all india aakash test series for jee (advanced)-2020 mock ... · all india aakash test series for...
Post on 28-Mar-2020
78 Views
Preview:
TRANSCRIPT
All India Aakash Test Series for JEE (Advanced)-2020
Test Date : 19/01/2020
ANSWERS
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
MOCK TEST - 1 (Paper-1) - Code-A
Mock Test - 1 (Paper-1) (Code-A)_(Answers) All India Aakash Test Series for JEE (Advanced)-2020
1/10
PHYSICS CHEMISTRY MATHEMATICS
1. (D)
2. (C)
3. (D)
4. (B)
5. (A)
6. (A, C)
7. (A, B, C)
8. (A, C, D)
9. (B, C)
10. (A, D)
11. (A, D)
12. (A, C)
13. (A, C)
14. (A, C)
15. (A, C)
16. A → (P)
B → (Q, S)
C → (R, T)
D → (Q, S)
17. A → (P, S)
B → (Q, T)
C → (P, S)
D → (Q, R)
18. (27)
19. (11)
20. (36)
21. (D)
22. (A)
23. (C)
24. (C)
25. (A)
26. (A, C, D)
27. (A, B, D)
28. (A, B, C)
29. (A, B, C, D)
30. (A, B, C, D)
31. (B)
32. (A)
33. (A, B, C)
34. (B)
35. (A, C, D)
36. A → (Q, T)
B → (P, R)
C → (P, R, S)
D → (P, Q)
37. A → (P, T)
B → (P, Q)
C → (S, T)
D → (S, T)
38. (06)
39. (12)
40. (09)
41. (A)
42. (A)
43. (C)
44. (D)
45. (B)
46. (C, D)
47. (B, D)
48. (B, C, D)
49. (B, C)
50. (A, D)
51. (B)
52. (A, B, C)
53. (A, C, D)
54. (D)
55. (B, C)
56. A → (P, Q, R)
B → (P, Q, R, S)
C → (P, Q, R, S)
D → (P)
57. A → (P, Q, R)
B → (T)
C → (P, Q, R, S, T)
D → (R)
58. (05)
59. (04)
60. (49)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
All India Aakash Test Series for JEE (Advanced)-2020 Mock Test-1 (Paper-1) (Code-A)_(Hints & Solutions)
2/10
PART - I (PHYSICS)
1. Answer (D)
Hint : aAB = aCD = g/3
Solution :
aAB = aCD = g/3
2
rel
1
2a t l=
rel
2
3
ga =
2
3
gt l=
3l
tg
=
2. Answer (C)
Hint : Linear momentum and kinetic energy of
system remains conserved.
Solution :
pB cos = p
pB sin = pA
2 2 2
2 2 2 2
B Ap p p
m m m+ =
2 2 2 2
2sec tan
2 1
p pp
+ =
2 2
23tan
2 2
p p =
1
tan3
=
2
sec3
=
3. Answer (D)
Hint : T2 R3
Solution :
( )
2 312 3
2 4
T R
T R=
2
1
8T
T= T2 = 16 hr
( )2 2 1 1
rel2 1
2 24 –– 16 2
– 3
R RR R
R R R
= =
rel 2 3 6
R
R
= =
rad/hr
4. Answer (B)
Hint : 2 2 sm
Tk
= =
Solution :
( ) ( )sin , sin2
AX A t A t= =
6
t
=
24 6 2
T Tt
= +
4 6
T Tt = +
5
s6
t =
5. Answer (A)
Hint : P1V1 = P2V2
Solution :
0 1
99
100
lP l P
=
1 0
100
99P P=
0 0
100 2–
99
TP P
r=
0 2
99
P T
r=
0
198
P rT =
6. Answer (A, C)
Hint : –KA dT
QL dx
=
Solution :
A BA B
A B
K A K AT T
l l =
A BA B
A B
K KT T
l l =
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Mock Test-1 (Paper-1) (Code-A)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
3/10
7. Answer (A, B, C)
Hint : Based on question of relative velocity of
approach and separation.
Solution :
Separation is maximum or minimum when
( )
( )1 20 – 0
d rr v v
dt= =
For 1 2
v v= separation will not change.
8. Answer (A, C, D)
Hint : T
v =
Solution :
= 1 m
= 16 Hz.
2
4 3 8sin x
=
2
3x
=
6
x
=
0
3 1– , 2 16
2 6 3X t
= =
0
1s
96t =
9. Answer (B, C)
Hint : Apply work energy theorem.
Solution :
2
2 0 00 0
1–
2 9 3
X XK X mg X
= +
0
3
KX
mg =
Also, 0 –3
KXma mg= a = 0
10. Answer (A, D)
Hint : Fl
lAY
=
Solution :
Stress in both the rod is same.
Stress
StrainY=
Strain is more in aluminium rod.
11. Answer (A, D)
12. Answer (A, C)
13. Answer (A, C)
Hint for Q. no. 11 to 13
Velocity of point of contact is zero and
mechanical energy is conserved.
Solution for Q. no. 11 to 13
21
2 Pmg R I=
2 21 142
2 5
mm R mgR
+ =
2 212
5
RmgR
=
5
12
g
R =
(mg + m2 R) 224
5R mR=
55
8512
24 288
gg
g
R R
+
= =
14. Answer (A, C)
15. Answer (A, C)
Hint for Q. no. 14 and 15
dutotal + dW = 0
Solution for Q. no. 14 and 15
3 5
– 2 200 1 200 02 2
W R+ + =
W = 1100R
1
5
2
RC =
5
1 2 02
RC + =
5
–4
RC =
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
All India Aakash Test Series for JEE (Advanced)-2020 Mock Test-1 (Paper-1) (Code-A)_(Hints & Solutions)
4/10
16. Answer A(P); B(Q, S); C(R, T); D(Q, S)
Hint : Pressure depends on height of liquid
column.
Solution :
Pressure depends on height of liquid column.
More pressure means more normal force by
bottom on liquid.
17. Answer A(P, S); B(Q, T); C(P, S); D(Q, R)
Hint : –p
dyv v
dx=
Solution :
–p
yv v
x
=
–y
p Bx
=
18. Answer (27)
Hint : 1 1
2 2
x
v= +
Solution :
( )
2
1v
x=
+
( ) ( )2
2 –2
1 1
dva v
dx x x= =
+ +
24– m/s
27a =
19. Answer (11)
Hint : It is based on principle of super position.
Solution :
8–3
2
P
GMGM
VRR
= +
1
– 1–12
GM
R
=
3 8–2
2
O
GMGM
VRR
= +
11
15
P
O
V
V=
20. Answer (36)
Hint : Finally pressure will be same.
Solution :
0 0 0 0 0 0 0
0 0 0 0
3
3 3 3
P l P l P l Pl
T T T T+ + =
05
3
PP =
0 00
5
3 3
P lx P =
0
5
lx =
Displacement = 0 0 02
–3 5 15
l l l=
= 12 cm
PART - II (CHEMISTRY)
21. Answer (D)
Hint : Q = oxygen
Solution : Group number = 2 (atomic number)
It means element may be O
22. Answer (A)
Hint : Resonance energy (per ring) of benzene >
naphthalene.
Solution : Resonance energy of naphthalene is
more than benzene but resonance energy per
ring of benzene is more than naphthalene.
23. Answer (C)
Hint : E+ at 2nd position of pyrrole and 3rd position
of indole.
Solution :
24. Answer (C)
Hint : It is an electrophilic aromatic substitution.
Solution :
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Mock Test-1 (Paper-1) (Code-A)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
5/10
25. Answer (A)
Hint : It is the preparation of borax from the ore
2CaO 3B2O3.
Solution :
26. Answer (A, C, D)
Hint : For irreversible process,
system surroundingS S 0 +
Solution :
qsystem = PV
= 1 × 10 = 1013 J
Ssurrounding = 1013
300− = –3.377 JK–1
Ssystem = 1 × 8.314 × ln2 = 5.76 JK–1
27. Answer (A, B, D)
Hint : In N3H and COS no p–d bonding takes
place.
Solution :
28. Answer (A, B, C)
Hint : 2 2 3Cl 3F 2ClF+ ⎯⎯→
Solution : If no limiting reagent in the reaction,
then final compound is ClF3(g). If ClF3 is formed,
then mole of gas become half, it means pressure
decreases to half.
29. Answer (A, B, C, D)
Hint : On increasing P, equilibrium shift in that
direction which contain less number of gaseous
mole.
Solution : If O2 is added, then reaction-II and III
move in forward direction.
30. Answer (A, B, C, D)
Hint : Buffer = wA + salt of wA with strong base
= wB + salt of wB with strong acid
Solution :
4 2 4 3 2 4(NH ) SO 2NaOH 2NH Na SO+ ⎯⎯→ +
If mol of 4 2 4(NH ) SO mol of NaOH
4 2 4 4(NH ) SO NH OH Basic buffer+ ⎯⎯→
31. Answer (B)
Hint : M is 80
32. Answer (A)
Hint : mol of Na2CO3 = mol of NaHCO3
= mol of NaOH
33. Answer (A, B, C)
Hint : x = 20, x1 = 20
z = 20, z1 = 10
y = 20, y1 = 30
Solution for Q.Nos. 31 to 33 :
x + y + z = 60
x1 + y1 + z1 = 60
From experiment-I,
y + z = 40
x + y = 40
So, x = 20, y = 20, z = 20
From experiment-II,
x1 + z1 = 40
It means y1 = 20
From experiment-III,
x + 2y + z = M 80
From experiment-IV,
2x1 + y1 + z1 = 90
20 + 2x1 + z1 = 90
2x1 + z1 = 70
x1 = 30
z1 = 10
34. Answer (B)
Hint : M is Li.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
All India Aakash Test Series for JEE (Advanced)-2020 Mock Test-1 (Paper-1) (Code-A)_(Hints & Solutions)
6/10
35. Answer (A, C, D)
Hint : Z
1n
= so velocity of H-atom in ground state
and X1 state of M+q is same.
Solution for Q.Nos. 34 & 35 :
X2 = 1 and n2 – – 1 = 0.
So, n2 = 2
If energy of X1 = ground state energy of H-atom,
it means 1
Z1
n= .
From the velocity of e– in X2,
6 6 Z
3.3 10 2.2 102
=
Z = 3
It means M+q = Li2+
36. Answer A(Q, T); B(P, R); C(P, R, S); D(P, Q)
Hint : Hybridisation : B2H6 → sp3, Si(OH)4 → sp3,
H3BO3 → sp2
Solution :
3 4 2 6 3BF LiAlH B H LiF AlF+ ⎯⎯→ + +
Heat
2 6 3 3 3 6 2B H NH B N H 12H+ ⎯⎯⎯→ +
4 2 4SiCl H O Si(OH) HCl+ ⎯⎯→ +
2 2 7 2 3 3Na B O HCl H O NaCl H BO+ + ⎯⎯→ +
37. Answer A(P, T); B(P, Q); C(S, T); D(S, T)
Hint : Solubility of carbonate of 2nd group
decreases top to bottom while for 1st group, it
increases top to bottom.
Solution : Solubility =
BeCO3 > MgCO3 > CaCO3 > SrCO3
Li2CO3 < Na2CO3 < K2CO3 < Cs2CO3
LiF < NaF < KF < CsF
MgF2 < CaF2 < SrF2 < BaF2 < BeF2
38. Answer (06)
Hint : -Cyano, malonic acid, -keto acid and
--unsaturated salt or acid show very high rate
of reaction.
Solution :
39. Answer (12)
Hint : X = 7, Y = 3, Z = 2
Solution :
40. Answer (09)
Hint : Write all the possible arrangement of
p2 configuration and then choose the
configuration which have total spin zero.
Solution :
There are nine possible arrangement with total
spin zero.
PART - III (MATHEMATICS)
41. Answer (A)
Hint : Modulus.
Solution :
1
1+
=
nz
z
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Mock Test-1 (Paper-1) (Code-A)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
7/10
|z + 1| = |z|
z lies on right bisector of the line segment
joining the points (–1, 0) and (0, 0).
42. Answer (A)
Hint : Point of intersection of tangents on
parabola.
Solution :
Locus of P.O.I of tangents to the parabola
y2 = 4ax include angle is
(y2 – 4ax) = tan2 (x + a)2
3
=
y2 = 4ax + 3(x + a)2
43. Answer (C)
Hint : Point of intersection of tangent on ellipse.
Solution :
cos sin 112 8
x y + = …(i)
sin cos 112 8
x y− + = …(ii)
By (i)2 + (ii)2 2 2
212 8
x y+ =
44. Answer (D)
Hint : Companion of C.O.C
Solution :
Equation of C.O.C T = 0
xh – yk = 9 …(i)
x = 9 …(ii)
Comparing 1
1 0 1
h k= =
h = 1; k = 0
Now equation of pair of tangents SS1 = T2
(x2 – y2 – 9) (–8) = (x – 9)2
–8x2 + 8y2 + 72 = x2 – 18x + 81
9x2 – 8y2 – 18x + 9 = 0
45. Answer (B)
Hint : ab + bc + ca – abc
= 1 + (1 – a) (1 – b) (1 – c)
Solution :
s = 1
So, s – a, s – b, s – c are positive numbers.
ab + bc + ca – abc
= 1 + (1 – a) (1 – b) (1 – c)
( ) ( ) ( )– – –
3
s a s b s c+ +
( )( )( )( )1
3– – –s a s b s c
( )( )( )1
1– 1– 1–27
a b c
Also, (1 – a) (1 – b) (1 – c) > 0
So, 28
– 1,27
ab bc ca abc
+ +
46. Answer (C, D)
Hint : The greatest number selected should be a
multiple of perfect square.
Solution :
Let the numbers be a, ar, ar2.
All the three numbers are natural numbers.
So, the largest number out of these three should
be a multiple of perfect square.
The greatest number may be
4, 8, 12, 20, 24, 28, 40, 44 (multiple of 22)
or 9, 18, 27, 45 (multiple of 32)
or 16, 32, 48 (multiple of 42)
or 25, 50, 36, 49 (multiple of 52, 62 and 72)
If the largest number is a multiple of k2, then
common ratio may be 1 2 3 1
, , , ...,k
k k k k
−
So, n = 8 × 1 + 4 × 2 + 3 × 3 + 2 × 4 + 1 × 5 + 1 ×
6
= 8 + 8 + 9 + 8 + 5 + 6 = 44
47. Answer ( B, D)
Hint : Find length of perpendicular from origin on
AB.
Solution :
Let equation of circle C : x2 + y2 = r2 (b < r < a)
Equation of common tangent, y = mx + c
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
All India Aakash Test Series for JEE (Advanced)-2020 Mock Test-1 (Paper-1) (Code-A)_(Hints & Solutions)
8/10
where 2 2 2 21c r m a m b= + = +
2 2
2 2
r bm
a r
−=
−
Equation of AB, y = mx – aem
Length of perpendicular from origin to AB
21
aem
m=
+
2 2
2 2
2 2
r bae r b
a b
−= = −
−
So, 2 2 22 ( ) 2AB r r b b= − − =
Area of triangle OAB = 2 2b r b−
48. Answer (B, C, D)
Hint : Position of two circles.
Solution :
C1(0, 0); r1 = 2 ; ( )2 2
3 3, 3 ; 4C r =
1 2
27 9 6C C = + =
C1C2 = r1 + r2
Equation transverse common tangent
6 3 6 24x y+ =
3 4x y+ =
3 1
22 2
x y+ =
3 1
cos ; sin2 2
= =
6
=
49. Answer (B, C)
Hint : 1 1
11
nIn n
= + −+
.
Solution :
100
1
1
( 1)rr
r
I−
=
−
= I1 – I2 + I3 – I4 + ... + I99 – I100
2 2 2 2 2 2
1 1 1 1 1 11 1 1 – ...
1 2 2 3 3 4
= + + − + + + + +
2 2
1 1
1 101= −
2
11
101= −
Also, 2 2
1 11
( 1)nI
n n= + +
+
2 1
( 1)
n n
n n
+ +=
+
1 1
1 –1n n
= ++
100
1 2 3 1001
...r
r
I I I I I=
= + + +
1 1 1 1 1 1
1 – 1 – ... 1 –1 2 2 3 100 101
= + + + + + +
1
101–101
=
50. Answer (A, D)
Hint : Let the slope of OP is m, then slope of OQ
will be 1
–m
. Now, find OP and OQ in terms of
m.
Solution :
Let equation of OP : y = mx. On solving with
hyperbola,
2 2 2
2 2– 1
x m x
a b=
2 2
2
2 2 2–
a bx
b a m=
2 2 2OP x m x= +
( )2 2 2
2
2 2 2
1
–
a b mOP
b a m
+=
If we put 1
–m
in place of m, we will get
( )2 2 2
2
2 2 2
1
–
a b mOQ
b m a
+=
Now, ( )( )
( )
2 2 2
2 2 2 2 2
1 1 – 1
1
b a m
OP OQ a b m
++ =
+
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Mock Test-1 (Paper-1) (Code-A)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
9/10
2 2
1 1–
a b=
( )( )
( )
2 2 2
2 2 2 2 2
1 1 1––
1
a b m
OP OQ a b m
+=
+
2
2 2 2
1 1 1–
1
m
a b m
= +
+
2
2
1–0, 1
1
m
m
+
So, 2 2 2 2
max
1 1 1–
a
OP OQ a b= +
51. Answer (B)
Hint : Remainder Theorem.
Solution :
100
118 18 205
− = =
option (B) is correct
52. Answer (A, B, C)
Hint : Divisibility concept.
Solution :
In option (D) 1
a bnk
d
−= Also a – b = nk2
2 21
1
nk knk d
d k= = need not N – {1}
53. Answer (A, C, D)
Hint : Remainder Theorem.
Solution :
5x – 2 = 7k
7 2
5
kx
+=
Now verify each and every option.
54. Answer (D)
Hint : Geometrical Probability.
Solution :
D 0 P2 – (P + 2) 0 P2 – P – 2 0
4 2
6 3P = =
55. Answer (B, C)
Hint : Geometrical Probability.
Solution :
T.C = × 102 = 100
F.C = 100 – 25 = 75
3
4m =
option (B, C) are correct.
56. Answer A(P, Q, R); B(P, Q, R, S); C(P, Q, R, S);
D(P)
Hint : Use –1
–1
n n
r r
nC C
r= and
0
2n
n n
rr
C=
=
Solution :
(A) ( ) ( )0
0 1 2 ... 1 2 3n n
rC n C+ + + + + + +
( )2
.... ) 2 3 .... ....nn C n+ + + + + + +
( ) ( )
0
1 – 1–
2 2
nn
rr
n n r rC
=
+=
( ) ( )
–21 – 12 – 2
2 2
n nn n n n+=
( )–32 3 5nn n= +
(B) ( ) ( )2 –2
–22 2
– – 1n n
n n
r rr r
N r r C n n C= =
= =
( ) –2– 1 2nn n=
(C) (C0 + C1) + (C0 + C2) + …. + (C0 + Cn)
+ (C1 + C2) + (C1 + C3) + …. + (C1 + Cn)
+ (C2 + C3) + (C2 + C4) +.…+ (C2 + Cn)
……………………………………………..
……………………………………………..
+ (Cn –1 + Cn)
= n(C0 + C1 + C2 + … + Cn) = n2n
(D) [nnC0 + 6(1 + 2 + 3 + …. + n)] +
[(n – 1) nC1 + 6(2 + 3 + 4 + ….+ n)] +
[(n – 2) nC2 + 6(3 + 4 + 5 + ….+ n] + ….
= n2n–1 + 6[12 + 22 + 32 + …. + n2]
= n2n–1 + n(n + 1) (2n + 1)
57. Answer A(P, Q, R); B(T); C(P, Q, R, S, T); D(R)
Hint : Property of modulus of complex number.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
All India Aakash Test Series for JEE (Advanced)-2020 Mock Test-1 (Paper-1) (Code-A)_(Hints & Solutions)
10/10
Solution :
(A) |iz + 3| = |z – 3i| |z – 2i| + 4
Also |z – 3i| 0
A → P, Q, R
(B) ( )( )( )
4
1 2 3
24
nn n n nC
− − −=
greatest term = middle term
n = 8
B → T
(C) Minimum value occurs at z = 1 + i
minimum value = 5 + 5 + 2 = 12
C → P, Q, R, S, T
(D) For (1 + i)n to be real sin 04
n =
n = 4, 8, 12
3
12P =
1
4P =
1
4P
=
D → R
58. Answer (05)
Hint : Assume the centre of circle on the normal of parabola at point P.
Solution :
Let normal at one end of latus rectum (1, 2) cuts the parabola at point P, then coordinates of P will be
(aT2, 2aT), where 2
– – –3T tt
= =
P(9, –6)
Equation of common normal at P,
y = 3x – (6 + 27) 3x – 33
Consider the centre of required circle C(a, 3a – 33)
CP = CS (S being the focus)
(a – 9)2 + (3a – 27)2 = (a – 1)2 + (3a – 33)2
10(a2 – 18a + 81) = 10a2 – 200a + 1090
20a = 280
a = 14
So, 10 – 9 5 10r a= =
59. Answer (04)
Hint : General Term of H.P
Solution :
1 1 3
– , – , – ,....are in A.P.10 5 10
–10n
nT =
General term of H.P. is 10
–n
10
–n
will be an integer, if n = 1, 2, 5, 10
60. Answer (49)
Hint : General Term.
Solution :
1 2
–1
2 – 12
3 2 2 ....to terms 2 12 – 1
2 2 – 1 2 1
n
n
n n n n
nt
++ + +
= = =+
–1
12
2n n
t = +
11–
22 2
11–
2
n
nS n n
= + =
All India Aakash Test Series for JEE (Advanced)-2020
Test Date : 19/01/2020
ANSWERS
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
MOCK TEST - 1 (Paper-1) - Code-B
Mock Test - 1 (Paper-1) (Code-B)_(Answers) All India Aakash Test Series for JEE (Advanced)-2020
1/10
PHYSICS CHEMISTRY MATHEMATICS
1. (A)
2. (B)
3. (D)
4. (C)
5. (D)
6. (A, D)
7. (B, C)
8. (A, C, D)
9. (A, B, C)
10. (A, C)
11. (A, D)
12. (A, C)
13. (A, C)
14. (A, C)
15. (A, C)
16. A → (P, S)
B → (Q, T)
C → (P, S)
D → (Q, R)
17. A → (P)
B → (Q, S)
C → (R, T)
D → (Q, S)
18. (36)
19. (11)
20. (27)
21. (A)
22. (C)
23. (C)
24. (A)
25. (D)
26. (A, B, C, D)
27. (A, B, C, D)
28. (A, B, C)
29. (A, B, D)
30. (A, C, D)
31. (B)
32. (A)
33. (A, B, C)
34. (B)
35. (A, C, D)
36. A → (P, T)
B → (P, Q)
C → (S, T)
D → (S, T)
37. A → (Q, T)
B → (P, R)
C → (P, R, S)
D → (P, Q)
38. (09)
39. (12)
40. (06)
41. (B)
42. (D)
43. (C)
44. (A)
45. (A)
46. (A, D)
47. (B, C)
48. (B, C, D)
49. (B, D)
50. (C, D)
51. (B)
52. (A, B, C)
53. (A, C, D)
54. (D)
55. (B, C)
56. A → (P, Q, R)
B → (T)
C → (P, Q, R, S, T)
D → (R)
57. A → (P, Q, R)
B → (P, Q, R, S)
C → (P, Q, R, S)
D → (P)
58. (49)
59. (04)
60. (05)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
All India Aakash Test Series for JEE (Advanced)-2020 Mock Test-1 (Paper-1) (Code-B)_(Hints & Solutions)
2/10
PART - I (PHYSICS)
1. Answer (A)
Hint : P1V1 = P2V2
Solution :
0 1
99
100
lP l P
=
1 0
100
99P P=
0 0
100 2–
99
TP P
r=
0 2
99
P T
r=
0
198
P rT =
2. Answer (B)
Hint : 2 2 sm
Tk
= =
Solution :
( ) ( )sin , sin2
AX A t A t= =
6
t
=
24 6 2
T Tt
= +
4 6
T Tt = +
5
s6
t =
3. Answer (D)
Hint : T2 R3
Solution :
( )
2 312 3
2 4
T R
T R=
2
1
8T
T= T2 = 16 hr
( )2 2 1 1
rel2 1
2 24 –– 16 2
– 3
R RR R
R R R
= =
rel 2 3 6
R
R
= =
rad/hr
4. Answer (C)
Hint : Linear momentum and kinetic energy of
system remains conserved.
Solution :
pB cos = p
pB sin = pA
2 2 2
2 2 2 2
B Ap p p
m m m+ =
2 2 2 2
2sec tan
2 1
p pp
+ =
2 2
23tan
2 2
p p =
1
tan3
=
2
sec3
=
5. Answer (D)
Hint : aAB = aCD = g/3
Solution :
aAB = aCD = g/3
2
rel
1
2a t l=
rel
2
3
ga =
2
3
gt l=
3l
tg
=
6. Answer (A, D)
Hint : Fl
lAY
=
Solution :
Stress in both the rod is same.
Stress
StrainY=
Strain is more in aluminium rod.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Mock Test-1 (Paper-1) (Code-B)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
3/10
7. Answer (B, C)
Hint : Apply work energy theorem.
Solution :
2
2 0 00 0
1–
2 9 3
X XK X mg X
= +
0
3
KX
mg =
Also, 0 –3
KXma mg= a = 0
8. Answer (A, C, D)
Hint : T
v =
Solution :
= 1 m
= 16 Hz.
2
4 3 8sin x
=
2
3x
=
6
x
=
0
3 1– , 2 16
2 6 3X t
= =
0
1s
96t =
9. Answer (A, B, C)
Hint : Based on question of relative velocity of
approach and separation.
Solution :
Separation is maximum or minimum when
( )
( )1 20 – 0
d rr v v
dt= =
For 1 2
v v= separation will not change.
10. Answer (A, C)
Hint : –KA dT
QL dx
=
Solution :
A BA B
A B
K A K AT T
l l =
A BA B
A B
K KT T
l l =
11. Answer (A, D)
12. Answer (A, C)
13. Answer (A, C)
Hint for Q. no. 11 to 13
Velocity of point of contact is zero and
mechanical energy is conserved.
Solution for Q. no. 11 to 13
21
2 Pmg R I=
2 21 142
2 5
mm R mgR
+ =
2 212
5
RmgR
=
5
12
g
R =
(mg + m2 R) 224
5R mR=
55
8512
24 288
gg
g
R R
+
= =
14. Answer (A, C)
15. Answer (A, C)
Hint for Q. no. 14 and 15
dutotal + dW = 0
Solution for Q. no. 14 and 15
3 5
– 2 200 1 200 02 2
W R+ + =
W = 1100R
1
5
2
RC =
5
1 2 02
RC + =
5
–4
RC =
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
All India Aakash Test Series for JEE (Advanced)-2020 Mock Test-1 (Paper-1) (Code-B)_(Hints & Solutions)
4/10
16. Answer A(P, S); B(Q, T); C(P, S); D(Q, R)
Hint : –p
dyv v
dx=
Solution :
–p
yv v
x
=
–y
p Bx
=
17. Answer A(P); B(Q, S); C(R, T); D(Q, S)
Hint : Pressure depends on height of liquid
column.
Solution :
Pressure depends on height of liquid column.
More pressure means more normal force by
bottom on liquid.
18. Answer (36)
Hint : Finally pressure will be same.
Solution :
0 0 0 0 0 0 0
0 0 0 0
3
3 3 3
P l P l P l Pl
T T T T+ + =
05
3
PP =
0 00
5
3 3
P lx P =
0
5
lx =
Displacement = 0 0 02
–3 5 15
l l l=
= 12 cm
19. Answer (11)
Hint : It is based on principle of super position.
Solution :
8–3
2
P
GMGM
VRR
= +
1
– 1–12
GM
R
=
3 8–2
2
O
GMGM
VRR
= +
11
15
P
O
V
V=
20. Answer (27)
Hint : 1 1
2 2
x
v= +
Solution :
( )
2
1v
x=
+
( ) ( )2
2 –2
1 1
dva v
dx x x= =
+ +
24– m/s
27a =
PART - II (CHEMISTRY)
21. Answer (A)
Hint : It is the preparation of borax from the ore
2CaO 3B2O3.
Solution :
22. Answer (C)
Hint : It is an electrophilic aromatic substitution.
Solution :
23. Answer (C)
Hint : E+ at 2nd position of pyrrole and 3rd position
of indole.
Solution :
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Mock Test-1 (Paper-1) (Code-B)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
5/10
24. Answer (A)
Hint : Resonance energy (per ring) of benzene >
naphthalene.
Solution : Resonance energy of naphthalene is
more than benzene but resonance energy per
ring of benzene is more than naphthalene.
25. Answer (D)
Hint : Q = oxygen
Solution : Group number = 2 (atomic number)
It means element may be O
26. Answer (A, B, C, D)
Hint : Buffer = wA + salt of wA with strong base
= wB + salt of wB with strong acid
Solution :
4 2 4 3 2 4(NH ) SO 2NaOH 2NH Na SO+ ⎯⎯→ +
If mol of 4 2 4(NH ) SO mol of NaOH
4 2 4 4(NH ) SO NH OH Basic buffer+ ⎯⎯→
27. Answer (A, B, C, D)
Hint : On increasing P, equilibrium shift in that
direction which contain less number of gaseous
mole.
Solution : If O2 is added, then reaction-II and III
move in forward direction.
28. Answer (A, B, C)
Hint : 2 2 3Cl 3F 2ClF+ ⎯⎯→
Solution : If no limiting reagent in the reaction,
then final compound is ClF3(g). If ClF3 is formed,
then mole of gas become half, it means pressure
decreases to half.
29. Answer (A, B, D)
Hint : In N3H and COS no p–d bonding takes
place.
Solution :
30. Answer (A, C, D)
Hint : For irreversible process,
system surroundingS S 0 +
Solution :
qsystem = PV
= 1 × 10 = 1013 J
Ssurrounding = 1013
300− = –3.377 JK–1
Ssystem = 1 × 8.314 × ln2 = 5.76 JK–1
31. Answer (B)
Hint : M is 80
32. Answer (A)
Hint : mol of Na2CO3 = mol of NaHCO3
= mol of NaOH
33. Answer (A, B, C)
Hint : x = 20, x1 = 20
z = 20, z1 = 10
y = 20, y1 = 30
Solution for Q.Nos. 31 to 33 :
x + y + z = 60
x1 + y1 + z1 = 60
From experiment-I,
y + z = 40
x + y = 40
So, x = 20, y = 20, z = 20
From experiment-II,
x1 + z1 = 40
It means y1 = 20
From experiment-III,
x + 2y + z = M 80
From experiment-IV,
2x1 + y1 + z1 = 90
20 + 2x1 + z1 = 90
2x1 + z1 = 70
x1 = 30
z1 = 10
34. Answer (B)
Hint : M is Li.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
All India Aakash Test Series for JEE (Advanced)-2020 Mock Test-1 (Paper-1) (Code-B)_(Hints & Solutions)
6/10
35. Answer (A, C, D)
Hint : Z
1n
= so velocity of H-atom in ground state
and X1 state of M+q is same.
Solution for Q.Nos. 34 & 35 :
X2 = 1 and n2 – – 1 = 0.
So, n2 = 2
If energy of X1 = ground state energy of H-atom,
it means 1
Z1
n= .
From the velocity of e– in X2,
6 6 Z
3.3 10 2.2 102
=
Z = 3
It means M+q = Li2+
36. Answer A(P, T); B(P, Q); C(S, T); D(S, T)
Hint : Solubility of carbonate of 2nd group
decreases top to bottom while for 1st group, it
increases top to bottom.
Solution : Solubility =
BeCO3 > MgCO3 > CaCO3 > SrCO3
Li2CO3 < Na2CO3 < K2CO3 < Cs2CO3
LiF < NaF < KF < CsF
MgF2 < CaF2 < SrF2 < BaF2 < BeF2
37. Answer A(Q, T); B(P, R); C(P, R, S); D(P, Q)
Hint : Hybridisation : B2H6 → sp3, Si(OH)4 → sp3,
H3BO3 → sp2
Solution :
3 4 2 6 3BF LiAlH B H LiF AlF+ ⎯⎯→ + +
Heat
2 6 3 3 3 6 2B H NH B N H 12H+ ⎯⎯⎯→ +
4 2 4SiCl H O Si(OH) HCl+ ⎯⎯→ +
2 2 7 2 3 3Na B O HCl H O NaCl H BO+ + ⎯⎯→ +
38. Answer (09)
Hint : Write all the possible arrangement of
p2 configuration and then choose the
configuration which have total spin zero.
Solution :
There are nine possible arrangement with total
spin zero.
39. Answer (12)
Hint : X = 7, Y = 3, Z = 2
Solution :
40. Answer (06)
Hint : -Cyano, malonic acid, -keto acid and
--unsaturated salt or acid show very high rate
of reaction.
Solution :
PART - III (MATHEMATICS)
41. Answer (B)
Hint : ab + bc + ca – abc
= 1 + (1 – a) (1 – b) (1 – c)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Mock Test-1 (Paper-1) (Code-B)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
7/10
Solution :
s = 1
So, s – a, s – b, s – c are positive numbers.
ab + bc + ca – abc
= 1 + (1 – a) (1 – b) (1 – c)
( ) ( ) ( )– – –
3
s a s b s c+ +
( )( )( )( )1
3– – –s a s b s c
( )( )( )1
1– 1– 1–27
a b c
Also, (1 – a) (1 – b) (1 – c) > 0
So, 28
– 1,27
ab bc ca abc
+ +
42. Answer (D)
Hint : Companion of C.O.C
Solution :
Equation of C.O.C T = 0
xh – yk = 9 …(i)
x = 9 …(ii)
Comparing 1
1 0 1
h k= =
h = 1; k = 0
Now equation of pair of tangents SS1 = T2
(x2 – y2 – 9) (–8) = (x – 9)2
–8x2 + 8y2 + 72 = x2 – 18x + 81
9x2 – 8y2 – 18x + 9 = 0
43. Answer (C)
Hint : Point of intersection of tangent on ellipse.
Solution :
cos sin 112 8
x y + = …(i)
sin cos 112 8
x y− + = …(ii)
By (i)2 + (ii)2 2 2
212 8
x y+ =
44. Answer (A)
Hint : Point of intersection of tangents on
parabola.
Solution :
Locus of P.O.I of tangents to the parabola
y2 = 4ax include angle is
(y2 – 4ax) = tan2 (x + a)2
3
=
y2 = 4ax + 3(x + a)2
45. Answer (A)
Hint : Modulus.
Solution :
1
1+
=
nz
z
|z + 1| = |z|
z lies on right bisector of the line segment
joining the points (–1, 0) and (0, 0).
46. Answer (A, D)
Hint : Let the slope of OP is m, then slope of OQ
will be 1
–m
. Now, find OP and OQ in terms of
m.
Solution :
Let equation of OP : y = mx. On solving with
hyperbola,
2 2 2
2 2– 1
x m x
a b=
2 2
2
2 2 2–
a bx
b a m=
2 2 2OP x m x= +
( )2 2 2
2
2 2 2
1
–
a b mOP
b a m
+=
If we put 1
–m
in place of m, we will get
( )2 2 2
2
2 2 2
1
–
a b mOQ
b m a
+=
Now, ( )( )
( )
2 2 2
2 2 2 2 2
1 1 – 1
1
b a m
OP OQ a b m
++ =
+
2 2
1 1–
a b=
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
All India Aakash Test Series for JEE (Advanced)-2020 Mock Test-1 (Paper-1) (Code-B)_(Hints & Solutions)
8/10
( )( )
( )
2 2 2
2 2 2 2 2
1 1 1––
1
a b m
OP OQ a b m
+=
+
2
2 2 2
1 1 1–
1
m
a b m
= +
+
2
2
1–0, 1
1
m
m
+
So, 2 2 2 2
max
1 1 1–
a
OP OQ a b= +
47. Answer (B, C)
Hint : 1 1
11
nIn n
= + −+
.
Solution :
100
1
1
( 1)rr
r
I−
=
−
= I1 – I2 + I3 – I4 + ... + I99 – I100
2 2 2 2 2 2
1 1 1 1 1 11 1 1 – ...
1 2 2 3 3 4
= + + − + + + + +
2 2
1 1
1 101= −
2
11
101= −
Also, 2 2
1 11
( 1)nI
n n= + +
+
2 1
( 1)
n n
n n
+ +=
+
1 1
1 –1n n
= ++
100
1 2 3 1001
...r
r
I I I I I=
= + + +
1 1 1 1 1 1
1 – 1 – ... 1 –1 2 2 3 100 101
= + + + + + +
1
101–101
=
48. Answer (B, C, D)
Hint : Position of two circles.
Solution :
C1(0, 0); r1 = 2 ; ( )2 2
3 3, 3 ; 4C r =
1 2
27 9 6C C = + =
C1C2 = r1 + r2
Equation transverse common tangent
6 3 6 24x y+ =
3 4x y+ =
3 1
22 2
x y+ =
3 1
cos ; sin2 2
= =
6
=
49. Answer ( B, D)
Hint : Find length of perpendicular from origin on
AB.
Solution :
Let equation of circle C : x2 + y2 = r2 (b < r < a)
Equation of common tangent, y = mx + c
where 2 2 2 21c r m a m b= + = +
2 2
2 2
r bm
a r
−=
−
Equation of AB, y = mx – aem
Length of perpendicular from origin to AB
21
aem
m=
+
2 2
2 2
2 2
r bae r b
a b
−= = −
−
So, 2 2 22 ( ) 2AB r r b b= − − =
Area of triangle OAB = 2 2b r b−
50. Answer (C, D)
Hint : The greatest number selected should be a
multiple of perfect square.
Solution :
Let the numbers be a, ar, ar2.
All the three numbers are natural numbers.
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
Mock Test-1 (Paper-1) (Code-B)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020
9/10
So, the largest number out of these three should
be a multiple of perfect square.
The greatest number may be
4, 8, 12, 20, 24, 28, 40, 44 (multiple of 22)
or 9, 18, 27, 45 (multiple of 32)
or 16, 32, 48 (multiple of 42)
or 25, 50, 36, 49 (multiple of 52, 62 and 72)
If the largest number is a multiple of k2, then
common ratio may be 1 2 3 1
, , , ...,k
k k k k
−
So, n = 8 × 1 + 4 × 2 + 3 × 3 + 2 × 4 + 1 × 5 + 1 ×
6
= 8 + 8 + 9 + 8 + 5 + 6 = 44
51. Answer (B)
Hint : Remainder Theorem.
Solution :
100
118 18 205
− = =
option (B) is correct
52. Answer (A, B, C)
Hint : Divisibility concept.
Solution :
In option (D) 1
a bnk
d
−= Also a – b = nk2
2 21
1
nk knk d
d k= = need not N – {1}
53. Answer (A, C, D)
Hint : Remainder Theorem.
Solution :
5x – 2 = 7k
7 2
5
kx
+=
Now verify each and every option.
54. Answer (D)
Hint : Geometrical Probability.
Solution :
D 0 P2 – (P + 2) 0 P2 – P – 2 0
4 2
6 3P = =
55. Answer (B, C)
Hint : Geometrical Probability.
Solution :
T.C = × 102 = 100
F.C = 100 – 25 = 75
3
4m =
option (B, C) are correct.
56. Answer A(P, Q, R); B(T); C(P, Q, R, S, T); D(R)
Hint : Property of modulus of complex number.
Solution :
(A) |iz + 3| = |z – 3i| |z – 2i| + 4
Also |z – 3i| 0
A → P, Q, R
(B) ( )( )( )
4
1 2 3
24
nn n n nC
− − −=
greatest term = middle term
n = 8
B → T
(C) Minimum value occurs at z = 1 + i
minimum value = 5 + 5 + 2 = 12
C → P, Q, R, S, T
(D) For (1 + i)n to be real sin 04
n =
n = 4, 8, 12
3
12P =
1
4P =
1
4P
=
D → R
57. Answer A(P, Q, R); B(P, Q, R, S); C(P, Q, R, S);
D(P)
Hint : Use –1
–1
n n
r r
nC C
r= and
0
2n
n n
rr
C=
=
Solution :
(A) ( ) ( )0
0 1 2 ... 1 2 3n n
rC n C+ + + + + + +
( )2
.... ) 2 3 .... ....nn C n+ + + + + + +
( ) ( )
0
1 – 1–
2 2
nn
rr
n n r rC
=
+=
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
All India Aakash Test Series for JEE (Advanced)-2020 Mock Test-1 (Paper-1) (Code-B)_(Hints & Solutions)
10/10
( ) ( )
–21 – 12 – 2
2 2
n nn n n n+=
( )–32 3 5nn n= +
(B) ( ) ( )2 –2
–22 2
– – 1n n
n n
r rr r
N r r C n n C= =
= =
( ) –2– 1 2nn n=
(C) (C0 + C1) + (C0 + C2) + …. + (C0 + Cn)
+ (C1 + C2) + (C1 + C3) + …. + (C1 + Cn)
+ (C2 + C3) + (C2 + C4) +.…+ (C2 + Cn)
……………………………………………..
……………………………………………..
+ (Cn –1 + Cn)
= n(C0 + C1 + C2 + … + Cn) = n2n
(D) [nnC0 + 6(1 + 2 + 3 + …. + n)] +
[(n – 1) nC1 + 6(2 + 3 + 4 + ….+ n)] +
[(n – 2) nC2 + 6(3 + 4 + 5 + ….+ n] + ….
= n2n–1 + 6[12 + 22 + 32 + …. + n2]
= n2n–1 + n(n + 1) (2n + 1)
58. Answer (49)
Hint : General Term.
Solution :
1 2
–1
2 – 12
3 2 2 ....to terms 2 12 – 1
2 2 – 1 2 1
n
n
n n n n
nt
++ + +
= = =+
–1
12
2n n
t = +
11–
22 2
11–
2
n
nS n n
= + =
59. Answer (04)
Hint : General Term of H.P
Solution :
1 1 3
– , – , – ,....are in A.P.10 5 10
–10n
nT =
General term of H.P. is 10
–n
10
–n
will be an integer, if n = 1, 2, 5, 10
60. Answer (05)
Hint : Assume the centre of circle on the normal of parabola at point P.
Solution :
Let normal at one end of latus rectum (1, 2) cuts the parabola at point P, then coordinates of P will be
(aT2, 2aT), where 2
– – –3T tt
= =
P(9, –6)
Equation of common normal at P,
y = 3x – (6 + 27) 3x – 33
Consider the centre of required circle C(a, 3a – 33)
CP = CS (S being the focus)
(a – 9)2 + (3a – 27)2 = (a – 1)2 + (3a – 33)2
10(a2 – 18a + 81) = 10a2 – 200a + 1090
20a = 280
a = 14
So, 10 – 9 5 10r a= =
top related