test - 3 (code-a) (answers) all india aakash test series ... · all india aakash test series for...
TRANSCRIPT
Test - 3 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2019
1/16
1. (2)
2. (4)
3. (2)
4. (2)
5. (4)
6. (3)
7. (3)
8. (2)
9. (3)
10. (3)
11. (4)
12. (3)
13. (2)
14. (3)
15. (4)
16. (3)
17. (4)
18. (4)
19. (3)
20. (2)
21. (3)
22. (1)
23. (2)
24. (3)
25. (2)
26. (2)
27. (4)
28. (3)
29. (3)
30. (3)
PHYSICS CHEMISTRY MATHEMATICS
31. (2)
32. (1)
33. (1)
34. (2)
35. (3)
36. (2)
37. (3)
38. (4)
39. (4)
40. (2)
41. (4)
42. (1)
43. (1)
44. (4)
45. (2)
46. (3)
47. (4)
48. (2)
49. (2)
50. (1)
51. (3)
52. (2)
53. (4)
54. (4)
55. (4)
56. (1)
57. (3)
58. (4)
59. (3)
60. (2)
61. (2)
62. (3)
63. (1)
64. (2)
65. (4)
66. (2)
67. (2)
68. (3)
69. (4)
70. (1)
71. (3)
72. (3)
73. (4)
74. (2)
75. (2)
76. (3)
77. (4)
78. (2)
79. (1)
80. (4)
81. (2)
82. (2)
83. (3)
84. (2)
85. (1)
86. (2)
87. (4)
88. (3)
89. (4)
90. (1)
Test Date : 18/11/2018
ANSWERS
TEST - 3 - Code-A
All India Aakash Test Series for JEE (Main)-2019
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-A) (Hints & Solutions)
2/16
1. Answer (2)
Hint : Pressure due to radiation = I
C where I is
intensity
Sol. :0 0
2 sin,
sin
av
I A AF S
c
P =
22 sin av
F I
S c
2. Answer (4)
Hint : KE can vary from 0 to Kmax
.
Sol. : KE of ejected photoelectrons varies from 0 to
Kmax
and its variation is non-linear
3. Answer (2)
Hint : Cut-off wavelength is maximum wavelength
which can produce photoelectrons
Sol. :hc
= 0
0
hceV
...(i)
hcn
=
2
0
0
hcn eV
...(ii)
0 =
1n
n
⎛ ⎞⎜ ⎟⎝ ⎠
4. Answer (2)
Hint : Use equation of photoelectric effect
Sol. :1240
500 = +
1
2mv
1
2 ...(i)
1240
620 = +
1
2mv
2
2 ...(ii)
= 1.6 eV
5. Answer (4)
Hint : Revolving electron can be equivalent to a
current carrying loop.
Sol. : 2
c
vB
r
r n2
1v
n
PART - A (PHYSICS)
6. Answer (3)
Hint : For inelastic collision to occur, loss of energy
should be equal to energy difference of two
Bohr's orbits.
Sol. : (Kloss
)max
= 2
0
1 4
2 5
m mv
m
= 4( )
5i
K
For inelastic collision,
4( )
5i
K 10.2 4
Ki 51 eV
7. Answer (3)
Hint : Use equation of photoelectric effect
Sol. :
1
hc
= + K
2
hc
= + 3 K
2 >
1
3
8. Answer (2)
Hint : Use equation 2 2
1 2
1 1 1R
n n
⎛ ⎞ ⎜ ⎟ ⎝ ⎠
Sol. : Series is Paschen
2 2
min
1 1 1
3R⎛ ⎞ ⎜ ⎟ ⎝ ⎠
min
= 820 nm.
9. Answer (3)
Hint : In -decay mass decreases by 4 a.m.u and
Z by 2.
Sol. : No. of -particles = 238 206
4
= 8
No. of -particles = 8 2 = 16
Test - 3 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
3/16
10. Answer (3)
Hint : Q-value of a reaction.
Sol. : 64Cu decay gives +ve Q-value
11. Answer (4)
Hint : Higher the half life, slower will be decay rate.
Sol. : Higher the half life, slower will be decay rate
12. Answer (3)
Hint : Angular momentum is integral multiple of 2
h
and
2
2Z
En
Sol. : � = 2
hn
�
Be > �
Li
and E
2
2
Z
n |E
Li| > |E
Be|
13. Answer (2)
Hint : Q-value = Kinetic energy produced
Sol. : KE =
2 2
2
1
2 2
p h
m m
⎛ ⎞ ⎜ ⎟
⎝ ⎠
and Q = (KE )4
A
A
=
2
2
232
228 2
h
m
⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠
=
2
2
29
57
h
m
14. Answer (3)
Hint : Modulation index = m
c
A
A
Sol. :
c m
c m
A A
A A =
x
y
c
m
A
A =
x y
x y
m
c
A
A =
x y
x y
15. Answer (4)
Hint : Power gain = 2 ⎛ ⎞ ⎜ ⎟
⎝ ⎠
L
B
R
R
Sol. : Power gain = 2 ⎛ ⎞ ⎜ ⎟
⎝ ⎠
L
B
R
R
200 = 2 2
101
⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠
Vin =
4
20 = 0.2 V
16. Answer (3)
Hint : NAND gates can be used to form OR gate
Sol. : Given logic gate represents OR gate
17. Answer (4)
Hint : Use 1 1 1 u v f
Sol. : v1 =
15 1030 cm
(15 10)
v2 =
25 10 50cm
(25 10) 3
� = 50 40
30 cm3 3
18. Answer (4)
Hint : Refraction through curved surface
2 1 2 1 v u R
Sol. :1.2 (1.5 1.8) (1.2 1.5)
20 ( 20)f
f =
19. Answer (3)
Hint : Use image formation concept by ray diagram
Sol. :1 1 1
22.5 22.5 10x x
x = 7.5 cm
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-A) (Hints & Solutions)
4/16
20. Answer (2)
Hint : Wavelength in a medium = 0
µ
Sol. : 1 sin 60° = μ sin(45°)
μ = 3
2
= 0 450 2
367 nmµ 3
21. Answer (3)
Hint : For maximum magnification, the image should
be formed at least distance of clear vision.
Sol. : v = –25 cm
1 1 1
( 25) 10u
u = 50
cm7
22. Answer (1)
Hint : Interference can be analysed by vector algebra.
Sol. : By vector representation:
2E0
3E0
E0
Er =
02 E
as I (Er)2
23. Answer (2)
Hint : I = Imax
2
cos2
⎛ ⎞⎜ ⎟⎝ ⎠
and 2
( )
x
Sol. : I1 =
0
0
4
4
II
I0 =
2
0(4 )cos
2I
⎛ ⎞⎜ ⎟⎝ ⎠
= 2
3
2
3
⎛ ⎞⎜ ⎟⎝ ⎠
= 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠
y d
D
y = 1
0.50 mm3
⎛ ⎞ ⎜ ⎟⎝ ⎠
D
d
24. Answer (3)
Hint : Number of minima = No. of maxima
Sol. :
No. of minima = 5 4 = 20
25. Answer (2)
Hint : Length of tube = f0 + f
e
Sol. : � = f0 + f
e, m =
0
e
f
f
10 = 20
2 cme
e
ff
⇒
� = 20 + 2 = 22 cm.
26. Answer (2)
Hint : Consider three lenses are in contact
Sol. : R = 20 cm
2
1 2 3(1.3 1)
20 100f
⎛ ⎞ ⎜ ⎟⎝ ⎠
1 1 350 cm
20 100e
e
ff ⇒
27. Answer (4)
Hint : If two mirrors are at 90°, then ray reflects back
antiparallel.
Sol. : Ray becomes anti-parallel if and only if = 90°.
28. Answer (3)
Hint : Use
sin2µ
sin2
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
mA
A and i + i = + A
Sol. :
sin2µ
sin2
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
mA
A
μ = 2cos2
A⎛ ⎞⎜ ⎟⎝ ⎠
cos2
A⎛ ⎞⎜ ⎟⎝ ⎠
= µ
1 µ 22 ⇒
and, + A = i + i < 180° (
Test - 3 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
5/16
PART - B (CHEMISTRY)
A < 90°
μ > 2 cos90
2
⎛ ⎞⎜ ⎟⎝ ⎠
μ > 2
29. Answer (3)
Hint : Use fundamental concept to calculate zero error.
Sol. : 1 VSD = 9
mm10
Zero error = 6 (1 mm) – 6 9
mm10
⎛ ⎞⎜ ⎟⎝ ⎠
= +0.6 mm
31. Answer (2)
Hint : A =
CH3
C =
H C5 2
C
H
C C
CH3
H
Sol. :
CH3 CH3
H / t2P
(A) (B)
(C) (D)
H C5 2
C
H
C C
CH3
H
H / t2 P
CH —(CH ) —CH3 2 4 3
32. Answer (1)
Hint : (CH3)3CBr is 3° while other are 1° or 2° halide
Sol. : 3° halide show SN1 while other show S
N2
reaction.
33. Answer (1)
Hint : Ease of SN2 decrease as hindrance
increases.
Sol. : Correct order of ease of SN2 is IV > II > I > III
34. Answer (2)
Hint : When —NH2 is present on equatorial then
ring contract to five membered.
Sol. :
OH
NH2
HNO2
CHO
OH
NH2
NaNO2
C — H
O
HCl
35. Answer (3)
Hint : Same M.F but different functional group
Sol. : Aldehydes and ketones are functional isomers
36. Answer (2)
Hint : OH– is a good nucleophile whereas H2O is a
poor nucleophile
Sol. :
+Cl
H O2
S TypeN1
OH
SN2
OH
(A)
(B)
–
::
OH
(A) and (B) are positional isomers.
30. Answer (3)
Hint : Meter Bridge works on Wheatstone
balancing
Sol. :40 60
P Q
and
50
50
40 10 (60 10)
⎛ ⎞⎜ ⎟⎝ ⎠
Q
P Q
P = 50 50
,3 2
Q
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-A) (Hints & Solutions)
6/16
37. Answer (3)
Hint : Boiling point of isomeric alcohol varies as
1° > 2° > 3°.
Sol. : Alcohols have higher BP as compared to
ethers because of H bonding.
38. Answer (4)
Hint : Fact Based.
Sol. : Antibiotics have either cidal effect or a static
effect on microbes on this basis drugs are
divided into Bactericidal or Bacteriostatic.
39. Answer (4)
Hint : End Product of reaction (4) is phenol
Sol. :
OH
NO2
Fe
HCl
OH
NH2
OH
N Cl2
OH
H
NaNO2
HCl
H PO3 2
(Phenol)
+
40. Answer (2)
Hint : The reaction is SN1 type
Sol. :
CH — CH — CH — CH3 2 3
CH 3
CH — CH — CH — CH3 3
CH 3
Cl
Cl2
h
OH (alc)
PhCO H3
CH 3
CH 3
C CH — CH3
CH 3
CH 3
C CH
3
H O
H/ H O2
::
18
(S Type)N1
CH — C — CH — CH3 3
CH3
OHOH18
41. Answer (4)
Hint : Aldehydes are more reactive than ketones as
ketones are hindered.
Sol. : The attack happens by lone pair over sulphur.
The rate of addition is much greater for
aldehydes and can be used for purification/
separation bisulphite addition can be
reversed by acidic as well as basic
hydrolysis.
42. Answer (1)
Hint : Aldol condensation
Sol. :
C
43. Answer (1)
Hint : P Sucrose
R and S Glucose and fructose
Sol. : Sucrose is formed by formation of glycosidic
linkage between C1 of -D-glucose and C
2
and -D-fructose
Test - 3 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
7/16
44. Answer (4)
Hint : Carbene is neutral
Sol. : Amine on reaction with HNO2 form diazonium
salt
45. Answer (2)
Hint : hydroxy acid with H+/ gives , unsaturated acid
Sol. :
CH — CH — CH —CH — COOH3 2 2
CH — CH — CH = CH — COOH3 2
OH H +,
46. Answer (3)
Hint : Nitrene rearrangement
Sol. :
CH — CH3
CH2
HF, 0°C
CH3
CH
CH3
KMnO , H , 4
+
COOH
PCl5
C — Cl
O
N 3 OH , ,
(Curtius rearrangement)
NH2
47. Answer (4)
Hint : H of esters are acidic
Sol. :
CH — C — OEt3
O
CH — C — CH — C — OEt3 2
O O
EtONa
H(Claisen condensation)
CH – I3
EtO Na
CH — C — CH — C — OEt3 2
O O
CH — C — CH — C — OEt3
O O
—
CH3
CH — C — CH — CH3 2 3
O
H /+
48. Answer (2)
Hint : Ammoniolysis of amine
Sol. :
+
49. Answer (2)
Hint : The NaOBr formed will oxidise the Benzylic
Alcohol also
Sol. :
C
Br2
O
CH 2 NH
2
OH
NaOH
OHC NH2
50. Answer (1)
Hint : —CH3
C— C—Cl
——
CH3
CH3
— —
O
AlCl3 —CH3
C— C (+)
——
CH3
CH3
— —
O
—CH3
C (+)
——
CH3
CH3
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-A) (Hints & Solutions)
8/16
Sol. :
+ —CH3
C— C— Cl
——
CH3
CH3
— —
O
AlCl3
—
C—
——
CH3
CH3
CH3
51. Answer (3)
Hint : Rate of reaction will depend upon electrophilic
nature of the electrophile
Sol. : NO2 is EWG whereas OCH
3 is EDG so III is
the most reactive and I is the least
52. Answer (2)
Hint : Cu is used in Gatterman’s reaction
Sol. :
Cu
HCl
N2
Cl
53. Answer (4)
Hint : Novolac is used to make paints
Sol. : Phenol + formaldehyde forms Novolac in
presence of acid or base catalyst which
undergoes cross linking on further heating
with HCHO to form Bakelite.
54. Answer (4)
Hint : Neoprene is a polymer of chloroprene
Sol. : Terylene is polyester, Buna–S is a polymer of
butadiene and styrene.
55. Answer (4)
Hint : Phenol > alcohol > alkyne
Sol. : Phenoxide is resonance stabilised.
56. Answer (1)
Hint : a
1pK
acidic strength
Sol. :
OH
> CH OH3 > H O2 >CH CH CH OH3 2 2
Acidic strength
57. Answer (3)
Hint : On the basis of product stability
Sol. : III leads to formation of benzene.
58. Answer (4)
Hint : POCl3 remove H
2O by -elimination
Sol. : In -elimination H and OH must be present
on axial.
59. Answer (3)
Hint : DNA has double strand helix whereas RNA
has single strand.
Sol. :
Base + sugar Nucleoside
Nucleoside + phosphoric acid Nucleotide
RNA does not contain thymine.
60. Answer (2)
Hint : Methyl ketone on reaction with NaOH/Br2 give
haloform reaction while ester give carboxylic
acid.
Sol. :
Test - 3 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
9/16
PART - C (MATHEMATICS)
61. Answer (2)
Hint : Use condition for coplanarities.
Sol. : Since vectors are coplanar,
So, 0 1 1 0
4
a a c
c c b
4ab = c2
c2 – 4ab = 0 ...(i)
And
Discriminant of given equation is
= c2 – 4ab = 0, [from (i)]
Roots are real and equal
62. Answer (3)
Hint : If a�
and
�
b are non-collinear vectors such that
0 � �
xa yb x = 0, y = 0
Sol. : Here, (1 – tan)
�
a – 1 2 sin 0b � �
and a�
and b�
are non-collinear
1 = tan and 1 2 sin 0
4
or 5
4
and
5
4
or 7
4
Most general value of 5
24
n , n I
63. Answer (1)
Hint : Use concept of coplanarity of vectors
Sol. : Since given vectors are coplanar
So, 61�
p + 40�
q + 21�
r = 0�
�
p , �
q , �
r are coplanar.
�
q × �
r and �
r × �
p are collinear
(�
q × �
r ) × {(�
r × �
p ) = 0�
64. Answer (2)
Hint : Use formula for 'dot' and cross product of two
vectors
Sol. : Let be the angle between b�
and c�
,
Then,
5b c � �
b c� �
sin = 5
1sin =
2
3
cos =2
Now, given,
3( ) 0 � �
� � �
a b a c ( 3 ) 0 � �
� �
a b c
3 ||�
� �
b c a
3 �
� �
b c a
2 2
9 – 6 ·b c b c� � � �
= 2
2a�
25 + 36 – 6 cosb c � �
= 2·4
2361– 6 5.2· 4
2
⎛ ⎞ ⎜ ⎟⎜ ⎟
⎝ ⎠
2 61– 30 3
4
65. Answer (4)
Hint : Use formula for cross product of three vectors
Sol. : From given relation
2 · – ·a c b a b c
� � � � � �
3b c � �
1
·2
a c � �
and 3
· –2
a b � �
3
cos –2
a b � �
5
cos cos – cos6 6
⎛ ⎞ ⎜ ⎟⎝ ⎠
angle between a�
and b�
is 5
6
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-A) (Hints & Solutions)
10/16
66. Answer (2)
Hint :
Volume of parallelepiped
· · ·
· · ·
· ·
·
���
��� ��� ��� ��� ��
�� ��� ��� �� ��
��
� � � � � �
a a a b a c
b a b b b c
c a c cc b
Sol. : Volume of parallelepiped
· · ·
· · ·
· ·
·
a a a b a c
V b a b b b c
c a c cc b
���
��� ��� ��� ��� ��
�� ��� ��� �� ��
��
� � � � � �
4 3 4288 cu units
3 9 6
12 2 cu units4 6 16
67. Answer (2)
Hint : Mode = 3 Median – 2 Mean
Sol. : We have mode = 3 Median – 2 Mean
Mode – Median = 2(Median – Mean)
(Median – Mean) = 1Mode – Median
2
136 18
2
68. Answer (3)
Hint :sum of variables
Mean n
n
If n is odd, then median item
th1
2
n ⎛ ⎞ ⎜ ⎟⎝ ⎠
term
Sol. : Given,
3 3 3 31 2 3 ...
1900n
n
22
1
4·
n n
n
21 ·1900
4
n n
7600 = n(n + 1)2
19 × (20)2 = n(n + 1)2
n = 19
Median of 1, 2, 3 ..., 18, 19 is
th19 1
2
⎛ ⎞⎜ ⎟⎝ ⎠
item = 10
69. Answer (4)
Hint : 9!
2! 3!n S
7 6
3 3
6!– 5!
2! n E C C
Sol. : 9! 9 8 7 6 5 4 3!
3! 2! 3! 2!
9 8 7 6 5 2 9 8 7 6 10
n S
To find n(E)
Case I.
position for T is denoted by ×
× A × E × N × I × O × N ×
No two T's come together but N may or may not
Total words fomed 3
6! 7! 6 5 4!7
2! 3! 4! 2C
7 6 5 4 3 2 542 300 12600
2
Case. II
× A × E × NN × I × O ×
No two T's come together but two N come together
['×' is the position for T]
In this case total words formed
6
3
6 5 4 3!= × 5! =
3! 3!
C × 5 × 4 × 3 × 2 × 1
= 20 × 120 = 2400
n(E) = 12600 – 2400 = 10200
Required probability
n E
n S
10200
9 8 7 6 10
85 85
18 14 252
70. Answer (1)
Hint : Use
2
22
1 2
1 2
( )
( ) ( )
RP E P
EEP
R R RP E P P P E
E E
⎛ ⎞ ⎜ ⎟
⎛ ⎞ ⎝ ⎠⎜ ⎟ ⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Test - 3 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
11/16
Sol. : Let
E1 the event when die A is used
E2 the event when die B is used
R when red face appears
1
3
5
n
RP
E
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
and
2
2
5
n
RP
E
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
,
P(E1) =
1
2, P(E
2) =
1
2
2
1 2
162 5
973 2 1
5 5 2
n
n n
EP
R
⎛ ⎞ ⎜ ⎟⎛ ⎞ ⎝ ⎠ ⎜ ⎟ ⎡ ⎤⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
2 16
973 2
n
n n
n = 4
71. Answer (3)
Hint : Use Bayes Theorem
Sol. :
3
33
1 2 3
1 2 3
·
· · ·
BP B P
BBP
B B B BP B P P B P P B P
B B B
⎛ ⎞⎜ ⎟
⎛ ⎞ ⎝ ⎠⎜ ⎟⎛ ⎞⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2
2
3
2 2 2
2 2 2
1 – 4 7
3 – 4 10
1 – 4 8 – 4 9 – 4 7
3 – 4 10 – 4 10 – 4 10
⎛ ⎞ ⎜ ⎟⎜ ⎟⎛ ⎞ ⎝ ⎠⎜ ⎟ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
a a
a aBP
B a a a a a a
a a a a a a
2
3
22
– 4 7 1 1 1–
3 33 – 4 8 – 2 4
B a aP
B a a a
⎡ ⎤⎛ ⎞⎢ ⎥ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦
1 1 1–
3 12 4
72. Answer (3)
Hint : n(S) = 66, n(E) = 6
6 5
6 1( ) = 1 – 1 –
6 6P E
Sol. : n(S) = 66, n(E) = 6C1 × 1 × 1 × 1 × 1 × 1
Required probability
6
61 –
6
5
5 5 5
1 6 –1 77751 –
6 6 6
73. Answer (4)
Hint :
5 2 4 2 3
5 1 3 2 1
5! 5!5!
2! 2! 2!
n S C C C C C
2 3
2 1
5!
2! 2!n E C C
Sol. : To find n(S)
Letters to given word are
K 2
O 1
L 1
A 2
T 1
Case I.
When all 5 letters are distinct
Required words formed = 5C5 × 5! = 1 × 120 = 120
Case II.
When 2 alike letters and 3 different letters are
taken
Required words formed 2 4
1 3
5!= × ×
2!C C
5 4 3 2!2 4 480
2!
Case III.
When 2 alike, 2 alike and 1 different letter are
taken
Required words formed 2 3
2 1
5!= × ×
2! 2!C C
5 4 3 2!1 3 90
2! 2
n(S) = 120 + 480 + 90 = 690
n(E) = total number of words formed with all
the repeated letters used = 90
Required probability
90 3
690 23
n E
n S
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-A) (Hints & Solutions)
12/16
74. Answer (2)
Hint : Probability
10 11 11 11 11
11
1 1 1 1 1 7
2 2 2 2 2 2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Sol. :
H or T (appearing)
10 times
H H....... H �����
�������
10 101 1
1 1 1 1 12 2
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
T
H or T (appearing)
10 times
H H....... H ���
�������
10 111 1 1
1 1 1 12 2 2
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
H or T (appearing)
10 times
T H.........HH���
�������
10 111 1 1
1 1 1 12 2 2
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Required probability
10 11 11 11 11 111 1 1 1 1 1
2 2 2 2 2 2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
10 11 11
11
1 1 1 75 2 5
2 2 2 2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
75. Answer (2)
Hint : Required probability
6 5 4
6 9 9 6
n
n n
⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
Sol. : Bag 1 Bag 2
6W 4W
nB 5B
16
6P W
n
2
4
9P W
16
nP B
n
2
5
9P B
P(one white, one black)
6 5 4
6 9 9 6
n
n n
⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
5030 4
(given)9 6 99
n
n
15 2 25
6 11
n
n
150 + 25n = 165 + 22n
3n = 165 – 150 = 15
n = 5
76. Answer (3)
Hint :
P
A
5
B
O
a�
a bOP
a b
⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠
� �
��
� �
Sol. :
A
B C
Dˆ ˆ–i k
ˆ ˆ( )i j
ˆ ˆ( )k iˆ ˆ( )j kˆ ˆ( – )i j
ˆ ˆ ˆ ˆ ˆ ˆ ˆ– – 2 – –
2 2 2
⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠
���� i k i j i j kBD
1 64 1 1 3
2 2BD ����
Unit vector along BD
ˆ ˆ ˆ2 – –
6 i j k
77. Answer (4)
Hint : Volume of tetrahedron
= 1
3(Area of base) × height
Test - 3 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
13/16
Sol. : Volume of tetrahedron, is
D
h
Q
A
B
C
1
3V
⎛ ⎞ ⎜ ⎟⎝ ⎠
(Area of base) (Height)
1120 20
3h
120 3
20h
h = 18 units
78. Answer (2)
Hint : 2a + 3b + 10c = 0
2a – 3b – 7 = 0
where a, b, c are the d.r' s of the line of
intersection of planes
Sol. : Planes are
2x + 3y + 10z = 8 ... (i)
2x – 3y – 7z = 2 ... (ii)
Let a, b, c be the d.r's of the line of intersection
of planes (i) and (ii)
2a + 3b + 10c = 0
2a – 3b – 7c = 0
–
–21 30 –14 – 20 –6 – 6
a b c
9 34 –12
a b c i.e., a : b : c = 9 : 34 : –12
79. Answer (1)
Hint :
2 1–
n
a a b
S
b
��� ��� �
�
Sol. : Here lines are parallel and they are
ˆ ˆ ˆ ˆ ˆ ˆ3 2 2 2r i j k i j k �
and
ˆ ˆ ˆ ˆ ˆ ˆ4 3 2 2r i j k i j k �
Distance between them
2 1–a a b
b
��� ��� �
�
ˆ ˆ ˆ ˆ ˆ ˆ( – 2 ) (2 2 )
4 1 4
i j k i j k
�
�
2 1 2
1 –1 2 4 – 2 – 3
4 1 4 9
i j k
i j k
� �
� �
16 4 9 29
39
80. Answer (4)
Hint : Use concept of the distance of any point from
line parallel to any plane
Sol. :
P(5, 4, 3)
Q R
2 3 4
3 4 5
x y z
x – y + z =2 3 0
– – –
Coordinates of R are (3 + 2, 4 + 3, 5+ 4)
and are satisfying the equation x – 2y + 3z = 0
i,e, (3 + 2) – 8 – 6 + 15 + 12 = 0
10 = –8
= 4
–5
Coordinates of R are 2 1
– , – , 05 5
⎛ ⎞⎜ ⎟⎝ ⎠
Here equation of line PQ is
– 5 – 4 – 3
3 4 5
x y zsay
Coordinates of Q are (3 + 5, 4 + 4, 5 + 3)
and the point Q lies on the plane
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-A) (Hints & Solutions)
14/16
x – 2y + 3z = 0
(3 + 5) – (8 + 8) + (15+ 9) = 0
10= –6 = –3
5
Coordinates of Q are 16 8
, , 05 5
⎛ ⎞⎜ ⎟⎝ ⎠
Required distance
= 9
5QR
81. Answer (2)
Hint : Take image of given point A(1, 2, 3) due to
plane then find the distance between image
point and the point B.
Sol. :
D
C
AB (5, 8 , 11) (1, 2, 3)
(–1, –2, –3)
plane
x y z + 2 + 3 = 0
Image of A is D (x, y, z) and co-ordinates of D
can be obtained as
–2 1 4 9–1 – 2 – 3
1 2 3 1 4 9
x y z
–1 – 2 – 3
– 2
1 2 3
x y z
x – 1 = – 2, y – 2 = –4, z – 3 = –6
x = –1, y = –2, z = –3
D (–1, –2, – 3)
Minimum length of AC + CB
= 36 100 196 332BD
82. Answer (2)
Hint : Put x + y = v –1dy dx
dx dn
Sol. : From given equation,
cos ,dy
x ydx
Put, x y v
1dy dv
dx dx
From given equation, dv
dx – 1 = cosv
21 cos 2cos
2
dv vv
dx
21sec
2 2∫ ∫
vdv dx
tan2
vx c
tan2
x yx c
⎛ ⎞ ⎜ ⎟⎝ ⎠
83. Answer (3)
Hint : Solve by equation by method of inspection
Sol. : Here,
–
x ydy
dx x
xdy + y dx + xdx = 0
d(xy) + xdx = 0
2
2
xxy c ... (i)
If (i) passes through the point (2, 1), then
42
2c
c = 4
Put c = 4 in equation (i)
2xy + y2 = 2 × 4 = 8
84. Answer (2)
Hint : Differentiate solution and then use method of
separation of variables
Sol. : Differentiating both sides w.r.t.x
x(1 – x) f(x) + 0
1–
x
t f t xdt x f x∫
Test - 3 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
15/16
2
0
1–
x
t f t dt x f x∫
Again differentiating w.r.t. x we get
(1 – x)f(x) = 2xf(x) + x2f'(x)
2' 1– 3x f x x f x
2
' 1 3–
f x
f x xx
ln f(x) = 1
– – 3lnx c
x
(integrating both sides)
1ln ( ) = – – 3 ln f x x + c
x...(i)
Put x = 1 and f(1) = 1, in (i) we get
0 –1– 0 c 1c
1ln – – 3 ln 1f x x
x
3
1 1 11 – ln 1 –
3
1x x xf x e e
x
1
1 –2
12
8 8
ef e
85. Answer (1)
Hint : Convert equation into linear form
Sol. : From given differential equation, we have
2 3
1,
dy y
dx x x
1–
IF = xe
Solution of equation (i) is
1–
·
xy e =
1–x
e +
1–x
ec
x
Put x = 1 and y (1) = 1, we get c = –1
e
1
11 –
xey
x e
2
11 2 – 3 –
2
⎛ ⎞ ⎜ ⎟⎝ ⎠
ey e
e
86. Answer (2)
Hint : Put x – 2 = X, y – 2 = Y
dY dy
dX dx and put Y = vX
Sol. : From given differential equation
2 2– 2 – 2
– 2 – 2
x ydy
dx x y
...(i)
Let x – 2 = X & y – 2 = Y dY dy
dX dx
From equation (i)
2 2dY X Y
dX XY
...(ii)
(homogeneous differential equation)
Put Y = VX, then dY dV
V XdX dX
Substituting these values in (ii) and then solving
we get,
2
22 In | |
YX c
X
2
2
– 2
– 2
y
x
= 2 ln |x – 2| + c
87. Answer (4)
Hint : Differentiate both sides w.r.t. x and eliminate
arbitrary constants
Sol. :
5
2y cx c , dyc
dx
5
2dy dyy x
dx dx
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2 5
–
dy dyy x
dx dx
⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
Degree of differential equation = 5
88. Answer (3)
Hint : Length of subsnormal dy
ydx
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-A) (Hints & Solutions)
16/16
Sol. : Given length of subnormal dy
ydx
= k
dy
y kdx
2
2
ykx c
Which passes through the origin O(0, 0)
0 = 0 + c
c = 0
Curve is 22y kx
89. Answer (4)
Hint : Here 5a + 3b + 3c = 0 and 2a + 3b + 4c = 0
and solve for a, b, c
Sol. : DRs of normal to the plane containing lines
5 2 3
x y z and 2 3 4 x y z
are a, b, c, then
5a + 2b + 3c = 0 and 2a + 3b + 4c = 0 and
so d.r.'s of nomral to the plane are –1, –14, 11
Also plane contains the line 1 2 3
x y z
Required equation of plane will be given by
1 2 3 0
–1 –14 11
x y z
32x – 7y – 6z = 0
90. Answer (1)
Hint : Use condition for coplanarity for two line
Sol. : Given lines are
– 5
0 4 – –1
x y z
and –
0 –2 2 –
x y z are coplanar
5 – 0 0
0 4 – –1 0
0 –2 2 –
3 – 112 + 36 – 30 = 0
sum of all possible values of
1 2 3
– 11
11
1
� � �
Test - 3 (Code-B) (Answers) All India Aakash Test Series for JEE (Main)-2019
1/16
1. (3)
2. (3)
3. (3)
4. (4)
5. (2)
6. (2)
7. (3)
8. (2)
9. (1)
10. (3)
11. (2)
12. (3)
13. (4)
14. (4)
15. (3)
16. (4)
17. (3)
18. (2)
19. (3)
20. (4)
21. (3)
22. (3)
23. (2)
24. (3)
25. (3)
26. (4)
27. (2)
28. (2)
29. (4)
30. (2)
PHYSICS CHEMISTRY MATHEMATICS
31. (2)
32. (3)
33. (4)
34. (3)
35. (1)
36. (4)
37. (4)
38. (4)
39. (2)
40. (3)
41. (1)
42. (2)
43. (2)
44. (4)
45. (3)
46. (2)
47. (4)
48. (1)
49. (1)
50. (4)
51. (2)
52. (4)
53. (4)
54. (3)
55. (2)
56. (3)
57. (2)
58. (1)
59. (1)
60. (2)
61. (1)
62. (4)
63. (3)
64. (4)
65. (2)
66. (1)
67. (2)
68. (3)
69. (2)
70. (2)
71. (4)
72. (1)
73. (2)
74. (4)
75. (3)
76. (2)
77. (2)
78. (4)
79. (3)
80. (3)
81. (1)
82. (4)
83. (3)
84. (2)
85. (2)
86. (4)
87. (2)
88. (1)
89. (3)
90. (2)
Test Date : 18/11/2018
ANSWERS
TEST - 3 - Code-B
All India Aakash Test Series for JEE (Main)-2019
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-B) (Hints & Solutions)
2/16
1. Answer (3)
Hint : Meter Bridge works on Wheatstone
balancing
Sol. :40 60
P Q
and
50
50
40 10 (60 10)
⎛ ⎞⎜ ⎟⎝ ⎠
Q
P Q
P = 50 50
,3 2
Q
2. Answer (3)
Hint : Use fundamental concept to calculate zero error.
Sol. : 1 VSD = 9
mm10
Zero error = 6 (1 mm) – 6 9
mm10
⎛ ⎞⎜ ⎟⎝ ⎠
= +0.6 mm
3. Answer (3)
Hint : Use
sin2µ
sin2
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
mA
A and i + i = + A
Sol. :
sin2µ
sin2
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
mA
A
μ = 2cos2
A⎛ ⎞⎜ ⎟⎝ ⎠
cos2
A⎛ ⎞⎜ ⎟⎝ ⎠
= µ
1 µ 22 ⇒
and, + A = i + i < 180° (
A < 90°
μ > 2 cos90
2
⎛ ⎞⎜ ⎟⎝ ⎠
μ > 2
PART - A (PHYSICS)
4. Answer (4)
Hint : If two mirrors are at 90°, then ray reflects back
antiparallel.
Sol. : Ray becomes anti-parallel if and only if = 90°.
5. Answer (2)
Hint : Consider three lenses are in contact
Sol. : R = 20 cm
2
1 2 3(1.3 1)
20 100f
⎛ ⎞ ⎜ ⎟⎝ ⎠
1 1 350 cm
20 100e
e
ff ⇒
6. Answer (2)
Hint : Length of tube = f0 + f
e
Sol. : � = f0 + f
e, m =
0
e
f
f
10 = 20
2 cme
e
ff
⇒
� = 20 + 2 = 22 cm.
7. Answer (3)
Hint : Number of minima = No. of maxima
Sol. :
No. of minima = 5 4 = 20
8. Answer (2)
Hint : I = Imax
2
cos2
⎛ ⎞⎜ ⎟⎝ ⎠
and 2
( )
x
Sol. : I1 =
0
0
4
4
II
I0 =
2
0(4 )cos
2I
⎛ ⎞⎜ ⎟⎝ ⎠
= 2
3
2
3
⎛ ⎞⎜ ⎟⎝ ⎠
= 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠
y d
D
y = 1
0.50 mm3
⎛ ⎞ ⎜ ⎟⎝ ⎠
D
d
Test - 3 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
3/16
9. Answer (1)
Hint : Interference can be analysed by vector algebra.
Sol. : By vector representation:
2E0
3E0
E0
Er =
02 E
as I (Er)2
10. Answer (3)
Hint : For maximum magnification, the image should
be formed at least distance of clear vision.
Sol. : v = –25 cm
1 1 1
( 25) 10u
u = 50
cm7
11. Answer (2)
Hint : Wavelength in a medium = 0
µ
Sol. : 1 sin 60° = μ sin(45°)
μ = 3
2
= 0 450 2
367 nmµ 3
12. Answer (3)
Hint : Use image formation concept by ray diagram
Sol. :1 1 1
22.5 22.5 10x x
x = 7.5 cm
13. Answer (4)
Hint : Refraction through curved surface
2 1 2 1 v u R
Sol. :1.2 (1.5 1.8) (1.2 1.5)
20 ( 20)f
f =
14. Answer (4)
Hint : Use 1 1 1 u v f
Sol. : v1 =
15 1030 cm
(15 10)
v2 =
25 10 50cm
(25 10) 3
� = 50 40
30 cm3 3
15. Answer (3)
Hint : NAND gates can be used to form OR gate
Sol. : Given logic gate represents OR gate
16. Answer (4)
Hint : Power gain = 2 ⎛ ⎞ ⎜ ⎟
⎝ ⎠
L
B
R
R
Sol. : Power gain = 2 ⎛ ⎞ ⎜ ⎟
⎝ ⎠
L
B
R
R
200 = 2 2
101
⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠
Vin =
4
20 = 0.2 V
17. Answer (3)
Hint : Modulation index = m
c
A
A
Sol. :
c m
c m
A A
A A =
x
y
c
m
A
A =
x y
x y
m
c
A
A =
x y
x y
18. Answer (2)
Hint : Q-value = Kinetic energy produced
Sol. : KE =
2 2
2
1
2 2
p h
m m
⎛ ⎞ ⎜ ⎟
⎝ ⎠
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-B) (Hints & Solutions)
4/16
and Q = (KE )4
A
A
=
2
2
232
228 2
h
m
⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠
=
2
2
29
57
h
m
19. Answer (3)
Hint : Angular momentum is integral multiple of 2
h
and
2
2Z
En
Sol. : � = 2
hn
�
Be > �
Li
and E
2
2
Z
n |E
Li| > |E
Be|
20. Answer (4)
Hint : Higher the half life, slower will be decay rate.
Sol. : Higher the half life, slower will be decay rate
21. Answer (3)
Hint : Q-value of a reaction.
Sol. : 64Cu decay gives +ve Q-value
22. Answer (3)
Hint : In -decay mass decreases by 4 a.m.u and
Z by 2.
Sol. : No. of -particles = 238 206
4
= 8
No. of -particles = 8 2 = 16
23. Answer (2)
Hint : Use equation 2 2
1 2
1 1 1R
n n
⎛ ⎞ ⎜ ⎟ ⎝ ⎠
Sol. : Series is Paschen
2 2
min
1 1 1
3R⎛ ⎞ ⎜ ⎟ ⎝ ⎠
min
= 820 nm.
24. Answer (3)
Hint : Use equation of photoelectric effect
Sol. :
1
hc
= + K
2
hc
= + 3 K
2 >
1
3
25. Answer (3)
Hint : For inelastic collision to occur, loss of energy
should be equal to energy difference of two
Bohr's orbits.
Sol. : (Kloss
)max
= 2
0
1 4
2 5
m mv
m
= 4( )
5i
K
For inelastic collision,
4( )
5i
K 10.2 4
Ki 51 eV
26. Answer (4)
Hint : Revolving electron can be equivalent to a
current carrying loop.
Sol. : 2
c
vB
r
r n2
1vn
27. Answer (2)
Hint : Use equation of photoelectric effect
Sol. :1240
500 = +
1
2mv
1
2 ...(i)
1240
620 = +
1
2mv
2
2 ...(ii)
= 1.6 eV
28. Answer (2)
Hint : Cut-off wavelength is maximum wavelength
which can produce photoelectrons
Test - 3 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
5/16
PART - B (CHEMISTRY)
31. Answer (2)
Hint : Methyl ketone on reaction with NaOH/Br2 give
haloform reaction while ester give carboxylic
acid.
Sol. :
32. Answer (3)
Hint : DNA has double strand helix whereas RNA
has single strand.
Sol. :
Base + sugar Nucleoside
Nucleoside + phosphoric acid Nucleotide
RNA does not contain thymine.
33. Answer (4)
Hint : POCl3 remove H
2O by -elimination
Sol. : In -elimination H and OH must be present
on axial.
34. Answer (3)
Hint : On the basis of product stability
Sol. : III leads to formation of benzene.
35. Answer (1)
Hint : a
1pK
acidic strength
Sol. :
OH
> CH OH3 > H O2 >CH CH CH OH3 2 2
Acidic strength
36. Answer (4)
Hint : Phenol > alcohol > alkyne
Sol. : Phenoxide is resonance stabilised.
37. Answer (4)
Hint : Neoprene is a polymer of chloroprene
Sol. : Terylene is polyester, Buna–S is a polymer of
butadiene and styrene.
Sol. :hc
= 0
0
hceV
...(i)
hcn
=
2
0
0
hcn eV
...(ii)
0 =
1n
n
⎛ ⎞⎜ ⎟⎝ ⎠
29. Answer (4)
Hint : KE can vary from 0 to Kmax
.
Sol. : KE of ejected photoelectrons varies from 0 to
Kmax
and its variation is non-linear
30. Answer (2)
Hint : Pressure due to radiation = I
C where I is
intensity
Sol. :0 0
2 sin,
sin
av
I A AF S
c
P =
22 sin av
F I
S c
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-B) (Hints & Solutions)
6/16
38. Answer (4)
Hint : Novolac is used to make paints
Sol. : Phenol + formaldehyde forms Novolac in
presence of acid or base catalyst which
undergoes cross linking on further heating
with HCHO to form Bakelite.
39. Answer (2)
Hint : Cu is used in Gatterman’s reaction
Sol. :
Cu
HCl
N2
Cl
40. Answer (3)
Hint : Rate of reaction will depend upon electrophilic
nature of the electrophile
Sol. : NO2 is EWG whereas OCH
3 is EDG so III is
the most reactive and I is the least
41. Answer (1)
Hint : —CH3
C— C—Cl
——
CH3
CH3
— —
O
AlCl3 —CH3
C— C (+)
——
CH3
CH3
— —
O
—CH3
C (+)
——
CH3
CH3
Sol. :
+ —CH3
C— C— Cl
——
CH3
CH3
— —
O
AlCl3
—
C—
——
CH3
CH3
CH3
42. Answer (2)
Hint : The NaOBr formed will oxidise the Benzylic
Alcohol also
Sol. :
C
Br2
O
CH 2 NH
2
OH
NaOH
OHC NH2
43. Answer (2)
Hint : Ammoniolysis of amine
Sol. :
+
44. Answer (4)
Hint : H of esters are acidic
Sol. :
CH — C — OEt3
O
CH — C — CH — C — OEt3 2
O O
EtONa
H(Claisen condensation)
CH – I3
EtO Na
CH — C — CH — C — OEt3 2
O O
CH — C — CH — C — OEt3
O O
—
CH3
CH — C — CH — CH3 2 3
O
H /+
Test - 3 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
7/16
45. Answer (3)
Hint : Nitrene rearrangement
Sol. :
CH — CH3
CH2
HF, 0°C
CH3
CH
CH3
KMnO , H , 4
+
COOH
PCl5
C — Cl
O
N 3 OH , ,
(Curtius rearrangement)
NH2
46. Answer (2)
Hint : hydroxy acid with H+/ gives , unsaturated acid
Sol. :
CH — CH — CH —CH — COOH3 2 2
CH — CH — CH = CH — COOH3 2
OH H +,
47. Answer (4)
Hint : Carbene is neutral
Sol. : Amine on reaction with HNO2 form diazonium
salt
48. Answer (1)
Hint : P Sucrose
R and S Glucose and fructose
Sol. : Sucrose is formed by formation of glycosidic
linkage between C1 of -D-glucose and C
2
and -D-fructose
49. Answer (1)
Hint : Aldol condensation
Sol. :
C
50. Answer (4)
Hint : Aldehydes are more reactive than ketones as
ketones are hindered.
Sol. : The attack happens by lone pair over sulphur.
The rate of addition is much greater for
aldehydes and can be used for purification/
separation bisulphite addition can be reversed
by acidic as well as basic hydrolysis.
51. Answer (2)
Hint : The reaction is SN1 type
Sol. :
CH — CH — CH — CH3 2 3
CH 3
CH — CH — CH — CH3 3
CH 3
Cl
Cl2
h
OH (alc)
PhCO H3
CH 3
CH 3
C CH — CH3
CH 3
CH 3
C CH
3
H O
H/ H O2
::
18
(S Type)N1
CH — C — CH — CH3 3
CH3
OHOH18
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-B) (Hints & Solutions)
8/16
52. Answer (4)
Hint : End Product of reaction (4) is phenol
Sol. :
OH
NO2
Fe
HCl
OH
NH2
OH
N Cl2
OH
H
NaNO2
HCl
H PO3 2
(Phenol)
+
53. Answer (4)
Hint : Fact Based.
Sol. : Antibiotics have either cidal effect or a static
effect on microbes on this basis drugs are
divided into Bactericidal or Bacteriostatic.
54. Answer (3)
Hint : Boiling point of isomeric alcohol varies as
1° > 2° > 3°.
Sol. : Alcohols have higher BP as compared to
ethers because of H bonding.
55. Answer (2)
Hint : OH– is a good nucleophile whereas H2O is a
poor nucleophile
Sol. :
+Cl
H O2
S TypeN1
OH
SN2
OH
(A)
(B)
–
::
OH
(A) and (B) are positional isomers.
56. Answer (3)
Hint : Same M.F but different functional group
Sol. : Aldehydes and ketones are functional isomers
57. Answer (2)
Hint : When —NH2 is present on equatorial then
ring contract to five membered.
Sol. :
OH
NH2
HNO2
CHO
OH
NH2
NaNO2
C — H
O
HCl
58. Answer (1)
Hint : Ease of SN2 decrease as hindrance
increases.
Sol. : Correct order of ease of SN2 is IV > II > I > III
59. Answer (1)
Hint : (CH3)3CBr is 3° while other are 1° or 2° halide
Sol. : 3° halide show SN1 while other show S
N2
reaction.
60. Answer (2)
Hint : A =
CH3
C =
H C5 2
C
H
C C
CH3
H
Sol. :
CH3 CH3
H / t2P
(A) (B)
(C) (D)
H C5 2
C
H
C C
CH3
H
H / t2 P
CH —(CH ) —CH3 2 4 3
Test - 3 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
9/16
PART - C (MATHEMATICS)
61. Answer (1)
Hint : Use condition for coplanarity for two line
Sol. : Given lines are
– 5
0 4 – –1
x y z
and –
0 –2 2 –
x y z are coplanar
5 – 0 0
0 4 – –1 0
0 –2 2 –
3 – 112 + 36 – 30 = 0
sum of all possible values of
1 2 3
– 11
11
1
62. Answer (4)
Hint : Here 5a + 3b + 3c = 0 and 2a + 3b + 4c = 0
and solve for a, b, c
Sol. : DRs of normal to the plane containing lines
5 2 3
x y z and 2 3 4 x y z
are a, b, c, then
5a + 2b + 3c = 0 and 2a + 3b + 4c = 0 and
so d.r.'s of nomral to the plane are –1, –14, 11
Also plane contains the line 1 2 3
x y z
Required equation of plane will be given by
1 2 3 0
–1 –14 11
x y z
32x – 7y – 6z = 0
63. Answer (3)
Hint : Length of subsnormal dy
ydx
Sol. : Given length of subnormal dy
ydx
= k
dy
y kdx
2
2
ykx c
Which passes through the origin O(0, 0)
0 = 0 + c
c = 0
Curve is 22y kx
64. Answer (4)
Hint : Differentiate both sides w.r.t. x and eliminate
arbitrary constants
Sol. :
5
2y cx c , dyc
dx
5
2dy dyy x
dx dx
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2 5
–
dy dyy x
dx dx
⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
Degree of differential equation = 5
65. Answer (2)
Hint : Put x – 2 = X, y – 2 = Y
dY dy
dX dx and put Y = vX
Sol. : From given differential equation
2 2– 2 – 2
– 2 – 2
x ydy
dx x y
...(i)
Let x – 2 = X & y – 2 = Y dY dy
dX dx
From equation (i)
2 2dY X Y
dX XY
...(ii)
(homogeneous differential equation)
Put Y = VX, then dY dV
V XdX dX
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-B) (Hints & Solutions)
10/16
Substituting these values in (ii) and then solving
we get,
2
22 In | |
YX c
X
2
2
– 2
– 2
y
x
= 2 ln |x – 2| + c
66. Answer (1)
Hint : Convert equation into linear form
Sol. : From given differential equation, we have
2 3
1,
dy y
dx x x
1–
IF = xe
Solution of equation (i) is
1–
·
xy e =
1–x
e +
1–x
ec
x
Put x = 1 and y (1) = 1, we get c = –1
e
1
11 –
xey
x e
2
11 2 – 3 –
2
⎛ ⎞ ⎜ ⎟⎝ ⎠
ey e
e
67. Answer (2)
Hint : Differentiate solution and then use method of
separation of variables
Sol. : Differentiating both sides w.r.t. x
x(1 – x) f(x) + 0
1–
x
t f t xdt x f x∫
2
0
1–
x
t f t dt x f x∫
Again differentiating w.r.t. x we get
(1 – x)f(x) = 2xf(x) + x2f'(x)
2' 1– 3x f x x f x
2
' 1 3–
f x
f x xx
ln f(x) = 1
– – 3ln x c
x
(integrating both sides)
1ln ( ) = – – 3 ln f x x + c
x...(i)
Put x = 1 and f(1) = 1, in (i) we get
0 –1– 0 c 1c
1ln – – 3 ln 1f x x
x
3
1 1 11 – ln 1 –
3
1x x xf x e e
x
1
1 –2
12
8 8
ef e
68. Answer (3)
Hint : Solve by equation by method of inspection
Sol. : Here,
–
x ydy
dx x
xdy + y dx + xdx = 0
d(xy) + xdx = 0
2
2
xxy c ... (i)
If (i) passes through the point (2, 1), then
42
2c
c = 4
Put c = 4 in equation (i)
2xy + y2 = 2 × 4 = 8
69. Answer (2)
Hint : Put x + y = v –1dy dx
dx dn
Sol. : From given equation,
cos ,dy
x ydx
Put, x y v
1dy dv
dx dx
Test - 3 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
11/16
From given equation, dv
dx – 1 = cosv
21 cos 2cos
2
dv vv
dx
21sec
2 2∫ ∫
vdv dx
tan2
vx c
tan2
x yx c
⎛ ⎞ ⎜ ⎟⎝ ⎠
70. Answer (2)
Hint : Take image of given point A(1, 2, 3) due to
plane then find the distance between image
point and the point B.
Sol. :
D
C
AB (5, 8 , 11) (1, 2, 3)
(–1, –2, –3)
plane
x y z + 2 + 3 = 0
Image of A is D (x, y, z) and co-ordinates of D
can be obtained as
–2 1 4 9–1 – 2 – 3
1 2 3 1 4 9
x y z
–1 – 2 – 3
– 2
1 2 3
x y z
x – 1 = – 2, y – 2 = –4, z – 3 = –6
x = –1, y = –2, z = –3
D (–1, –2, – 3)
Minimum length of AC + CB
= 36 100 196 332BD
71. Answer (4)
Hint : Use concept of the distance of any point from
line parallel to any plane
Sol. :
P(5, 4, 3)
Q R
2 3 4
3 4 5
x y z
x – y + z =2 3 0
– – –
Coordinates of R are (3 + 2, 4 + 3, 5+ 4)
and are satisfying the equation x – 2y + 3z = 0
i,e, (3 + 2) – 8 – 6 + 15 + 12 = 0
10 = –8
= 4
–5
Coordinates of R are 2 1
– , – , 05 5
⎛ ⎞⎜ ⎟⎝ ⎠
Here equation of line PQ is
– 5 – 4 – 3
3 4 5
x y zsay
Coordinates of Q are (3 + 5, 4 + 4, 5 + 3)
and the point Q lies on the plane
x – 2y + 3z = 0
(3 + 5) – (8 + 8) + (15+ 9) = 0
10= –6 = –3
5
Coordinates of Q are 16 8
, , 05 5
⎛ ⎞⎜ ⎟⎝ ⎠
Required distance = 9
5QR
72. Answer (1)
Hint :
2 1–
n
a a b
S
b
��� ��� �
�
Sol. : Here lines are parallel and they are
ˆ ˆ ˆ ˆ ˆ ˆ3 2 2 2r i j k i j k �
and
ˆ ˆ ˆ ˆ ˆ ˆ4 3 2 2r i j k i j k �
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-B) (Hints & Solutions)
12/16
Distance between them
2 1–a a b
b
��� ��� �
�
ˆ ˆ ˆ ˆ ˆ ˆ( – 2 ) (2 2 )
4 1 4
i j k i j k
�
�
2 1 2
1 –1 2 4 – 2 – 3
4 1 4 9
i j k
i j k
� �
� �
16 4 9 29
39
73. Answer (2)
Hint : 2a + 3b + 10c = 0
2a – 3b – 7 = 0
where a, b, c are the d.r' s of the line of
intersection of planes
Sol. : Planes are
2x + 3y + 10z = 8 ... (i)
2x – 3y – 7z = 2 ... (ii)
Let a, b, c be the d.r's of the line of intersection
of planes (i) and (ii)
2a + 3b + 10c = 0
2a – 3b – 7c = 0
–
–21 30 –14 – 20 –6 – 6
a b c
9 34 –12
a b c i.e., a : b : c = 9 : 34 : –12
74. Answer (4)
Hint : Volume of tetrahedron
= 1
3(Area of base) × height
Sol. : Volume of tetrahedron, is
D
h
Q
A
B
C
1
3V
⎛ ⎞ ⎜ ⎟⎝ ⎠
(Area of base) (Height)
1120 20
3h
120 3
20h
h = 18 units
75. Answer (3)
Hint :
P
A
5
B
O
a�
a bOP
a b
⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠
� �
��
� �
Sol. :
A
B C
Dˆ ˆ–i k
ˆ ˆ( )i j
ˆ ˆ( )k iˆ ˆ( )j kˆ ˆ( – )i j
ˆ ˆ ˆ ˆ ˆ ˆ ˆ– – 2 – –
2 2 2
⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠
���� i k i j i j kBD
1 64 1 1 3
2 2BD ����
Unit vector along BD
ˆ ˆ ˆ2 – –
6 i j k
76. Answer (2)
Hint : Required probability
6 5 4
6 9 9 6
n
n n
⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
Sol. : Bag 1 Bag 2
6W 4W
nB 5B
16
6P W
n
2
4
9P W
16
nP B
n
2
5
9P B
Test - 3 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
13/16
P(one white, one black)
6 5 4
6 9 9 6
n
n n
⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
5030 4
(given)9 6 99
n
n
15 2 25
6 11
n
n
150 + 25n = 165 + 22n
3n = 165 – 150 = 15
n = 5
77. Answer (2)
Hint : Probability
10 11 11 11 11
11
1 1 1 1 1 7
2 2 2 2 2 2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Sol. :
H or T (appearing)
10 times
H H....... H �����
�������
10 101 1
1 1 1 1 12 2
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
T
H or T (appearing)
10 times
H H....... H ���
�������
10 111 1 1
1 1 1 12 2 2
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
H or T (appearing)
10 times
T H.........HH���
�������
10 111 1 1
1 1 1 12 2 2
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Required probability
10 11 11 11 11 111 1 1 1 1 1
2 2 2 2 2 2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
10 11 11
11
1 1 1 75 2 5
2 2 2 2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
78. Answer (4)
Hint :
5 2 4 2 3
5 1 3 2 1
5! 5!5!
2! 2! 2!
n S C C C C C
2 3
2 1
5!
2! 2!n E C C
Sol. : To find n(S)
Letters to given word are
K 2
O 1
L 1
A 2
T 1
Case I.
When all 5 letters are distinct
Required words formed = 5C5 × 5! = 1 × 120 = 120
Case II.
When 2 alike letters and 3 different letters are
taken
Required words formed 2 4
1 3
5!= × ×
2!C C
5 4 3 2!2 4 480
2!
Case III.
When 2 alike, 2 alike and 1 different letter are
taken
Required words formed 2 3
2 1
5!= × ×
2! 2!C C
5 4 3 2!1 3 90
2! 2
n(S) = 120 + 480 + 90 = 690
n(E) = total number of words formed with all
the repeated letters used = 90
Required probability
90 3
690 23
n E
n S
79. Answer (3)
Hint : n(S) = 66, n(E) = 6
6 5
6 1( ) = 1 – 1 –
6 6P E
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-B) (Hints & Solutions)
14/16
Sol. : n(S) = 66, n(E) = 6C1 × 1 × 1 × 1 × 1 × 1
Required probability 6
61 –
6
5
5 5 5
1 6 –1 77751 –
6 6 6
80. Answer (3)
Hint : Use Bayes Theorem
Sol. :
3
33
1 2 3
1 2 3
·
· · ·
BP B P
BBP
B B B BP B P P B P P B P
B B B
⎛ ⎞⎜ ⎟
⎛ ⎞ ⎝ ⎠⎜ ⎟⎛ ⎞⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2
2
3
2 2 2
2 2 2
1 – 4 7
3 – 4 10
1 – 4 8 – 4 9 – 4 7
3 – 4 10 – 4 10 – 4 10
⎛ ⎞ ⎜ ⎟⎜ ⎟⎛ ⎞ ⎝ ⎠⎜ ⎟ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
a a
a aBP
B a a a a a a
a a a a a a
2
3
22
– 4 7 1 1 1–
3 33 – 4 8 – 2 4
B a aP
B a a a
⎡ ⎤⎛ ⎞⎢ ⎥ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦
1 1 1–
3 12 4
81. Answer (1)
Hint : Use
2
22
1 2
1 2
( )
( ) ( )
RP E P
EEP
R R RP E P P P E
E E
⎛ ⎞ ⎜ ⎟
⎛ ⎞ ⎝ ⎠⎜ ⎟ ⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Sol. : Let
E1 the event when die A is used
E2 the event when die B is used
R when red face appears
1
3
5
n
RP
E
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
and
2
2
5
n
RP
E
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
,
P(E1) =
1
2, P(E
2) =
1
2
2
1 2
162 5
973 2 1
5 5 2
n
n n
EP
R
⎛ ⎞ ⎜ ⎟⎛ ⎞ ⎝ ⎠ ⎜ ⎟ ⎡ ⎤⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
2 16
973 2
n
n n
n = 4
82. Answer (4)
Hint : 9!
2! 3!n S
7 6
3 3
6!– 5!
2! n E C C
Sol. : 9! 9 8 7 6 5 4 3!
3! 2! 3! 2!
9 8 7 6 5 2 9 8 7 6 10
n S
To find n(E)
Case I.
position for T is denoted by ×
× A × E × N × I × O × N ×
No two T's come together but N may or may not
Total words fomed 3
6! 7! 6 5 4!7
2! 3! 4! 2C
7 6 5 4 3 2 542 300 12600
2
Case. II
× A × E × NN × I × O ×
No two T's come together but two N come together
['×' is the position for T]
In this case total words formed
6
3
6 5 4 3!= × 5! =
3! 3!
C × 5 × 4 × 3 × 2 × 1
= 20 × 120 = 2400
n(E) = 12600 – 2400 = 10200
Required probability
n E
n S
10200
9 8 7 6 10
85 85
18 14 252
83. Answer (3)
Hint :sum of variables
Mean n
n
If n is odd, then median item
th1
2
n ⎛ ⎞ ⎜ ⎟⎝ ⎠
term
Test - 3 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019
15/16
Sol. : Given,
3 3 3 31 2 3 ...
1900n
n
22
1
4·
n n
n
21 ·1900
4
n n
7600 = n(n + 1)2
19 × (20)2 = n(n + 1)2
n = 19
Median of 1, 2, 3 ..., 18, 19 is
th19 1
2
⎛ ⎞⎜ ⎟⎝ ⎠
item = 10
84. Answer (2)
Hint : Mode = 3 Median – 2 Mean
Sol. : We have mode = 3 Median – 2 Mean
Mode – Median = 2(Median – Mean)
(Median – Mean) = 1Mode – Median
2
136 18
2
85. Answer (2)
Hint :
Volume of parallelepiped
· · ·
· · ·
· ·
·
���
��� ��� ��� ��� ��
�� ��� ��� �� ��
��
� � � � � �
a a a b a c
b a b b b c
c a c cc b
Sol. : Volume of parallelepiped
· · ·
· · ·
· ·
·
a a a b a c
V b a b b b c
c a c cc b
���
��� ��� ��� ��� ��
�� ��� ��� �� ��
��
� � � � � �
4 3 4288 cu units
3 9 6
12 2 cu units4 6 16
86. Answer (4)
Hint : Use formula for cross product of three vectors
Sol. : From given relation
2 · – ·a c b a b c
� � � � � �
3b c � �
1
·2
a c � �
and 3
· –2
a b � �
3
cos –2
a b � �
5
cos cos – cos6 6
⎛ ⎞ ⎜ ⎟⎝ ⎠
angle between a�
and b�
is 5
6
87. Answer (2)
Hint : Use formula for 'dot' and cross product of two
vectors
Sol. : Let be the angle between b�
and c�
,
Then,
5b c � �
b c� �
sin = 5
1sin =
2
3
cos =2
Now, given,
3( ) 0 � �
� � �
a b a c ( 3 ) 0 � �
� �
a b c
3 ||�
� �
b c a
3 �
� �
b c a
2 2
9 – 6 ·b c b c� � � �
= 2
2a�
25 + 36 – 6 cosb c � �
= 2·4
2361– 6 5.2· 4
2
⎛ ⎞ ⎜ ⎟⎜ ⎟
⎝ ⎠
2 61– 30 3
4
88. Answer (1)
Hint : Use concept of coplanarity of vectors
Sol. : Since given vectors are coplanar
So, 61�
p + 40�
q + 21�
r = 0�
�
p , �
q , �
r are coplanar.
All India Aakash Test Series for JEE (Main)-2019 Test - 3 (Code-B) (Hints & Solutions)
16/16
� � �
�
q × �
r and �
r × �
p are collinear
(�
q × �
r ) × {(�
r × �
p ) = 0�
89. Answer (3)
Hint : If a�
and
�
b are non-collinear vectors such that
0 � �
xa yb x = 0, y = 0
Sol. : Here, (1 – tan)
�
a – 1 2 sin 0b � �
and a
�
and b�
are non-collinear
1 = tan and 1 2 sin 0
4
or 5
4
and
5
4
or 7
4
Most general value of 5
24
n , n I
90. Answer (2)
Hint : Use condition for coplanarities.
Sol. : Since vectors are coplanar,
So, 0 1 1 0
4
a a c
c c b
4ab = c2
c2 – 4ab = 0 ...(i)
And
Discriminant of given equation is
= c2 – 4ab = 0, [from (i)]
Roots are real and equal