Test - 6 (Code-C) (Answers) All India Aakash Test Series for JEE (Main)-2021
All India Aakash Test Series for JEE (Main)-2021
Test Date : 16/02/2020
ANSWERS
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
TEST-6 - Code-C
1/10
PHYSICS CHEMISTRY MATHEMATICS
1. (3)
2. (1)
3. (2)
4. (3)
5. (2)
6. (3)
7. (4)
8. (4)
9. (4)
10. (3)
11. (2)
12. (1)
13. (4)
14. (2)
15. (2)
16. (1)
17. (3)
18. (2)
19. (4)
20. (2)
21. (10)
22. (66)
23. (25)
24. (30)
25. (25)
26. (2)
27. (1)
28. (4)
29. (3)
30. (3)
31. (1)
32. (2)
33. (2)
34. (1)
35. (4)
36. (1)
37. (2)
38. (3)
39. (1)
40. (2)
41. (4)
42. (2)
43. (3)
44. (4)
45. (2)
46. (01)
47. (05)
48. (28)
49. (10)
50. (06)
51. (4)
52. (2)
53. (2)
54. (3)
55. (2)
56. (3)
57. (2)
58. (4)
59. (2)
60. (4)
61. (3)
62. (3)
63. (4)
64. (4)
65. (1)
66. (2)
67. (2)
68. (4)
69. (3)
70. (2)
71. (06)
72. (03)
73. (02)
74. (04)
75. (00)
All India Aakash Test Series for JEE (Main)-2021 Test - 6 (Code-C) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
2/10
HINTS & SOLUTIONS
PART - A (PHYSICS)
1. Answer (3)
Hint: 8
22
= =
Sol.: red 22 2 s
8
MT
k= = =
cm
4 4 16m/s , V 4 m/s
3 3 3ACV = → = + =
4 8
4 m/s3 3
PxV = − = →
KE in C frame
2 21 16 8
3 62 3 3
= +
1 256 6 64
2 3 3 3
= +
128 64
643
+= =
sinA t =
2 s6 12
t t
= =
2164 8 4 m
2A A= =
2. Answer (1)
Hint: 2
3 2
2
42
3
d xR R xg
dt = −
Sol.: 2
2
6 605
4 4
d x xg
R Rdt
− = = =
3. Answer (2)
Hint: 0 0.1mF
Ak
= =
Sol.: x = 0.1 – 0.1 cos(t – 0.2)
= 0.1[1 – cos5(t – 0.2)]
4. Answer (3)
Hint: 3 31 1 2 2
2
k xk x k x= =
Sol.: 3 3 3 31 23 3
1 22 4 4
k x k xx xx x x
k k
+= + = + +
2 3 3 1 1 23
1 2
( 4 )
4
k k k k k kx x
k k
+ +=
1 23
2 3 3 1 1 2
4
4
k k xx
k k k k k k=
+ +
2
1 2 3
22 3 3 1 1 2
4
( 4 )
k k k xd xM
k k k k k kdt= −
+ +
1 2 3
2 3 3 1 1 2
4
( 4 )
k k k
M k k k k k k =
+ +
5. Answer (2)
Hint: 2
cos3
dx t= −
Sol.: 3
cos2 2 3
t t
= − = +
13
t
=
12M
T tk
= +
2
3
M M
k k
= +
5
3
M
k
=
6. Answer (3)
Hint: y(x, t) = 5 sin(kx – t)
Sol.: y(x, t = 0) = 5 mm sinkx
y(x, t) = y(x – vt, t = 0) = 5 mm sink(x – vt)
= 5 mm sin(kx – t)
5 mm cos( )y
kx tt
= − −
At 3
0, cos2
t kx= =
3 ˆ5 mm 100 250 3 mm/s
2PV j= − = −
Test - 6 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2021
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
3/10
7. Answer (4)
Hint: 2
3 2 3
GMrm d r GMrma
R dt R= − = −
Sol.: 3
3, 2
GM RT
GMR = =
3
4 2
= =
T Rt
GM
8. Answer (4)
Hint:
210
2 20 0
btt
mA A e A e
−
− −
= =
Sol.: 35 10 31
ln10 5 1010
te t−− − = =
t = 200 ln10 s = 461 s
9. Answer (4)
Hint: 3
0cm 2
0
2 2
33
L
L
x kx dxL L
xL
kx dx
= = =
Sol.: 4
2
04
= =L
kLI kx dx x
4
2
3 22 2
4 334 2
2 2
kL LT
gkL gL
= =
10. Answer (3)
Hint: 2sindU
F a ax a xdx
= − = − = −
Sol.: 2 2
2
2
d x a am a x
mdt m= − = =
2
T ma
=
11. Answer (2)
Hint: 0
2( , 0) siny x t a x
= =
Sol.: = 4b
12. Answer (1)
Hint: 202 sin
2
dT rd r
=
Sol.: 2 20T r =
rel 0
Tv r= =
v = 20r
13. Answer (4)
Hint: v xg=
Sol.: 1
22
dv ga xg g
dt x= = =
14. Answer (2)
Hint: 2 2
cos ( )P
y vv A vt x
dt
= = −
Sol.: , max wave
2,Pv Av v v
= =
2
2A A
= =
15. Answer (2)
Hint: 330 330
4000 4250330 330
v
v
+ =
−
Sol.: 16(330 + v) = 17(330 – v)
33v = 330 v = 10 m/s
16. Answer (1)
Hint: 110 10 1 10 0
0
20 10log 2 log logI
I II
= = −
Sol.: 210 10 2 10 0
0
30 10log 3 log logI
I II
= = −
210 2 1
1
1 log 10= =I
I II
17. Answer (3)
Hint: 1 0 2 0,4 4
u uT T = + = −
Sol.:
00
01
02 00
21 214 5 2020[19 ] 19
4 5
u
uu
u
+
= = =
−
05uT
=
All India Aakash Test Series for JEE (Main)-2021 Test - 6 (Code-C) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
4/10
18. Answer (2)
Hint: nf0 = 225, (n + 1)f0 = 375
Sol.: f0 = 150 Hz
n is not an integer
(2n – 1)f0 = 225, (2n + 1)f0 = 375
0 0
225 3751 1
f f+ = −
00
1502 75 Hzf
f= =
n = 2
75 1.1m4 300
v vL
L= = =
19. Answer (4)
Hint: 24
PI
r=
Sol.: 20
02 224 2
pP VPI p
PVr r
= = =
P0 = BkS0
00 2 22
p V VPS
Bk V r
= =
2
1
2
P
V r=
51 20m 1.4 10 m
6000 1.3 340 2 3.14
−= =
20. Answer (2)
Hint: 2 280 18 6400 324+ = +
6724 82 m= =
Sol.: 2 m 4 m2
x
= = =
340
85 Hz4
f = =
21. Answer (10)
Hint: 1 2 20 0
1 1,
2
D D at t
v f v f v= = + −
Sol.: 02 1 2 2
0 0 0
21
2 2
f v aaT t t
f f v f v
− = − = − =
2 2
0 0
0 0
2 101
2 10 5
f v f vf
T f v a f v a = = =
− −
22. Answer (66)
Hint: 100
2 5 cm/s 10 m/sr = =
Sol.: max min
330 3301088
320 340
− = −
f f
= 2
1088 3332 34
= 66 Hz
23. Answer (25)
Hint: 340 3
3 85 Hz 255 Hz4
f
= = =
Sol.: 1 260.1 0.4
2550.8 m
=
260.1 0.4
255 255 0.8 0.8m
=
3260.1 0.410 2.5 g
255 255 0.8 0.8m
= =
10m = 25
24. Answer (30)
Hint: cos 602 4
= =
Sol.: 3 3 300 md
d DD
= = =
30 m10
d=
25. Answer (25)
Hint: 2
50 cm25
= =
Sol.: min 25 cm2
L
= =
PART - B (CHEMISTRY)
26. Answer (2)
Hint : B.P. surface area
Sol. : Boiling point for the isomers same
number of C-atoms follows the order of
normal > iso > neo.
Test - 6 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2021
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
5/10
27. Answer (1)
Hint : In propagation step R is formed
Sol. : R X R X+ ⎯⎯→ − is a termination step
28. Answer (4)
Hint : This is the competition between the
reactivity and stability Cl and Br
Sol. : Br is more stable, hence less reactive
majorly attacks at 3° carbon position
that why S is formed in very less amount.
29. Answer (3)
Hint :
Sol. : Molar mass of CH2O is in lesser but
carbon is in higher oxidation state as
compared to X.
30. Answer (3)
Hint : To show geometrical isomerism C = C
bond must have different groups.
Sol. :
No geometrical isomerism
Can show geometrical isomerism
31. Answer (1)
Hint : -anti elimination of fluoroalkanes give
kinetically controlled product in which
transition state of intermediate is more
stable (i.e. carbanion)
Sol. :
Only two hyperconjugable H’s
32. Answer (2)
Hint : NBS cause substitution of Br at allylic
position
Sol. : P2 is
33. Answer (2)
Hint : Lindlar’s catalyst form cis product
Sol. :
34. Answer (1)
Hint : In presence of Pd and BaSO4, partial
addition takes place
Sol. :
35. Answer (4)
Hint & Sol. :
Benzene do not give test for unsaturation
36. Answer (1)
Hint : General mechanism of ozonolysis is
Sol. :
All India Aakash Test Series for JEE (Main)-2021 Test - 6 (Code-C) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
6/10
37. Answer (2)
Hint : Cl2 in water exist as
HOCl (HO– + Cl+)
Sol. :
38. Answer (3)
Hint : AlCl3 is a Lewis acid
Sol. :
39. Answer (1)
Hint :
Sol. : x is an aromatic compound and Cl is ortho-para directing
40. Answer (2)
Hint :
Sol.
41. Answer (4)
Hint : –OCH3 is ortho-para directing
Sol. :
42. Answer (2)
Hint : Dehydrohalogenation is elimination type
of reaction.
Sol. :
43. Answer (3)
Hint : 4KMnO cause the addition of –OH to
both the double bonded (atoms)
Sol. :
44. Answer (4)
Hint : Free radical substitution reaction takes
place
Sol. :
45. Answer (2)
Hint : HBr/ROOR cause anti-Markovnikoff
addition
Sol. :
Test - 6 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2021
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
7/10
46. Answer (01)
Hint : The hydro carbon with molar weight
72 g/mol and 12 primary hydrogens is
C5H12
Sol. : Only one isomer is possible.
47. Answer (05)
Hint :
⎯⎯⎯⎯→ +alc.KOH
4 7 4 6C H Br C H HBr
Sol. : Possible isomers with molar formula C4H6
48. Answer (28)
Hint : In Kolbe’s electrolysis method, reaction
goes via free radical intermediate.
Sol. :
49. Answer (10)
Hint : Naphthalene has 10 e’s
Sol. :
50. Answer (06)
Hint : The site which is less hindered and more
electron dense will participate most in
EAS.
Sol. :
is electron withdrawing group
and –NH – is electron donating in
nature.
So EAS will occur according to –NH–
group.
PART - C (MATHEMATICS)
51. Answer (4)
Hint: Range of both sides.
Sol.: L.H.S. 5 and R.H.S. 5
equation holds when
2 29sec 4cosec 5y y+ =
2 2 1tan cos(2 )
3 5y y= =
52. Answer (2)
Hint: P(A) + P(B) = P(A B) + P(A B)
Sol.: P(A B) = 0.2, P(A B) = 0.6
P(A) + P(B) = 0.8
( ) ( ) 2 0.8 1.2+ = − =P A P B
53. Answer (2)
Hint: Modulus Property.
Sol.: For maximum value cos2x = 0
max value ( 3 1)= −
54. Answer (3)
Hint: Probability few cases
total case=
Sol.: Few cases = {(1, 1), (1, 2), (2, 1), (1, 4),
(4, 1), (2, 3), (3, 2), (3, 4), (4, 3), (1, 6),
(6, 1), (2, 5), (5, 2), (6, 5), (5, 6)}
15 5
Pr36 12
= =
55. Answer (2)
Hint: Replace r → n – r.
Sol.: Let 1000
1
sin( )cos(1001 )r
S rx r x=
= − …(1)
Replace r → 1001 – r
1000
1
sin(1001 ) cos( )r
S r x rx=
= − …(2)
By (1) and (2)
2 1000 sin(1001 )S x=
500 sin(1001 )S x=
All India Aakash Test Series for JEE (Main)-2021 Test - 6 (Code-C) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
8/10
56. Answer (3)
Hint: Simplify.
Sol.: 2 2 4 2 2
2 2 2
4sin cos 4sin 4sin cos 1
94 4sin 4sin cos
x x x x x
x x x
+ −=
− −
4
4
sin 1 1tan
9 6cos 3
= = =
xx x n
x
57. Answer (2)
Hint: 21 cos 2cos2
xx+ =
21 cos 2sin2
xx− =
Sol.: cos sin
1 cos 1 cos 2 2
1 cos 1 coscos sin
2 2
++ + −
=+ − −
−
x x
x x
x xx x
( )tan cot2 4 4 2
x x = − − = +
58. Answer (4)
Hint: tan tanA A = .
Sol.: tanA tanC = 2 and tanB tanC = 18
Let tanC = x
2 18
tan and tanA Bx x
= =
tanA + tanB + tanC = tanA tanB × tanC
x = 4 = tanC
2
2
1 tan 1 16 15cos2
1 17 171 tan
CC
C
− − −= = =
++
59. Answer (2)
Hint: 10cot 45 10 =
Sol.: 10cot 45 10 =
60. Answer (4)
Hint: cos3 = 4cos3 – 3cos
Sol.: 10 10
3
0 0
1cos 3cos cos( )
3 4 3r r
r rr
= =
= +
10 11cos sin
3 16 6
4 4sin
6
= +
1
8
−=
61. Answer (3)
Hint: 5 1
cos364
+ =
Sol.: ( ) 1 cos 1 sin4 4
F kk k
= + +
1 sin 1 cos4 4k k
− −
2 21 cos 1 sin4 4k k
= − −
21
sin4 2k
=
1 3 5
(5) (1 cos36 )8 32
F−
= − =
62. Answer (3)
Hint: A.M. G.M.
Sol.:
1 1 2 2( )
cos cos6 6
f xa b ab
x x
+
− +
2( ) 0,
31 cos(2 )
4 2
f x xx
+
max
4( )
3 =f x
63. Answer (4)
Hint: Number of relation = 2mn,
Number of function = mn
Sol.:
44
12
3 3Pr
82
= =
Test - 6 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2021
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
9/10
64. Answer (4)
Hint: 4 sin sin sin2 2
A Cr R B=
Sol.: 8 8 4 sin sin sin2 2 2
A B CR r R
= =
1
2sin sin sin2 2 2 16
=A B C
1
sin 1 sin2 2 16
C C − =
1
sin2 4
C=
65. Answer (1)
Hint: 2 2 2
cos2
b c aA
bc
+ −=
Sol.: 2 2 2 2 cos
3AC OA OC AOOC
= + −
2 2d r=
66. Answer (2)
Hint: Mean.
Sol.: 1 2 1
( 2 )2 1 2
nx a a nd a nd
n
+ = + + = + +
Mean deviation from mean is
2
0
1( ) ( )
2 1
n
r
a rd a ndn =
+ − ++
2
0
1
2 1
n
r
r n dn =
= −+
( 1)
2 1
n nd
n
+=
+
67. Answer (2)
Hint: 2var( ) var( )ax a x=
Sol.: 2var( ) var( )ax b a x+ =
2 2
2
2 2var var( )
ax b a ax
c c c
+ = =
Standard deviation 2
2
2
a a
cc= =
68. Answer (4)
Hint: Probability by classical theory.
Sol.: T.C = 25
Set of F.C = {2, 3, 5, 7, 11, 13, 17, 19, 23}
9
25P =
69. Answer (3)
Hint: Probability P and C.
Sol.: T.C = 30C3
F.C = apart from 9 : 25 one number is to
be chosen from 10, 11, …… 24 which are
selected in 15 ways
30
3
15 15 6 3
30 29 28 812
= = =
P
C
70. Answer (2)
Hint: Required probability 1
2=
Sol.: Required probability 12 1
22
−
= =n
n
71. Answer (06)
Hint: A.M H.M
Sol.: a + b + c = 2s
Given expression is a b c
s a s b s c+ +
− − −
3s s s
s a s b s c= − + + +
− − −
A.M H.M
3
3
+ +− − −
− − −+ +
s s s
s a s b s cs a s b s c
s s s
9s s s
s a s b s c+ +
− − −
Minimum value of given expression is 6
All India Aakash Test Series for JEE (Main)-2021 Test - 6 (Code-C) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
10/10
72. Answer (03)
Hint: Distance of circumcenter from sides.
Sol.: Distance of circumcenter from side
AC = R cosB
Distance of orthocenter from side AC =
2R cosA cosC
2cos cos cos( )A C A C= − +
tan tan 3A C =
73. Answer (02)
Hint: Cosine rule in cyclic quadrilateral.
Sol.:
In PQR PR2 = 4 + 25 – 20cos60° = 19
In PRS PR2 = 9 + x2 + 3x = 19
(x + 5) (x – 2) = 0 x = 2
74. Answer (04)
Hint: Cosine rule.
Sol.: 2 2
1 22 2cos( )a b+ = + +
minimum value = 0; maximum value = 4
75. Answer (00)
Hint: Compound angle formulas.
Sol.: sin( )cos( ) sin cos
sin( )cos( ) sin cos
+ − − +=
− + + −
A B C A B C C B
A B C A B C B C
Apply componendo and dividendo
( )sin 2( ) sin( ) sin(2 )sin( ) 0B C B C A B C − + + − =
sin( ) 2cos( )sin( ) sin2 0B C B C B C A − − + + =
sin2 sin2 sin2 0 ( )A B C B C + + =
Test - 6 (Code-D) (Answers) All India Aakash Test Series for JEE (Main)-2021
All India Aakash Test Series for JEE (Main)-2021
Test Date : 16/02/2020
ANSWERS
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
TEST-6 - Code-D
1/10
PHYSICS CHEMISTRY MATHEMATICS
1. (2)
2. (4)
3. (2)
4. (3)
5. (1)
6. (2)
7. (2)
8. (4)
9. (1)
10. (2)
11. (3)
12. (4)
13. (4)
14. (4)
15. (3)
16. (2)
17. (3)
18. (2)
19. (1)
20. (3)
21. (25)
22. (30)
23. (25)
24. (66)
25. (10)
26. (2)
27. (4)
28. (3)
29. (2)
30. (4)
31. (2)
32. (1)
33. (3)
34. (2)
35. (1)
36. (4)
37. (1)
38. (2)
39. (2)
40. (1)
41. (3)
42. (3)
43. (4)
44. (1)
45. (2)
46. (06)
47. (10)
48. (28)
49. (05)
50. (01)
51. (2)
52. (3)
53. (4)
54. (2)
55. (2)
56. (1)
57. (4)
58. (4)
59. (3)
60. (3)
61. (4)
62. (2)
63. (4)
64. (2)
65. (3)
66. (2)
67. (3)
68. (2)
69. (2)
70. (4)
71. (00)
72. (04)
73. (02)
74. (03)
75. (06)
All India Aakash Test Series for JEE (Main)-2021 Test - 6 (Code-D) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
2/10
HINTS & SOLUTIONS
PART - A (PHYSICS)
1. Answer (2)
Hint: 2 280 18 6400 324+ = +
6724 82 m= =
Sol.: 2 m 4 m2
x
= = =
340
85 Hz4
f = =
2. Answer (4)
Hint: 24
PI
r=
Sol.: 20
02 224 2
pP VPI p
PVr r
= = =
P0 = BkS0
00 2 22
p V VPS
Bk V r
= =
2
1
2
P
V r=
51 20m 1.4 10 m
6000 1.3 340 2 3.14
−= =
3. Answer (2)
Hint: nf0 = 225, (n + 1)f0 = 375
Sol.: f0 = 150 Hz
n is not an integer
(2n – 1)f0 = 225, (2n + 1)f0 = 375
0 0
225 3751 1
f f+ = −
00
1502 75 Hzf
f= =
n = 2
75 1.1m4 300
v vL
L= = =
4. Answer (3)
Hint: 1 0 2 0,4 4
u uT T = + = −
Sol.:
00
01
02 00
21 214 5 2020[19 ] 19
4 5
u
uu
u
+
= = =
−
05uT
=
5. Answer (1)
Hint: 110 10 1 10 0
0
20 10log 2 log logI
I II
= = −
Sol.: 210 10 2 10 0
0
30 10log 3 log logI
I II
= = −
210 2 1
1
1 log 10= =I
I II
6. Answer (2)
Hint: 330 330
4000 4250330 330
v
v
+ =
−
Sol.: 16(330 + v) = 17(330 – v)
33v = 330 v = 10 m/s
7. Answer (2)
Hint: 2 2
cos ( )P
y vv A vt x
dt
= = −
Sol.: , max wave
2,Pv Av v v
= =
2
2A A
= =
8. Answer (4)
Hint: v xg=
Sol.: 1
22
dv ga xg g
dt x= = =
9. Answer (1)
Hint: 202 sin
2
dT rd r
=
Sol.: 2 2
0T r =
rel 0
Tv r= =
v = 20r
Test - 6 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2021
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
3/10
10. Answer (2)
Hint: 0
2( , 0) siny x t a x
= =
Sol.: = 4b
11. Answer (3)
Hint: 2sindU
F a ax a xdx
= − = − = −
Sol.: 2 2
2
2
d x a am a x
mdt m= − = =
2
T ma
=
12. Answer (4)
Hint: 3
0cm 2
0
2 2
33
L
L
x kx dxL L
xL
kx dx
= = =
Sol.: 4
2
04
= =L
kLI kx dx x
4
2
3 22 2
4 334 2
2 2
kL LT
gkL gL
= =
13. Answer (4)
Hint:
210
2 20 0
btt
mA A e A e
−
− −
= =
Sol.: 35 10 31
ln10 5 1010
te t−− − = =
t = 200 ln10 s = 461 s
14. Answer (4)
Hint: 2
3 2 3
GMrm d r GMrma
R dt R= − = −
Sol.: 3
3, 2
GM RT
GMR = =
3
4 2
= =
T Rt
GM
15. Answer (3)
Hint: y(x, t) = 5 sin(kx – t)
Sol.: y(x, t = 0) = 5 mm sinkx
y(x, t) = y(x – vt, t = 0) = 5 mm sink(x – vt)
= 5 mm sin(kx – t)
5 mm cos( )y
kx tt
= − −
At 3
0, cos2
t kx= =
3 ˆ5 mm 100 250 3 mm/s
2PV j= − = −
16. Answer (2)
Hint: 2
cos3
dx t= −
Sol.: 3
cos2 2 3
t t
= − = +
13
t
=
12M
T tk
= +
2
3
M M
k k
= +
5
3
M
k
=
17. Answer (3)
Hint: 3 31 1 2 2
2
k xk x k x= =
Sol.: 3 3 3 31 23 3
1 22 4 4
k x k xx xx x x
k k
+= + = + +
2 3 3 1 1 23
1 2
( 4 )
4
k k k k k kx x
k k
+ +=
1 23
2 3 3 1 1 2
4
4
k k xx
k k k k k k=
+ +
2
1 2 3
22 3 3 1 1 2
4
( 4 )
k k k xd xM
k k k k k kdt= −
+ +
1 2 3
2 3 3 1 1 2
4
( 4 )
k k k
M k k k k k k =
+ +
All India Aakash Test Series for JEE (Main)-2021 Test - 6 (Code-D) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
4/10
18. Answer (2)
Hint: 0 0.1mF
Ak
= =
Sol.: x = 0.1 – 0.1 cos(t – 0.2)
= 0.1[1 – cos5(t – 0.2)]
19. Answer (1)
Hint: 2
3 2
2
42
3
d xR R xg
dt = −
Sol.: 2
2
6 605
4 4
d x xg
R Rdt
− = = =
20. Answer (3)
Hint: 8
22
= =
Sol.: red 22 2 s
8
MT
k= = =
cm
4 4 16m/s , V 4 m/s
3 3 3ACV = → = + =
4 8
4 m/s3 3
PxV = − = →
KE in C frame
2 21 16 8
3 62 3 3
= +
1 256 6 64
2 3 3 3
= +
128 64
643
+= =
sinA t =
2 s6 12
t t
= =
2164 8 4 m
2A A= =
21. Answer (25)
Hint: 2
50 cm25
= =
Sol.: min 25 cm2
L
= =
22. Answer (30)
Hint: cos 602 4
= =
Sol.: 3 3 300 md
d DD
= = =
30 m10
d=
23. Answer (25)
Hint: 340 3
3 85 Hz 255 Hz4
f
= = =
Sol.: 1 260.1 0.4
2550.8 m
=
260.1 0.4
255 255 0.8 0.8m
=
3260.1 0.410 2.5 g
255 255 0.8 0.8m
= =
10m = 25
24. Answer (66)
Hint: 100
2 5 cm/s 10 m/sr = =
Sol.: max min
330 3301088
320 340
− = −
f f
= 2
1088 3332 34
= 66 Hz
25. Answer (10)
Hint: 1 2 20 0
1 1,
2
D D at t
v f v f v= = + −
Sol.: 02 1 2 2
0 0 0
21
2 2
f v aaT t t
f f v f v
− = − = − =
2 2
0 0
0 0
2 101
2 10 5
f v f vf
T f v a f v a = = =
− −
Test - 6 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2021
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
5/10
PART - B (CHEMISTRY)
26. Answer (2)
Hint : HBr/ROOR cause anti-Markovnikoff
addition
Sol. :
27. Answer (4)
Hint : Free radical substitution reaction takes
place
Sol. :
28. Answer (3)
Hint : 4KMnO cause the addition of –OH to
both the double bonded (atoms)
Sol. :
29. Answer (2)
Hint : Dehydrohalogenation is elimination type
of reaction.
Sol. :
30. Answer (4)
Hint : –OCH3 is ortho-para directing
Sol. :
31. Answer (2)
Hint :
Sol.
32. Answer (1)
Hint :
Sol. : x is an aromatic compound and Cl is
ortho-para directing
33. Answer (3)
Hint : AlCl3 is a Lewis acid
Sol. :
All India Aakash Test Series for JEE (Main)-2021 Test - 6 (Code-D) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
6/10
34. Answer (2)
Hint : Cl2 in water exist as
HOCl (HO– + Cl+)
Sol. :
35. Answer (1)
Hint : General mechanism of ozonolysis is
Sol. :
36. Answer (4)
Hint & Sol. :
Benzene do not give test for unsaturation
37. Answer (1)
Hint : In presence of Pd and BaSO4, partial
addition takes place
Sol. :
38. Answer (2)
Hint : Lindlar’s catalyst form cis product
Sol. :
39. Answer (2)
Hint : NBS cause substitution of Br at allylic
position
Sol. : P2 is
40. Answer (1)
Hint : -anti elimination of fluoroalkanes give
kinetically controlled product in which
transition state of intermediate is more
stable (i.e. carbanion)
Sol. :
Only two hyperconjugable H’s
41. Answer (3)
Hint : To show geometrical isomerism C = C
bond must have different groups.
Sol. :
No geometrical isomerism
Can show geometrical isomerism
42. Answer (3)
Hint :
Sol. : Molar mass of CH2O is in lesser but
carbon is in higher oxidation state as
compared to X.
Test - 6 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2021
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
7/10
43. Answer (4)
Hint : This is the competition between the
reactivity and stability Cl and Br
Sol. : Br is more stable, hence less reactive
majorly attacks at 3° carbon position
that why S is formed in very less amount.
44. Answer (1)
Hint : In propagation step R is formed
Sol. : R X R X+ ⎯⎯→ − is a termination step
45. Answer (2)
Hint : B.P. surface area
Sol. : Boiling point for the isomers same
number of C-atoms follows the order of
normal > iso > neo.
46. Answer (06)
Hint : The site which is less hindered and more
electron dense will participate most in
EAS.
Sol. :
is electron withdrawing group
and –NH – is electron donating in
nature.
So EAS will occur according to –NH–
group.
47. Answer (10)
Hint : Naphthalene has 10 e’s
Sol. :
48. Answer (28)
Hint : In Kolbe’s electrolysis method, reaction
goes via free radical intermediate.
Sol. :
49. Answer (05)
Hint :
⎯⎯⎯⎯→ +alc.KOH
4 7 4 6C H Br C H HBr
Sol. : Possible isomers with molar formula C4H6
50. Answer (01)
Hint : The hydro carbon with molar weight
72 g/mol and 12 primary hydrogens is
C5H12
Sol. : Only one isomer is possible.
PART - C (MATHEMATICS)
51. Answer (2)
Hint: Required probability 1
2=
Sol.: Required probability 12 1
22
−
= =n
n
52. Answer (3)
Hint: Probability P and C.
Sol.: T.C = 30C3
F.C = apart from 9 : 25 one number is to
be chosen from 10, 11, …… 24 which are
selected in 15 ways
30
3
15 15 6 3
30 29 28 812
= = =
P
C
53. Answer (4)
Hint: Probability by classical theory.
Sol.: T.C = 25
Set of F.C = {2, 3, 5, 7, 11, 13, 17, 19, 23}
9
25P =
All India Aakash Test Series for JEE (Main)-2021 Test - 6 (Code-D) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
8/10
54. Answer (2)
Hint: 2var( ) var( )ax a x=
Sol.: 2var( ) var( )ax b a x+ =
2 2
2
2 2var var( )
ax b a ax
c c c
+ = =
Standard deviation 2
2
2
a a
cc= =
55. Answer (2)
Hint: Mean.
Sol.: 1 2 1
( 2 )2 1 2
nx a a nd a nd
n
+ = + + = + +
Mean deviation from mean is
2
0
1( ) ( )
2 1
n
r
a rd a ndn =
+ − ++
2
0
1
2 1
n
r
r n dn =
= −+
( 1)
2 1
n nd
n
+=
+
56. Answer (1)
Hint: 2 2 2
cos2
b c aA
bc
+ −=
Sol.: 2 2 2 2 cos
3AC OA OC AOOC
= + −
2 2d r=
57. Answer (4)
Hint: 4 sin sin sin2 2
A Cr R B=
Sol.: 8 8 4 sin sin sin2 2 2
A B CR r R
= =
1
2sin sin sin2 2 2 16
=A B C
1
sin 1 sin2 2 16
C C − =
1
sin2 4
C=
58. Answer (4)
Hint: Number of relation = 2mn,
Number of function = mn
Sol.:
44
12
3 3Pr
82
= =
59. Answer (3)
Hint: A.M. G.M.
Sol.:
1 1 2 2( )
cos cos6 6
f xa b ab
x x
+
− +
2( ) 0,
31 cos(2 )
4 2
f x xx
+
max
4( )
3 =f x
60. Answer (3)
Hint: 5 1
cos364
+ =
Sol.: ( ) 1 cos 1 sin4 4
F kk k
= + +
1 sin 1 cos4 4k k
− −
2 21 cos 1 sin4 4k k
= − −
21
sin4 2k
=
1 3 5
(5) (1 cos36 )8 32
F−
= − =
61. Answer (4)
Hint: cos3 = 4cos3 – 3cos
Sol.: 10 10
3
0 0
1cos 3cos cos( )
3 4 3r r
r rr
= =
= +
10 11cos sin
3 16 6
4 4sin
6
= +
1
8
−=
Test - 6 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2021
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
9/10
62. Answer (2)
Hint: 10cot 45 10 =
Sol.: 10cot 45 10 =
63. Answer (4)
Hint: tan tanA A = .
Sol.: tanA tanC = 2 and tanB tanC = 18
Let tanC = x
2 18
tan and tanA Bx x
= =
tanA + tanB + tanC = tanA tanB × tanC
x = 4 = tanC
2
2
1 tan 1 16 15cos2
1 17 171 tan
CC
C
− − −= = =
++
64. Answer (2)
Hint: 21 cos 2cos2
xx+ =
21 cos 2sin2
xx− =
Sol.: cos sin
1 cos 1 cos 2 2
1 cos 1 coscos sin
2 2
++ + −
=+ − −
−
x x
x x
x xx x
( )tan cot2 4 4 2
x x = − − = +
65. Answer (3)
Hint: Simplify.
Sol.: 2 2 4 2 2
2 2 2
4sin cos 4sin 4sin cos 1
94 4sin 4sin cos
x x x x x
x x x
+ −=
− −
4
4
sin 1 1tan
9 6cos 3
= = =
xx x n
x
66. Answer (2)
Hint: Replace r → n – r.
Sol.: Let 1000
1
sin( )cos(1001 )r
S rx r x=
= − …(1)
Replace r → 1001 – r
1000
1
sin(1001 ) cos( )r
S r x rx=
= − …(2)
By (1) and (2)
2 1000 sin(1001 )S x=
500 sin(1001 )S x=
67. Answer (3)
Hint: Probability few cases
total case=
Sol.: Few cases = {(1, 1), (1, 2), (2, 1), (1, 4),
(4, 1), (2, 3), (3, 2), (3, 4), (4, 3), (1, 6),
(6, 1), (2, 5), (5, 2), (6, 5), (5, 6)}
15 5
Pr36 12
= =
68. Answer (2)
Hint: Modulus Property.
Sol.: For maximum value cos2x = 0
max value ( 3 1)= −
69. Answer (2)
Hint: P(A) + P(B) = P(A B) + P(A B)
Sol.: P(A B) = 0.2, P(A B) = 0.6
P(A) + P(B) = 0.8
( ) ( ) 2 0.8 1.2+ = − =P A P B
70. Answer (4)
Hint: Range of both sides.
Sol.: L.H.S. 5 and R.H.S. 5
equation holds when
2 29sec 4cosec 5y y+ =
2 2 1tan cos(2 )
3 5y y= =
All India Aakash Test Series for JEE (Main)-2021 Test - 6 (Code-D) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
10/10
71. Answer (00)
Hint: Compound angle formulas.
Sol.: sin( )cos( ) sin cos
sin( )cos( ) sin cos
+ − − +=
− + + −
A B C A B C C B
A B C A B C B C
Apply componendo and dividendo
( )sin 2( ) sin( ) sin(2 )sin( ) 0B C B C A B C − + + − =
sin( ) 2cos( )sin( ) sin2 0B C B C B C A − − + + =
sin2 sin2 sin2 0 ( )A B C B C + + =
72. Answer (04)
Hint: Cosine rule.
Sol.: 2 2
1 22 2cos( )a b+ = + +
minimum value = 0; maximum value = 4
73. Answer (02)
Hint: Cosine rule in cyclic quadrilateral.
Sol.:
In PQR PR2 = 4 + 25 – 20cos60° = 19
In PRS PR2 = 9 + x2 + 3x = 19
(x + 5) (x – 2) = 0 x = 2
74. Answer (03)
Hint: Distance of circumcenter from sides.
Sol.: Distance of circumcenter from side
AC = R cosB
Distance of orthocenter from side AC =
2R cosA cosC
2cos cos cos( )A C A C= − +
tan tan 3A C =
75. Answer (06)
Hint: A.M H.M
Sol.: a + b + c = 2s
Given expression is a b c
s a s b s c+ +
− − −
3s s s
s a s b s c= − + + +
− − −
A.M H.M
3
3
+ +− − −
− − −+ +
s s s
s a s b s cs a s b s c
s s s
9s s s
s a s b s c+ +
− − −
Minimum value of given expression is 6
Top Related