Test - 4 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2020
All India Aakash Test Series for JEE (Main)-2020
Test Date : 17/11/2019
ANSWERS
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
TEST - 4 - Code-A
1/11
PHYSICS CHEMISTRY MATHEMATICS
1. (4)
2. (1)
3. (2)
4. (2)
5. (1)
6. (4)
7. (2)
8. (1)
9. (3)
10. (1)
11. (3)
12. (1)
13. (2)
14. (3)
15. (4)
16. (1)
17. (3)
18. (1)
19. (2)
20. (4)
21. (16)
22. (10)
23. (14)
24. (07)
25. (88)
26. (3)
27. (3)
28. (1)
29. (3)
30. (1)
31. (2)
32. (3)
33. (2)
34. (3)
35. (2)
36. (3)
37. (3)
38. (1)
39. (2)
40. (3)
41. (4)
42. (1)
43. (2)
44. (2)
45. (1)
46. (02)
47. (09)
48. (02)
49. (12)
50. (20)
51. (4)
52. (2)
53. (1)
54. (2)
55. (4)
56. (4)
57. (2)
58. (1)
59. (2)
60. (3)
61. (1)
62. (4)
63. (2)
64. (4)
65. (1)
66. (3)
67. (1)
68. (4)
69. (4)
70. (2)
71. (24)
72. (01)
73. (00)
74. (22)
75. (04)
All India Aakash Test Series for JEE (Main)-2020 Test - 4 (Code-A) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
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1. Answer (4)
Hint : A = λN
Sol. : 3λ1NA = λ2NB
1 2
3 2 1T T×
=
12 6
TT =
2. Answer (1)
Hint : hv = φ + eV0
Sol. : φ = 5.2 eV
E = 6.2 eV
V0 = 1 V
01
4ne
r=
πε
9 19
29 10 1.6 10 1
2 10n −
−
× × × ×=
×
10 210 2 109 1.6
n−× ×
=×
7 71.38 10 1.4 10= × ≈ ×
3. Answer (2)
Hint : 2
21 602
k emvd
=
Sol. : 2
2 120( ) ke mmvd
=
2120dhmke
λ =
4. Answer (2)
Hint : Fringes formed will be symmetric on screen
Sol. : Number of Fringes are 5d = λ .
5. Answer (1)
Hint : 2
1En
∝
Sol. : 2
1nE
n∝
1 2 21 1
( 1)n nE E kn n−
− = − −
31hc
n=
λ
3nλ ∝
6. Answer (4)
Hint : ( ) ( )Y A B A B= + ⋅ +
Sol. : ( ) ( ) 1Y A B A B= + + + =
7. Answer (2)
Hint : 24 cos2oI I φ
=
Sol. : 3π
φ =
1 6Dyd
λ=
2 6Dyd
λ= −
8. Answer (1)
Hint : Effective phase difference must be 2nπ
Sol. : 2 sec2
t r x λµ − =
2 tan sinx t r= θ
2 2 sin sincos cos 2
t t rr r
µ θ λ− =
22 sincos 2
tr
θ λµ − =
µ
2 2
2 2
2 [ sin ]2sin
tµ µ − θ λ⇒ =
µ µ − θ
2 24 sint λ
⇒ =µ − θ
PART - A (PHYSICS)
Test - 4 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
3/11
9. Answer (3)
Hint : For polarizing angle reflected and refracted rays are mutually perpendicular.
Sol. : sincos
p
p
θ= µ
θ
tan pθ = µ
1sin cθ =µ
tan sin 1p cθ θ =
10. Answer (1)
Hint : 00 00.7 tN N e−λ=
Sol. : 00.7 te−λ=
03(1 )tx e− λ= −
31 (0.7)x = −
1 0.49 0.7x = − ×
65.7%x =
11. Answer (3)
Hint : 2
2hc P
m= φ +
λ
Sol. : 2
21
35 2
hc hc hm
= +λ λ λ
2
1
25 2hc h
m=
λ λ
221
54
hhmc
λλ =
21
54
hmc
λλ =
11 52
hmc
λλ =
12. Answer (1)
Hint : 24PId
=π
Sol. : 24PId
=π
2Pressure4
I PL d c
= =π
13. Answer (2)
Hint : 22 21 2
1 1 1Rzn n
= − λ
Sol. : 22
1 1 1 164 16 3
RzRz
= − ⇒ λ = λ
2
least
1 1116
Rz = − λ
2
least
1 1516
Rz=
λ
least 216
15Rzλ =
14. Answer (3)
Hint : 1/2ln2t =λ
Sol. : 0tN N e−λ=
ln22
0N N eλ
−λ=
0
2N
N =
00
0
2 12Decayed2
NN
N
−−
= =
15. Answer (4)
Hint : duFdr
= −
Sol. : 2 2
3mv ke
r r=
2
2K.E.2ke
r=
2
22keU
r= −
Total energy = zero
All India Aakash Test Series for JEE (Main)-2020 Test - 4 (Code-A) (Hints & Solutions)
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4/11
16. Answer (1)
Hint : hp
λ =
Sol. : 1 2
fh hp = −λ λ
1 2
h h h= −
λ λ λ
1 2
2 1
λ λλ =
λ − λ
17. Answer (3)
Hint : E = (B.E)l – (B.E)R
Sol. : E = E1 – 2E2
18. Answer (1)
Hint : Apply KVL and KCL
Sol. : 210 4 1.2 0I− − =
2 5 mAI =
14 0I− =
1 4 mAI =
2 1 1mA= − =zI I I
19. Answer (2)
Hint and Sol. : max min
max min
13
v vm
v v−
= =+
20. Answer (4)
Hint : Reading = MSR + n × LC
Sol. : 0.5LC 0.005 mm100
= =
Reading 3 40 0.005= + ×
= 3.200 mm
21. Answer (16)
Hint : C
B
II
β =
Sol. : 0C C
b b i
I R VI R V
=
250b
iC
RV
R×
=×
Vi = 16 mV
22. Answer (10)
Hint : It will vary from fc – fm to fc + fm
Sol. : 2 mf f∆ = 10 kHz=
23. Answer (14)
Hint : For intensity to be 1 th4
223
n πφ = π ±
Sol. : 5 cos 53λ λ θ = λ −
145 cos3
λλ θ =
1 14cos15
− θ =
24. Answer (07)
Hint : Path difference will be 3λ corresponding to that point.
Sol. : 4 cos 3λ θ = λ
1 3cos4
− θ =
tan yD
θ =
73
y=
25. Answer (88)
Hint : Analyse the circuit for half of the cycle
Sol. : 2
01 1 42 2 2 11
VPR
= × +
2019
88VR
=
Test - 4 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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26. Answer (3)
Hint : A strong acid can produce weak acid.
Sol. : Acidic strength of carbolic acid (Phenol) is less than H2CO3 so, it not produce CO2.
27. Answer (3)
Hint : Hemiacetal can reduce Tollen’s reagent
Sol. : If A give positive iodoform test so it must contain
group and A does not reduce
Tollen’s reagent it means A not contain
or hemiacetal
28. Answer (1)
Hint :
Sol. :
29. Answer (3)
Hint : Carbylamine test is given by primary amines.
Sol. : Carbylamine test is not given by secondary amines.
30. Answer (1)
Hint : Gabriel phthalimide synthesis Sol. :
31. Answer (2)
Hint : A and D → α
Sol. : CH2OH at C–5 and OH at C–1 decide the α or β nomenclature.
CH2OH and OH same side → β-form
CH2OH and OH opposite side → α-form
32. Answer (3)
Hint : β–Keto acid show decarboxylation on heating.
Sol. :
33. Answer (2)
Hint : B not cleaved with HI
Sol. :
34. Answer (3)
Hint : 4-methylpent-3-en-2-one
Sol. :
35. Answer (2)
Hint : p-chlorophenol is most acidic
Sol. :
36. Answer (3)
Hint : Reaction is intermolecular, it is a Cannizzaro reaction.
PART - B (CHEMISTRY)
All India Aakash Test Series for JEE (Main)-2020 Test - 4 (Code-A) (Hints & Solutions)
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Sol . :
37. Answer (3)
Hint :
Sol. :
38. Answer (1)
Hint :
Sol. :
39. Answer (2)
Hint : B.P of isomeric amine 1°> 2° > 3°
Sol. : A → 1° amine
B → 2° amine
C → 3° amine
Due to less extent of H–Bonding in 3° amine,
B.P. of 3° amine is minimum.
40. Answer (3)
Hint : (iv), (v), (vi) are not Sandmeyer’s reaction
Sol. :
41. Answer (4)
Hint : As positive charge on carbonyl carbon decrease, rate of reaction decrease.
Sol. : sp2 ‘N’ in 3 membered ring and bridge head is not possible.
42. Answer (1)
Hint : When NH2 and OH are at axial position then form compound P.
Sol. :
43. Answer (2)
Hint : HNO3Glucose Saccharic acid→
Sol. : Aspartic acid
Test - 4 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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44. Answer (2)
Hint : Melamine and formaldehyde
Sol. :
45. Answer (1)
Hint : Teflon
Sol. :
46. Answer (02)
Hint : PCC oxidises alcohol to carbonyl group.
Sol. : MnO2 oxidises allylic alcohol.
47. Answer (09)
Hint : CH3MgBr react with acidic hydrogen
Sol. :
48. Answer (02)
Hint : Glyptal is used in manufacture of paints and lacquers.
Sol. : Polyester is a condensation polymer. Two statements are incorrect (ii) & (vii)
49. Answer (12)
Hint : β - D ribose has 4 chiral centre.
Sol. :
xy = 4 × 3 = 12
50. Answer (20)
Hint :
Sol. : Chloramphenicol has two –OH groups, two chiral carbon and zero –COOH.
5(2 + 0 + 2) = 20
51. Answer (4)
Hint : Find the d.r. of normal to the plane
Sol. : 1 2 1
1 2 3 02 1 5
x y z− + −=
−
13( 1) ( 2) 5( 1) 0x y z⇒ − + + − − =
13 5 6x y z⇒ + − =
52. Answer (2)
Hint : ( 2 ) 0a b c+ ⋅ =
Sol. : 3a b=
ˆ ˆ ˆ ˆ ˆ ˆ2 3 3 3 3p i j k i q j k⇒ + + = − +
23,3
p q⇒ = = −
ˆ ˆ ˆ( 2 ) ( 2) (2 2 ) (3 2)a b p i q j k+ = + + − + +
2ˆ ˆ ˆ5 53
i j k= + +
( 2 ) 0a b c+ ⋅ =
PART - C (MATHEMATICS)
All India Aakash Test Series for JEE (Main)-2020 Test - 4 (Code-A) (Hints & Solutions)
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139
r −⇒ =
2 13( , , ) 3, ,3 9− − =
p q r
53. Answer (1)
Hint : Use section formula
Sol. : A = (2, 4, 5), B = (–2, 0, 1)
2 2 4 5, ,1 1 1
P − λ + λ + λ + λ + λ +
33 2 2⇒ λ + = λ +
31⇒ λ =
⇒ 31 : 1
54. Answer (2)
Hint : Find the probability of complementary event
Sol. : Required probability
2 2 3 3 313 3 4 4 4
= − × × × =
55. Answer (4)
Hint : d.r. of line segment is same as that of normal to the plane
Sol. : Equation of plane
( 5)2 ( 0)6 ( 2)10 0x x y− + − + − =
3 5 0x y z⇒ + + =
56. Answer (4)
Hint : Proceed with parametric form of point
Sol. : Point Q on the line = (–5k – 9, k + 2, 2k + 5)
Q lies on plane 5 9 2 2 5 2k k k⇒ − − + + + + =
2k⇒ = −
(1, 0,1)Q∴ =
Distance, 2 2 25 1 2 30PQ = + + =
57. Answer (2)
Hint : variance = 22
i ix xN N
∑ ∑ −
Sol. : 5
120 5 100i
ix
== × =∑
6
1100 100 0i
ix
=⇒ = − =∑
52
21 (20) 165
ii
x= − =∑
5
2
12080i
ix
=⇒ =∑
6
2 2
12080 ( 100) 12080i
ix
=∴ = + − =∑
Variance
26 62
1 16 6
i ii i
x x= =
= −
∑ ∑
12080 60406 3
= =
58. Answer (1)
Hint : Make cases
Sol. : Required probability
5 5 5 5
2 2 2 35 7
2 3
( )C C C CC C
+ +=
×
10 10 10 635 7
+ += =
59. Answer (2)
Hint : | || | cosa b a b⋅ = θ
Sol. : | | , | |a a b b= =
ab(1 + cosθ) = 12
ab(1 – cosθ) = 6
1 cos 21 cos
+ θ⇒ =
− θ
1cos and 93
ab⇒ θ = =
| | | sin |a b ab∴ × = θ
∴ option (2) can be a b×
Test - 4 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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60. Answer (3)
Hint : Use the concept of family of planes
Sol. : Required plane is
( 3) (3 2 5) 0x y z x y z+ + − + λ − + + =
Parallel to xz plane 0(1 3 ) 1(1 2 ) 0(1 ) 0⇒ + λ + − λ + + λ =
12
⇒ λ =
∴ Plane:
1( 3) (3 2 5) 02
x y z x y z+ + − + − + + =
5 3 1 0x z⇒ + − =
61. Answer (1)
Hint : Find the d.r. of line.
Sol. : Any point on line is (–λ + 2, –3λ – 1, 2λ + 3) which also passes through (1, 1, –2)
∴ d.r. of line = (–λ + 1, –3λ – 2, 2λ + 5)
This line is parallel to the given plane.
⇒ (–λ + 1)(–2) + (–3λ – 2)(1) + (2λ + 5)2 = 0
⇒ 3λ + 6 = 0 ⇒ λ = –2
∴ d.r. = (3, 4, 1)
∴ Equation of line: 1 1 23 4 1
x y z− − += =
62. Answer (4)
Hint : Find the d.r. of lines
Sol. : First line is 10 32 1
x zy− −= =
− and second
line is 5 23 7
x y z− −= =
− as
2 ( 3) 1 7 ( 1) 1 0× − + × + − × =
⇒ L1 is perpendicular to L2
63. Answer (2) Hint : Break 5 into 3 non-zero parts
Sol. : 5 3 5 31 2( ) 3 (2 2)n E C C= − − −
5( ) 3n S =
( ) 50( ) 81
n EPn S
= =
64. Answer (4)
Hint : Use total probability theorem
Sol. : Required probability
( ) ( ) = ⋅ + ⋅ W WP W P P B PW B
3 4 2 3 18 35 6 5 6 30 5
= × + × = =
65. Answer (1)
Hint : Find the angle between ( )OA OB×
and
( )AB AC×
.
Sol. : 1n OA OB= ×
ˆ ˆ ˆ
ˆ ˆ ˆ2 1 1 3 51 3 2
i j ki j k= = − − +
2n AB AC= ×
ˆ ˆ ˆˆ ˆ ˆ1 2 1 2 2 6
3 0 1
i j ki j k= − = − +
−
1 2
1 2cos
| || |n nn n
⋅θ =
2 6 30
1 9 25 4 4 36− + +
=+ + + +
34 17
35 44 385= =
66. Answer (3)
Hint : Use triangle rule of vector addition
Sol. :
All India Aakash Test Series for JEE (Main)-2020 Test - 4 (Code-A) (Hints & Solutions)
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AB BM AM+ =
…(1)
AC CM AM+ =
…(2)
(1) + (2)
⇒ 2AB BC AM+ =
67. Answer (1)
Hint : , ,a b c
do not make a triangle.
Sol. : 1 | | 7a b c≤ + + ≤
68. Answer (4)
Hint : P1 + λP2 = 0
Sol. : Intersection point of
1 2 31 2 3
x y z− − −= = and xy plane
is P(λ + 1, 2λ + 2, 3λ + 3) where 3λ + 3 = 0
⇒ P(0, 0, 0)
Plane passing through intersection of
P1 = 0 and P2 = 0 is
P1 + tP2 = 0
⇒ (1 + t)x + (1 – t)y + (1 + t)z + t – 1 = 0
Passes through P(0, 0, 0) ⇒ 1t =
⇒ required plane is 2x + 2z = 0
0x z⇒ + =
69. Answer (4)
Hint : 1 2n b b= ×
Sol. : ˆ ˆ ˆ
ˆ ˆ1 4 5 40 81 4 5
i j kn i k= = −
−
70. Answer (2)
Hint : Use Bayes theorem
Sol. : Let Bi be the number of black balls transferred (i = 0, 1, 2, 3). B is the event of drawing a black ball. Therefore,
0 15 30( ) , ( ) ,
70 70P B P B= =
2 330 5( ) , ( ) ,70 70
P B P B= =
Also 0 1
10, ,4
B BP PB B
= =
2 3
2 3, ,4 4
B BP PB B
= =
∴ By Bayes theorem
3
5 370 4
5 30 1 30 2 5 3070 70 4 70 4 70 4
BP
B
× =
× + × + × + ×
17
=
71. Answer (24)
Hint : Find P(x = 1) + P(x = 2)
Sol. : ( 2) ( 1) ( 2)P X P X P X≤ = = + =
12 40 12
1 1 252 52
2 2
C C CC C×
= +
6 91 751 26 17
×= =
×
72. Answer (01)
Hint : 22
2 i ix xN N
∑ ∑ σ = −
Sol. : 2(2 1) 21ix n+ =∑
4 4 29A n B n⇒ + + =
7A B n⇒ + =
Where 2
1 1,
n n
i ii i
A x B x= =
= =∑ ∑
2(3 1) 34ix n− =∑
9 6 34A n B n⇒ + − =
3 2 11A B n⇒ − =
5 , 2A n B n∴ = =
2
2 21 1i ix x
n n σ = − ∑ ∑
2
2 5 4 1A Bn n
= − = − =
Test - 4 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
11/11
73. Answer (00)
Hint : 0 (acute), 0 (obtuse)a b a b⋅ > ⋅ <
Sol. : x2 – x – x > 0 …(1)
–x < 0 …(2)
2
cos1 1
x
x
−θ =
+ + …(3)
( 2) 0, 0 2x x x x− > > ⇒ > …(a)
52 6π π
< θ <
50 cos cos6π
> θ >
2
3 02 2
x
x
− −⇒ < <
+
2
32 2
x
x
− −⇒ <
+
2
3 6 62 2
x xx
⇒ > ⇒ − < <+
…(b)
and2
0 {0}2
x x Rx
−< ⇒ ∈ −
+ …(c)
From (a), (b) and (c), 2 6x< <
74. Answer (22)
Hint : Make cases
Sol. : 5 6 5 3 6 3
1 1 1 1 1 114
2( ) C C C C C C
P EC
⋅ + ⋅ + ⋅=
(30 15 18) 63 913 14 13 7 13
2
+ += = =
⋅ ⋅
75. Answer (04)
Hint : 22 1 ( )ix xn
σ = −∑
Sol. : 1 21
... nx x xxn
+ + +=
2 2 2
2 1 1 1 2 1( ) ( ) ... ( )nx x x x x xn
− + − + + −σ =
1 22 1
... 4ny y yx xn
+ + += =
2 2 2
2 2 1 2 2 2( ) ( ) ... ( )( ) − + − + + −′σ = nx y x y x yn
2 2 2
1 1 1 2 1( ) ( ) ... ( )16 nx x x x x xn
− + − + + −=
2 216 4′ ′σ = σ ⇒ σ = σ
Test - 4 (Code-B) (Answers) All India Aakash Test Series for JEE (Main)-2020
All India Aakash Test Series for JEE (Main)-2020
Test Date : 17/11/2019
ANSWERS
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
TEST - 4 - Code-B
1/11
PHYSICS CHEMISTRY MATHEMATICS
1. (4)
2. (2)
3. (1)
4. (3)
5. (1)
6. (4)
7. (3)
8. (2)
9. (1)
10. (3)
11. (1)
12. (3)
13. (1)
14. (2)
15. (4)
16. (1)
17. (2)
18. (2)
19. (1)
20. (4)
21. (88)
22. (07)
23. (14)
24. (10)
25. (16)
26. (1)
27. (2)
28. (2)
29. (1)
30. (4)
31. (3)
32. (2)
33. (1)
34. (3)
35. (3)
36. (2)
37. (3)
38. (2)
39. (3)
40. (2)
41. (1)
42. (3)
43. (1)
44. (3)
45. (3)
46. (20)
47. (12)
48. (02)
49. (09)
50. (02)
51. (2)
52. (4)
53. (4)
54. (1)
55. (3)
56. (1)
57. (4)
58. (2)
59. (4)
60. (1)
61. (3)
62. (2)
63. (1)
64. (2)
65. (4)
66. (4)
67. (2)
68. (1)
69. (2)
70. (4)
71. (04)
72. (22)
73. (00)
74. (01)
75. (24)
All India Aakash Test Series for JEE (Main)-2020 Test - 4 (Code-B) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
2/11
1. Answer (4)
Hint : Reading = MSR + n × LC
Sol. : 0.5LC 0.005 mm100
= =
Reading 3 40 0.005= + ×
= 3.200 mm
2. Answer (2)
Hint and Sol. : max min
max min
13
v vm
v v−
= =+
3. Answer (1)
Hint : Apply KVL and KCL
Sol. : 210 4 1.2 0I− − =
2 5 mAI =
14 0I− =
1 4 mAI =
2 1 1mA= − =zI I I
4. Answer (3)
Hint : E = (B.E)l – (B.E)R
Sol. : E = E1 – 2E2
5. Answer (1)
Hint : hp
λ =
Sol. : 1 2
fh hp = −λ λ
1 2
h h h= −
λ λ λ
1 2
2 1
λ λλ =
λ − λ
6. Answer (4)
Hint : duFdr
= −
Sol. : 2 2
3mv ke
r r=
2
2K.E.2ke
r=
2
22keU
r= −
Total energy = zero
7. Answer (3)
Hint : 1/2ln2t =λ
Sol. : 0tN N e−λ=
ln22
0N N eλ
−λ=
0
2N
N =
00
0
2 12Decayed2
NN
N
−−
= =
8. Answer (2)
Hint : 22 21 2
1 1 1Rzn n
= − λ
Sol. : 22
1 1 1 164 16 3
RzRz
= − ⇒ λ = λ
2
least
1 1116
Rz = − λ
2
least
1 1516
Rz=
λ
least 216
15Rzλ =
9. Answer (1)
Hint : 24PId
=π
Sol. : 24PId
=π
2Pressure4
I PL d c
= =π
PART - A (PHYSICS)
Test - 4 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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10. Answer (3)
Hint : 2
2hc P
m= φ +
λ
Sol. : 2
21
35 2
hc hc hm
= +λ λ λ
2
1
25 2hc h
m=
λ λ
221
54
hhmc
λλ =
21
54
hmc
λλ =
11 52
hmc
λλ =
11. Answer (1)
Hint : 00 00.7 tN N e−λ=
Sol. : 00.7 te−λ=
03(1 )tx e− λ= −
31 (0.7)x = −
1 0.49 0.7x = − ×
65.7%x =
12. Answer (3)
Hint : For polarizing angle reflected and refracted rays are mutually perpendicular.
Sol. : sincos
p
p
θ= µ
θ
tan pθ = µ
1sin cθ =µ
tan sin 1p cθ θ =
13. Answer (1)
Hint : Effective phase difference must be 2nπ
Sol. : 2 sec2
t r x λµ − =
2 tan sinx t r= θ
2 2 sin sincos cos 2
t t rr r
µ θ λ− =
22 sincos 2
tr
θ λµ − =
µ
2 2
2 2
2 [ sin ]2sin
tµ µ − θ λ⇒ =
µ µ − θ
2 24 sint λ
⇒ =µ − θ
14. Answer (2)
Hint : 24 cos2oI I φ
=
Sol. : 3π
φ =
1 6Dyd
λ=
2 6Dyd
λ= −
15. Answer (4)
Hint : ( ) ( )Y A B A B= + ⋅ +
Sol. : ( ) ( ) 1Y A B A B= + + + =
16. Answer (1)
Hint : 2
1En
∝
Sol. : 2
1nE
n∝
1 2 21 1
( 1)n nE E kn n−
− = − −
31hc
n=
λ
3nλ ∝
All India Aakash Test Series for JEE (Main)-2020 Test - 4 (Code-B) (Hints & Solutions)
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17. Answer (2)
Hint : Fringes formed will be symmetric on screen
Sol. : Number of Fringes are 5d = λ .
18. Answer (2)
Hint : 2
21 602
k emvd
=
Sol. : 2
2 120( ) ke mmvd
=
2120dhmke
λ =
19. Answer (1)
Hint : hv = φ + eV0
Sol. : φ = 5.2 eV
E = 6.2 eV
V0 = 1 V
01
4ne
r=
πε
9 19
29 10 1.6 10 1
2 10n −
−
× × × ×=
×
10 210 2 109 1.6
n−× ×
=×
7 71.38 10 1.4 10= × ≈ ×
20. Answer (4)
Hint : A = λN
Sol. : 3λ1NA = λ2NB
1 2
3 2 1T T×
=
12 6
TT =
21. Answer (88)
Hint : Analyse the circuit for half of the cycle
Sol. : 2
01 1 42 2 2 11
VPR
= × +
2019
88VR
=
22. Answer (07)
Hint : Path difference will be 3λ corresponding to that point.
Sol. : 4 cos 3λ θ = λ
1 3cos4
− θ =
tan yD
θ =
73
y=
23. Answer (14)
Hint : For intensity to be 1 th4
223
n πφ = π ±
Sol. : 5 cos 53λ λ θ = λ −
145 cos3
λλ θ =
1 14cos15
− θ =
24. Answer (10)
Hint : It will vary from fc – fm to fc + fm
Sol. : 2 mf f∆ = 10 kHz=
25. Answer (16)
Hint : C
B
II
β =
Sol. : 0C C
b b i
I R VI R V
=
250b
iC
RV
R×
=×
Vi = 16 mV
Test - 4 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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26. Answer (1)
Hint : Teflon
Sol. :
27. Answer (2)
Hint : Melamine and formaldehyde
Sol. :
28. Answer (2)
Hint : HNO3Glucose Saccharic acid→
Sol. : Aspartic acid
29. Answer (1)
Hint : When NH2 and OH are at axial position then form compound P.
Sol. :
30. Answer (4)
Hint : As positive charge on carbonyl carbon decrease, rate of reaction decrease.
Sol. : sp2 ‘N’ in 3 membered ring and bridge head is not possible.
31. Answer (3)
Hint : (iv), (v), (vi) are not Sandmeyer’s reaction
Sol. :
32. Answer (2)
Hint : B.P of isomeric amine 1°> 2° > 3°
Sol. : A → 1° amine
B → 2° amine
C → 3° amine
Due to less extent of H–Bonding in 3° amine,
B.P. of 3° amine is minimum.
33. Answer (1)
Hint :
Sol. :
PART - B (CHEMISTRY)
All India Aakash Test Series for JEE (Main)-2020 Test - 4 (Code-B) (Hints & Solutions)
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34. Answer (3)
Hint :
Sol. :
35. Answer (3)
Hint : Reaction is intermolecular, it is a Cannizzaro reaction.
Sol . :
36. Answer (2)
Hint : p-chlorophenol is most acidic
Sol. :
37. Answer (3)
Hint : 4-methylpent-3-en-2-one
Sol. :
38. Answer (2)
Hint : B not cleaved with HI
Sol. :
39. Answer (3)
Hint : β–Keto acid show decarboxylation on heating.
Sol. :
40. Answer (2)
Hint : A and D → α
Sol. : CH2OH at C–5 and OH at C–1 decide the α or β nomenclature.
CH2OH and OH same side → β-form
CH2OH and OH opposite side → α-form
41. Answer (1)
Hint : Gabriel phthalimide synthesis
Sol. :
42. Answer (3)
Hint : Carbylamine test is given by primary amines.
Sol. : Carbylamine test is not given by secondary amines.
43. Answer (1)
Hint :
Test - 4 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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Sol. :
44. Answer (3)
Hint : Hemiacetal can reduce Tollen’s reagent
Sol. : If A give positive iodoform test so it must contain
group and A does not reduce
Tollen’s reagent it means A not contain
or hemiacetal
45. Answer (3)
Hint : A strong acid can produce weak acid.
Sol. : Acidic strength of carbolic acid (Phenol) is less than H2CO3 so, it not produce CO2.
46. Answer (20)
Hint :
Sol. : Chloramphenicol has two –OH groups, two chiral carbon and zero –COOH.
5(2 + 0 + 2) = 20
47. Answer (12)
Hint : β - D ribose has 4 chiral centre. Sol. :
xy = 4 × 3 = 12
48. Answer (02) Hint : Glyptal is used in manufacture of paints and lacquers. Sol. : Polyester is a condensation polymer. Two statements are incorrect (ii) & (vii)
49. Answer (09)
Hint : CH3MgBr react with acidic hydrogen
Sol. :
50. Answer (02)
Hint : PCC oxidises alcohol to carbonyl group.
Sol. : MnO2 oxidises allylic alcohol.
51. Answer (2)
Hint : Use Bayes theorem
Sol. : Let Bi be the number of black balls transferred (i = 0, 1, 2, 3). B is the event of drawing a black ball. Therefore,
0 15 30( ) , ( ) ,
70 70P B P B= =
2 330 5( ) , ( ) ,70 70
P B P B= =
Also 0 1
10, ,4
B BP PB B
= =
2 3
2 3, ,4 4
B BP PB B
= =
∴ By Bayes theorem
PART - C (MATHEMATICS)
All India Aakash Test Series for JEE (Main)-2020 Test - 4 (Code-B) (Hints & Solutions)
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3
5 370 4
5 30 1 30 2 5 3070 70 4 70 4 70 4
BP
B
× =
× + × + × + ×
17
=
52. Answer (4)
Hint : 1 2n b b= ×
Sol. : ˆ ˆ ˆ
ˆ ˆ1 4 5 40 81 4 5
i j kn i k= = −
−
53. Answer (4)
Hint : P1 + λP2 = 0
Sol. : Intersection point of
1 2 31 2 3
x y z− − −= = and xy plane
is P(λ + 1, 2λ + 2, 3λ + 3) where 3λ + 3 = 0
⇒ P(0, 0, 0)
Plane passing through intersection of
P1 = 0 and P2 = 0 is
P1 + tP2 = 0
⇒ (1 + t)x + (1 – t)y + (1 + t)z + t – 1 = 0
Passes through P(0, 0, 0) ⇒ 1t =
⇒ required plane is 2x + 2z = 0
0x z⇒ + =
54. Answer (1)
Hint : , ,a b c
do not make a triangle.
Sol. : 1 | | 7a b c≤ + + ≤
55. Answer (3)
Hint : Use triangle rule of vector addition
Sol. :
AB BM AM+ =
…(1)
AC CM AM+ =
…(2)
(1) + (2)
⇒ 2AB BC AM+ =
56. Answer (1)
Hint : Find the angle between ( )OA OB×
and
( )AB AC×
.
Sol. : 1n OA OB= ×
ˆ ˆ ˆ
ˆ ˆ ˆ2 1 1 3 51 3 2
i j ki j k= = − − +
2n AB AC= ×
ˆ ˆ ˆ
ˆ ˆ ˆ1 2 1 2 2 63 0 1
i j ki j k= − = − +
−
1 2
1 2cos
| || |n nn n
⋅θ =
2 6 301 9 25 4 4 36
− + +=
+ + + +
34 1735 44 385
= =
57. Answer (4)
Hint : Use total probability theorem
Sol. : Required probability
( ) ( ) = ⋅ + ⋅ W WP W P P B PW B
3 4 2 3 18 35 6 5 6 30 5
= × + × = =
Test - 4 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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58. Answer (2) Hint : Break 5 into 3 non-zero parts
Sol. : 5 3 5 31 2( ) 3 (2 2)n E C C= − − −
5( ) 3n S =
( ) 50( ) 81
n EPn S
= =
59. Answer (4)
Hint : Find the d.r. of lines
Sol. : First line is 10 32 1
x zy− −= =
− and second
line is 5 23 7
x y z− −= =
− as
2 ( 3) 1 7 ( 1) 1 0× − + × + − × =
⇒ L1 is perpendicular to L2
60. Answer (1)
Hint : Find the d.r. of line.
Sol. : Any point on line is (–λ + 2, –3λ – 1, 2λ + 3) which also passes through (1, 1, –2)
∴ d.r. of line = (–λ + 1, –3λ – 2, 2λ + 5)
This line is parallel to the given plane.
⇒ (–λ + 1)(–2) + (–3λ – 2)(1) + (2λ + 5)2 = 0
⇒ 3λ + 6 = 0 ⇒ λ = –2
∴ d.r. = (3, 4, 1)
∴ Equation of line: 1 1 23 4 1
x y z− − += =
61. Answer (3)
Hint : Use the concept of family of planes
Sol. : Required plane is
( 3) (3 2 5) 0x y z x y z+ + − + λ − + + =
Parallel to xz plane 0(1 3 ) 1(1 2 ) 0(1 ) 0⇒ + λ + − λ + + λ =
12
⇒ λ =
∴ Plane:
1( 3) (3 2 5) 02
x y z x y z+ + − + − + + =
5 3 1 0x z⇒ + − =
62. Answer (2)
Hint : | || | cosa b a b⋅ = θ
Sol. : | | , | |a a b b= =
ab(1 + cosθ) = 12
ab(1 – cosθ) = 6
1 cos 21 cos
+ θ⇒ =
− θ
1cos and 93
ab⇒ θ = =
| | | sin |a b ab∴ × = θ
∴ option (2) can be a b×
63. Answer (1)
Hint : Make cases
Sol. : Required probability
5 5 5 5
2 2 2 35 7
2 3
( )C C C CC C
+ +=
×
10 10 10 635 7
+ += =
64. Answer (2)
Hint : variance = 22
i ix xN N
∑ ∑ −
Sol. : 5
120 5 100i
ix
== × =∑
6
1100 100 0i
ix
=⇒ = − =∑
52
21 (20) 165
ii
x= − =∑
5
2
12080i
ix
=⇒ =∑
6
2 2
12080 ( 100) 12080i
ix
=∴ = + − =∑
All India Aakash Test Series for JEE (Main)-2020 Test - 4 (Code-B) (Hints & Solutions)
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Variance
26 62
1 16 6
i ii i
x x= =
= −
∑ ∑
12080 60406 3
= =
65. Answer (4)
Hint : Proceed with parametric form of point
Sol. : Point Q on the line = (–5k – 9, k + 2, 2k + 5)
Q lies on plane 5 9 2 2 5 2k k k⇒ − − + + + + =
2k⇒ = −
(1, 0,1)Q∴ =
Distance, 2 2 25 1 2 30PQ = + + =
66. Answer (4)
Hint : d.r. of line segment is same as that of normal to the plane
Sol. : Equation of plane
( 5)2 ( 0)6 ( 2)10 0x x y− + − + − =
3 5 0x y z⇒ + + =
67. Answer (2)
Hint : Find the probability of complementary event
Sol. : Required probability
2 2 3 3 313 3 4 4 4
= − × × × =
68. Answer (1)
Hint : Use section formula
Sol. : A = (2, 4, 5), B = (–2, 0, 1)
2 2 4 5, ,1 1 1
P − λ + λ + λ + λ + λ +
33 2 2⇒ λ + = λ +
31⇒ λ =
⇒ 31 : 1
69. Answer (2)
Hint : ( 2 ) 0a b c+ ⋅ =
Sol. : 3a b=
ˆ ˆ ˆ ˆ ˆ ˆ2 3 3 3 3p i j k i q j k⇒ + + = − +
23,3
p q⇒ = = −
ˆ ˆ ˆ( 2 ) ( 2) (2 2 ) (3 2)a b p i q j k+ = + + − + +
2ˆ ˆ ˆ5 53
i j k= + +
( 2 ) 0a b c+ ⋅ =
139
r −⇒ =
2 13( , , ) 3, ,3 9
p q r − − =
70. Answer (4)
Hint : Find the d.r. of normal to the plane
Sol. : 1 2 1
1 2 3 02 1 5
x y z− + −=
−
13( 1) ( 2) 5( 1) 0x y z⇒ − + + − − =
13 5 6x y z⇒ + − =
71. Answer (04)
Hint : 22 1 ( )ix xn
σ = −∑
Sol. : 1 21
... nx x xxn
+ + +=
2 2 2
2 1 1 1 2 1( ) ( ) ... ( )nx x x x x xn
− + − + + −σ =
1 22 1
... 4ny y yx xn
+ + += =
2 2 2
2 2 1 2 2 2( ) ( ) ... ( )( ) − + − + + −′σ = nx y x y x yn
2 2 2
1 1 1 2 1( ) ( ) ... ( )16 nx x x x x xn
− + − + + −=
2 216 4′ ′σ = σ ⇒ σ = σ
Test - 4 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
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72. Answer (22)
Hint : Make cases
Sol. : 5 6 5 3 6 3
1 1 1 1 1 114
2( ) C C C C C C
P EC
⋅ + ⋅ + ⋅=
(30 15 18) 63 913 14 13 7 13
2
+ += = =
⋅ ⋅
73. Answer (00)
Hint : 0 (acute), 0 (obtuse)a b a b⋅ > ⋅ <
Sol. : x2 – x – x > 0 …(1)
–x < 0 …(2)
2
cos1 1
x
x
−θ =
+ + …(3)
( 2) 0, 0 2x x x x− > > ⇒ > …(a)
52 6π π
< θ <
50 cos cos6π
> θ >
2
3 02 2
x
x
− −⇒ < <
+
2
32 2
x
x
− −⇒ <
+
2
3 6 62 2
x xx
⇒ > ⇒ − < <+
…(b)
and2
0 {0}2
x x Rx
−< ⇒ ∈ −
+ …(c)
From (a), (b) and (c), 2 6x< <
74. Answer (01)
Hint : 22
2 i ix xN N
∑ ∑ σ = −
Sol. : 2(2 1) 21ix n+ =∑
4 4 29A n B n⇒ + + =
7A B n⇒ + =
Where 2
1 1,
n n
i ii i
A x B x= =
= =∑ ∑
2(3 1) 34ix n− =∑
9 6 34A n B n⇒ + − =
3 2 11A B n⇒ − =
5 , 2A n B n∴ = =
2
2 21 1i ix x
n n σ = − ∑ ∑
2
2 5 4 1A Bn n
= − = − =
75. Answer (24)
Hint : Find P(x = 1) + P(x = 2)
Sol. : ( 2) ( 1) ( 2)P X P X P X≤ = = + =
12 40 12
1 1 252 52
2 2
C C CC C×
= +
6 91 751 26 17
×= =
×