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Test - 5 (Code-C) (Answers) All India Aakash Test Series for JEE (Main)-2021
All India Aakash Test Series for JEE (Main)-2021
Test Date : 02/02/2020
ANSWERS
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
TEST-5 - Code-C
1/10
PHYSICS CHEMISTRY MATHEMATICS
1. (3)
2. (1)
3. (2)
4. (3)
5. (3)
6. (4)
7. (2)
8. (1)
9. (2)
10. (1)
11. (2)
12. (4)
13. (1)
14. (4)
15. (4)
16. (3)
17. (3)
18. (1)
19. (4)
20. (2)
21. (24)
22. (04)
23. (25)
24. (31)
25. (07)
26. (2)
27. (1)
28. (2)
29. (3)
30. (3)
31. (4)
32. (3)
33. (2)
34. (2)
35. (1)
36. (3)
37. (2)
38. (3)
39. (3)
40. (1)
41. (2)
42. (4)
43. (1)
44. (3)
45. (1)
46. (02)
47. (06)
48. (06)
49. (60)
50. (56)
51. (3)
52. (3)
53. (3)
54. (3)
55. (2)
56. (4)
57. (4)
58. (1)
59. (2)
60. (1)
61. (3)
62. (3)
63. (4)
64. (2)
65. (1)
66. (3)
67. (4)
68. (4)
69. (3)
70. (3)
71. (65)
72. (09)
73. (61)
74. (10)
75. (36)
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All India Aakash Test Series for JEE (Main)-2021 Test - 5 (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: Let R = thermal resistance of each rod
of y
then 2R = thermal resistance of each rod of x.
Sol.:
1 16A BT T T RI = − =
2 14B ET T T RI = − =
1 12
3 3100 60 C
2 5
TT
T
= = =
2 40 = T C
60 60 20 C= − = − = B A B B CT T T T T
TC = 40°C
2. Answer (1)
Hint: 140 z = 100°C
Sol.: 14 7
1 C10 5
z = =
1 1 14200
4200 J kg J kg7
5
ws k
z
− − −= =
= 3000 J kg–1 z–1
3. Answer (2)
Hint: 2
12
mlI =
Sol.: 2
2I l
TI l
= =
4. Answer (3)
Hint: V = VT
Sol.: 1 dV
V dT =
3nRT
PV nRT KV
= =
ln( ) 3 ln ln ln+ = +nR T K V
3 1 dV
T V dT= =
5. Answer (3)
Hint: W = Area under P-V graph
Sol.: In general W may be positive or
negative.
6. Answer (4)
Hint: dT dt
H KA aATdx dx
= − = −
Sol.: 0
0
2 20 0
4 0
16
2
T l
T
T THdx HlTdT
aA aA
− −= − =
2
015
2
TaAH
l=
2 2 2
0 016 15
2 2 2
T T TaA l
l aA
− −=
2
2 2 00
1516
2
TT T= −
2
017
2
T=
017
2T T=
7. Answer (2)
Hint: PV = RT
Sol.: 22 2
Constant ConstantP
PVP V
= =
3
2 2 1 2
R R RC = − =
−
02
RQ C T T = =
8. Answer (1)
Hint: dQ = nCdT
Sol.: dQ = nCvdT + PdV
PdV = nTdT
nRT
dV n TdTV
=
-
Test - 5 (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
dV dT
V R
=
ln lnT
k VR
+ =
=
T
RkV e
Constant
−
=
T
RVe
9. Answer (2)
Hint: 4 4[ ]sP A T T= −
Sol.: P = 5 × 10–4 × 6.0 × 10–8 [6004 – 3004]
= 3 × 10–11 × 108 [1296 – 81] = 3 × 10–3 ×
1215 W
= 3.645 W
10. Answer (1)
Hint: dQ = nCdT
Sol.: nCdT = nCvdT + nVdT
PdV = nVdT
nRT
PV nRT PV
= =
dV
nRT n VdTV
=
2
=
dT RdV
T V
1
ln( )
= −
kT RV
R
VkT e−
=
constant =
R
VTe
11. Answer (2)
Hint: U = nCvT
Sol.: 15
2i
RU n T=
1 2 25 3
( ) 2 2 22 2
= − + fR R
U n n T n T
1 2 2( )5 6n n RT n RT= − +
1 12 2
5 5
2 2
= − = + = +
f i
n RT nU U U n RT n RT
12. Answer (4)
Hint: The decrease in the KE due to ordered
motion is equal to increase in internal
energy.
Sol.: 2
2 21 1
2 2 4
= − =
VU nm V nmV
1
v
RnC T n T= =
−
2( 1)
4
− =
mVT
R
13. Answer (1)
Hint: p = nRT
Sol.: 1 1 1 2
2 2 1 2
p n m M
p n M m= =
2 1 2
1 2 1
4 2
3 3
M M
M M
= =
1
2
1
2
M
M=
14. Answer (4)
Hint: TV = constant PV2 = constant
Sol.: 22 2
k kP
v m = =
15. Answer (4)
Hint: SlopeP
V
= −
Sol.:
12 22 1
2
f
f
f f f
+ + = = = +
16. Answer (3)
Hint: dp = –gdh
Sol.: nRT mRT RT
PV MV M
= = =
RdT dT Mg
gdhM dh R
−= − =
dT Mg
dh R=
-
All India Aakash Test Series for JEE (Main)-2021 Test - 5 (Code-C) (Hints & Solutions)
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17. Answer (3)
Hint: A = A0[1 + T]
Sol.: 3 5[1 ] 1 12 2
= + + = +
A lb T T lb T
5
2 =
18. Answer (1)
Hint: =
FLY
A L
Sol.: L F F
T Y TL A T AY
= = =
19. Answer (4)
Hint: extracted
Wn
Q=
Sol.: 0 02W PV=
0 0 0 033
2 2AB
P V P Vn RQ
nR= =
0 0 0 045
102
BC
P Vn RQ P V
nR= =
0 0
0 0
2 42 100%
23 23
P Vn
P V= =
20. Answer (2)
Hint: 2Q
W =
Sol.: 2 2
1 2 1 2
279 93
24 8
Q T
Q Q T T = = = =
− −
= 11.6
21. Answer (24)
Hint: 21
0.6 1002
m
Sol.: 20.6
1002
ms T m =
0.3 104
24 C125
T
= =
22. Answer (04)
Hint: isothermal isothermalQ W =
Sol.: ln = CBCB
VQ nRT
V
23. Answer (25)
Hint: Maximum temperature is attained in
process AB
Sol.: 0
0
= − +P
P V CV
00 0 00
4 5= − + =P
P V C C PV
0 00
5P
P V PV
= − +
20 00
5P
PV RT V P V RTV
= − + =
0 000
2 55 0
2
P VV P V
V− + = =
2
0 0 0 0max
0
25 251
4 2
= − +
P V P VT
R V
0 025
4
P V
R=
24. Answer (31)
Hint: Q = nCPT
Sol.: 5
1252
RT=
125 2
3 6 K5 25
= =
T
25. Answer (07)
Hint: 5
2
RU n T =
Sol.: 7
2
RQ n T =
W = nRT
-
Test - 5 (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
PART - B (CHEMISTRY)
26. Answer (2)
Hint :
possible tautomers
Sol. : A is poly hydroxy aromatic compound.
B is poly keto non-aromatic compound
27. Answer (1)
Hint :
Sol. : Acidic strength order is Hb > Ha > Hc
28. Answer (2)
Hint :
According to lowest sum rule correct path
of numbering is given above
Sol. : Correct IUPAC name is
4-Amino-4-hydroxy oct-2-en-5-yne-1,
8-dioic acid.
29. Answer (3)
Hint : To show geometrical isomerism molecule
must have a group which can be oriented
in different spatial arrangements.
Sol. :
Two geometrical isomers
30. Answer (3)
Hint : Alcohols and ethers are functional
isomers
Sol. :
and CH3CH2 – O – CH3
are the two compounds which have same
molecular formula with different functional
groups.
31. Answer (4)
Hint & Sol. :
1, 2 and 3 are conjugated system.
32. Answer (3)
Hint : Energy of structure I and V and II and IV
are same
Sol. : III is least stable
33. Answer (2)
Hint : Lesser is the double bond character
lower would be the energy required for
rotation.
Sol. : Both ring becomes aromatic after
dissociation of -bond hence have least
double bond order.
34. Answer (2)
Hint : Greater the electron density on nitrogen
greater will be the basic strength
-
All India Aakash Test Series for JEE (Main)-2021 Test - 5 (Code-C) (Hints & Solutions)
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Sol. :
Lone pair of nitrogen is in conjugation
with benzene rings
35. Answer (1)
Hint & Sol. :
Hyperconjugation participate with in
toluene and increases the resonance
energy.
36. Answer (3)
Hint : Greater the stability of conjugate base
lesser will be strength of conjugate base
Sol. :
Most stable conjugate base due to
identical resonating structure.
37. Answer (2)
Hint : The species which have complete octet
cannot act as an electrophile.
Sol. : NH3 has lone pair of electron.
38. Answer (3)
Hint : Negative charge on more electronegative
atom is more stable.
Sol. :
Negative charge on N is more stable than
at carbon.
39. Answer (3)
Hint :
Sol. : Lesser the charge density on cation
greater will be the stability.
40. Answer (1)
Hint : Dipole moment is a vector quantity and to
calculate the net dipole moment, vector
sum formula can be used.
Sol. : Dipole moment would be maximum when
the angle between the two vectors is
180°.
41. Answer (2)
Hint & Sol. :
2 46Fe CN [Fe(CN) ]
+ − −+ ⎯⎯→
Some of the Fe2+ is converted into Fe3+
when heated with conc. H2SO4
3 4
6 4 6 3Prussian blue
Fe [Fe(CN) ] Fe [Fe(CN) ]+ −+ ⎯⎯→
42. Answer (4)
Hint & Sol. :
S2– + Pb2+ ⎯⎯→ PbS black ppt.
43. Answer (1)
Hint : Equivalent of BaSO4 = Equivalents of S
in organic compound
Sol. : 0.9625 g of BaSO4 contains
= 32 0.9625
g233
of sulphur
% of sulphur = 32 0.9625 100
233 0.314
42%
44. Answer (3)
Hint : In compound P the degree of
unsaturation is 2 where as in Q is 4.
Sol. : Both have 4 primary carbon atoms.
45. Answer (1)
Hint : Consider the involvement of lone pair in
resonance.
Sol. :
-
Test - 5 (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 (02)
Hint : To show geometrical isomerism along a
C=C, both C-atom should posses
different groups
Sol. :
47. Answer (06)
Hint : Except 3-chlorocyclopropene all have
DBE 3.
Sol. : DBE = 2.
48. Answer (06)
Hint : Check all the possibilities
Sol. :
49. Answer (60)
Hint : Let the given compound is CxHy
x y 2 2 2
4.4g2g
C H O CO H O+ ⎯⎯→ +
Sol. : 4.4 g of CO2 1.2 g of C
Composition of C = 1.2
1002
= 60%
50. Answer (56)
Hint : Equivalents of H2SO4 acid = eq. of NH3 +
eq. of NaOH
Sol. : 100 × 0.25 × 2 = eq. of NH3 + 60 × 0.5
eq. of NH3 = 20 m eq. = 20 m moles
% of N = 314 20 10 100
56%0.5
− =
PART - C (MATHEMATICS)
51. Answer (3)
Hint: Use chain rule.
Sol.: y = f(x3)
3 2( ) 3= dy
f x xdx
6 2 13 6 3 9 3 9== + = =xx x
52. Answer (3)
Hint: Use binomial theorem.
Sol.: 5
01 5
5(1 )5 ... 5 4
+= + + +
CxC C
x x
5
41 2 5
(1 ) 15 5 ... 5
+ = + + + +
C C C
5(1 ) 1+ −
53. Answer (3)
Hint: Make truth table.
Sol.:
54. Answer (3)
Hint: Express equation as factors.
Sol.: P i(2i) = –2
z4 + 5z3 + 18z2 – 4z2 + 5 = (z – z1)
(z – z2) (z – z3) (z – z4)
16 – 40 + 72 – 16 + 5 = (–2 – z1)
(–2 – z2) (–2 – z3) (–2 – z4)
37 = ((PA) (PC))2
-
All India Aakash Test Series for JEE (Main)-2021 Test - 5 (Code-C) (Hints & Solutions)
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8/10
55. Answer (2)
Hint: G.P of infinite series.
Sol.: 1
11 2 1
12 12
=
= = −
rr
56. Answer (4)
Hint: Section formula.
Sol.:
9 10 4
6 3 41 3
+= = − = −
+
4(1) 1( 3)
133 131 1
3
z− + −
−= = =
−−
57. Answer (4)
Hint: Rationalise.
Sol.:
2 2
2 21
(2 3 4) ( 6 2)lim
( 1)( 2 3 4 6 2)→
+ + − + +
− + + + + +x
x x x x
x x x x x
1
( 2)( 1) 1lim
( 1)(3 3) 6x
x x
x→
− −= −
− +
58. Answer (1)
Hint: Locus.
Sol.: |z1| = |z2| = |z1 – z2|
2 21 2 1 2z z z z+ =
1 2
2 1
1z z
z z+ =
59. Answer (2)
Hint: Condition of line intersection.
Sol.: z lies on line or segment joining z1 & z2
and also on line or segment joining 1z
and z2. Hence only 1 possible solution.
60. Answer (1)
Hint: Chain rule.
Sol.: f(x) = f(2 – x)
f (x) = –f (2 – x)
f (x) + f (2 – x) = 0
Put x = 3 f (3) + f (–1) = 0
61. Answer (3)
Hint: Negation of statement.
Sol.: ~( (~ )) ~(~ )→ = q q p q q p
( ~ ) ~q q p q p= =
62. Answer (3)
Hint: Truth table.
Sol.:
63. Answer (4)
Hint: 1 form.
Sol.:
1 1 1 1
2 3 5 7 4lim 2
4→
+ + + −
x x x x
xx
e
( ) ( ) ( ) ( )1 1 1 12 1 3 1 5 1 7 1lim
12
→
− + − + − + −
x x x x
x
xe
1ln(2 3 5 7)
2 210e
=
64. Answer (2)
Hint: Create Indeterminacy.
Sol.: 2 2 ( ) (1 )
lim 31x
x ax b a x b
x→
− + + + −=
−
a = 1 and b – a = 3 b = 2
-
Test - 5 (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
65. Answer (1)
Hint: Locus is a straight line.
Sol.: 1
| 1|2021
z z− = −
Locus
11
101120212 2021
x+
= = =
66. Answer (3)
Hint: Section formula.
Sol.: Let B divides AC in ratio : 1, then for
y-coordinates
( 2) 1(4)
21
− +=
+
–2 + 4 = 2 + 2
1
4 22
= =
5(1) ( )
61
a+ =
+
5 9 82
aa+ = =
67. Answer (4)
Hint: Locus based.
Sol.: 21
(3 ) (4 )2
+ − −k i i
21
1 22
i − +
5 5
,2 2
k
68. Answer (4)
Hint: 1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2
cosa a b b c c
a b c a b c
+ + =
+ + + +
Sol.: For perpendicular lines
2( 6) 5( ) 6( 1) 0 − + + − =
3 26 11 6 0 1, 2, 3 − + − = =
69. Answer (3)
Hint: Put z = x + iy
Sol.: 2 2 2 4x iy x y i+ − + = − −
24 and 16 2y x x= + − + = −
2 24 4 16x x x+ + = +
x = 3
14
3 4 Arg( ) tan3
z i z −
= + =
70. Answer (3)
Hint: Check L.H.L. and R.H.L.
Sol.: R.H.L. = e–1 and L.H.L. = 1
Limit does not exist
71. Answer (65)
Hint: | x | < a –a < x < a
Sol.: | | 13 2z −
2 | | 13 2z− −
11 | | 15z
72. Answer (09)
Hint: 0
form0
Sol.:
23
3
2
3
2 3
1 1tan
3 7
1
1lim
7 81
→−
+ + − = =
+ +
x
xxx x
xL
xx x
22
1
L=
73. Answer (61)
Hint: u
v rule of differentiation
-
All India Aakash Test Series for JEE (Main)-2021 Test - 5 (Code-C) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
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Sol.: 22
( )( 1) (3 1)
xf x
x x= −
+ −
2
2 2
2 (3 2 )( )
( 1) (3 1)
x xf x
x x
− − = −
+ −
2 8 50 72 122
(2)9 25 225 225
f− − −
= − − = =
74. Answer (10)
Hint: Locus based
Sol.: |z – 4| = |z – 8| x = 6
and 3|z – 12| = 5|z – 8i|
9(x2 + y2 – 24x + 144) = 25(x2 + y2 – 16y
+ 64)
y = 8, 17
| z |min = 10
75. Answer (36)
Hint: V = abc
Sol.: a = 3 b = 6 c = 2
V = abc = 3 6 2 = 36
-
Test - 5 (Code-D) (Answers) All India Aakash Test Series for JEE (Main)-2021
All India Aakash Test Series for JEE (Main)-2021
Test Date : 02/02/2020
ANSWERS
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
TEST-5 - Code-D
1/10
PHYSICS CHEMISTRY MATHEMATICS
1. (2)
2. (4)
3. (1)
4. (3)
5. (3)
6. (4)
7. (4)
8. (1)
9. (4)
10. (2)
11. (1)
12. (2)
13. (1)
14. (2)
15. (4)
16. (3)
17. (3)
18. (2)
19. (1)
20. (3)
21. (07)
22. (31)
23. (25)
24. (04)
25. (24)
26. (1)
27. (3)
28. (1)
29. (4)
30. (2)
31. (1)
32. (3)
33. (3)
34. (2)
35. (3)
36. (1)
37. (2)
38. (2)
39. (3)
40. (4)
41. (3)
42. (3)
43. (2)
44. (1)
45. (2)
46. (56)
47. (60)
48. (06)
49. (06)
50. (02)
51. (3)
52. (3)
53. (4)
54. (4)
55. (3)
56. (1)
57. (2)
58. (4)
59. (3)
60. (3)
61. (1)
62. (2)
63. (1)
64. (4)
65. (4)
66. (2)
67. (3)
68. (3)
69. (3)
70. (3)
71. (36)
72. (10)
73. (61)
74. (09)
75. (65)
-
All India Aakash Test Series for JEE (Main)-2021 Test - 5 (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: 2Q
W =
Sol.: 2 2
1 2 1 2
279 93
24 8
Q T
Q Q T T = = = =
− −
= 11.6
2. Answer (4)
Hint: extracted
Wn
Q=
Sol.: 0 02W PV=
0 0 0 033
2 2AB
P V P Vn RQ
nR= =
0 0 0 045
102
BC
P Vn RQ P V
nR= =
0 0
0 0
2 42 100%
23 23
P Vn
P V= =
3. Answer (1)
Hint: =
FLY
A L
Sol.: L F F
T Y TL A T AY
= = =
4. Answer (3)
Hint: A = A0[1 + T]
Sol.: 3 5[1 ] 1 12 2
= + + = +
A lb T T lb T
5
2 =
5. Answer (3)
Hint: dp = –gdh
Sol.: nRT mRT RT
PV MV M
= = =
RdT dT Mg
gdhM dh R
−= − =
dT Mg
dh R=
6. Answer (4)
Hint: SlopeP
V
= −
Sol.:
12 22 1
2
f
f
f f f
+ + = = = +
7. Answer (4)
Hint: TV = constant PV2 = constant
Sol.: 2
2 2
k kP
v m = =
8. Answer (1)
Hint: p = nRT
Sol.: 1 1 1 2
2 2 1 2
p n m M
p n M m= =
2 1 2
1 2 1
4 2
3 3
M M
M M
= =
1
2
1
2
M
M=
9. Answer (4)
Hint: The decrease in the KE due to ordered
motion is equal to increase in internal
energy.
Sol.: 2
2 21 1
2 2 4
= − =
VU nm V nmV
1
v
RnC T n T= =
−
2( 1)
4
− =
mVT
R
-
Test - 5 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2021
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3/10
10. Answer (2)
Hint: U = nCvT
Sol.: 15
2i
RU n T=
1 2 25 3
( ) 2 2 22 2
= − + fR R
U n n T n T
1 2 2( )5 6n n RT n RT= − +
1 12 2
5 5
2 2
= − = + = +
f i
n RT nU U U n RT n RT
11. Answer (1)
Hint: dQ = nCdT
Sol.: nCdT = nCvdT + nVdT
PdV = nVdT
nRT
PV nRT PV
= =
dV
nRT n VdTV
=
2
=
dT RdV
T V
1
ln( )
= −
kT RV
R
VkT e−
=
constant =
R
VTe
12. Answer (2)
Hint: 4 4[ ]sP A T T= −
Sol.: P = 5 × 10–4 × 6.0 × 10–8 [6004 – 3004]
= 3 × 10–11 × 108 [1296 – 81] = 3 × 10–3 ×
1215 W
= 3.645 W
13. Answer (1)
Hint: dQ = nCdT
Sol.: dQ = nCvdT + PdV
PdV = nTdT
nRT
dV n TdTV
=
dV dT
V R
=
ln lnT
k VR
+ =
=
T
RkV e
Constant
−
=
T
RVe
14. Answer (2)
Hint: PV = RT
Sol.: 22 2
Constant ConstantP
PVP V
= =
3
2 2 1 2
R R RC = − =
−
02
RQ C T T = =
15. Answer (4)
Hint: dT dt
H KA aATdx dx
= − = −
Sol.: 0
0
2 20 0
4 0
16
2
T l
T
T THdx HlTdT
aA aA
− −= − =
2
015
2
TaAH
l=
2 2 2
0 016 15
2 2 2
T T TaA l
l aA
− −=
2
2 2 00
1516
2
TT T= −
2
017
2
T=
017
2T T=
16. Answer (3)
Hint: W = Area under P-V graph
Sol.: In general W may be positive or
negative.
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17. Answer (3)
Hint: V = VT
Sol.: 1 dV
V dT =
3nRT
PV nRT KV
= =
ln( ) 3 ln ln ln+ = +nR T K V
3 1 dV
T V dT= =
18. Answer (2)
Hint: 2
12
mlI =
Sol.: 2
2I l
TI l
= =
19. Answer (1)
Hint: 140 z = 100°C
Sol.: 14 7
1 C10 5
z = =
1 1 14200
4200 J kg J kg7
5
ws k
z
− − −= =
= 3000 J kg–1 z–1
20. Answer (3)
Hint: Let R = thermal resistance of each rod
of y
then 2R = thermal resistance of each rod of x.
Sol.:
1 16A BT T T RI = − =
2 14B ET T T RI = − =
1 12
3 3100 60 C
2 5
TT
T
= = =
2 40 = T C
60 60 20 C= − = − = B A B B CT T T T T
TC = 40°C
21. Answer (07)
Hint: 5
2
RU n T =
Sol.: 7
2
RQ n T =
W = nRT
22. Answer (31)
Hint: Q = nCPT
Sol.: 5
1252
RT=
125 2
3 6 K5 25
= =
T
23. Answer (25)
Hint: Maximum temperature is attained in
process AB
Sol.: 0
0
= − +P
P V CV
00 0 00
4 5= − + =P
P V C C PV
0 00
5P
P V PV
= − +
20
00
5P
PV RT V P V RTV
= − + =
0 000
2 55 0
2
P VV P V
V− + = =
2
0 0 0 0max
0
25 251
4 2
= − +
P V P VT
R V
0 025
4
P V
R=
24. Answer (04)
Hint: isothermal isothermalQ W =
Sol.: ln = CBCB
VQ nRT
V
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25. Answer (24)
Hint: 21
0.6 1002
m
Sol.: 20.6
1002
ms T m =
0.3 104
24 C125
T
= =
PART - B (CHEMISTRY)
26. Answer (1)
Hint : Consider the involvement of lone pair in
resonance.
Sol. :
27. Answer (3)
Hint : In compound P the degree of
unsaturation is 2 where as in Q is 4.
Sol. : Both have 4 primary carbon atoms.
28. Answer (1)
Hint : Equivalent of BaSO4 = Equivalents of S
in organic compound
Sol. : 0.9625 g of BaSO4 contains
= 32 0.9625
g233
of sulphur
% of sulphur = 32 0.9625 100
233 0.314
42%
29. Answer (4)
Hint & Sol. :
S2– + Pb2+ ⎯⎯→ PbS black ppt.
30. Answer (2)
Hint & Sol. :
2 46Fe CN [Fe(CN) ]
+ − −+ ⎯⎯→
Some of the Fe2+ is converted into Fe3+
when heated with conc. H2SO4
3 4
6 4 6 3Prussian blue
Fe [Fe(CN) ] Fe [Fe(CN) ]+ −+ ⎯⎯→
31. Answer (1)
Hint : Dipole moment is a vector quantity and to
calculate the net dipole moment, vector
sum formula can be used.
Sol. : Dipole moment would be maximum when
the angle between the two vectors is
180°.
32. Answer (3)
Hint :
Sol. : Lesser the charge density on cation
greater will be the stability.
33. Answer (3)
Hint : Negative charge on more electronegative
atom is more stable.
Sol. :
Negative charge on N is more stable than
at carbon.
34. Answer (2)
Hint : The species which have complete octet
cannot act as an electrophile.
Sol. : NH3 has lone pair of electron.
35. Answer (3)
Hint : Greater the stability of conjugate base
lesser will be strength of conjugate base
Sol. :
Most stable conjugate base due to
identical resonating structure.
36. Answer (1)
Hint & Sol. :
Hyperconjugation participate with in
toluene and increases the resonance
energy.
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37. Answer (2)
Hint : Greater the electron density on
nitrogen greater will be the basic
strength
Sol. :
Lone pair of nitrogen is in conjugation
with benzene rings
38. Answer (2)
Hint : Lesser is the double bond character
lower would be the energy required
for rotation.
Sol. : Both ring becomes aromatic after
dissociation of -bond hence have
least double bond order.
39. Answer (3)
Hint : Energy of structure I and V and II and
IV are same
Sol. : III is least stable
40. Answer (4)
Hint & Sol. :
1, 2 and 3 are conjugated system.
41. Answer (3)
Hint : Alcohols and ethers are functional
isomers
Sol. :
and CH3CH2 – O – CH3
are the two compounds which have
same molecular formula with different
functional groups.
42. Answer (3)
Hint : To show geometrical isomerism molecule
must have a group which can be oriented
in different spatial arrangements.
Sol. :
Two geometrical isomers
43. Answer (2)
Hint :
According to lowest sum rule correct path
of numbering is given above
Sol. : Correct IUPAC name is
4-Amino-4-hydroxy oct-2-en-5-yne-1,
8-dioic acid.
44. Answer (1)
Hint :
Sol. : Acidic strength order is Hb > Ha > Hc
45. Answer (2)
Hint :
possible tautomers
Sol. : A is poly hydroxy aromatic compound.
B is poly keto non-aromatic compound
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46. Answer (56)
Hint : Equivalents of H2SO4 acid = eq. of NH3 +
eq. of NaOH
Sol. : 100 × 0.25 × 2 = eq. of NH3 + 60 × 0.5
eq. of NH3 = 20 m eq. = 20 m moles
% of N = 314 20 10 100
56%0.5
− =
47. Answer (60)
Hint : Let the given compound is CxHy
x y 2 2 2
4.4g2g
C H O CO H O+ ⎯⎯→ +
Sol. : 4.4 g of CO2 1.2 g of C
Composition of C = 1.2
1002
= 60%
48. Answer (06)
Hint : Check all the possibilities
Sol. :
49. Answer (06)
Hint : Except 3-chlorocyclopropene all have
DBE 3.
Sol. : DBE = 2.
50. Answer (02)
Hint : To show geometrical isomerism along a
C=C, both C-atom should posses
different groups
Sol. :
PART - C (MATHEMATICS)
51. Answer (3)
Hint: Check L.H.L. and R.H.L.
Sol.: R.H.L. = e–1 and L.H.L. = 1
Limit does not exist
52. Answer (3)
Hint: Put z = x + iy
Sol.: 2 2 2 4x iy x y i+ − + = − −
24 and 16 2y x x= + − + = −
2 24 4 16x x x+ + = +
x = 3
14
3 4 Arg( ) tan3
z i z −
= + =
53. Answer (4)
Hint: 1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2
cosa a b b c c
a b c a b c
+ + =
+ + + +
Sol.: For perpendicular lines
2( 6) 5( ) 6( 1) 0 − + + − =
3 26 11 6 0 1, 2, 3 − + − = =
54. Answer (4)
Hint: Locus based.
Sol.: 21
(3 ) (4 )2
+ − −k i i
21
1 22
i − +
5 5
,2 2
k
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55. Answer (3)
Hint: Section formula.
Sol.: Let B divides AC in ratio : 1, then for
y-coordinates
( 2) 1(4)
21
− +=
+
–2 + 4 = 2 + 2
1
4 22
= =
5(1) ( )
61
a+ =
+
5 9 82
aa+ = =
56. Answer (1)
Hint: Locus is a straight line.
Sol.: 1
| 1|2021
z z− = −
Locus
11
101120212 2021
x+
= = =
57. Answer (2)
Hint: Create Indeterminacy.
Sol.: 2 2 ( ) (1 )
lim 31x
x ax b a x b
x→
− + + + −=
−
a = 1 and b – a = 3 b = 2
58. Answer (4)
Hint: 1 form.
Sol.:
1 1 1 1
2 3 5 7 4lim 2
4→
+ + + −
x x x x
xx
e
( ) ( ) ( ) ( )1 1 1 12 1 3 1 5 1 7 1lim
12
→
− + − + − + −
x x x x
x
xe
1ln(2 3 5 7)
2 210e
=
59. Answer (3)
Hint: Truth table.
Sol.:
60. Answer (3)
Hint: Negation of statement.
Sol.: ~( (~ )) ~(~ )→ = q q p q q p
( ~ ) ~q q p q p= =
61. Answer (1)
Hint: Chain rule.
Sol.: f(x) = f(2 – x)
f (x) = –f (2 – x)
f (x) + f (2 – x) = 0
Put x = 3 f (3) + f (–1) = 0
62. Answer (2)
Hint: Condition of line intersection.
Sol.: z lies on line or segment joining z1 & z2
and also on line or segment joining 1z
and z2. Hence only 1 possible solution.
63. Answer (1)
Hint: Locus.
Sol.: |z1| = |z2| = |z1 – z2|
2 21 2 1 2z z z z+ =
1 2
2 1
1z z
z z+ =
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64. Answer (4)
Hint: Rationalise.
Sol.:
2 2
2 21
(2 3 4) ( 6 2)lim
( 1)( 2 3 4 6 2)→
+ + − + +
− + + + + +x
x x x x
x x x x x
1
( 2)( 1) 1lim
( 1)(3 3) 6x
x x
x→
− −= −
− +
65. Answer (4)
Hint: Section formula.
Sol.:
9 10 4
6 3 41 3
+= = − = −
+
4(1) 1( 3)
133 131 1
3
z− + −
−= = =
−−
66. Answer (2)
Hint: G.P of infinite series.
Sol.: 1
11 2 1
12 12
=
= = −
rr
67. Answer (3)
Hint: Express equation as factors.
Sol.: P i(2i) = –2
z4 + 5z3 + 18z2 – 4z2 + 5 = (z – z1)
(z – z2) (z – z3) (z – z4)
16 – 40 + 72 – 16 + 5 = (–2 – z1)
(–2 – z2) (–2 – z3) (–2 – z4)
37 = ((PA) (PC))2
68. Answer (3)
Hint: Make truth table.
Sol.:
69. Answer (3)
Hint: Use binomial theorem.
Sol.: 5
01 5
5(1 )5 ... 5 4
+= + + +
CxC C
x x
5
41 2 5
(1 ) 15 5 ... 5
+ = + + + +
C C C
5(1 ) 1+ −
70. Answer (3)
Hint: Use chain rule.
Sol.: y = f(x3)
3 2( ) 3= dy
f x xdx
6 2 13 6 3 9 3 9== + = =xx x
71. Answer (36)
Hint: V = abc
Sol.: a = 3 b = 6 c = 2
V = abc = 3 6 2 = 36
72. Answer (10)
Hint: Locus based
Sol.: |z – 4| = |z – 8| x = 6
and 3|z – 12| = 5|z – 8i|
9(x2 + y2 – 24x + 144) = 25(x2 + y2 – 16y
+ 64)
y = 8, 17
| z |min = 10
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73. Answer (61)
Hint: u
v rule of differentiation
Sol.: 22
( )( 1) (3 1)
xf x
x x= −
+ −
2
2 2
2 (3 2 )( )
( 1) (3 1)
x xf x
x x
− − = −
+ −
2 8 50 72 122
(2)9 25 225 225
f− − −
= − − = =
74. Answer (09)
Hint: 0
form0
Sol.:
23
3
2
3
2 3
1 1tan
3 7
1
1lim
7 81
→−
+ + − = =
+ +
x
xxx x
xL
xx x
22
1
L=
75. Answer (65)
Hint: | x | < a –a < x < a
Sol.: | | 13 2z −
2 | | 13 2z− −
11 | | 15z
AIATS_JEE(M)2021_Test-5_(Code-C)_02-02-2020_SolAIATS_JEE(M)2021_Test-5_(Code-D)_02-02-2020_Sol
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