Solutions - AIATS JEE (Adv) - 2020 (XI Studying) - Test-3A (Code-E & F) - Paper-1 - (24-02-2019) PDF
Solutions - AIATS JEE (Adv) - 2020 (XI Studying) - Test-3A (Code-E & F) - Paper-1 - (24-02-2019) PDF
Solutions - AIATS JEE (Adv) - 2020 (XI Studying) - Test-3A (Code-E & F) - Paper-1 - (24-02-2019) PDF
ANSWERS
PHYSICS CHEMISTRY MATHEMATICS
1. (A) 21. (B) 41. (C)
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All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper - 1) (Code-E) (Hints & Solutions)
M 2 M 2 0 (v ) vs
I = 12 16 = (v v cos )
s
cot
7M 2
I= v sin vs
48
vs = v cos
2
7M
2 0
T=
48 Mg sin2
4
4 3
sin = , cos =
2 7 5 5
=
2 3g
4. Answer (C)
Hint:
7
=
3g Calculate the effective phase difference at point P.
Solution:
2. Answer (C)
Hint: 2 2
3 cos 3 6 P
Add the total no. of mole by
Pdv 2
6 cos 6
using : dn RT 3 S1 S2
Solution: 1
3 cos 3
3
T T0
T = T0 L x
L 8
cos
PA dx = dn RT 9
= cos–1 (8/9)
T0 n
PA dx
R T T
dn
Y
L 0 tan
D
PV TL Y 17
n = (T T )R ln T
L 0 0 D 8
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Test - 3A (Paper - 1) (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
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All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper - 1) (Code-E) (Hints & Solutions)
x = A sin
2 2
A0 A B
= Solution :
2
N
2
18. Answer (03)
N=8
Hint:
9 nodes
0 = 2
PART - II (CHEMISTRY)
21. Answer (B) Solution:
Hint:
O Non-aromatic
R Reductive R
C = CH2 ozonolysis C =O & H – C – H H
R R
Solution: B
CH 2 CH 2 Antiaromatic
B
O3 H
CH 2 6HCHO
Zn–H2 O
CH 2
Aromatic
CH 2 CH 2
22. Answer (D)
Hint:
Aromatic
is non-aromatic compound as it is non-planar
molecule.
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Test - 3A (Paper - 1) (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
One canonical structure will not explain all the HBr & HCl
properties of molecule. both follow Br2
Markownikoff
26. Answer (B) with carbonium ion Br
rearrangement
Hint:
H Ring 30. Answer (A)
opening
CH 2 – CH = CH 2
Hint:
Solution:
H
carbonium ion
Br2/high temperature will do allylic substitution
rearrangment
Solution:
CH2 – CH = CH 2 CH2 – CH = CH2
H2 O O O–H O–H
O
–H
H H CH2 = CH – CH 2 – OH
H
27. Answer (A, C, D)
Hint: Carbonium ion rearrangement.
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All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper - 1) (Code-E) (Hints & Solutions)
OH OH OH 34. Answer (A, B, C)
Hint:
Ring Deprotonation
Expansion
Electrophilic aromatic substitution
(–H +)
Solution:
(P)
E
NBS CH2 O
OH More CH2
activating
(major)
Br (Q)
31. Answer (B) E
Hint: bonds which are not the part of aromatic ring, 35. Answer (B, C)
can undergo addition reaction. Hint:
Solution: Friedel Craft alkylation reaction
OH OH OH
Solution:
alc KOH Br 2
CH3 Me
Hint:
36. Answer A(P, T); B(P, Q, R, S, T); C(P), D(P, Q, R, S)
HO Hint: Alkene decolourise Br2/CCl4.
is aromatic compound.
Solution:
Solution:
+ H 2/Ni CH3 – CH 2 – CH3
OH
It is aromatic
cyclic and stable
Compound R is Has acidic , , CH2 &
H-atom
CH
33. Answer (A, B, C) are unsaturated compound can give bromine water
Hint: test
CH4 + CH3 – C C Mg Br
– + –
Electrophilic aromatic substitution CH3 – C CH + CH 3MgBr
Solution:
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Test - 3A (Paper - 1) (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
37. Answer A(Q, R, T); B(P, S); C(Q, R, T); D(Q, R, T) 39. Answer (06)
Hint: Hint: Decarboxylation reaction
Reaction intermediate singlet carbene is sp 2 Solution:
hybridised.
Solution: O O
COOH COONa
R1 NaOH
2 14
= 7
4 ,
i.e 7 OH NH2
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All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper - 1) (Code-E) (Hints & Solutions)
Solution: 2 – 4 0
48. Answer (A, C)
Total no. of ways = (7)4
Hint:
7
Now no. of favourable case C2 (2 2) 21 14 4
2 – 4 > 0
49. Answer (B)
21 14 6
Now probability of required event
(7)4 49 Hint:
44. Answer (C) z –
log l lim – sin t log(t 2 (1 t 2 ))
t 0
2 4
z z
2
2log(t ) log(1 t 2 )
lim
t 0 cosect
2 4 2 1
z
2 sin2 t 2 2 4
lim =0l=1
t 0 t cos t 50. Answer (B)
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Test - 3A (Paper - 1) (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
3 3 zz
sin( A B ) , tan( A B )
5 4
2 1 | z |2
2i
( A B) zz
2 tan 1
2 3
zz
AB 4 2i ( z z ) (1 )
1 tan2 1 | z |2
2
z z i
AB 1 2 1 | z |2
tan
2 3 56. Answer A(Q, S); B(Q, R, S, T); C(R S, T); D(R, S, T)
AB z
∵ is an acute angle
2 Hint:
|z|
= unimodular
ab C 1 Solution:
cot
ab 2 3 z z
(A) 2i 2i
C 90 |z| |z|
2 2 x2 y 2
Hint: z
2
Equation of ellipse 1
(1 )2 7 16
(D) Iz – 2 – 3i| = 1
Solution:
( x 2)2 ( y 3)2 1
2 2 2 x 2 cos , y 3 sin
| z |2 2
2
(1 ) (1 ) 1
x [1, 3], y [2, 4]
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All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper - 1) (Code-E) (Hints & Solutions)
57. Answer A(R, T); B(Q, R, T); C(Q, R, T); D(Q, R, S, T) 59. Answer (05)
Hint: x – 1 x Hint:
Solution:
sin x cos x 1 z1 C is
lim 10
(A) lim(sin x cos x )cosecx e x 0 sin x
x 0
=e 3
z2 C is
n 10
4 n 10
(B) L lim (1 form)
5
n
Solution:
1
lim z1 C is
Le n 4 (10) n 5 n 10
5
3
z2 C is
1 10
ln(10)
L e 5
1 3 3
z1 z2 cos cos + i sin sin
L 10 5 10 10 10 10
L5
2 10 2 5 10 2 5 1
5
z1 z2 i
xx 1 4 2
(C) xlim
0
2 L= 2
3 5
| z1 z2 | 2
5 x 2
(D) L lim
x x
3 60. Answer (02)
Hint:
x x x
3 1 3 3 |a – d|, |a|, |a + d| be three distances
Solution:
5 x 5 x 5 x
1
x 3 x 3 x 3 As perpendicular distances are in A.P.
lim
Let |a – d|, |a|, |a + d| be these distances
5 x 5
x
x 3 3 so the co-ordinates of point P are
1 a = ±2, d = ±1
cos –
2 so points P (±1, ±2, ±3)
Number of values of [–2, 2] is 4 Now = 1, = 2, = 3
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Test - 3A (Paper - 1) (Code-F) (Answers) All India Aakash Test Series for JEE (Advanced)-2020
ANSWERS
PHYSICS CHEMISTRY MATHEMATICS
1. (A) 21. (B) 41. (C)
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All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
(b r ) Solution:
Req
4k br vs
0 (v )
dT b(T1 T2 ) = (v v cos )
s
dr (b r )r
2. Answer (A) cot
Hint: v sin vs
Temperature of each body will change based on their
vs = v cos
heat capacity.
Solution: 0
dT kA sin2
ms 1 (T2 T1 )
dt
4 3
dT kA sin = , cos =
2ms 2 (T2 T1 ) 5 5
dt
5. Answer (C)
2ms ⎛ 4 ⎞
t n⎜ ⎟ Hint:
3kA ⎝3⎠
Add the total no. of mole by
3. Answer (C)
Hint: Pdv
Calculate the effective phase difference at point P.
using : ∫ dn ∫ RT
Solution: Solution:
⎡ 2 2 ⎤ ⎛ T T0 ⎞
⎢ 3 cos 3 ⎥ 6 P T = T0 ⎜ L ⎟x
⎣ ⎦ ⎝ L ⎠
2 PA dx = dn RT
6 cos 6
3 S1 S2
T n
1 PA 0 dx
R T∫ T
3 cos 3 ∫ dn
3 0
L
⎛8⎞
cos ⎜ ⎟
⎝9⎠ PV ⎛ TL ⎞
n = (T T )R ln ⎜ T ⎟
= cos–1 (8/9) L 0 ⎝ 0⎠
Y
tan 6. Answer (A)
D
Y 17 I
Hint: T = 2
D 8 Mgd
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Test - 3A (Paper - 1) (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
Solution: Solution:
Hint: ⎛P ⎞
vp = v s ⎜ ⎟
Draw the corresponding PV diagram and calculate ⎝ B0 ⎠
corresponding quantities.
v s ⎛ ⎞
Solution: ap = B ⎜ t ⎟
0 ⎝ ⎠
P = kV
v s P0
RT = B x vs
kV 0 0
V
v s2 p0
RT = kV 2 ap
B0 x 0
2nV = nT
16. Answer A(Q, T), B(R), C(P), D(Q, S, T)
dV dT Hint:
2
V T Distance will be maximum when velocity of each is
dV 1 same.
2
VdT T Solution:
10. Answer (A, C) This has to be solved using
11. Answer (B, C) Phasor diagram
12. Answer (A, D) 2Asin = 3A
Hint and Solution of Q. Nos. 10 to 12 A
=
Hint: 3
Process for system will be adiabatic. More over work = 2 or 2 – 2
done by external force will be used to raise the internal
2 4 B
energy. = or
3 3
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All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
Hint: 31
4 2
2 2
1 =
Solution : 3
N 1 = 600 Hz
2
PART - II (CHEMISTRY)
21. Answer (B)
Na/
Hint: CH2 – (CH2)4 – CH2 dry ether
H Ring
Br Br
opening
CH 2 – CH = CH 2
Solution:
Solution:
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Test - 3A (Paper - 1) (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
CH 2 CH 2
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All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
OH OH OH 34. Answer (A, B, C)
Hint:
Ring Deprotonation
Expansion (–H +) Electrophilic aromatic substitution
(P) Solution:
NBS
E
CH2 O
OH
More CH2
activating
Br (major)
(Q)
31. Answer (B)
E
Hint: bonds which are not the part of aromatic ring,
can undergo addition reaction. 35. Answer (B, C)
Solution: Hint:
OH OH OH
Friedel Craft alkylation reaction
alc KOH Br 2
Solution:
CH3 Me
Br Br
Br
& has greater electron
(Q) (R) (S)
CH3 Me Me
32. Answer (A, C)
Hint:
density than CH3
HO
is aromatic compound. 36. Answer A(Q, R, T); B(P, S); C(Q, R, T); D(Q, R, T)
Solution: Hint:
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Test - 3A (Paper - 1) (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
O O
COOH COONa
NaOH
, , CH2 &
HOOC COOH NaOOC COONa
CH COOH COONa
are unsaturated compound can give bromine water (salt)
test electrolysis
+ 4 CO 2 (m = 4)
hydrogen atom OH
CH3 – C CH
(DBE = 4)
H
Hydrogen is
2m d
acidic due to
aromaticity
6
aromatic 2
40. Answer (07)
38. Answer (04)
Hint: Kjeldahl method
Hint: RMgX RH
Solution: m.e of H2SO4 used for NH3
Solution: = (25 × 2) – 30 × 0.5 × 2 = 50 – 30 = 20
C6H5 SO3H, C 6H5OH No. of m.e of NH3 = 20
NO 2 OH % mass of nitrogen in organic compound
(20)(14) 100
=
, (1000)(4)
OH NH2
1.4
= 20
4
39. Answer (06)
2 14
Hint: Decarboxylation reaction = 7
4
Solution: i.e 7
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All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
44. Answer (B) 49. Answer (B)
4
Hint : Number of exhaustive cases = 7 Hint:
Solution: z –
Total no. of ways = (7)4 Solution for Q. Nos. 47 to 49
z2 + z + 1 = 0
Now no. of favourable case 7 C2 (24 2) 21 14
2 4
z
21 14 6 2
Now probability of required event 4
(7) 49 Case 1
45. Answer (A) –2 < < 2
3 2 2 – 4 < 0
Hint: Area = a
4
i 4 2
Solution: z
2
|z1 + z2|2 + |z1 – z2|2 = 2(|z1|2 + |z2|2)
x ⇒ y 2 1 x2
2
| – z3|2 + |z1 – z2|2 = 2(2)
x2 + y2 = 1
|z1 – z2|2 = 3 ⇒ | z1 z2 | 3 Case 2
>2
Similarly | z2 z3 |2 | z3 z1 | 3
2 – 4 > 0
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Test - 3A (Paper - 1) (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
3 3 zz
⇒ sin( A B ) , tan( A B )
5 4
2 1 | z |2
2i
zz
⎡ ( A B) ⎤ 1
2 tan ⎢ ⎥
⇒ ⎣ 2 ⎦ 3 zz
⎛ AB⎞ 4 2i ( z z ) (1 )
1 tan2 ⎜ ⎟ 1 | z |2
⎝ 2 ⎠
z z i
⎛ AB⎞ 1 2 1 | z |2
⇒ tan ⎜ ⎟
⎝ 2 ⎠ 3
56. Answer A(R, T); B(Q, R, T); C(Q, R, T); D(Q, R, S, T)
⎧ AB Hint: x – 1 x
⎨∵ is an acute angle
⎩ 2 Solution:
sin x cos x 1
ab C 1 (A) lim(sin x cos x )cosecx e x 0
lim
⇒ cot sin x
ab 2 3 x 0
=e
⇒ C 90
n
So triangle ABC is right angle triangle ⎛ 4 n 10 ⎞
(B) L lim ⎜ (1 form)
n ⎝ 5 ⎟⎠
1
Area of ABC 6 3 = 9 sq. units 1
2 lim
⎛ ⎞
n ⎜ 4 (10) n 5
⎟n
Le ⎜ ⎟
5
⎜⎝ ⎟⎠
r
9
s ⎛363 5 ⎞ 2
3
3 5
⎜⎜ ⎟⎟ 1
ln(10)
⎝ 2 ⎠ L e5
1
1 3 5 L 10 5
Radius of the circumcircle = × hypotenuse
2 2 L5
2
53. Answer (C) 5
54. Answer (A) (C) xlim xx 1
0
2 2 2 5 ⎡x⎤
Hint: z (D) L lim
(1 )2 x x ⎢⎣ 3 ⎥⎦
Solution:
⎛ x ⎞ ⎡x⎤ x
⎜⎝ 3 1⎟⎠ ⎢ 3 ⎥ 3
⎣ ⎦
2 2 2
| z |2
(1 ) 2
(1 )2 1 5 ⎛ x ⎞ 5 ⎡x⎤ 5 x
1
x ⎜⎝ 3 ⎟⎠ x ⎢⎣ 3 ⎥⎦ x 3
1 1
Now | z |2 1 ⇒ 1 | z |2 lim 5 ⎡x⎤ 5
1 1 1 x
x ⎢⎣ 3 ⎥⎦ 3
| z |2
57. Answer A(Q, S); B(Q, R, S, T); C(R S, T); D(R, S, T)
1 | z |2
z
zz Hint: = unimodular
⇒ 2 ( z z ) (1 ) |z|
1 | z |2
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All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper - 1) (Code-F) (Hints & Solutions)
Solution: Solving (i), (ii) and (iii)
z z we get
(A) 2i 2i
|z| |z| a = ±2, d = ±1
z so points P (±1, ±2, ±3)
unimodular complex no.
|z| Now = 1, = 2, = 3
and lies on perpendicular bisector of 2i and –2i
59. Answer (05)
z
1 ⇒ z | z | ⇒ a real number Hint:
|z|
Im (z) = 0
z1 C is
(B) = ei 10
1 2 1 e 2i ⎛ 3 ⎞
z 2cos z2 C is ⎜ ⎟
ei ⎝ 10 ⎠
(C) 2b = 8 Solution:
2be = 6 a2 = 7
z1 C is
3 10
e=
4
x2 y 2 ⎛ 3 ⎞
z2 C is ⎜ ⎟
Equation of ellipse 1 ⎝ 10 ⎠
7 16
(D) |z – 2 – 3i| = 1
⎛ 3 ⎞ ⎛ 3 ⎞
( x 2)2 ( y 3)2 1 z1 z2 ⎜ cos cos ⎟ + i ⎜ sin sin ⎟
⎝ 10 10 ⎠ ⎝ 10 10 ⎠
x 2 cos , y 3 sin
⎛ 10 2 5 10 2 5 ⎞ 1
x [1, 3], y [2, 4] z1 z2 ⎜ ⎟ i
⎜⎝ 4 ⎟⎠ 2
58. Answer (02)
Hint: 3 5
| z1 z2 | 2
|a – d|, |a|, |a + d| be three distances 2
Solution: 60. Answer (04)
Let |a – d|, |a|, |a + d| be these distances sin sin 2 sin 3 0 sin2 (1 + 2cos) = 0
so the co-ordinates of point P are ..(1)
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