15-04-2023 - SR C 120 - Incoming - Jee-Mains - WTM-01 - Key & Sol's

SRI CHAITANYA IIT ACADEMY, INDIA 15-04-2023_ Sr C 120 _ Incoming __Jee-Mains_WTM-01_KEY & Sol’S
Sri Chaitanya IIT Academy., India.

A.P, TELANGANA, KARNATAKA, TAMILNADU, MAHARASHTRA, DELHI, RANCHI
A right Choice for the Real Aspirant
ICON Central Office – Madhapur – Hyderabad
Sec: Sr C 120 Incoming Sr's Jee Main Date: 15-04-23
Time: 09.00Am to 12.00Noon WTM-01 Max. Marks: 300
KEY SHEET
PHYSICS
1 4 2 3 3 1 4 2 5 1
6 3 7 4 8 1 9 4 10 2
11 3 12 1 13 3 14 3 15 3
16 3 17 2 18 2 19 2 20 1
21 81 22 20 23 2 24 2 25 4
26 120 27 5 28 335 29 9 30 48
CHEMISTRY
31 3 32 3 33 3 34 2 35 1
36 2 37 4 38 3 39 4 40 3
41 2 42 1 43 4 44 2 45 3
46 2 47 4 48 1 49 2 50 2
51 0 52 4 53 6 54 6 55 6
56 5 57 2 58 6 59 12 60 4
MATHEMATICS
61 1 62 1 63 3 64 2 65 2
66 3 67 2 68 3 69 3 70 2
71 2 72 2 73 2 74 1 75 4
76 3 77 3 78 3 79 2 80 2
81 1 82 4 83 8 84 0 85 0
86 3 87 5 88 0 89 3 90 0
Sec: Sr.C 120 Page 1

SOLUTIONS
PHYSICS
1. 2h 1
t  sec Frequency  7  8  56Hz
g 7
2. 1
n  500 Hz;T  S
500
T 1
 S
2 1000
3. w
V
K
4. 0.06
Amplitude, A   0.03
2
2
k

  2 f
5. d   t
6. Frequency of first over tone of closed pipe= frequency of first over tone of open pipe
3v v 3 P 1 P  P
     v  
4 L1 L2 4 L1 1 L2 2   
4 L1 1 4 L 1
 L2  
3 2 3 2
7. Frequency of vibration in tight string
p T n T 1
n  n T      4%   2%
2l m n 2T 2
2 2
 Number of beats  n  n   100  2
100 100
8. Vmax T  M  m g m
 B   1
Vmin TA Mg M

9. T  Mg  mg sin 300
m
2
M
10. Given
V p  4Vw
 A  4  n
2 fy0  4 f 
y
 0
2
11. The wave 1 and 3 reach out of phase. Hence resultant phase difference between them is
.
Resultant amplitude of 1 and 3  10  7  3 m

This wave has phase difference of with 4  m
2
Resultant amplitude  32  42  5 m
12. Sound waves require material medium to travel. As there is no atmosphere (vacuum) on
the surface of moon, therefore the sound waves cannot reach from one person to another
13.   2  l2  l1   2 100  200cm  2m
14. x
Given y  sin cos 40 t
3
2 vt 2 x
Comparing with y  2a cos sin    6cm
 

The separation between adjacent nodes 
 3cm
2
15. The apparent frequency observed by detector in the situation described in question is
expressed as

 v  v0 
f ' f  
 v  vs 
 340  20 
 180  f 
 324  20 
180  360
 f 
320
 f  2025Hz
16. For a vibrating string
n1l1  n2l2  n3l3......  cons tan t  k  say   nl
Also l1  l2  l3  l4  .......  1
k k k k k 1 1 1 1
    ......       ...
n1 n2 n3 n4 n n n1 n2 n3
17. V V
nc  ; n0 
4lc 2l0
18. T
velocity v  Where T= weight of part of rope hanging below the point under

M 
 M    xg
consideration    L
 xg v   xg
 L  M 
 
 L 
19. l1  l2  l3  110cm and n1l1  n2l2  n3l3

n1 : n2 : n3 :1: 2 : 3
n1 1 l2 l n 1 l l
   l2  1 and 1   3  l3  1
n2 2 l1 l2 n3 3 l1 3
l1 l1
l1    110 sol1  60cm, l2  30cm, l3  20cm.
2 3
20. n 2Vs

n V
n 2  4

240 320
n  6

21. P T
n
2l 
P T  cons tan t
1
T 2
P
2 2
T2  P1   6  9
    
T1  P2   4  4
9
T2   36  81N
4
22. v T
v1 T 1.44T1
  2   1.2
v2 T1 T1
 v2  1.2v1
v v v
  100%  2 1  100%
v v1
1.2v1  v1
  100%  20%
v1
23. T
V

24. V  T
25. On comparing with P  P0 sin t  kx  , we have
  1000rad / s, K  3m 1
 1000
v0    333.3m / s
k 3
v1 T
 1
v2 T2
333.3 273  0 t 
or ,   or Vt  V0 1  
336 273  t  546 
0
t  4.3 C
t  40 C

26. V
nc 
4lc
V
n0 
2l0
V
n
4  lc  l0 
27. W
V
K
28. n  n0  n
29.  2 
  x
  
30. T 
 
 K
CHEMISTRY
31. PHBV is a polyester
32. Glyptal is polymer of ethylene glycol & Thallic acid
33. Sucrose with anomeric–OH groups in glycosidic linkage & thus non-Reducing
34. It contains different types of amino acids.
35. Cysteine is sulphur containing amino acid.
36. Conceptual
37. Conceptual
38. Natural rubber is cis-1, 4-polyisoprene.
All trans-1, 4-polyisoprene also occurs, and is known as Gutta Percha.

39. Teflon, polystyrene and PVC are formed by addition polymerization
40. Both highly inflammable and Non-inflammable
41. ‘NO’ odd electron molecule
42. Absence of vacant d-oribital
43. NH 4 NO2  N 2  2 H 2O
NH 4 NO3  NO2  2 H 2O
1
44.
H 3 PO2
45. 0
P4  3 NaOH  3H 2O 3 PH 3  3NaH 2 1PO4
46. SO3 ; two  p  d  bonds

47. Bond order of H 2O2  1, O3  1.5, O2  2
Hence O  O bond length order is
H 2O2  O3  O2
48. S  Se  Te  O
49. E.N  Acidic character 
50. S
HO S OH
O
51. x  0; y  0; z  0
52. O2 , S8
53. O O
H O S S S S O H
O O
54. H 2CO3 , H 2 SO4 , H 3 PO3 , H 2 S2O7 , H 2CrO4 , H 2 SO3

55. Based on structure
56. Conceptual
57. (I) –ve
(II) +ve
(III) +ve
(IV) -ve
58.
Iso electric point  Pi  
 pK a1  pK a 2    9.69  2.31
2 2
59.
MN 
 30  20000    40  3000    30  60000 
30  40  30
 36000
MW 
 30  20000    40  3000   30  60000 
2 2 2
 30  20000    40  3000    30  60000 

 43333.33

MW
Poly disparity index (PDI) 
MN
43333.333

36000
 1.20
 12.0 101
x  12
60. LDPE, HDPE, PAN, Telfon are chain growth polymers
MATHEMATICS
61. 2  1
  0  a  1  4  a  
 4
62. 2elog k  1  7  K  4  S .O.R  3K  12
63. x 2  2 x sec  1  0
 
 x  sec  tan  and     
6 12
    
 sec     sec  sec   
 6  12 
   
 sec    sec  sec  
6  12 
    
and tan     tan   tan   
 6  12 
   
 tan     tan   tan  
6  12 
 ,  are the root of x 2  2 x sec  1  0 and 1  1
 1  sec  tan  and 1  sec  tan 
  2  2 are the root of x 2  2 x tan   1  0 and  2   2
 2   tan   sec ,  2   tan   sec
Here 1   2  2 tan 
64. f  3  0   2  7  10  0  2    5
65. x2  2x  9
2
 y  1  y  x 2  1  y  2 x  9 1  y   0;
x  2x  9
1 
  0  2 y 2  5 y  2  0  y   , 2
2 

2
66.  x  m  1  0  x  m  1, m  1
Romgiven condition, m  1  2 and m  1  4
 m  1 and m  3
67. For real roots q 2  4 pr  0
2
 pr
   4 pr  0  p, q, r are in A.P 
 2 
p2 p
 p 2  r 2  14 pr  0  2
 14  1  0
r r
p   p 
   7   48  0    7  4 3
r   r 
68. Put x  1  t  x  t  1
2
3 t  1   t  1  1 3t 2  7t  5 3 7 5
  2 3 4
t4 t4 t t t
A  0, B  3, C  7, D  5
A B C  D 1
69. Put x 2  1  t  x 2  t  1
5  t  1  9 5t  4 5 4 5 4
    
t5 t5 t 4 t 5  x 2  14  x 2  15
No. of partial fractions=2.

70. f  x   x 4  ax3  13x 2  bx  4  0
 
roots are 2 5, 2  5,  ,   2  5 2  5  .  4 
  2    1
   13   2  8  12  0    2,   6    2 
from 1 &  2     2, a  2  5  2  5  2  2  a  0
f  2   0  b  20

71. 1 5
f ' x   0  x  ,
2 2
 1 
f  0
 2 
fx
72. 3 21
0
73. For x 2  3x  5  0, the roots are imaginary
x 2  3 x  5  0 and ax 2  bx  c  0
Have same roots
a b c
  
1 3 5
 a : b : c  1: 3: 5
74. f  x   x 4  3x3  9 x 2  27 x  81
Remainder  f  3  81  81  81  81  81  81
75. x4
is an improper fraction
 x  1 x  2 
x 2
 x  2  x 4  x 2  x  3
x 4  x3  2 x 2
 x3  2 x 2
 x3  x 2  2 x
3x 2  2 x
3x 2  3 x  6
5 x  6
Q  x   x2  x  3  k  3
76. x  A 1  x 2    Bx  C  3  2 x 
Equating x 2 coefficients  A  2 B  0
Equating x coefficients  3B  2C  1
Equating constants  A  3C  0

2
Solving  C  
13
2
77.  x  1  A  x 2  1   Bx  C  x
Equating x 2 coefficients  A  B  1
Equating x coefficients  C  2
Equating constants  A  1
 A 1 
B  0 cos 1    cos 1   
C 2 3
78. 1 2
x 2  A  x  1 x  1   x  1  B  x  1
2
Equating x 2 coefficients  A  B  1
1 
Equating constants   A   B  0  A B  
2 2
3 1
A  ,B 
4 4
| 3i  j | 9  1  10
79. x1  x2  3, x1x2  A
x3  x412, x3 x4  B
Take x1  a, x2  ar , x3  ar 2 , x4  ar 3
a 1  r   3, ar 2 1  r   12
 r 2  4  r  2, a  1
A  a 2 r  2, B  a 2r 5  32
80. f  x  2
81.   3a  1
Let roots be p,2 p 3 p 
a 2  5a  3
2
 2p  22 2

3a  1  1
 a  5a  3 9  a 2  5a  3
 39q  2b, a  2 / 3

82.
  
 ,  are roots of 5  2 x 2 4  5 x  8  2  0 
Harmonic mean 
2

2 8 5
4
 
  4 5
83. 2  4  4
x  1  1  
2
n  1 is a perfect square
  8,15,24,35,48,63,80,99.... 8 
84. Sum of positive terms can never be zero
85. 12  1 11
max. of RHS   LHS  0  No sol
4 4
86. Let f  x   4 x 2  16 x    0
  0     ,16 
f 1  0    8 f  2   0
   16  f  3   0
   12    12,16  ;   13,14,15
87. Let f  x   x 3  10 x 2  11x  100  0
10  133
f 1
x  0  x 
3
10  133
 7.16 f  8   0, f  9   0, f 10   0, f 11  0, f 12   0    11,12      11
3
10  133
  3

10  133 O
3
88. x 2  x  1  0 are non-real then equation ax 2  bx  a  0 has both roots common with
x2  x  1  0

a b a
 
1 1 1
a b  0
89.  2  6  2  0and  2  6  2  1
  10  6 9  2 8  0and  10  6 9   8  0
     10   10   6  9   9   2  8   8   0
 10  2a8  6a9
a10  2a8
 3
2a9
90. 1 1 1 2  2  2
  
 2 2  2 2  2  2  2  2 2
2

       2      
2
 
4  2  3 1
    0.12  0.125
16 8


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SRI CHAITANYA IIT ACADEMY, INDIA 15-04-2023_ Sr C 120 _ Incoming __Jee-Mains_WTM-01_KEY & Sol’S

Sri Chaitanya IIT Academy., India.

Sec: Sr.C 120 Page 1

Sec: Sr.C 120 Page 2

Resultant amplitude of 1 and 3  10  7  3 m

Sec: Sr.C 120 Page 3

n1l1  n2l2  n3l3......  cons tan t  k  say   nl

19. l1  l2  l3  110cm and n1l1  n2l2  n3l3

Sec: Sr.C 120 Page 4

Sec: Sr.C 120 Page 5

All trans-1, 4-polyisoprene also occurs, and is known as Gutta Percha.

Sec: Sr.C 120 Page 6

47. Bond order of H 2O2  1, O3  1.5, O2  2

Hence O  O bond length order is

54. H 2CO3 , H 2 SO4 , H 3 PO3 , H 2 S2O7 , H 2CrO4 , H 2 SO3

 30  20000    40  3000    30  60000 

Sec: Sr.C 120 Page 7

Sec: Sr.C 120 Page 8

Romgiven condition, m  1  2 and m  1  4

No. of partial fractions=2.

from 1 &  2     2, a  2  5  2  5  2  2  a  0

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Have same roots

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