRY
37.
Conceptual
38.
From PV=nRT, total moles =0.6.
Let C x H 8 be ‘a’ mole therefore moles of C x H12   0.6  a  mole
Mass of ‘c’ in ‘a’ mole of C x H 8  12axg .
Mass of ‘c’ in 0.6  a mole of C x H12  12  0.6  a  g .
Therefore total mass of c in mixture = 41.4g and x=5.
39.
U rms 
u12 N1  u22 N 2
U 2 N  U 22 N 2
2
U rms
 1 1
N1  N 2
N1  N 2
25 
40.
4  72  422
5.4
 u2 
 3 ms 1
10
6
v,n are constant
T
Pi Pf
 109 

 Pf  f  Pi  
 Pi
Ti T f
Ti
 100 
P increases P  Pf  Pi 
“1” pressure increases 
9
Pi
100
Pf
Pc
 100 
9 Pi
 100  % X  9  1
100 Pi
X=9.
41.
mol.wt of aceylated  M .wt of organic compound ACompound
42
 No.of amino groups=
390  180
 5.
42
42.
a,d,e,f,g can form stable carbanion.
43.
Two different amides give two different amines.
44.
HOOC
COOH



D
H
CH 3
CH 3
CH 3
HOOC
H
H
D
CH 3
CH 3
45.
Dehydration products of carboxylic acids.
46.
Mechanism of HVZ reaction.
Sec : Sr.Elite(C-120, C-IPL, IPL-IC)

H
H
COOH  CO2
D
CH 3
Page 9
Sri Chaitanya IIT Academy
47.
08-10-23_Sr.Elite(C-120, C-IPL, IPL-IC)_Jee Adv(2022-P2)_RPTA-08_Paper 2_Key & Sol's
Decarboxylation is facilitated by the presence of any carbonyl group at the   position,
including that of carboxyl group (or) ester. The mechanism followed through a cyclic,
six members. Transition state involving rearrangement of 3e  pairs gives the enol form
of a carboxylic acid, which is tautomerised to the carboxylic acid.
3RT
PMw
3
d
.
RT 1mole  U rms 
Mw
RT
2
48.
KE 
49.
U mp  U av  U rms
2 RT
M
U mp 
2
U mp

2 RT
M
3 2
3
mv mp  RT
4
2
50.
r1
v
m2 32
44
 m2 / m1 1 


 x  27.3ml
r2
v2
m1
x
32
Similarly volume of SO2 will be 22.62 ml .
51.
Conceptual.
52.
Rate of effusion=
PA
2 RTm
A=Area of cross section of orifice.
53.
Conceptual
54.
O
O
||
||
C  Cl  H
NaOH
OH 

C O
HNO3 / H 2 SO4
HONO


C O
Sn / HCl
NH 2a 
||
O
C O
NO2
C O
||
O

CaCl
N 2  C l   


||
O
Sec : Sr.Elite(C-120, C-IPL, IPL-IC)

Cl
C O
||
O
C 
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