SRI CHAITANYA IIT ACADEMY, INDIA 01-10-22_ Sr.

IPL-IC_Jee-Main-WTM-17_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.IPL-IC JEE-MAIN Date:01-10-2022
Time: 09.00Am to 12.00Pm WTM-17 Max. Marks: 300

KEY SHEET
PHYSICS
1 4 2 4 3 3 4 2 5 3
6 4 7 3 8 4 9 2 10 2
11 1 12 4 13 1 14 3 15 3
16 4 17 3 18 2 19 3 20 1
21 9 22 2 23 6 24 7 25 6
26 3 27 2 28 80 29 21 30 300

CHEMISTRY
31 3 32 2 33 4 34 2 35 2
36 3 37 1 38 4 39 2 40 2
41 1 42 1 43 1 44 2 45 3
46 2 47 2 48 2 49 1 50 4
51 4 52 5 53 9 54 19 55 11
56 4 57 4 58 4 59 1 60 4

MATHEMATICS
61 2 62 1 63 1 64 2 65 2
66 3 67 3 68 1 69 4 70 3
71 3 72 2 73 4 74 3 75 2
76 1 77 2 78 2 79 3 80 1
81 12 82 4 83 3 84 4 85 4
86 50 87 1 88 15 89 0 90 5

Sec: Sr.IPL-IC Page 1

 SRI CHAITANYA IIT ACADEMY, INDIA 01-10-22_ Sr.IPL-IC_Jee-Main-WTM-17_KEY & Sol’S

SOLUTIONS
PHYSICS
1. Fermi energy is maximum possible allowable energy value for an electron at 0 K.

2/3
h2  3 N 
Fermi energy E F   
8m   V 
2. Conceptual
3. In a semiconductor energy gap of valence and conduction band is small.

Few electrons near top of valence band can acquire enough energy to reach the
conduction band. The Fermi energy level is so, in middle of gap.
4. A transistor may be a p–n–p or n–p–n type.

 its equivalent to 2 p–n junctions.

5. In a transistor, first layer E–B is forward biased  Depletion layer thickness decreases
due to biasing.

Sec: Sr.IPL-IC Page 2

 SRI CHAITANYA IIT ACADEMY, INDIA 01-10-22_ Sr.IPL-IC_Jee-Main-WTM-17_KEY & Sol’S

The base – collector region is in reverse bias, hence depletion layer thickness increases
due to biasing
6. 90 100 100
IC   IE  IE   IC   10  11mA
100 90 90
 I B  IE  IC  11  10  1 mA

7. All of charge carriers entering base region will be neutralized.

8. I  I 50
  C  50    C  0.98
IB 1   IB 51

9. Given logic circuit is

Let we draw truth table for this,

Clearly output resembles output resembles output of an AND gate.

10. Two NOR gates in series are

This is equivalent to

Sec: Sr.IPL-IC Page 3

 SRI CHAITANYA IIT ACADEMY, INDIA 01-10-22_ Sr.IPL-IC_Jee-Main-WTM-17_KEY & Sol’S

Truth table for above is

Two NOR gates in series are equivalent to an OR gate.

11. Truth table for given logic circuit is

Clearly above circuit is equivalent to AND gate

12. Inputs and outputs are as follows

Clearly, Y  A. A  B

13.

Sec: Sr.IPL-IC Page 4

 SRI CHAITANYA IIT ACADEMY, INDIA 01-10-22_ Sr.IPL-IC_Jee-Main-WTM-17_KEY & Sol’S

14. Signal from out cell phone travels through roof tops. Plants, vehicles etc. to the nearest
service providers tower by ground wave propagation mode.
15. 3  108 ms 1
Using f .  c,  
6
 10 2  100kms
3  10 Hz
So, only in two wavelengths signal can be reaching the space.
16. Out put of G1 : Y1  A  B Output of G 2 : Y2   Y1  C  '   A  B '.C'  A'.B'.C'

Find output  A '.B'.C'

17. Standard circuit diagram
18. Truth table for given pulse

A B C

0 0 0

1 1 1

0 1 1

1 0 1

Which is OR gate
19. Principle of superposition
20. Conceptual
21. V V
IS  S & IL  L  I Z  IS  IL
RS RL
22. D1 & D 2 are F.B and D3 is in R.B
23. In the given condition diode is in reverse biasing so it acts as open circuit. Hence P.d
bet A & B is 6V
24. I 7 
  C   
I B 8 1 
25. 80 80
ic  ie  24  i  ie  30mA & i b  i c  ic  6mA
100 100 e
26. V R
Voltage gain = current gain & Resistance gain 0  . 0
Vi Ri

27. 20 
Power gain  .voltage gain      & IC  .I6
21 1 
Sec: Sr.IPL-IC Page 5
 SRI CHAITANYA IIT ACADEMY, INDIA 01-10-22_ Sr.IPL-IC_Jee-Main-WTM-17_KEY & Sol’S

28. dT  2RhT
29. d  2Rh  A  d 2  A  2R 2h
30.   C / 100.
CHEMISTRY
31. Paracetamol acts as both antipyretic and analgesic
32. Bithional is added to soaps to impart antiseptic properties
33. From Definitions
34. Diphenylhydramine is employed as antihistamine drug
35. Penicillin ‘G’ has a narrow spectrum. Chloramphenicol is a board spectrum antibiotic
Ampicillin and amoxicillin are synthetic modifications of penicillins. These have board
spectrum.
36. Contain a mixture of estrogen and progesterone
37. Conceptual
38. Conceptual
39. Conceptual
40. Wt of AgBr A.wt of Br
%Br    100
mol.vol.of .AgBr Wt.of .org.comp
41. Adsorption process is

G   ve
H   ve
S   ve
42. On increasing the temperature, adsorption decreases
43. Hardness of water is due to carbonate or sulphate of Ca 2 and Mg 2  ions. When
sodium aluminium silicate is added Na  replaces Ca 2 and Mg 2  by adsorption, and
hence water becomes soft.
44. x 1 x 1
 P n or  K n
m m
Taking log on both sides

x 1
log  log K  log P
m n
45. Temp  extent of adsorption 
46. Tyndall effect is observed for colloidal solution

Sec: Sr.IPL-IC Page 6

 SRI CHAITANYA IIT ACADEMY, INDIA 01-10-22_ Sr.IPL-IC_Jee-Main-WTM-17_KEY & Sol’S

47. Ni is used for catalytic hydrogenation of oil

48. Smaller the gold number of protective colloid, greater is the protective power
49. AS2S3 is a negative colloid hence Al3 will be host effective for its coagulation
50. Statement D is wrong

Correct statement is :–Above CMC, surfactant molecules undergo aggregation and

micelles are formed

Micelles are generally formed by the aggregation of several ions or molecules with
lyophilic as well as lyophobic ports. They act as detergents, emulsifiers and dispersants.
They are used in petroleum recovery. They cleansing action of soap is due to the fact
that soap molecules form micelles around the oil droplets.
51. A,B,C,E
52. 1,2,3,5,8
53. O O O
|| || ||
HO  C  CH 2 CH  C  NH CH  C  OCH 3
| |
2 2
sp NH 2 sp CH 2 sp 2

6 sp 2

No.of sp2  carbons=9

54. Pressure of N 2  758  aqueous tension (in mm) of Hg

758  14 30  10 3
From PV=nRTMoles of N 2    1.246  103 mol.
760 0.0821  287
mass of N 2
Mass of N 2  1.246  103  28 % of nitrogen =  100
Total mass

1.246  28  10 3
  100  18.9%  19%
0.184

Sec: Sr.IPL-IC Page 7

 SRI CHAITANYA IIT ACADEMY, INDIA 01-10-22_ Sr.IPL-IC_Jee-Main-WTM-17_KEY & Sol’S

55.  y y
Cx H y   x 
  O 2  xCO 2  H 2O. Volume of CO2  16x  48  x  3
4 2

Volume of reactant = volume of product + contraction

16mL  VO2 (excess)  16  3mL  VO2  left  +48mL

 y
VO2  excess  VO2  left   80mL 16  x    80mL
 4

 y  8 Compound is C3H8  3  8  11

56. Hemoglobin is positively charged sol. Hence the sol, with negative charge can
coagulant hemoglobin, i.e., starch, clay AS2S3 ,cds
57. Silicic acid is negatively charged sol. Hence the sol with positive charge can coagulate
silicic acid, Fe  OH 3 ,Ca  OH 2 ,Al  OH3 ,basic dye.
58. 2
2mL of 1M Nacl contains Nacl   2m mol
1000
Thus 500mL of AS2S3 sol require Nacl Fc6 complete coagulation  2m mol

Hence, 1L, i.e., 1000mL of the sol require Nacl for complete coagulation = 4m mol

Therefore, Hocculation value of Nacl = 4

59. Gold number of gelatin  0.01 or 0.01 mg gelatin required to be added to 10mL of
gold sol completely prevent coagulation of 1mL of 10% Nacl solution.

Therefore gelatin added to 1000mL of gold sol to prevent coagulation

0.01  1000
  1mg
10
60. Mass of HCL acid adsorbed by log charcoal


 526.3  103  0.5  0.4  38  2 MW of HCl  38g mol1 
x 2
The amount of adsorption  4
m 0.5

Sec: Sr.IPL-IC Page 8

 SRI CHAITANYA IIT ACADEMY, INDIA 01-10-22_ Sr.IPL-IC_Jee-Main-WTM-17_KEY & Sol’S

MATHEMATICS
61. 1 dx 1 1 1 1 dx dt dt 1
  1 Put  t   t  1
 dy y x x 2 dy dy dy y
x x
1
  dy 1 11 1 1 1
I.F  e y  G.S,   dy  c   log y  c   ylog y  cy
y
y yx xy x

62.  dy 
xy  dy log x
 dx 
The given equation is reduced to x  e log x  xy
dx  ydy  
x
dx
2
y 2  log x  c
 
2 2 2
63. dy dx 1
 y  x2
 log y   c y  e1/x .ec
x

64. dy  e2x 1 1
  c  dy   e2x dx   cdx y  e2x  cx  d.
dx 2 2 4
65. dy y2 d
y  c  y dy   c dx  cx  y 2  2cx  d  y 2  2ax  b
dx 2 2

66. dy a2 dy dt dt a 2  t 2 t2
 Put x  y  t  1   a 2  t 2 dt   dx
dx  x  y  2 dx dx dx t2

 t  x  y
t  a tan 1    x  c y  a tan 1  c
 a  a 

67. dy
y  x  ydy   x dx y2  c  x 2 x 2  y2  c.
dx
68. dy dt dt
Put x  y  1  t  1  t 2  1 tan 1 t  x  c tan 1  x  y  1  x  c
dx dx dx
69. dy dy
xy  c x  y  0  x  y  0 x 2  y2  c
dx dx

Sec: Sr.IPL-IC Page 9

 SRI CHAITANYA IIT ACADEMY, INDIA 01-10-22_ Sr.IPL-IC_Jee-Main-WTM-17_KEY & Sol’S

70. dy sin x  x cos ydy  xdy  ycos xdx  sin ydx  ydx  0

  xdy  ydx    ycos x dx  sin x dy    x cos ydy  sin ydx   0

 d  xy  d  ysin x   d  xsiny  0  xy  ysin x  x sin y  k.

71. 4  5334  3  
365
 4  5  3 2  i 3
72. A   0,1 ,B   0, 1  K  AB  K  2

73. z1  3  4z  z1  3  4z  z1  3  4z  z1  3  4 2

74. z 2  2 z1 and

B  z2 


A  z1   3,4
4
O

z2 z 2 i 4    1 i
 e  z 2  2z1  cos  isin   2  3  4i   2  1  7i
z1 z1  4 4  2 

75. x n 1   x  1  x  1  x   2  ......... x   n 1 

Put x  3
76. 3 z
Area  z
4
77. x 2  10x  13 32  10  3  13
C  4
 x  1 x  2 x  3  3  13  2
78. x 2  1  2x 1 2
  2  A  1,C  2

x x2  1  x x 1

79. Let r  x   ax  b r 1  a  b  2  1 r  2  2a  b  5   2

Solving 1 &  2 ,a  3,b  1

Sec: Sr.IPL-IC Page 10

 SRI CHAITANYA IIT ACADEMY, INDIA 01-10-22_ Sr.IPL-IC_Jee-Main-WTM-17_KEY & Sol’S

80. 2  i 1 i 1  i
Point   Line y  x Then image 
3 i 2 2
81. dy
x cos   ysin   5   cot 
dx
82. dy dy 2dt
Let y  t   y  y'  t    2y  2e2t I.F  e  e2t G.S, ye2t  2t  c
dt dt
2t  2 4
 y  0  2  c  2  ye 2t  2t  2 y  y 1  2
t
e e
83. 2 dy
y  c  x  c  1  2c  x  c   2 From 1 &  2
dx
dy 2y
 x  c  2y dy
dx x  c
dx
84. dy
dx
 x  y dy  x dx y2  x 2  8 f  x   x 2  8 f 1  3,f 2 2  4  
85.
dy x  x 2  4y dy x  x 2  4y
 
dx 2 dx 2
86. z1  z 2  z1  z 2  25  6  31  Maximum
z1  z 2  z1  z 2  25  6  19  Minimum

87. W 5  1  W 4.W1  1and 1  W  W 2  W 3  W 4  0

1
log 2 1  W  W 2  W3   log 2  W 4  W 4  log 2 2W 4  log 22  1
W
88. cis4  cos 4  isin 4
n  4
89.  3,3  A but not B.
90. 2
3x 2  x  1 3 y  1   y  1  1 3 7 5
Put x  1  y     A  0,B  3,C  7,D  5
 x  1 4
y 4 y2 y3 y4

Sec: Sr.IPL-IC Page 11