GTM 49 Key

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.Super-60 & All Jee-Main Date:26-08-2021
Time: 02.30Pm to 05.30 GTM-49 Max.Marks:300
Key Sheet
PHYSICS
1 4 2 4 3 1 4 1 5 2
6 3 7 3 8 2 9 4 10 3
11 1 12 2 13 3 14 3 15 3
16 3 17 3 18 3 19 2 20 2
21 9 22 0.57 23 12 24 1.36 25 1
26 101 27 476 28 51 29 10 30 3
CHEMISTRY
31 2 32 1 33 3 34 4 35 3
36 4 37 1 38 3 39 3 40 4
41 4 42 1 43 4 44 4 45 4
46 2 47 2 48 4 49 2 50 1
51 20.00 52 4.80 to 5.00 53 0.33 to 0.34 54 0.16 to 0.17 55 4.00
56 5.00 57 1.00 58 3.00 59 2.50 60 4
MATHEMATICS
61 3 62 4 63 1 64 1 65 1
66 2 67 1 68 4 69 3 70 2
71 1 72 3 73 4 74 3 75 2
76 4 77 4 78 2 79 3 80 4
81 7.00 82 1.00 83 3.00 84 78.00 85 18
86 4 87 2 88 10 89 130.00 90 2.00
Sri Chaitanya IIT Academy 26-08-2021_Sr.Super60 & ICON All_Jee-Main_GTM-49_Key & Sol’s
SOLUTIONS
PHYSICS
1. Use the formula for angular magnification.
2. Use the formula for fringe width  D
d
3. T  m 2
v N
4. Use 1  1
v2 N 2
2  x   10
5.  and 
x 100   2 90  
 90   90
So      45 cm
100     10 110
h
6.  Since v is increasing in case (i), but is it is not changing in case (ii). Hence, in
mv
the first case de-Broglie wavelength will change, but it second case, it remain the same
N Pb 1
7. 
N Ur 2
N Ur
2
N Pb
NUr 2

N0 3
2 1

3 2n
3
 2n
2
3
n    n n 2
2
3 t
n    n 2
2 T
n  3 / 2 
tT
n  2 
8. Photons per area per second at a distance r are 5.00 1018 / 4 r 2 . Photons per second
entering the eye, radius R is then this times R 2 . Set this product equal to 500 per
second and solve for r. The result is [B].
9. The P-N junctions will conduct only when it is forward biased i.e. when – 5V is fed
to it, so it will conduct only for 3rd quarter part of signal shown and when it conducts
potential drop 5 volt will be across both the resistors, so output voltage across R 2 is
2.5 V.
 V0  2.5V
10.
Sec: Sr.Super-60 & All Page 2

200  T T  20

R 2R
400  2T  T  20
T  140C
11. BH  Bcentre
 0 Ni 4107 100  2
 
2r 2  0.1
 4 10 4 T
12. Speed max  7  105 m / s
Kinetic energy of e   1 mv 2  1 m   7  105 

2
2 2
1 31 10 21
  9.1 10  49  10  9.1 10  49  0.5 J
2
Given incident frequency  8 1014 Hz
 From photoelectric equation k max  h  h 0
 h 0  h  k max
 6.6 1034  8 1014  91 49  5 1023
  6.6  8 1020  91 49  5 1023 
 1021  66  8  91 49  5 102 
 528  223 1021  6.6 1034  0
305 30.5
0   1021  1034   1014   0
6.6 6.6
 4.64  1014 Hz
k
13. As we know,  
I
3k  1 
 2 
 I rod  m 2 
m  3 
Tension when it passes through the mean position,

 3k  k2
 m220  m 2 20  0
3 m 3 
vu 1
14. Using equation, a  and S  ut  at 2
t 2
1  45  2
Distance travelled by car in 15 sec  15
2 15
675
 m
2
Distance travelled by scooter in 15 seconds  30 15  450
( distance = speed  time)
Difference between distance travelled by car and scooter
in 15 sec, 450 – 337.5 = 112.5 m
675
Let car catches scooter in time t ;  45  t  15   30 t
2
337.5  45t  675  30 t  15t  337.5
 t  22.5 sec
lateral strain   
15. Poisson’s ratio,  
longitudinal strain   
For material like copper,   0.33
And, Y  3k 1  2  
9 1 3
Also,  
Y k 
Y  2 1   
Hence,   Y  k
16. For minimum density of liquid, solid sphere has to float (completely immersed) in
the liquid.
mg  FB  also Vimmersed  Vtotal 
4
or  dV  R 3
3
  r2  
   r    0 1  2 
0  r  R given 
  R  
R 2
 r  4
   0 4 1  2   r 2 dr  R 3
0  R  3
R
 r3 r5  4
 4 0   2   R 3 
 3 5R  0 3
3
40 R 2 4 3
  R 
3 5 3
2 0
  
5
1
17. Velocity of EM wave is given by v 


Velocity in air  C
k

Velocity in medium  C
2
Here, 1   2  1 as medium is non-magnetic
1
r1 C r1 1
  2  
1 C r2 4
 
r 2
2
18. Ratio of AM signal Bandwidths

15200  200 15000
  6.
2700  200 2500
19. i DC   / R
i AC   / Z
KP sin 45 4K p
20. E0  3
2 
 a  a3
 
 2
EA  E 2
KP
 2 2
a
E0 4
  2 2
EA 2
21.
 q1 max  1 6  6
 q 2 max  2  4  8
q   q1  max ,  q 2  max
 q6
q 6
V   9 KV
Ceq 2
3

I
22. T  2
B
23. Pv  nRT
2  P / nT  Constant
P2 / n 2T2  P1 / n1T1
P2  P2 n 2 T2 / n1T1
If initial no of moles = 1  final no of moles = 0.5
T1  27C  273  27  300 k
T2  300  50  350 k
put in above eq n
P2  20  0.5  350 / 1 300 
 11.66 atm
h
24. r
2
, d  2r
r d  1
25. See figure. Potential difference across R L
 8V  VCE
 8V  4V  4V
Now IC R L  4V
4
RL   103   1k 
4 103
26. Given : Radius of capillary tube, r  0.015 cm  15 10 5 mm
h  15 cm  15  10 2 mm
2T cos 
Using, h  cos   cos 0  1
gr
rhg 15 105 15 102  900 10
Surface tension, T    101 milli newton m 1
2 2
27. Given,
Distance between an object and screen, D = 100 cm
Distance between the two position of lens, d = 40 cm
Focal length of lens,

D2  d 2 1002  40 2 100  40 100  40 
f    21 cm
4D 4 100  4 100 
Power, P  1  100  N
f 21 100
 N  476.19 .
28.
From linear momentum conservation, Li  Lf

v
mV  0  0  5mV '  V ' 
5
2
1 1 v
Loss of KE  KE i  KE f  mv 2   5m   
2 2 5
2
1  1  4  mv 
 mv 2 1     
2  5 5 2 
4
 KE i  10.2 eV
5
[ Energy in first excited state of atom = 10.2 eV]
N
KE i  12.75 eV   N  51
4
The value of N = 51.
Li 80
w  100 
cw 1  10C
29.  mix  
2 2
30. He should move away from 224 Hz and towards 220 Hz.
CHEMISTRY
31. Stability of carbocation Aromatic > non aromatic > Anti aromatic
32.
33. Conjugation is preferred & Benzyl carbanion is stable

H    
H2 O 
O Et   O  Et O Et   O  C2 H 5OH 2
  
C2 H 5OH  H 
34.
35. Electron with drawing groups increases reactivity but electron donating groups
decreases reactivity
36.  -Keto acids undergo decarboxylation simply by heating.
37. In basic medium O is stronger donar than  NH 2
38. Acidic amino acids will have low Isoelectric points. at pH=7 it exists as anion
39.
2
2
40. This is an example of Azeotrope

41. Rentadine is also anti histamine in stomach wall.
42.
N0
N N  final amount
2n
N0  initial amount
n  no of half lifes
43. Molar conductance of NO3 = molar conductance of Cl 

As long as reaction takes place Cl  is replaced by NO3 . Hence conductance remains
unchanged .
After completion of reaction Ag  also present in solution
Hence conductance increases
 RT wt 1000
44.      RT
 C  GMwt gMwt  vol of solution in ml
1000
 C  RT
GMwt
Slope 1000 RT  4.65  103
GMwt
1000 RT
GMwt 
4.65  10 3

0.0821 293 1000

4.65 10 3
82.1  293  1000

4.65
 5.17  106
45. Zn  OH  2  NH 4 OH   Zn  NH 3  4   OH 2
Zn  OH 2 is soluble in alkali ( NaOH ) as well as in ammonia
46. NO is paramagnetic.
47.
G 0  nFE
G 0  G products  Greac tan ts
48. When electron density is pushed from metal atom into  -bond, the CO bond is
weaken as electrons enter into anti bonding orbital of CO. With 2 unit negative charge
on metal atom  -back bonding from metal to CO increases maximum electron
density.
49. Syn addition of H2 takes place. Hence (b) is only optically active
50. Marconicovs addition of water
51. H   103 M
1
So Ca 2    103 M
2
103
wt.of Ca 2   40
2
1
 103  40  106
So, weight of Ca 2 6
ions in 10 mL hard water  2  20
103
52. work done = ng RT
53. x millimoles of Na2CO3 and y millimoles of NaOH are present in initial mixture
x  y  20 x 5
2 x  y  25 y  15
x 1
 
y 3
54. It will undergo disproportionation
55. Pb, Sn, Al , Zn
56.  N 2( g )  3H 2( g )
2 NH 3( g ) 
 3
1
2 2
Totalmoles  1  
PV  nRT
20V 1 R 300

50V n R 600
n 1.25
  0.25

%  25 i.e. x  5
57. Normality = 10 1.78
5.6
Milii equivalents of H 2O2 = milliequivalents of KMnO4
1.78x 112 = 0.04 x 5 x volume in ml
58.
2-D or sheet Silicate
59. Na2 S 4O6

60. 4.00
MATHEMATICS
61. Ans: 3
cos x  sin x
Sol:  dx
8  sin 2x
cos x  sin x
 dx
2
9   sin x  cos x 
Let sin x  cos x  t
dt 1 t
 9  t 2  sin 3  c
 sin x  cos x 
 sin 1  c
 3 
So a = 1, b = 3.
62. Ans: 4
Sol: (i)  2  i  z   2  i  z
x
y
2
(ii)  2  i  z   i  z  z  4i  0
x  2y  2
(iii) iz  z  1  i  0
Equation of tangent x  y  1  0
Solving (i) and (ii)
1
x  1, y 
2
1
1 1
Now, p  r  2 r
2

3
r
2 2
63. Ans: 1
1 n n n 
Sol: lim   2
 2
 ...  2
n  n
  n  1  n  2   2n  1 
n 1 n 1
n n
 lim  2
 lim  2 2
r 0  n  r  r 0 n  2nr  r
n  n 
1 n 1 1
 lim
n  n
 2
r 0  r / n   2  r / n   1
1 1
dx  1  1
 2
  
0  x  1   x  1  0 2
64. Ans: 1
2 7 12 17
Sol: S  1      ....
3 32 33 34
S 1 2 7 12 17
      ....
3 3 32 33 34 35
2S 1 5 5 5
 1   2  3  4  .... 
3 3 3 3 3
13
S
4
65. Ans: 1
Sol:
6
sin 30   x  12
x
66. Ans: 2
10!
Sol: Total number of arrangements =
3!2!
The number of arrangements in which two
9!
L’s appear adjacently =
3!
 Total number of arrangements
10! 9!
 
3!2! 3!

9!
  4  6  8!
3!
67. Ans: 1
x 3   sin   x 2   cos   x  1
  x    x    x   
put x = 1
2   sin   cos    1   1   1   
max = 2  2
68. Ans: 4
 sin x   1 and  cos y   1
sin x  1 of cos y  1
 
x  ,  and y  0, 
2 2
69. Ans: 3
D  0  m2  3

 m ,  3    3,  ……(1) 
b
1  2  m  1,2  ……(2)
2a
f 1  0  m  2 and f  2   0
7
m ……(3)
4
 7
from (1), (2), (3) m   3, 
 4
70. Ans: 2
6x 8  5x 6  3x 2
Sol:  dx
x6  x4  1
1 12x11  10x 9  6x 5
  dx
2 x12  x10  x 6
1

2
2 x12  x10  x 6  c 
 x3 x 6  x 4  1  c

71. Ans: 1
p q pq p ~ q  p  q    p ~ q  ~ pq
(1)  (2) pq
(1) (2)
T T T T T T T T
T F F T T F F F
F T F F F T F F
F F F T T T T T
72. Ans: 3
Sol:  A,A   R because it is not necessary that each matrix is inverse of itself.
Let  A,A   R  A & B are inverse of each other.
 AB  I  BA
 BA  AB   B,A   R
Now Let  A, B  R  AB  I  BA ….. (1)
 B,C   R  BC  I  CB …..(2)
From (1) & (2) (AB) (BC) = I
AB2C  I
It is clear that AB2C  I is AC  I if, B2  I
which not possible of each matrix.
73. Ans: 4
Sol: x 6  6x  8 is non differentiable at x = 2, 4
at x = 2,
at x = 2, sin   x  2     0
x  4sin   x  2     0
So x 2  6x  8 sin   x  2    is differentiable everywhere
Similarly, ex  1 sin x is differentiable every where and x 2 is differentiable
everywhere
74. Ans: 3
1
8
1 cos  x  5x  dx
1 5
Sol:

8
1
8

2I  1  dx I 
8

8
75. Ans: 2
Sol: Let A = (2, 3, 4) and B = (5, 6, 7).
Then AB = 3 3 .
Plane : 2x + y + z = 1
Projection of AB on vector 2iˆ  ˆj  kˆ which is normal to plane.
Normal to plane,
B
 
3iˆ  3jˆ  3kˆ . 2iˆ  ˆj  kˆ
12
2 6
6 6
 Projection of AB on plane,
2 2
AC  3 3    2 6   3
76. Ans: 4
Sol: Area of parallelogram formed by line y  m1x  c1 , y  m1x  c 2 , y  m 2 x  d1 ,
y  m 2 x  d 2 is given by
Area of parallelogram = 1 2
 c  c    d1  d 2 
m1  m 2
x 5 x 15 c
For the given lines y    ; y    and y  3x  10; y  3x  ,
2 2 2 2 2
c
5.  10
2
Area of parallelogram = 1
5
2
(Given)
 c  20  1  21 or c  19
77. Ans: 4
 
Sol: Image of  2 ,2 in the line mirror x  y  1  0 is
2

h   2 k  2 2   2  1
 

1 1 2

 h  1  2 …….(1)
and k   2  1 ……(2)
2
 h 1
Putting value of  from (1) and (2), we get k  1   
 2 
2
or 4  y  1   x  1 , which is the required locus
78. Ans: 2
 1 
  1   r  1  r
Sol: Tr  tan 1  2   tan   tan 1  

 1  r  r  1   2  2
 2 2 
 1
S   tan 1  
2 2
79. Ans: 3
Sol: Normal vector:
ˆi ˆj kˆ
3 1 2  11iˆ  ˆj  17kˆ
2 5 1
So drs of normal to the required plane is <11, 1, 17>
Plane passes through (1, 2, -3)
So equation of plane :
11 x  1  1 y  2   17  z  3  0
11x  y  17z  38  0
80. Key 4
3 2 k
Sol:   2 4 2  0
1 2 1
 24  2  0   k  8  0  k  3
10 2 3
 x  6 4 2
5m 2 1
 10  8  2  10m  6   312  20m 
 8  4  5m 
3 10 3
y  2 6 2
1 5m 1
 3  6  10m   10  0   3 10m  6 
=0
3 2 10
 z  2 4 6
1 2 5m
 3  20m  12   2  6  10m   10 8 
 40m  32  8  5m  4 
For inconsistent
4
k  2&m 
5
81. Ans: 7.00
2
Sol: Parabola is  x  2   8  y  1
1
Now equation of normal having slope  is
2
1  2 
 y  1    x  2   4   
2  1 / 4  
 x  2y  28
82 Ans: 1.00
Minimum value of z  i = distance of point (0, 1) from the line y – x = 0

1

2
83. Ans: 3.00

Sol. a  k

b  2k

c  3k
  
a  2b  3c  21
     
a 2  4b 2  9c2  2  2a  b  6b  c  3c  a   21
2 2  4k 2 36k 2 9k 2 
2
k  16k  81k  2      21
 2 2 2 
1
147k 2  21  k  
7
3
c  3k  [magnitude is always +ve]
7
84. Ans: 78.00
A  5  3,7  , 2   
B  3  3,3  2,6  4 
direction of
AB  2  3  3, 4    2,  8    4 
AB is perpendicular to L 2
11  29  18  0 …..(1)
2  3  3 4    2 8    4
Also  
2 7 5
   m  2  0 …..(2)
from (1) and (2)   1
  1
 A(2,8, -3) and B(0, 1, 2)
2
 AB  4  49  25  78
85. Ans: 18
f x
 4t 3dt
6 0 
Sol: lim  form 
x 2 x  2  0 
D’L rule
3
4  f  x    f ' x   0
lim
x 2 1
1
 4.63   18
48
86. Ans: 4
1 9 1 1
Sol: 1  n    n
2 10 10 2
n
 2  10  n = 4
87. Ans: 2
2
Sol:  2  2        2
2
 3     2 2   
  2  9  6  4  2
  2  4  5  For least value   2
88. Ans: 10
5
Sol: 4x 2  9x  5  0  x  1,
4
5 t t
Now given  p q , t  t p t q where
4 2
 1
r 1
5  1 p1  1 q 1 
t r  16    so  8        
 2 4  2   2  
pq 2
 1 pq2 8
1  256     2 pq2   1 2
 2
hence p + q = 10
89. Ans: 130.00
P(B) = 1 if the event ‘A’ occurs atleast twice
P(B) = 0 if the event ‘A’ does not occurs
1
P(B) = if ‘A’ occurs once
2
3 2 2 3 1 4
4  1  2   1   4  1   2  4  1   2  4  1  
P  B   C1        C2      C3      C4    1
 3  3   2   3  3 3  3  3  
49
=
81
90. Ans: 2.00
3  10  10  4  7  10  5
Sol: x  7
7
xi d1  x i  x
3 4
10 3
10 3
4 3
7 0
10 3
5 2
18
18
MD   2.57
7

GTM 49 Key

Uploaded by

Copyright:

Available Formats

GTM 49 Key

Uploaded by

Document Information

Original Description:

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

GTM 49 Key

Uploaded by

Copyright:

Available Formats

Sri Chaitanya IIT Academy., India.

A.P, TELANGANA, KARNATAKA, TAMILNADU, MAHARASHTRA, DELHI, RANCHI

A right Choice for the Real Aspirant

Sec: Sr.Super-60 & All Page 2

Kinetic energy of e   1 mv 2  1 m   7  105 

Tension when it passes through the mean position,

Sec: Sr.Super-60 & All Page 3

Sec: Sr.Super-60 & All Page 4

18. Ratio of AM signal Bandwidths

Sec: Sr.Super-60 & All Page 5

25. See figure. Potential difference across R L

Sec: Sr.Super-60 & All Page 6

From linear momentum conservation, Li  Lf

33. Conjugation is preferred & Benzyl carbanion is stable

Sec: Sr.Super-60 & All Page 7

40. This is an example of Azeotrope

43. Molar conductance of NO3 = molar conductance of Cl 

Sec: Sr.Super-60 & All Page 8

Sec: Sr.Super-60 & All Page 9

2-D or sheet Silicate

59. Na2 S 4O6

Sec: Sr.Super-60 & All Page 10

Sec: Sr.Super-60 & All Page 11

Sec: Sr.Super-60 & All Page 12

Sec: Sr.Super-60 & All Page 14

Minimum value of z  i = distance of point (0, 1) from the line y – x = 0

Sec: Sr.Super-60 & All Page 18

You might also like