18-05-24 - SR - Super60 (Incoming) - STERLING BT - Jee-Main - WTM-07 - Key & Sol's

Sri Chaitanya IIT Academy, India 18-05-24_Sr.
Super60(Incoming)_STERLING BT_Jee-Main_WTM-07 _Key & Sol’s
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
SEC: Sr.Super60(Incoming)_STERLING BT JEE-MAIN Date: 18-05-2024
Time: 09:00AM to 12:00PM WTM-07 Max. Marks: 300
KEY SHEET
PHYSICS
1) 2 2) 4 3) 3 4) 4 5) 2
6) 1 7) 1 8) 1 9) 3 10) 2
11) 2 12) 1 13) 3 14) 3 15) 1
16) 4 17) 2 18) 1 19) 2 20) 3
21) 12 22) 24 23) 9 24) 3 25) 8
26) 2 27) 6 28) 90 29) 12 30) 2
CHEMISTRY
31) 4 32) 2 33) 4 34) 4 35) 3

36) 1 37) 1 38) 2 39) 4 40) 3
41) 1 42) 4 43) 1 44) 3 45) 1
46) 2 47) 3 48) 2 49) 3 50) 3
51) 3 52) 3 53) 46 54) 4 55) 3
56) 3 57) 3 58) 5 59) 6 60) 2
MATHEMATICS
61) 2 62) 2 63) 1 64) 2 65) 3

66) 1 67) 2 68) 3 69) 3 70) 2
71) 3 72) 3 73) 1 74) 1 75) 1
76) 4 77) 3 78) 2 79) 2 80) 3
81) 108 82) 7 83) 6 84) 576 85) 933
86) 666 87) 10 88) 369 89) 720 90) 553
SEC: Sr.Super60(Incoming)_STERLING BT Page 1

Sri Chaitanya IIT Academy, India 18-05-24_Sr.Super60(Incoming)_STERLING BT_Jee-Main_WTM-07 _Key & Sol’s
SOLUTIONS
PHYSICS
 1
1. Shift due to a slab  1   t is in the direction of ray of light. Hence, Ram appears
 
nearer to Anoop by that much distance.
2. i  45
For   2 or 1.414   C  45
 They get TIR on face AC.
d
3. d app. 

 of red is least. So, d app. for red is maximum. So, they appear to be raised least.
 A  m 
sin  
4.   2 
sin  A / 2 
 A
Given,   cot  
2
Solving, we get sin  180  2A
d
5. d app 
 n1 / n2 
n2
d  d app
n1
n 
 d   2   d app    2  x
d n d

dt n1  dt   n1 
 d  x R n2
2
dv
 A  d  
dt  dt  n1

6. (a) and (b)
2
v vu
1
as 1   2
i.e v and u are of same sign.
Or they are on same side of plane surface. From plane surface, if object is real, image is
virtual and vice-versa.
R 6 / 5 4
7. sin C   
D 3 / 2 5
r1  0  r2  B  90  
Now, r2   c
or sin r2  sin  c
4
 sin  90    
5
4
or cos 
5
4
cos37 
5
From Eq.(i), we see that   37
2 1 2  1
8. Apply  
v v R
1 2 1 2
We get  
v 10 10
Solving we get, v  10 cm
22
9. Using, sin  max  1 22  12  1
12
 2 
or  max  sin  1
2
 1 
 12 
 
 2 
For T1R,  sin 1  2
 1 
 12 
 A

10. DBM  45
 DM  h tan 45  h  32 cm
sin i

sin r
sin i sin 45 3
sin r   
 4/3 4 2
sin r sin r
 tan r  
cos r 1  sin 2 r
3
3
 tan r  4 2 
9 23
1
32
3
 CM  h tan r  32   20 cm
23
 CM  DM  CM  32  20  12 cm
11. Applying Snell’s Law at the interface separating two media, we get
1 sin i   2 sin r
From the figure, we get
 3  a   c 
     2  
 2   a 2  b2   c d 
2 2
Since ai  b j and ci  d j are unit vectors, so we get
a 2  b2  c2  d 2  1
a 4
Substituting in equation, we get 
c 3

12. From figure, we get
x x
 and  
OC OC '
 OC ' 1
  
 OC 
For small angles, we have
4
       6   8  Bending angle      2
3
13. For combination to produce dispersion without deviation.
 n1  1 A1   n2  1 A2
 n 1   1.54  1  3
 A2   1  A1  A2    4  A2   4   A2  3
 n2  1   1.72  1  4
14.
sin i
   at P 
sin r1
and r1  C  A
 sin i   sin  A  C   sin i    sin A cos C  cos A sin C 
 1 1  1
 sin i    sin A 1  2  cos A  sin C  
    
 i  sin 1  
 2  1 sin A  cos A
15. In first case, the apparent depth of the liquid is  b  a 
 Real depth    b  a 
In second case, apparent depth   d  c 
 Real depth    d  c 
The difference in depth of liquid is
  d  c    b  a 
Here we use   d  c     b  a 
From experimental data, the difference is equal to  d  b 
   d  c    b  a   d  b
d b

a  d  c  b
 1
16. Lateral displacement, x  t sin i 1    t
 
17. Conceptual
18. Conceptual
19. After refraction at two parallel faces of a glass slab, a ray of light emerges in a
direction parallel to the direction of incidence of white light on the slab. As rays of all
colours emerge in the same direction (of incidence of white light), hence there is no
dispersion, but only lateral displacement.
20. Here nose of the boy is the object and fish is observer. Using refraction formula, we
have
2 1 2  1 4
  where u   R, 1  1, 2  , R   R
v u R 3
4
1
4 1 3
    v  2 R
3v R  R
Thus, the image of child’s nose will appear at a distance 3R from the centre of the
bowl. The ray diagram is shown in below figure.

21.
2 1 2  1
Applying   we get,
v u R
1.0 1.5 1.0  1.5
 
1.0 u 2.0
Solving we get, u  1.2 cm
22.
sin i 4 4 / 16  h 2
Applying,   we get 
sin r 3 2 / 4  h2
Solving this equation we get, h  2.4 cm

23.
dy
Given,   4 cm/s
dt
4
Distance of bird as observed by fish Z  y   x  y  x
3
dZ dy 4  dx 
    
dt dt 3  dt 
dZ
Given,  16 cm/s
dt
dx
Substituting in Eq. (i) we get  9 cm/s
dt
24. 2 sin 30  1.sin r  r  45
PR d0 d
 cos i  PQ   0
PQ cos30 3
QS  d  diameter of refracted light
d d
 cos r  PQ   2d
PQ cos 45
2d 0 2
 2d  d  d0
3 3
The refracted beam has lesser diameter. Intensity of refracted beam
2
 d0 
I 0   2
 2  d0 
I 2
 I 0 . 
d  d 
 
2
3
 I  I0
2

25.  Total   Refraction  2 Reflection   Refraction
or    i  r   2 180  2r    i  r 
 sin i 
 360  2i  6r  360  2i  6sin 1  
  
d
For deviation to be minimum, 0
di
By putting first derivative of  (w.r.t. i) equal to zero, we get the desired result.
26. For the convex spherical refracting surface i.e., air-oil interface
7
u  24 cm, v  ?, u1  1, 2  and R  6 cm
4
 1 2 2  1
 
u v R
7
1 7 4 4 1
  
 24  v 6
 v  21 cm
This image will not as object for the water-oil interface
7 4
u  21 cm, v  v ', 1  , 2  and R  
4 3
7 4
4  3 0
21 V '
V '  16 cm
Therefore the distance of the image from the bottom of the tank  18  16  2 cm

27. In the figure, C  critical angle
OM R
In POM , tan C  
PM 8
1 3 3
 sin C    tan C 
 5 4
R 3 3
   R   8  6 cm
8 4 4
28. In the figure, QR is the reflected ray and QS is refracted ray. CQ is normal.
Apply Snell’s law at P
1
1sin 60  3 sin r  sin r   r  30
2
From geometry, CP  CQ
 r  30
Again apply Snell’s law at Q,
3
3 sin r '  1sin e   sin e  e  60
2
From geometry
r '   e  180 (As angles lies on a straight line)
 30    60  180    90
29. The necessary and sufficient condition for all the rays to pass around the arc is that the
ray with least angle of incidence should get internally reflected i.e., should suffer TIR.
From the figure, it becomes obvious that the ray with least angle of incidence is the one
which is incident almost grazingly with the inner wall.

For this ray, let  be the angle of incidence, then we observe that
Rd
sin   where d is the diameter of tube
R
For TIR,  C
 sin   sin C
Rd 1
 
R 
d
R
 1
Since,   1.5 and d  4 cm, so we get R  12 cm
Hence, the least required is 12 cm.
30. Since,   90  i
dy
 tan   cot i   cot i
dx
According to Snell’s Law at O and P, we have 0 sin i0   p sin i p
Since   1  ay
 At y  0,   1  sin  90    
1  ay sin i
y x
1 dy dy y

ay 0
 sin i   cot i  ay   dx x2
1  ay dx 0
a
Substituting, y  2 m and a  2  106 m 1 , we get xmax  2000 m  2 km

CHEMISTRY
31. The COH bond angle in alcohols is lower than the tetrahedral angle.
32. Chloro group is ortho para directing group
33. Hydration of alcohols in acid medium proceeds through the formation of carbocation
intermediate
34. Inductive effect is distance dependent hence molecule A is more acidic than B
35. on reaction with HCl gives a carbocation which undergoes hydride shift
to form a more stable carbocation.
36. Boiling point of 1,2-dichlorobenzene is higher than 1,4-dichlorobenzene. Solubility of
isomeric alcohols increases with branching.
37. Syn hydroxylation
38. Ethyl alcohol and n-propyl alcohol are distinguished by Iodoform reaction as only ethyl
alcohol gives iodoform reaction
39. Tertairy alcohols with copper at 573K gives alkene
40. Conceptual
41. Arylhalides do not react with AgNO3
42. Reactions of alcohols with Grignard reagent involves cleavage of O-H bond
43. In case of acid catalyzed hydration of alkenes carbocation intermediate is formed
which undergoes rearrangement. OMDM follows markonikov’s rule. Hydroboration
oxidation follows anti markonikov’s rule.
44. Presence of electron withdrawing groups increases the reactivity of aryl halides
towards aromatic nucleophilic substitution reactions
45.

46.
47. Formic acid esters gives secondary alcohols with Grignard reagent
48. NaBH 4 can’t reduce esters
49. Retention in the configuration with thionyl chloride if pyridine in not used
i
( SN - mechanism) and inversion in the configuration with PCl3 ( SN 2 -
mechanism)
50. Electron releasing groups stabilizes carbocation’s
51. Compounds 1 and 2 are primary alcohols and 6th compound cannot form carbocation
according to Bredt’s rule or doesn’t undergo SN 2 reaction due to steric hinderance to
form chloride.
52. Iodoform reaction is given by 2-ol ‘s and ethyl alcohol
53. One mole of mono hydric alcohol reacts with one mole of ethyl magnesium bromide to
form 22.4 litres of ethane gas
54. Periodic acid is an oxidized agent to oxidize vicinal diols
55. A,B,C reagents can bring about the given conversion
56. Fluorobenzene is prepared by Balz-schiemann reaction
57. Reagents I, III, IV gives visible change with butan-2-ol
58. 1 mole of ethyl alcohol on reaction with one mole of sodium gives 11.2 litres of
hydrogen gas
59.
60. In acetylation replacement of H of O  H occurs by CH 3CO  group
i.e., H atom of mass 1 amu is lost and an acetyl group of mass 43 amu is added. Thus,
there is a net gain of 43  1  42 amu for every acetyl group introduced. Mass of
difference of final product and original compound is 190  106  84. Hence, number of
84
OH groups   2.
42

MATHEMATICS
61. Number g required ways 2  4!  48
62. 8  P  Q  r  12
P  Q  r  8,9,10,11,12
P , Q, r  1
Required no of ways 7C2  8C2  9C2  10C2  11C2  185
63. x1 P.P x2 P.P x3
x1  x2  x3  6 x1 x3  0 x2  1
64. 5 digit no’s 5 P5  120
4 digit no’s greater than 7000 mus begin with 7,8,9
3  4 P3  72
120  72  192
65. yI
y  7,6,...0,1,..6,7
Total no. of integrals points 12
Total no. lines formed  12C2  66
Total no. lines 66  12  54
66. No of 4 digit no’s  7 4  2401
4 digit no’s repeation allowed  7 P 4  840
The numbers that can be formed when at least one digit is repeated  2401  840  1561
67. 7 m ends with 7,9,3,1
7 n ends with 7,9,3,1

7 m ends with 7 then 7 n ends with 3
7 m ends with 7 in 25 cases
7 n ends with 3 in 25 cases
7 m  7 n is divisible by 5 in 625 cases
Similarly other 3 cases
Total 4  625  2500

68. FORTUNE
VOWELS OUE
CONSONANTS FRTN
x  7c3  4!
y  7c4  1  3!
z  7c3  1  1!
7c3  4! 7c4  3!
 30
7c3
69. 6P4  3
6  5 4  3 3
70. 18c3  18c2  18c2  17c1  17c0
19c3  18c2  18c1
19c3  19c2   20c3
 n  20
71. 18cx 1  2.10cx x  10
x 1  0 x 1
72. Numbers  1000 and less than 4,000
3 5 5 5  375
375  1  374
1,000 
73. 15 members 3 particular persons
 12  treating as one unit
Total  13
13 members around a circle  13  1!
 12!
3 particular persons can be managed in 2 ways
 12! 2!
74. Total no of books a  2b  3c  d
No of arrangements
 a  2b  3c  d !
a ! b!  c !
2 3

5  5  5  ... 20
75. Each position can be filled in 5 ways. So total 20 positions 5
20
76. 2 places for vowels which can be selected in 4C2 ways this 2 places can be filled by 25
ways (repetition) allowed the remaining 2 places can be filled by 212 ways.
Total no of ways 212  52  4C2
 150  212
77. Exponent of 5
125  125  125  125 
 5    52    53    54 
25  5  1  31
31 Zero’s
78. 20!  2 .3 .5r.7 1.11 2.13 3.17 4.19 5
20!  218.38.54.7 2.111.131.17.19
 214.38.7 2.11.13.17.19
1
  ends with 4  ends with 1 ends with 9  ends with 1 ends with 3 ends with 7  ends with 9 
108  63  ends with 4

79. No of 4 digit no’s divisible by 5
6  4  10
0  3!  6
5  3!  2!  4
Total 4 digit no’s 4! 3!  18
Required cases 18  10  8
80. No’s from 000, 001, 002, …. 999. Since each digit has an equal likely wood of coming
and digits are used 3  1000  3000
0 appears 300 times
but 000,001,......099 have some unwanted zero’s
3  1  2  9  1  90  111
Total no of zeros 300  111  189
189  3  zero's in 1000 
 192

81. Divisible b 6 means divisible by 2 as well as 3
I without 0 -------2/4  48
II without 3 and 0 on units place ------0 = 24
III without 3 and 2 and 4 at place = 36
Total no of way 48  24  36  108
82. n  1c3  nc3  21
 n  1 n  n  1  n  n  1 n  2  21
1.23 1.2.3
n  n  1
 21
1.2
n  n  1  42  n  7
2n  1Pn1 7
83.  n3
2n  1Pn 10
3P3  3!  6
84. First 4 places are filled with 1, 2, 3, 4 in 4! ways. The last 4 places are filled with 5, 6,
7, 8 in 4! ways. No of ways  4! 4!  576
85. 1111   2  3  4  5   3!  93324
 N 
100   933.24  933
86. 36  33  22
No. of numbers between 2 and 2000 divisible by 2 is 1000
No. of numbers divisible by 2 or 3 is 1000  666  333  1333
No. of integers that are not divisible by 2 or 3 is 1999  1333  666
87. Divide x3  ax 2  bx  c with x 2  1
The remainder is  b  1 x  c  a  0 for any x
b 1  0 ca0 b 1 ca
10  1  1  10 ways

88. I 7      9 ways
4
II 4  8  93 ways
Total No of cases 94  32  93
 93  41
93  41
  369
92
89. d 7 6C3  3C3  20
d 8 7C3  4C3  140
d 9 8C3  5C3  560
20  140  560  720
90. Words starting with E = 360
Words starting with GE = 60
Words starting with GN = 60
Words starting with GTE = 24
Words starting with GTN = 24
Words starting with GTT = 24
GTWENTY – 1
Total – 553

18-05-24 - SR - Super60 (Incoming) - STERLING BT - Jee-Main - WTM-07 - Key & Sol's

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18-05-24 - SR - Super60 (Incoming) - STERLING BT - Jee-Main - WTM-07 - Key & Sol's

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Sri Chaitanya IIT Academy, India 18-05-24_Sr.

Super60(Incoming)_STERLING BT_Jee-Main_WTM-07 _Key & Sol’s

Sri Chaitanya IIT Academy.,India.

31) 4 32) 2 33) 4 34) 4 35) 3

61) 2 62) 2 63) 1 64) 2 65) 3

SEC: Sr.Super60(Incoming)_STERLING BT Page 1

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From the figure, we get

Since ai  b j and ci  d j are unit vectors, so we get

SEC: Sr.Super60(Incoming)_STERLING BT Page 4

From experimental data, the difference is equal to  d  b 

SEC: Sr.Super60(Incoming)_STERLING BT Page 6

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SEC: Sr.Super60(Incoming)_STERLING BT Page 9

SEC: Sr.Super60(Incoming)_STERLING BT Page 10

Substituting, y  2 m and a  2  106 m 1 , we get xmax  2000 m  2 km

SEC: Sr.Super60(Incoming)_STERLING BT Page 11

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7 n ends with 7,9,3,1

SEC: Sr.Super60(Incoming)_STERLING BT Page 14

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108  63  ends with 4

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