Test - 3A (Paper-2) (Code-C)_(Answers) All India Aakash Test Series for JEE (Advanced)-2020

All India Aakash Test Series for JEE (Advanced)-2020

TEST - 3A (Paper-2) - Code-C
Test Date : 06/10/2019

ANSWERS
PHYSICS CHEMISTRY MATHEMATICS
1. (C) 21. (D) 41. (B)
2. (A) 22. (A) 42. (A)
3. (A) 23. (C) 43. (B)
4. (C) 24. (D) 44. (A)
5. (A) 25. (B) 45. (A)
6. (A, D) 26. (A, C, D) 46. (C, D)
7. (A, C) 27. (A, C, D) 47. (A, B, C)
8. (A, B) 28. (B, C, D) 48. (A, D)
9. (A, C) 29. (A, B, D) 49. (C, D)
10. (C) 30. (A, B, C) 50. (A, B, D)
11. (A, D) 31. (D) 51. (A, C)
12. (A, C) 32. (D) 52. (C, D)
13. (A, C, D) 33. (A) 53. (A, B, C)
14. (A, C) 34. (A, B, D) 54. (A, B, C)
15. (A, C) 35. (A, B, C) 55. (C, D)
16. A → (Q) 36. A → (Q, R, S, T) 56. A → (P, S)
B → (P, T) B → (P, Q, R) B → (S)
C → (P, S) C → (P, Q, R) C → (Q, S, T)
D → (P, R, T) D → (P, R, T) D → (Q, S, T)
17. A → (Q, T) 37. A → (P, R, T) 57. A → (P, Q)
B → (R, S, T) B → (Q, T) B → (P, R, S, T)
C → (P, R, S) C → (P, S, T) C → (R, T)
D → (P, Q) D → (P, Q, T) D → (R, S, T)
18. (25) 38. (07) 58. (19)
19. (20) 39. (03) 59. (30)
20. (22) 40. (13) 60. (32)

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

1/12
All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper-2) (Code-C)_(Hints & Solutions)

PART - I (PHYSICS) 3. Answer (A)

1 1 1 1
1. Answer (C) Hint : = + +
Leq L1 L2 L3
Hint : Displacement method.
Solution : Solution :
1 1 1 1
m1 =
150 – x = + +
x Leff L1 L2 L3

x 2
m2 =  Leff = H
150 – x 3
itotal = 2 + 3 + 5 = 10 A
16 (150 – x )
2
m1
 = =
m2 1 x2 1 2 q2
Li =
2 2C
150 – x
 =4 2
x  q2 =  10  10  2  10–6
3
 x = 30 cm
20 20
 q=  10 –3 C = mC
1 1 1
 + = 3 3
120 30 f
4. Answer (C)
1 5 Hint : AC circuit Analysis.
 =
f 120
V
Solution : VA =
 f = 24 cm 2

2. Answer (A) VR  –  L
and VB = tan  =
R + ( L )
2 2 R
Hint : At the instant of sharp change, the flux
would remain same. For R → 0 VB = V
Solution : Just after changing flux would remain V
same  v =
2
 L For R →  VB = V
 L = i
R 3
V
 v =
3 2
 i = = 3l 0
R
At any time V
L di
Now,  – Ri – =0
3 dt
L di
 (  – Ri ) =
3 dt
t i
3dt di
  L =   – Ri
0 3i
V = VA – VB = VA2 + VB – 2 (VA )(VB )  cos 
0 2

–3Rt   – Ri 
 = ln 
L  –2   V 2 =
V2
+
V 2R 2
–
2 V
4 R 2 + ( L )2 2
3Rt
–
 Ri –  = 2e L VR

R
R + ( L ) R + ( L )
2 2 2 2
 
2Rt
–
 i = 1 + 2e L 
R
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

2/13
Test - 3A (Paper-2) (Code-C)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020

V2 f2
 V 2 = m=
4 f 2 – 2df + 2 x ( d – f )
V For m to be independent of x
 V =
2 df = 0  d=f
5. Answer (A)
f2
Hint : Induced electric field is non-conservative So, m = =1
f 2 – 2f 2
in nature.
Solution : Induced electric field are produced by 8. Answer (A, B)
changing magnetic field and they form a closed Hint : It could be real image or virtual image.
loop, –f –f
Solution : = 2 and = –2
So  E  dl  0 and potential can’t be defined. –f + u –f + u
6. Answer (A, D) f 3f
 u= and u =
Hint : Image by reflection. 2 2
Solution : As the object is not on the bisector, 9. Answer (A, C)
the polygon will be irregular. V
Hint : Z = ; P = irms  Vrms cos 
I
Solution :  = 100
 
V = 400 sin  t +  volt
 6
I = 600 sin(t) mA
7. Answer (A, C)
V 400  103 2000
1 1 1
Hint : − = ; m =
v  Z = = = 
I 600 3
v u f u
3
Power factor =
2
Solution : Let x be the object distance from lens
and d be the distance between lens and mirror. Average power dissipation:
1 1 1
+ = 400 600  10–3 3
Then for image by lens   = irms  Vrms cos 
v1 x f 2 2 2
1 x–f
=  P avg = 60 3 watt.

v1 xf
10. Answer (C)
v1 f x f Hint : L-R circuit.
 m1 = = v1 =
x x–f x–f 5L
Solution :  =
xf R
For 2nd time object distance is 2d –
x–f iL(t = 0) = 0
1 1 1 4
+ = iL(t =  ) =
v2 xf f R
2d –
x–f
4  
−tR

f iL = 1 − e 5L 
m2 = R 
xf  
2d – –f
x–f 11. Answer (A, D)
 f  f 12. Answer (A, C)
 m1. m2 =  .
x–f  xf 13. Answer (A, C, D)
2d – –f
x–f Hint for Q.Nos. 11 to 13 :
f (x – f )
2 1 1 1
+ =
=
( x – f ) ( 2 xd – 2df – xf – xf + f 2 ) v u f

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

3/13
All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper-2) (Code-C)_(Hints & Solutions)

Solution for Q.Nos. 11 to 13 : dx0 f f

Velocity of object = −  sin t = − 3
dt 2 4
4f 3 f 3 4 1
 Vrel = (v 2 − v1 ) = + = 3f  + 
9 4 9 4
25
At any time t, x =
f
cos t  vrel = 3 f
2 36
 object distance from mirror u = (2f – x ) Magnification is always negative so always real image
(towards left) will be formed. For the time when object lies between
2f and f from mirror it will produce magnified image.
Let v is the position of image w.r.t. mirror then
For rest of the time it will be diminished image.
1 1 2
+ =− 14. Answer (A, C)
v u R
15. Answer (A, C)
1 1 −1
 − = Hint for Q.Nos. 14 and 15 :
v 2f − x f
eE = m2  x
1 f − 2f + x
 = Solution for Q.Nos. 14 and 15 :
v f ( 2f − x )
f ( 2f − x )
 v=
x −f
 Distance of image from mirror
Because of pseudo force, free charge would try
 f 
f  2f − cos t  to shift outward since free charges are electrons.
 v =  2 
So Fout = m2x. Because of that electric field
f
cos t − f
2 would induce in a way that eE = m2  x
f 2 ( 4 − cos t )  2 f ( 4 − cos t ) m2
 v IM = =  E=  x (from A to B as electron has
2  f ( cos t − 2 ) 2 − cos t e
Now position of image w.r.t. origin is shifted towards B)
l +L
f ( 2f − x ) m2
rI = rIM + rM =
0 0 x −f
+ 2f VAB =  Edx =
e 
xdx
l

2f − fx + 2fx − 2f
2 2
f x m2 2 l +L m2 (
 rI = =  VAB = x l = L L + 2l )
0 x −f x −f 2e 2e
f
f  cos t m2
2 f 2 cos t As we see E =  x implies that the field
 rI = =
0 f f ( cos t − 2 ) e
cos t − f region (in rod and outside of it) would exist in x
2
direction as per the equation then there must be
f cos t
 rI = − (oscillating but not SHM) uniform charge distribution.
0 2 − cos t 16. Answer A(Q); B(P, T); C(P, S); D(P, R, T)
Now velocity of image is Hint : Req(t = 0) = 40 
80

d
( ) =v
rI
0
=
( 2 − cos t ) f  sin t + f cos t   sin t
Req(t = ) =
3


dt I
( 2 − cos t )2 transient LR circuit.
Solution :
f sin t ( 2 − cos t + cos t ) 2f sin t
vI = =
( 2 − cos t ) 2
( 2 − cos t )2
 2f 3 f 3  4
At t = vI = =
3  1
2 9
2 2 − 
 2 Req (t = 0) = 40 
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

4/13
Test - 3A (Paper-2) (Code-C)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020

80 19. Answer (20)

Req ( t =  ) = 
3 Hint : After first refraction from the plane surface
100 3
Clearly at t = 0 current i1 = = 2.5 A the image seems to be at L from plane surface.
40 2

5 Solution : After first refraction from the plane

and at t =  l1 = = 1.25 A
4 3
surface p seems to be at L from plane surface.
Current i2 at t = 0 is zero at inductor and it will 2
oppose the sudden change of current. 1  (1 −  )
Now for 2nd surface + =
10 v  3  −R
And i2 at t =  is = 2.5 A R + L
4  2 
Power delivered by battery p = i  v=
As i increases from 2.5 A to 3.75 A
  −1
So power delivered by battery increases from  =
3 R
250 watt to 375 watt. R+ L
2
17. Answer A(Q, T); B(R, S, T); C(P, R, S); D(P, Q)
 L = 20 cm
1 1 1
Hint : For lens − = 20. Answer (22)
v u f
2 1 2 − 1
1 1 1 Hint : − =
For mirror + = v u R
v u f
1 1 1 Solution :
Solution : For lens − =
v u f
1 1 1
For mirror + =
v u f
18. Answer (25)
7 
Hint :
1
=
1
+
1  − 1
= 4 
7 1
Z  5 − 5 j 10 j +
4v1 24 6
Solution : from water surface
4 7
 − 
= 3 4
4 7
−
3v 2 4v1 

4 1 3
 + =
3v 2 24 24
4 1
Let Z be the impedance across MN.  =
3v 2 12
1 1 1
Then = +  v2 = 16 cm
z  5 − 5 j 10 j
 Distance from bottom of tank is 38 – 16 = 22 cm.
1 10 j + 5 − 5 j 5+5j
 = =
Z 50 j + 50 50 + 50 j
PART - II (CHEMISTRY)
 Z = 10 
21. Answer (D)
1
 = 20  Hint : Spectator ligand will affect the C — O bond
C length. Order of ligand field strength of the given
1 ligand is
 f = = 25 Hz
2  20  C CO  PF3 > PCl3 > PAr3 > PMe3

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

5/13
All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper-2) (Code-C)_(Hints & Solutions)

Solution : The ligand PPh3 is a weaker  Fe(OH)3 is a brown residue, NaAlO2 is a

acceptor than CO. As a result d electrons of colourless solution and Na2CrO4 is a yellow
solution.
metal in such mixed carbonyl will be drawn more
towards CO than in pure metal carbonyl. If the Ph 24. Answer (D)
groups of PPh3 are replaced by more electron Hint : Cation present is Na+.
attracting Cl or F groups, the tendency of d
Solution : Fe(CH3COO)3 is red in colour.
electrons of metal to move towards P increases.
And if Ph groups of PPh3 are replaced by electron 25. Answer (B)
releasing Me groups, the tendency of d Hint : Dispersion of charge decreases the
electrons of metal to move towards P further energy.
decreases. As d electrons of metal move
Solution :
towards P more and less towards CO, the CO
bond Order will be more or CO bond length will
be less and vice versa.
22. Answer (A)
Hint : For the fast rate, back side of LG Should The general energy diagram of the above reaction
be less hindered. is
Solution : In an SN2 reaction, the leaving group
must be in an axial position in order to allow
backside attack to occur without steric
hinderance from the cyclohexane ring. When the
Br-atom is in axial position in the all cis-isomer,
both the methyl groups are in equatorial
positions, the structure (II) is most reactive as the
approaching nucleophile experiences least
On increasing polarity of the solvent, all charged
crowding. In the all trans isomer (IV) both the
species will get solvated. Thus their energies will be
ethyl groups are in axial positions providing the
lowered. In the above reaction the reactants do not
maximum crowding to the approaching
carry any charge and hence their energy remains
nucleophile. Thus structure (IV) is least reactive.
unaffected by the increase in polarity of the solvent.
Structure (I) and (III) are cis-trans type. Their
But the energy of transition state is lowered due to
reactivity lies between those of (II) and (IV),
its solvation. This results in decrease of energy
structure (I) is less reactive than (III)as it has
barrier and hence increase in the rate of reaction.
bulky ethyl group at the axial position
26. Answer (A, C, D)

Hint : CFSE  stability.

Solution :
1. In a transition group, stability increases down
the group due to increase in effective nuclear
charge.
23. Answer (C)
2. NO2– is stronger ligand than NH3.
Hint : Fe(OH)3 and Na2CrO4 will be formed.
3. Chelate complexes are more stable and as
Solution : Na2O2 + 2H2O → 2NaOH + H2O2 the number of cyclic rings increases, the
2FeSO4 + 4NaOH + H2O2 → 2Fe ( OH)3  + 2Na 2SO4 stability of the complex increases.
(Brown)
4. For the same metal ion, stability of the
complex increases with increase in oxidation
Al2 ( SO4 ) + NaOH → 2NaAlO2 + 3Na2SO4 + 4H2O state of the metal ion.
3

27. Answer (A, C, D)

Cr2 ( SO4 ) + 10NaOH + 3H2 O2 →
3
Hint : Fact based.
2Na2 CrO4 + 3Na2 SO4 + 8H2 O
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

6/13
Test - 3A (Paper-2) (Code-C)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
Solution : (B) Dehydrohalogenation occurs.
2+
2Zn + K 4 Fe ( CN)6  → Zn2 Fe ( CN)6 + 4K +
(C) It has 3 stereocentres which means 8
White ppt. optically active isomers
2+ (D) The given compound has 3 chiral centres and
2Cd + K 4 Fe ( CN)6  → Cd2 Fe ( CN)6 + 4K +
White ppt. disubstituted cyclic ring that show
2+ geometrical isomerism.
2Cu + K 4 Fe ( CN)6  → Cu2 Fe ( CN)6 + 4K +
Brown ppt.

Al3+ ion does not form any precipitate with

K4[Fe(CN)6].
28. Answer (B, C, D)
Hint : With Fe, NO exist in +1 O.S.
Solution :
31. Answer (D)
2+
(I) FeSO4 + 5H2O + NO → Fe (H2O ) NO 
 5  Hint :
( X)

Oxidation state of Fe changes from +2 to +1

due to transfer of an electron from NO to Fe+2.
Electronic configuration of Fe+ is 3d7. It has
three unpaired electrons and hence magnetic

moment of (X) is 15 BM .

(II) Na2 Fe ( CN)5 NO + Na2S → Na4 Fe ( CN)5 NOS
 (Y) 
Solution :
Oxidation state and hybridisation of Fe in the Compound (P) has two dissimilar chiral
C-atoms. It has two pair of enantiomers or four
reactant and product (Y) of reaction
pair of diastereomers.
(II) remains unchanged i.e. +2 and d2sp3 32. Answer (D)
respectively. Hint : Compound (Q) has two chiral centres.
29. Answer (A, B, D) Solution :
No. of optical Isomers = 2n
Hint : Less hindered more rate.
(for unsymmetrical molecule) = 2n = 4
Solution :
33. Answer (A)
(CH3 )3 C — Br  CH3CH2CH2CH2 — Br  CH3 — CH2 — Br  CH3 — Br
Hint : Saytzeff elimination.
Rate of SN 2 reaction
Solution : R is Ph — CD = CH — CH3
30. Answer (A, B, C) 34. Answer (A, B, D)
Hint : 35. Answer (A, B, C)
Hint : Sodium salt gives violet colour solution with
Less hindered double bond will be more reactive.
sodium nitroprusside.
Solution : Solution : P, Q and R are Na2SO3, Na2S2O3 and
Na2S Respectively
(A) Loss of Br (a) atom in dehydrobromination
reaction results in the formation of least
Na2SO3 + 2HCl → 2NaCl + H2O + SO2 ( g)
(P ) ( X)
stable alkene and hence most reactive
towards hydrogenation. Na2S2O3 + 2HCl → 2NaCl + SO2 ( g) + S  + H2O
( Q) (X)

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

7/13
All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper-2) (Code-C)_(Hints & Solutions)

Na2S+ 2HCl → 2NaCl + H2S ( g) Zn2+ + (NH4 ) Hg ( SCN)4  → Zn Hg ( SCN)4   + 2NH+4
2
(R ) ( Y)

37. Answer A(P, R, T); B(Q, T); C(P, S, T); D(P, Q, T)

SO2 + 2H2S → 3S + 2H2O
Hint :
2H2S + O2 → 2H2O + 2S The possible stereoisomer of the given
complexes are
H2S + 2FeCl3 → 2FeCl2 + 2HCl + S
(A)
H2S + ( CH3COO) Pb → PbS + 2CH3COOH
2 Black

Na2S + Na2 Fe ( CN)5 NO  → Na4 Fe ( CN)5 NOS 

Violet colour

36. Answer A(Q, R, S, T); B(P, Q, R); C(P, Q, R);

D(P, R, T)
Hint : Fact based
Solution :

Co2+ + 4KSCN → Co ( SCN)4  + 4K +

2–

Blue colour
(B)
Co(NO3 )2 + NaOH → Co ( OH) (NO3 )  +NaNO3
Blue

Co ( OH) (NO3 ) + NaOH ⎯⎯⎯⎯

Warm
→ Co ( OH)2 +NaNO3
Excess Pink

Co2+ + 7NO2– + 2H+ + 3K + → K3 Co (NO2 )   +NO  + H2O

 6
Yellow Solution :
(C)
2Cu2+ + K 4 Fe ( CN)6  → Cu2 Fe ( CN)6   +4K +
Chocolate brown

Cu2 + + 2NaOH → Cu ( OH)2  +2Na+

Bluish white

2Cu2+ + 4KSCN → 2Cu (SCN)2  → 2CuSCN  + (SCN)2

Black White

4Fe3 + + 3K 4 Fe ( CN)6  → Fe 4 Fe ( CN)6 3 + 12K +

Pr ussian Blue

Fe3+ + 3SCN– ⎯⎯→ Fe ( SCN)3

Deep red colour

Fe3 + + 3NaOH ⎯⎯→ Fe ( OH)3  + 3Na+

Brown

3Zn2+ + 2K 4 Fe ( CN)6  → K 2 Zn3 Fe ( CN)6 2 + 6K +

White
(D)
Zn2+ + 2NaOH → Zn ( OH)2  + 2Na+
White

Zn ( OH)2 + 2NaOH → Na  Zn (OH)4 

(Excess) Water soluble

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

8/13
Test - 3A (Paper-2) (Code-C)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
38. Answer (07) PART - III (MATHEMATICS)
Hint : D. B. E = 1. 41. Answer (B)
Solution : Hint : Put secx + tanx = t
Solution :
dt 1
 sec x dx = and sec x − tan x =
t t

4 1+ 2

 t (t + t )
2
 ( 2sec x )( 2sec x )
12 12
−1
I= dx = dt
0 1

39. Answer (03) 42. Answer (A)

Hint : PbBr2 → Colourless Hint :  +  = –t2 and  = –2t

MnS → Buff Solution :

Solution : As2S5, PbI2 and AgI are yellow

2
   2 + 2  1 1 
f (t ) =   x 2 +  2 2  x + 2 2 +  dx
coloured compound. −1       
40. Answer (13)
x 3  t − 2 ( −2t )  x 2  1
2
4
1
= +  + 2 − x
Hint : Pt form square planar complex. 3  4t 2  2  4t 2t
−1
Solution :
 8 1  t 1 3  1 1
2
=  +  +  +   +  2 − 3
 3 3   4 t  2  4t 2t 

3t 2 3
=3+ + 2
8 4t
6t 6
f  (t ) =
1
− =0t =  2 4
8 4t 3

3 2 3 3 2
f ( t )min = 3 + + =3+
8 4 2 4
43. Answer (B)
3
3 9  2
Hint : ( 9 − 16 x ) = x  2 − 16 
3
2 2

x 
Solution :
 x=7 dx
 3
 9  2
x  2 − 16 
3

x 

9
Put − 16 = t 2
x2
18
− dx = 2t dt
x3
1 2t dt
18  t 3
−

 y=6 1 x
= +c = +c
( )
1
9t 9 9 − 16 x 2 2
x + y = 13
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

9/13
All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper-2) (Code-C)_(Hints & Solutions)

44. Answer (A) 46. Answer (C, D)

Hint : Degree is power of highest differential Hint : Substitute x = sint
coefficient when expressed as polynomial in Solution :

differential coefficients 1 2

Solution :  f ( x ) dx =  f ( sin t ) cos tdt = I

0 0
2sin−1 x A 
y = − 2

1− x2 1− x2 =  f (cos t ) sin tdt = I

0
 y  1 − x 2 = 2sin−1 x − A 
2

y  ( −2 x ) 2
 2I =  ( f ( sin t ) cos t + f (cos t ) sin t ) dt
 y  1 − x 2 + = 0
1− x2 1− x2 
2

 ( )
y  1 − x 2 − 2xy  = 2  2I   1 dx
0
Hence degree = 1 
45. Answer (A) I
4
Hint : Area between f(x) and g(x) is

47. Answer (A, B, C)
 ( f ( x ) − g ( x ) ) dx ,

where  and  are point of Hint : Form linear differential equation
Solution :
intersection of the curves
Solution : ( −1
e tan y − x
dy
dx
)
= 1+ y 2
−1
dx e tan y − x
 =
dy 1+ y 2
−1
dx x e tan y
 + =
dy 1 + y 2
1+ y 2
dy
 1+ y 2 −1
I.F = e = etan y

−1

 d x e(tan −1
y
) =
e2 tan y
1+ y 2
dy

−1 1 2 tan−1 y
 x e tan y
= e +c
( )
1 0
A=  x dy =  − y + 2y + 3 − (1 − y ) dy
2 2
− 2 +1 − 2 +1 1 
  , 0
2 
( )
1
+ − y 2 + 2y + 3 − ( y + 1) dy −1

0  2x = etan y

1 0 1 
e 3
  4 − ( y − 1) dy +  ( y − 1) dy −  ( y + 1) dy
2
 x0 =
1− 2 1− 2 0 2
1 48. Answer (A, D)
 y −1 4  y − 1 
4 − ( y − 1) + sin−1 
2
   Hint : Multiply and divide by sec2x
 2 2  2   1− 2 Solution :

( ) sec 4 x
0 1 −11 −1
 y2   y2  I =  sin 3
x  cos 3
x  dx
+ −y − +y sec 4 x
 2  1−  2 0
2

 I=
(1 + tan x ) sec 2 2
x dx
 1  1 
( 0 ) + 2sin−1  −   + 0 +
11
 0 + 0 − − tan x 3

 2  2 
 Put tanx = t  sec2x dx = dt
 1+ 2 − 2 2 − 2 + 2 2   1   1 t2 
  −  + 1 − 0   I =   11 + 11  dt
 2  2  t 3 t 3 
    1 3   3 3
( tan x ) 3 − ( tan x ) 3 + c
−8 −2
 −2  −  +   −   − 1 sq. units =−
 4   
2 2 2  8 2
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

10/13
Test - 3A (Paper-2) (Code-C)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
49. Answer (C, D) 1 1 y 
2

 − = +c
Hint : Area between f(x) and g(x) is xy 2  x 

 4, −2 
 ( f ( x ) − g ( x ) ) dx ,

where  and  are point of
 c=0
intersection of f(x) and g(x)
f ( x )  y = ( −2x )
1
3
Solution :
Also g(x) = sin–1sinx·2sinxcosx
+ cos–1cosx 2cosx(–sinx) = 0
 g(x) = constant
 3  
Put x = we get g(x)  y =  x  0, 
4 16  2

 2


2
A1 =  ( sin x − cos x ) dx +

 sin x dx = − cos x − sin x

4 2 
4

 1 1
+ ( − cos x )  = −1 + + + 1− 0 = 2
2 2 2
5
5 4
4
A nd A1 + A2 =  ( sin x − cos x ) dx = − cos x − sin x

4 
4

2

=
1
2
+
1
2
+
1
2
+
1
2
=2 2 A=  ( g ( x ) − f ( x )) dx
0

 A2 = 2 and A1 = 2  A1A2 = 2 3 3
4
  3
2

A= 1 + 
50. Answer (A, B, D) 8  4 

Hint : If x  (0, 1)  x2 > x3 and if x  (1, 2) x3 > x2 54. Answer (A, B, C)

Solution : 55. Answer (C, D)

Hint for Q.Nos. 54 and 55 :
If x  ( 0, 1) ( 2020 )  ( 2020 )
x2 x3

Substitute expression in tan.

and if x  (1, 2 ) ( 2020 )  ( 2020 )
3 2
x x Solution for Q.Nos. 54 and Q.55 :

 I4 > I3 > I1 > I2 (sin 3

2
3
 + cos 2  d  )
51. Answer (A, C)  sin3  cos3  sin (  +  )
52. Answer (C, D)
3
sin 2  d 
53. Answer (A, B, C)   sin 3
2
 cos3   ( sin  cos  + cos  sin  )
Hint for Q.Nos. 51 to 53 :
xdy − ydx y  3
xdy + ydx = d(xy) and =d  cos 2  d 
x2 x +
cos 2  sin3  ( sin  cos  + cos  sin  )
3

Solution for Q.Nos. 51 to 53 :

xy dx + x 2dy y ( x dy − y dx ) d
3

=   cos 2
 ( tan  cos  + sin  )
x2 x2
I1
y y
  d ( xy ) =  x y
2 2

x  x 
d
d
+  sin 2
 ( cos  + cos  sin  )
d ( xy ) y y
  ( xy ) 2
=  d 
x x
I2

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

11/13
All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper-2) (Code-C)_(Hints & Solutions)

For I1 put tan cos + sin = t2 −5

3
2
( x + 5 )2
e dx + k  e 3 x −2 dx
( )2
(D) I =
 cos sec2 d = 2t dt −4 1
3
For I2 put cos + cot sin = 2
I1 + I2
 –cosec2·sin d = 2 d
In I1 put x + 5 = y and in I2 put 3x – 2 = –t
2t dt 2 d 
 cos   t −  sin  
0 0
 I =  e y dy +
k y2 (
e −dt ) = 0
3 1
2

 2sec  sin  + tan  cos  − 2cosec   k=3

cos  + cot  sin  + c 57. Answer A(P, Q); B(P, R, S, T); C(R, T); D(R, S, T)

 f() = sin + cos tan Hint : Applying properties of integrals

Solution :
and g() = cos + cot sin
1 1 1
f() = 2sin  (0, 2) (A) 5  sin  x dx = 5  sin x dx = 5 ( − cos x )
0 0
g() = 2cos 6

= 5 (1 − cos1)
(
 f (  ) + g (  )  2, 2 2 

5
56. Answer A(P, S); B(S); C(Q, S, T); D(Q, S, T) (B)  ( −1) dx
−5
= 10
(A) Hint : Convert all terms into sinx and cosx
1
r4 n
x5 1
( sin2 x + cos2 x ) dx (C) lim  = 0 x 4
dx = =
I= n →
r =1 n
5
5 5
sin4 x cos2 x

 cos2 x  (e 2x
−1 )  2x = lim
e2 x − 1 4 x 2
 n
=   sec 2 x + 2cosec 2 x +  dx
(D) lim
x →0 xn n → 2x x
 sin4 x 
 n  3 for limit to be finite
cot 3 x
= tan x − 2cot x − +k 58. Answer (19)
3
Hint : Integration by parts
1
A = 1, B = −2 and c = −
3 Solution :
1
 4A + B + 3C = 1 I n =  1
1
dx
(1 + x )
n
2
27 0
(B) Hint : Distribute x in both brackets

g ( x ) =  ( x 4 + x 5 + x 6 ) ( 6x 5 + 5x 4 + 4x 3 ) dx
6 1
1
x 2 x 2dx
= + n
(1 + x ) (1 + x )
n n +1
2 2
Put x6 + x5 + x4 = t 0
0

 (6x5 + 5x4 + 4x3)dx = dt

1 1 
1  dx dx 
 0 1 + x 2
1( 6 = + 2n −
x + x4 + x5 ) + c
7
= ( ) ( 
)
n n +1
2n 1+ x2
7  0


( x6 + x4 + x5 ) 7

As g(0) = 0 g (x) =
7 1
= + 2n In − 2n In +1
2n
37
 g (1) =
7 2n In + 1 = 2–n + (2n – 1)In

2
Put n = 10, we get
1
(C)  12sin3 x G x d x = 12  =3 20 I11 = 2–10 + 19 I10
0
4
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

12/13
Test - 3A (Paper-2) (Code-C)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
59. Answer (30) 3
2

Hint : Distance from origin  distance from line

 A= 
0
9 − 6 x dx

x=3 3
2

(9 − 6x ) 2
3
Solution :
=
3
 ( −6 )
2 0

(9)
3
2
27
=0+ = = 3 sq. units
9 9
 10A = 30
60. Answer (32)
a
Hint :  f ( x ) dx = 0
−a
if f(x) is odd.

Solution :

4
I=  sec
2
x dx
Given x + y  x −3
2 2
−
4


 x2 + y2  x2 – 6x + 9 4
= 2  sec 2 x dx
 y2  – 6x + 9 0


= 2 tan x 0 4 = 2
 3
 y 2  −6  x − 
 2  I5 = 25 = 32



Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

13/13
Test - 3A (Paper-2) (Code-D)_(Answers) All India Aakash Test Series for JEE (Advanced)-2020

All India Aakash Test Series for JEE (Advanced)-2020

TEST - 3A (Paper-2) - Code-D
Test Date : 06/10/2019

ANSWERS
PHYSICS CHEMISTRY MATHEMATICS
1. (A) 21. (B) 41. (A)
2. (C) 22. (D) 42. (A)
3. (A) 23. (C) 43. (B)
4. (A) 24. (A) 44. (A)
5. (C) 25. (D) 45. (B)
6. (C) 26. (A, B, C) 46. (A, B, D)
7. (A, C) 27. (A, B, D) 47. (C, D)
8. (A, B) 28. (B, C, D) 48. (A, D)
9. (A, C) 29. (A, C, D) 49. (A, B, C)
10. (A, D) 30. (A, C, D) 50. (C, D)
11. (A, D) 31. (D) 51. (A, C)
12. (A, C) 32. (D) 52. (C, D)
13. (A, C, D) 33. (A) 53. (A, B, C)
14. (A, C) 34. (A, B, D) 54. (A, B, C)
15. (A, C) 35. (A, B, C) 55. (C, D)
16. A → (Q, T) 36. A → (P, R, T) 56. A → (P, Q)
B → (R, S, T) B → (Q, T) B → (P, R, S, T)
C → (P, R, S) C → (P, S, T) C → (R, T)
D → (P, Q) D → (P, Q, T) D → (R, S, T)
17. A → (Q) 37. A → (Q, R, S, T) 57. A → (P, S)
B → (P, T) B → (P, Q, R) B → (S)
C → (P, S) C → (P, Q, R) C → (Q, S, T)
D → (P, R, T) D → (P, R, T) D → (Q, S, T)
18. (22) 38. (13) 58. (32)
19. (20) 39. (03) 59. (30)
20. (25) 40. (07) 60. (19)

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

1/12
All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper-2) (Code-D)_(Hints & Solutions)

PART - I (PHYSICS)  Leff =

2
H
1. Answer (A) 3
Hint : Induced electric field is non-conservative itotal = 2 + 3 + 5 = 10 amp
in nature. 1 2 q2
Solution : Induced electric field are produced by Li =
2 2C
changing magnetic field and they form a closed
2
loop,  q2 =  10  10  2  10–6
 E  dl  0
So and potential can’t be defined. 3

20 20
2. Answer (C)  q=  10 –3 C = mC
Hint : AC circuit Analysis. 3 3
V 4. Answer (A)
Solution : VA =
2 Hint : At the instant of sharp change, the flux
VR  –  L would remain same.
and VB = tan  =
R 2 + ( L ) R
2 Solution : Just after changing flux would remain
same
For R → 0 VB = V
 L
V  L = i
 v = R 3
2
For R →  VB = V 3
 i = = 3l 0
V R
 v =
2 L di
Now,  – Ri – =0
At any time V 3 dt
L di
 (  – Ri ) =
3 dt
t i
3dt di
  L
= 
 – Ri
0 3i 0

–3Rt   – Ri 
 = ln 
V = VA – VB = VA2 + VB – 2 (VA )(VB )  cos   –2 
2
 L
3Rt
 V2 V 2R 2 2 V –
V 2 = + –  Ri –  = 2e L
4 R 2 + ( L )2 2

 
2Rt
VR R –
  i = 1 + 2e L 
R 2 + ( L ) R 2 + ( L )
2 2 R
5. Answer (C)
V2
 V 2 =
4 Hint : Displacement method.
V Solution :
 V =
2 150 – x
m1 =
3. Answer (A) x
1 1 1 1
Hint : = + + x
Leq L1 L2 L3 m2 =
150 – x
Solution :
16 (150 – x )
2
m1
1 1 1 1  = =
= + + m2 1 x2
Leff L1 L2 L3

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

2/13
Test - 3A (Paper-2) (Code-D)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020

150 – x Solution : Let x be the object distance from lens

 =4 and d be the distance between lens and mirror.
x
1 1 1
 x = 30 cm Then for image by lens + =
v1 x f
1 1 1
 + =
1 x–f
120 30 f  =
v1 xf
1 5
 =
f 120 v1 f x f
 m1 = = v1 =
 f = 24 cm x x–f x–f

6. Answer (C) xf
For 2nd time object distance is 2d –
x–f
Hint : L-R circuit.
1 1 1
Solution :  =
5L + =
v2 xf f
R 2d –
x–f
iL(t = 0) = 0
f
4 m2 =
iL(t =  ) = xf
R 2d – –f
x–f
4  
−tR
iL = 1 − e 5L   f  f
R   m1. m2 =  .
  x–f  xf
2d – –f
7. Answer (A, C) x–f

V f 2 (x – f )
Hint : Z = ; P = irms  Vrms cos  =
I ( x – f ) ( 2 xd – 2df – xf – xf + f 2 )
Solution :  = 100
f2
 m=
 f 2 – 2df + 2 x ( d – f )
V = 400 sin  t +  volt
 6
For m to be independent of x
I = 600 sin(t) mA
df = 0  d=f
V 400  103 2000
 Z = = =  f2
I 600 3 So, m = =1
f 2 – 2f 2
3
Power factor = 10. Answer (A, D)
2
Average power dissipation: Hint : Image by reflection.
Solution : As the object is not on the bisector,
400 600  10–3 3
  = irms  Vrms cos  the polygon will be irregular.
2 2 2

 P avg = 60 3 watt.
8. Answer (A, B)
Hint : It could be real image or virtual image.
–f –f
Solution : = 2 and = –2 11. Answer (A, D)
–f + u –f + u
12. Answer (A, C)
f 3f
 u= and u = 13. Answer (A, C, D)
2 2
9. Answer (A, C) Hint for Q.Nos. 11 to 13 :
1 1 1 v 1 1 1
Hint : − = ;m = + =
v u f u v u f

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

3/13
All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper-2) (Code-D)_(Hints & Solutions)

Solution for Q.Nos. 11 to 13 : dx0 f f

Velocity of object = −  sin t = − 3
dt 2 4

4f 3 f 3 4 1
 Vrel = (v 2 − v1 ) = + = 3f  9 + 4 
9 4
f 25
At any time t, x = cos t  vrel = 3 f
2 36
 object distance from mirror u = (2f – x ) Magnification is always negative so always real image
(towards left) will be formed. For the time when object lies between
Let v is the position of image w.r.t. mirror then 2f and f from mirror it will produce magnified image.
1 1 2 For rest of the time it will be diminished image.
+ =−
v u R 14. Answer (A, C)
1 1 −1 15. Answer (A, C)
 − =
v 2f − x f Hint for Q.Nos. 14 and 15 :
1 f − 2f + x eE = m2  x
 =
v f ( 2f − x )
Solution for Q.Nos. 14 and 15 :
f ( 2f − x )
 v=
x −f
 Distance of image from mirror
 f 
f  2f − cos t 
v =  
2 Because of pseudo force, free charge would try

f to shift outward since free charges are electrons.
cos t − f
2 So Fout = m2x. Because of that electric field
f 2 ( 4 − cos t )  2 f ( 4 − cos t ) would induce in a way that eE = m2  x
 v IM = =
2  f ( cos t − 2 ) 2 − cos t m2
Now position of image w.r.t. origin is  E=  x (from A to B as electron has
e
f ( 2f − x ) shifted towards B)
rI = rIM + rM = + 2f
0 0 x −f l +L
m2
 rI =
2f − fx + 2fx − 2f
2 2
=
f x VAB =  Edx =
e 
xdx
l
0 x −f x −f
m2 2 l +L m2 (
f
f  cos t  VAB = x l = L L + 2l )
2 f 2 cos t 2e 2e
 rI = =
0 f f ( cos t − 2 )
cos t − f m2
2 As we see E =  x implies that the field
e
f cos t
 rI = − (oscillating but not SHM) region (in rod and outside of it) would exist in x
0 2 − cos t direction as per the equation then there must be
Now velocity of image is uniform charge distribution.
16. Answer A(Q, T); B(R, S, T); C(P, R, S); D(P, Q)

d
( ) =v
rI
0
=
( 2 − cos t ) f  sin t + f cos t   sin t
Hint : For lens
1 1 1
− =
dt I
( 2 − cos t ) 2 v u f
1 1 1
For mirror + =
f sin t ( 2 − cos t + cos t ) 2f sin t v u f
vI = =
( 2 − cos t ) 2
( 2 − cos t )2 1 1 1
Solution : For lens − =
v u f
 2f 3 f 3  4
At t = vI = =
3  1
2 9 For mirror
1 1 1
+ =
2 2 −  v u f
 2

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

4/13
Test - 3A (Paper-2) (Code-D)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
17. Answer A(Q); B(P, T); C(P, S); D(P, R, T) 19. Answer (20)
Hint : Req(t = 0) = 40  Hint : After first refraction from the plane surface
80 3
Req(t = ) =  the image seems to be at L from plane surface.
3 2
transient LR circuit. Solution : After first refraction from the plane
Solution : 3
surface p seems to be at L from plane surface.
2

1  (1 −  )
Now for 2nd surface + =
v  3  −R
R + L
 2 

Req (t = 0) = 40   v=
80   −1
Req ( t =  ) =   =
3 3 R
R+ L
100 2
Clearly at t = 0 current i1 = = 2.5 A
40  L = 20 cm
5
and at t =  l1 = = 1.25 A 20. Answer (25)
4
1 1 1
Current i2 at t = 0 is zero at inductor and it will Hint : = +
oppose the sudden change of current. 
z 5 − 5 j 10 j
10 Solution :
And i2 at t =  is = 2.5 A
4
Power delivered by battery p = i
As i increases from 2.5 A to 3.75 A
So power delivered by battery increases from
250 watt to 375 watt.
18. Answer (22)
2 1 2 − 1 Let Z be the impedance across MN.
Hint : − =
v u R
1 1 1
Solution : Then = +
Z  5 − 5 j 10 j

1 10 j + 5 − 5 j 5+5j
 = =
Z 50 j + 50 50 + 50 j

 Z = 10 
7 
 − 1 
1
= 20 
7
+
1
=  4 
C
4v1 24 6
from water surface 1
4 7  f = = 25 Hz
 −  2  20  C
= 3 4
4 7
−
3v 2 4v1  PART - II (CHEMISTRY)
4 1 3 21. Answer (B)
 + =
3v 2 24 24 Hint : Dispersion of charge decreases the
4 1 energy.
 =
3v 2 12 Solution :
 v2 = 16 cm
 Distance from bottom of tank is 38 – 16 = 22 cm.

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

5/13
All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper-2) (Code-D)_(Hints & Solutions)

The general energy diagram of the above reaction nucleophile. Thus structure (IV) is least reactive.
is Structure (I) and (III) are cis-trans type. Their
reactivity lies between those of (II) and (IV),
structure (I) is less reactive than (III)as it has
bulky ethyl group at the axial position

On increasing polarity of the solvent, all charged

species will get solvated. Thus their energies will be
lowered. In the above reaction the reactants do not 25. Answer (D)
carry any charge and hence their energy remains
Hint : Spectator ligand will affect the C — O bond
unaffected by the increase in polarity of the solvent.
length. Order of ligand field strength of the given
But the energy of transition state is lowered due to
ligand is
its solvation. This results in decrease of energy
barrier and hence increase in the rate of reaction. CO  PF3 > PCl3 > PAr3 > PMe3

22. Answer (D) Solution : The ligand PPh3 is a weaker 

acceptor than CO. As a result d electrons of
Hint : Cation present is Na+.
metal in such mixed carbonyl will be drawn more
Solution : Fe(CH3COO)3 is red in colour. towards CO than in pure metal carbonyl. If the Ph
23. Answer (C) groups of PPh3 are replaced by more electron
attracting Cl or F groups, the tendency of d
Hint : Fe(OH)3 and Na2CrO4 will be formed.
electrons of metal to move towards P increases.
Solution : Na2O2 + 2H2O → 2NaOH + H2O2 And if Ph groups of PPh3 are replaced by electron
2FeSO4 + 4NaOH + H2O2 → 2Fe ( OH)3  + 2Na 2SO4 releasing Me groups, the tendency of d
(Brown)
electrons of metal to move towards P further
decreases. As d electrons of metal move
Al2 ( SO4 ) + NaOH → 2NaAlO2 + 3Na2SO4 + 4H2O
3 towards P more and less towards CO, the CO
bond Order will be more or CO bond length will
Cr2 ( SO4 ) + 10NaOH + 3H2 O2 → be less and vice versa.
3

2Na2 CrO4 + 3Na2 SO4 + 8H2 O 26. Answer (A, B, C)

Hint :
Fe(OH)3 is a brown residue, NaAlO2 is a
colourless solution and Na2CrO4 is a yellow Less hindered double bond will be more reactive.
solution. Solution :
24. Answer (A) (A) Loss of Br (a) atom in dehydrobromination
reaction results in the formation of least
Hint : For the fast rate, back side of LG Should
stable alkene and hence most reactive
be less hindered.
towards hydrogenation.
Solution : In an SN2 reaction, the leaving group (B) Dehydrohalogenation occurs.
must be in an axial position in order to allow (C) It has 3 stereocentres which means 8
backside attack to occur without steric optically active isomers
hinderance from the cyclohexane ring. When the (D) The given compound has 3 chiral centres and
Br-atom is in axial position in the all cis-isomer, disubstituted cyclic ring that show
both the methyl groups are in equatorial geometrical isomerism.
positions, the structure (II) is most reactive as the
approaching nucleophile experiences least
crowding. In the all trans isomer (IV) both the
ethyl groups are in axial positions providing the
maximum crowding to the approaching
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

6/13
Test - 3A (Paper-2) (Code-D)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
27. Answer (A, B, D) 3. Chelate complexes are more stable and as
Hint : Less hindered more rate. the number of cyclic rings increases, the
stability of the complex increases.
Solution :
4. For the same metal ion, stability of the
(CH3 )3 C — Br  CH3CH2CH2CH2 — Br  CH3 — CH2 — Br  CH3 — Br
complex increases with increase in oxidation
Rate of SN 2 reaction
state of the metal ion.
28. Answer (B, C, D)
31. Answer (D)
Hint : With Fe, NO exist in +1 O.S.
Hint :
Solution :
2+
(I) FeSO4 + 5H2O + NO → Fe (H2O ) NO 
 5 
( X)

Oxidation state of Fe changes from +2 to +1

due to transfer of an electron from NO to Fe+2.
Electronic configuration of Fe+ is 3d7. It has
three unpaired electrons and hence magnetic
moment of (X) is 15 BM .
Solution :
(II) Na2 Fe ( CN)5 NO + Na2S → Na4 Fe ( CN)5 NOS Compound (P) has two dissimilar chiral
 (Y) 
C-atoms. It has two pair of enantiomers or four
Oxidation state and hybridisation of Fe in the pair of diastereomers.
reactant and product (Y) of reaction 32. Answer (D)
(II) remains unchanged i.e. +2 and d2sp3
Hint : Compound (Q) has two chiral centres.
respectively.
Solution :
29. Answer (A, C, D)
No. of optical Isomers = 2n
Hint : Fact based.
Solution : (for unsymmetrical molecule) = 2n = 4

2+ 33. Answer (A)

2Zn + K 4 Fe ( CN)6  → Zn2 Fe ( CN)6 + 4K +
White ppt. Hint : Saytzeff elimination.
Solution : R is Ph — CD = CH — CH3
2+
2Cd + K 4 Fe ( CN)6  → Cd2 Fe ( CN)6 + 4K + 34. Answer (A, B, D)
White ppt.
35. Answer (A, B, C)
2+
2Cu + K 4 Fe ( CN)6  → Cu2 Fe ( CN)6 + 4K +
Hint : Sodium salt gives violet colour solution with
Brown ppt.
sodium nitroprusside.
Al3+ ion does not form any precipitate with Solution : P, Q and R are Na2SO3, Na2S2O3 and
K4[Fe(CN)6]. Na2S Respectively
30. Answer (A, C, D)
Na2SO3 + 2HCl → 2NaCl + H2O + SO2 ( g)
(P ) ( X)
Hint : CFSE  stability.
Solution : Na2S2O3 + 2HCl → 2NaCl + SO2 ( g) + S  + H2O
( Q) (X)
1. In a transition group, stability increases down
the group due to increase in effective nuclear Na2S+ 2HCl → 2NaCl + H2S ( g)
(R ) ( Y)
charge.
2. NO2– is stronger ligand than NH3. SO2 + 2H2S → 3S + 2H2O

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

7/13
All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper-2) (Code-D)_(Hints & Solutions)

2H2S + O2 → 2H2O + 2S (D)

H2S + 2FeCl3 → 2FeCl2 + 2HCl + S

H2S + ( CH3COO) Pb → PbS + 2CH3COOH

2 Black

Na2S + Na2 Fe ( CN)5 NO  → Na4 Fe ( CN)5 NOS 

Violet colour

36. Answer A(P, R, T); B(Q, T); C(P, S, T); D(P, Q, T) 37. Answer A(Q, R, S, T); B(P, Q, R); C(P, Q, R);
Hint : D(P, R, T)
The possible stereoisomer of the given
complexes are Hint : Fact based
(A) Solution :

Co2+ + 4KSCN → Co ( SCN)4  + 4K +

2–

Blue colour

Co(NO3 )2 + NaOH → Co ( OH) (NO3 )  +NaNO3

Blue

Co ( OH) (NO3 ) + NaOH ⎯⎯⎯⎯

Warm
→ Co ( OH)2 +NaNO3
Excess Pink

Co2+ + 7NO2– + 2H+ + 3K + → K3 Co (NO2 )   +NO  + H2O

 6
Yellow

(B) 2Cu2+ + K 4 Fe ( CN)6  → Cu2 Fe ( CN)6   +4K +

Chocolate brown

Cu2 + + 2NaOH → Cu ( OH)2  +2Na+

Bluish white

2Cu2+ + 4KSCN → 2Cu (SCN)2  → 2CuSCN  + (SCN)2

Black White

Solution :
4Fe3 + + 3K 4 Fe ( CN)6  → Fe 4 Fe ( CN)6 3 + 12K +
(C)
Pr ussian Blue

Fe3+ + 3SCN– ⎯⎯→ Fe ( SCN)3

Deep red colour

Fe3 + + 3NaOH ⎯⎯→ Fe ( OH)3  + 3Na+

Brown

3Zn2+ + 2K 4 Fe ( CN)6  → K 2 Zn3 Fe ( CN)6 2 + 6K +

White

Zn2+ + 2NaOH → Zn ( OH)2  + 2Na+

White

Zn ( OH)2 + 2NaOH → Na  Zn (OH)4 

(Excess) Water soluble

Zn2+ + (NH4 ) Hg ( SCN)4  → Zn Hg ( SCN)4   + 2NH+4

2

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

8/13
Test - 3A (Paper-2) (Code-D)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
38. Answer (13) PART - III (MATHEMATICS)
Hint : Pt form square planar complex. 41. Answer (A)
Solution : Hint : Area between f(x) and g(x) is


 ( f ( x ) − g ( x ) ) dx ,

where  and  are point of

intersection of the curves

Solution :

( )
1 0
A=  x dy =  − y 2 + 2y + 3 − (1 − y ) dy
− 2 +1 − 2 +1
 x=7
( )
1
+ − y 2 + 2y + 3 − ( y + 1) dy
0
1 0 1
  4 − ( y − 1) dy +  ( y − 1) dy −  ( y + 1) dy
2

1− 2 1− 2 0

1
 y −1 4  y − 1 
4 − ( y − 1) + sin−1 
2
  
 2 2  2   1− 2
0 1
y 2
y 2
+ −y − +y
 2  1− 2  2 0

 y=6  1  1 
 0 + 0 − − ( 0 ) + 2sin−1  −   + 0 +
 2  2 
x + y = 13
 1+ 2 − 2 2 − 2 + 2 2   1 
39. Answer (03)   −  + 1 − 0 
 2  2 
Hint : PbBr2 → Colourless
    1 3  
MnS → Buff  −2  −  +   −   − 1 sq. units
 4   
2 2 2 
Solution : As2S5, PbI2 and AgI are yellow 42. Answer (A)
coloured compound. Hint : Degree is power of highest differential
coefficient when expressed as polynomial in
40. Answer (07) differential coefficients
Hint : D. B. E = 1. Solution :
2sin−1 x A
Solution : y = −
1− x 2
1− x2
 y  1 − x 2 = 2sin−1 x − A
y  ( −2 x ) 2
 y  1 − x 2 + =
1− x 2
1− x2
 ( )
y  1 − x 2 − 2xy  = 2
Hence degree = 1
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

9/13
All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper-2) (Code-D)_(Hints & Solutions)

43. Answer (B) 46. Answer (A, B, D)

3 Hint : If x  (0, 1)  x2 > x3 and if x  (1, 2) x3 > x2
 9  2
Hint : ( 9 − 16 x 2 ) 2 = x 3  2 − 16 
3

x  Solution :

If x  ( 0, 1) ( 2020 )  ( 2020 )
x2 x3
Solution :

and if x  (1, 2 ) ( 2020 )  ( 2020 )

x3 x2
dx
 3
 9  2
x 3  2 − 16   I4 > I3 > I1 > I2
 x 
47. Answer (C, D)
9
Put 2 − 16 = t 2 Hint : Area between f(x) and g(x) is
x 

18  ( f ( x ) − g ( x ) ) dx , where  and  are point of

− 3 dx = 2t dt 
x intersection of f(x) and g(x)
1 2t dt Solution :
18  t 3
−

1 x
= +c = +c
( )
1
9t 9 9 − 16 x 2 2

44. Answer (A)

Hint :  +  = –t2 and  = –2t 
 2


2
Solution : A1 =  ( sin x − cos x ) dx +

 sin x dx = − cos x − sin x

4 2 
2
  2 + 2  1 1 
4

f (t ) =   x 2 +  2 2  x + 2 2 +  dx
−1        
+ ( − cos x )  = −1 +
1
+
1
+ 1− 0 = 2
2 2 2
x 3  t − 2 ( −2t )  x 2  1
2
4
1 5
= +  + 2 − x 5 4
 4
 2  4t 2t 
2
3  4t
−1 A nd A1 + A2 =  ( sin x − cos x ) dx = − cos x − sin x

4 
4
 8 1  t 1 3  1 1
2
=  +  +  +   +  2 − 3 1 1 1 1
 3 3   4 t  2  4t 2t  = + + + =2 2
2 2 2 2
3t 2 3
=3+ + 2  A2 = 2 and A1 = 2  A1A2 = 2
8 4t
48. Answer (A, D)
6t 6
f  (t ) =
1
− 3 =0t =  2 4 Hint : Multiply and divide by sec2x
8 4t
Solution :
3 2 3 3 2
f ( t )min =3+ + =3+
( ) sec
4
−11 −1 x
8 4 2 4 I =  sin 3
x  cos 3
x  dx 4
sec x

(1 + tan x ) sec
45. Answer (B) 2 2
x dx
Hint : Put secx + tanx = t  I= 11
3
tan x
Solution :
 Put tanx = t  sec2x dx = dt
dt 1
 sec x dx = and sec x − tan x =  1 t2 
t t  I =   11 + 11  dt
t 3 t 3 

4 1+ 2

 t (t + t )
2
 ( 2sec x )( 2sec x )
12 12
−1
I= dx = dt =−
3 3
( tan x ) 3 − ( tan x ) 3 + c
−8 −2

0 1 8 2

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

10/13
Test - 3A (Paper-2) (Code-D)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
49. Answer (A, B, C) Solution for Q.Nos. 51 to 53 :

xy dx + x 2dy y ( x dy − y dx )
Hint : Form linear differential equation 3

=
Solution : x2 x2

(e tan−1 y
−x ) dy = 1+ y 2
  d ( xy ) =  x y
2 2 y y
x  x 
dx d
−1
dx e tan y − x
 = d ( xy ) y y
dy 1+ y 2   ( xy ) 2
=
x  x 
d
−1
dx x e tan y
 + = 2
dy 1 + y 2
1+ y 2 1 1 y 
 − =   +c
dy
xy 2  x 
 1+ y 2 −1
I.F = e = etan y
 4, −2 
−1

 d x e (
tan −1
y
) =
e2 tan y
1+ y 2
dy  c=0

f ( x )  y = ( −2x )
1
3
−1 1 −1
 x e tan y
= e 2 tan y + c
2 Also g(x) = sin–1sinx·2sinxcosx
1  + cos–1cosx 2cosx(–sinx) = 0
  , 0
2 
 g(x) = constant
−1
 2x = etan y

 3  
 Put x = we get g(x)  y =  x  0, 
e 3 4 16  2
 x0 =
2
50. Answer (C, D)
Hint : Substitute x = sint
Solution :

1 2

 f ( x ) dx =
0
 f ( sin t ) cos tdt = I
0


2
=  f (cos t ) sin tdt = I
0 
2


2
A=  ( g ( x ) − f ( x )) dx
 ( f ( sin t ) cos t + f (cos t ) sin t ) dt
0
 2I =
0
3 3
4
  3
2

 A= 1 + 
2 8  4 
 2I   1 dx
0 54. Answer (A, B, C)
 55. Answer (C, D)
I
4
Hint for Q.Nos. 54 and 55 :
51. Answer (A, C)
Substitute expression in tan.
52. Answer (C, D)
Solution for Q.Nos. 54 and Q.55 :
53. Answer (A, B, C)
Hint for Q.Nos. 51 to 53 : (sin 3
2
3
 + cos 2  d  )
xdy − ydx y   sin  cos  sin (  +  )
3 3
xdy + ydx = d(xy) and =d 
x 2
x
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

11/13
All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper-2) (Code-D)_(Hints & Solutions)
3
sin 2  d  (e 2x
−1 )  2x e2 x − 1 4 x 2
  sin 3
2
 cos3   ( sin  cos  + cos  sin  )
(D) lim
x →0 xn
= lim
n → 2x
 n
x

 n  3 for limit to be finite

3
cos 2  d 
+ 57. Answer A(P, S); B(S); C(Q, S, T); D(Q, S, T)
cos 2  sin3  ( sin  cos  + cos  sin  )
3

(A) Hint : Convert all terms into sinx and cosx

d
  cos 2
 ( tan  cos  + sin  ) I=
( sin2 x + cos2 x ) dx
sin4 x cos2 x
I1
 cos2 x 
d =   sec 2 x + 2cosec 2 x +  dx
+  sin 2
 ( cos  + cos  sin  )
 sin4 x 
I2 cot 3 x
= tan x − 2cot x − +k
3
For I1 put tan cos + sin = t2

 cos sec2 d = 2t dt 1
A = 1, B = −2 and c = −
3
For I2 put cos + cot sin = 2
 4A + B + 3C = 1
 –cosec2·sin d = 2 d
27
(B) Hint : Distribute x in both brackets
2t dt 2 d 
  cos   t −  sin  
g ( x ) =  ( x 4 + x 5 + x 6 ) ( 6x 5 + 5x 4 + 4x 3 ) dx
6

 2sec  sin  + tan  cos  − 2cosec 

Put x6 + x5 + x4 = t
cos  + cot  sin  + c  (6x5 + 5x4 + 4x3)dx = dt
 f() = sin + cos tan 1( 6
x + x4 + x5 ) + c
7
=
and g() = cos + cot sin 7

f() = 2sin  (0, 2) ( x 6 + x 4 + x 5 )7

As g(0) = 0 g (x) =
g() = 2cos 7

 f (  ) + g (  )  2, 2 2  (   g (1) =
37
7
56. Answer A(P, Q); B(P, R, S, T); C(R, T); D(R, S, T) 
2
1
 12sin x G x d x = 12  =3
3
Hint : Applying properties of integrals (C)
0
4
Solution :
3
−5 2
1
( x + 5 )2
e dx + k  e 3 x −2 dx
( )2
(D) I =
1 1

(A) 5  sin  x dx = 5  sin x dx = 5 ( − cos x ) −4 1

0 0 3
6

= 5 (1 − cos1)
I1 + I2

In I1 put x + 5 = y and in I2 put 3x – 2 = –t

5
(B)  ( −1) dx = 10 0 0
k y2 (
I =  e y dy + e −dt ) = 0
3 1
2
−5

1
1
n
r4 x5 1
(C) lim  5 =  x 4dx = = k=3
n →
r =1 n 0
5 5

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

12/13
Test - 3A (Paper-2) (Code-D)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
58. Answer (32)  3
a  y 2  −6  x − 
 2
Hint :  f ( x ) dx = 0
−a
if f(x) is odd.
3
2
Solution :

 A= 
0
9 − 6 x dx
4
I=  sec
2
x dx 3
2
−
(9 − 6x )
3
4 2

 =
3
 ( −6 )
4
= 2  sec 2 x dx 2 0
0

(9)
3
 2
27
= 2 tan x 0 4 = 2 =0+ = = 3 sq. units
9 9
 I5 = 25 = 32  10A = 30
59. Answer (30) 60. Answer (19)
Hint : Distance from origin  distance from line Hint : Integration by parts
x=3 Solution :
Solution : 1
1
I n =  1 dx
(1 + x )
n
2
0

1
1
x 2 x 2dx
= + n
(1 + x ) (1 + x )
n n +1
2 2
0
0

1 1 
1  dx dx 
= + 2n  −
 0 1+ x2
( ) ( )
n n n +1
2 1+ x 2
 0


1
= + 2n In − 2n In +1
2n
Given x2 + y 2  x − 3
2n In + 1 = 2–n + (2n – 1)In
 x2 + y2  x2 – 6x + 9 Put n = 10, we get
 y2  – 6x + 9 20 I11 = 2–10 + 19 I10



Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

13/13