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Test - 2A (Paper - 2) (Code-E) (Answers) All India Aakash Test Series for JEE (Advanced)-2019
PHYSICS
1. (A, C)
2. (A, B, C)
3. (B, C)
4. (A, B, D)
5. (A, C)
6. (A)
7. (C)
8. (D)
9. (C)
10. (B)
11. (A)
12. (C)
13. (C)
14. (A)
15. (C)
16. A (R, T)
B(R, T)
C(Q, S)
D(P, R)
17. A(R, S)
B (T)
C(P, S)
D(Q, S)
18. (05)
19. (09)
20. (02)
CHEMISTRY
21. (A, C)
22. (A, B, C, D)
23. (A, B, C)
24. (B, D)
25. (A, C)
26. (D)
27. (A)
28. (D)
29. (B)
30. (C)
31. (C)
32. (C)
33. (C)
34. (B)
35. (A)
36. A(P, S)
B(P, S)
C(P, Q, T)
D(P, S)
37. A(P, Q)
B(R, S)
C(T)
D(P)
38. (04)
39. (09)
40. (07)
MATHEMATICS
41. (A, C)
42. (C, D)
43. (B, C, D)
44. (B, C, D)
45. (A, C)
46. (A)
47. (C)
48. (A)
49. (C)
50. (D)
51. (D)
52. (A)
53. (A)
54. (B)
55. (C)
56. A(Q, S)
B(Q, S, T)
C(P, R)
D(Q, R, T)
57. A(P)
B(Q)
C(P, T)
D(P, Q, R)
58. (04)
59. (08)
60. (08)
Test Date : 25/11/2018
ANSWERS
TEST - 2A (Paper-2) - Code-E
All India Aakash Test Series for JEE (Advanced)-2019
1/9
All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper - 2) (Code-E) (Hints & Solutions)
2/9
PART - I (PHYSICS)
HINTS & SOLUTIONS
1. Answer (A, C)
Hint: x
y f tv
⎛ ⎞ ⎜ ⎟⎝ ⎠
v is wave speed
Solution:
2
0.8
4 – 42
yx
t
⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
v = 2 m/s
m
0.80.2 m
4 y
2. Answer (A, B, C)
Hint: PV2 = C
Solution:
3 3– –
2 2 –1 2 2
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
pr
R R R RC R
TC
V
0
22
TT
⎛ ⎞ ⎜ ⎟⎝ ⎠
dU = 0
3(decrement)
4RT
01
2 2
TRQ
⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
.... rejected by gas
Q = U + W
W = Q + U 01
1 2
TR ⎛ ⎞ ⎜ ⎟⎝ ⎠
0
2
RTW
⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠
3. Answer (B, C)
Hint: Req
= 20
402 A
20 I
Solution:
1 2
1A 1 A
4 2 2 I I
I I=
Radius of V3 = 20 volt
4. Answer (A, B, D)
Hint: Q = CV
0⎛ ⎞ ⎜ ⎟
⎝ ⎠
KAC
d
Solution:
1
11–⎛ ⎞ ⎜ ⎟⎝ ⎠
Q QK
0
4
KbL
d
0 0
0
11–
4
⎛ ⎞ ⎜ ⎟⎝ ⎠
Lb VV
K d
0
2
2
3
bL K
Qd
0 0
0
11–
4
V b LV
d
⎛ ⎞ ⎜ ⎟⎝ ⎠
0
1 2
VE E
d
5. Answer (A, C)
Hint: 1 2
1 2
r
rE E E
r r r
Solution:
If current in ammeter is i2 and current in voltmeter is i
1
then
10 = 11i1 + i
2and 80 = 10i
2 + i
2
it gives i1 =
20
109 A and i
2 =
870
109 A
Voltage across voltmeter V = 20 10 200
volt109 109
V = 200
volt109
and current in ammeter = 870
109 A
6. Answer (A)
7. Answer (C)
8. Answer (D)
Hint & Solution of Question Nos. 6 to 8
Hint: (VA – V
B) = 4(V
P – V
Q)
2
12
⎛ ⎞⎜ ⎟⎝ ⎠
RR
2
4 4
Test - 2A (Paper - 2) (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
3/9
Solution:
Rr = 2 + 2 = 4
9. Answer (C)
10. Answer (B)
11. Answer (A)
Solution of Question Nos. 9 to 11
P = 5
0
0
10 (2 )Kx
P xA
dw = 5 3
010 8 10 2PA dx x dx
∫ ∫
w =
0.12
0
2 1800 2 800 164 J
2 10 200
xx
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
Tf =
4 40
4 40 0
(20 1) 10 (32 10 ) 200
20 10 24 10
f f
PV T
PV
Tf = 280 K
T = 80 K
And U = NCv T =
0 0
0
380
2
⎛ ⎞ ⎜ ⎟
⎝ ⎠
PVR
RT
U =
4 420 10 24 10 3
80 288J200 2
RR
12. Answer (C)
Hint: T2 > T
1
U = nCv T
Solution:
TB > T
A
Q = U + W
Q1 = U
1 + W
1
Q2 = U
2 + W
2
W2 >W
1
13. Answer (C)
Hint: For maximum intensity path difference (x) = n
Solution:
x = 20 – 12 = 8
m
= 8 m
14. Answer (A)
Hint: ·V E r
��
Solution:
0 0
0 0
ˆ ˆ16 123 3
E OO i j⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
������
0
0
ˆ ˆ ˆ ˆ– 16 12 · – – 53
P QV V i j i j
⎛ ⎞ ⎜ ⎟⎝ ⎠
0 0
0 0
7616 60
3 3
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
15. Answer (C)
Hint: E inside shell is zero
Solution:
Vp =
2 2
kQ kQ kQ
R R R
16. Answer A(R, T); B(R, T); C(Q, S), D(P, R)
Hint : Principle of superposition
Solution :
14 2 sin 100 – 2y t x
24 2 cos 100 – 2y t x
(y1 + y
2) = y
R
8sin 100 – 24
t x⎛ ⎞ ⎜ ⎟
⎝ ⎠
yR
= (y1 + y
2 + y
3)
17. Answer A(R, S); B(T), C(P, S); D(Q, S)
Hint : VA – V
B =
B
AE dl∫
����
Solution :
(VA – V
B) =
2
10
4
∫
r
r
drr
= 2
0 1
ln4
r
r
⎛ ⎞⎜ ⎟ ⎝ ⎠
18. Answer (05)
Hint: 1 2
1 2
r
eq
E E E
r r r in parallel combination
Solution:
1 2
1 2
r
r
E E E
r r r
Er = 5 unit and r
eq =
2
3
1
5 152 A
2 175
3
I
All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper - 2) (Code-E) (Hints & Solutions)
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PART - II (CHEMISTRY)
19. Answer (09)
Hint: If PVn = C C
Pr = C
V +
1
R
nSolution:
PT = constant PV1/2 = C
5 9–12 2– 1
2
⎛ ⎞ ⎜ ⎟⎝ ⎠
pr
R R RC
n = 9
20. Answer (02)
Hint: B.F. =1 2
| |f f
Solution:
T2 = T
1 + 4
Hence, f2 = 404 Hz
21. Answer (A, C)
Hint : To find the total spin, we can add spin of each
atom.
Solution :
Ortho hydrogen � para hydrogen
Total spin of ortho hydrogen is 1 and para hydrogen is 0.
At lower temperature para hydrogen is present in
greater concentration.
22. Answer (A, B, C, D)
Hint : Dut to inert pair effect, Bi+5 is not stable.
Solution :
Melting point of Bi is low due to metallic behaviour.
23. Answer (A, B, C)
Hint : Factual.
Cu2S
Roasting
n limited i supply airof
Cu2S + Cu
2O
Heating in
absence of air
Cu + SO2
24. Answer (B, D)
Hint : M–L bond is stronger if L is a -acid ligand.
Solution :
Due to Pi-acid behaviour of –NO2, Co–Cl* bond would
be weaker than Co–Cl and the attack would take place
from the back side of the Co–Cl* bond, hence
pentagonal bipyramidal intermediate is formed.
25. Answer (A, C)
Hint : Surface tension first decreases and then
increases with the increase in concentration of the
surfactant.
Solution :
Correct curve for surface tension is
C
26. Answer (D)
Hint : Fact
Solution :
Observations from the curve.
27. Answer (A)
Hint : As given in paragraph, conversion of h.s. to l.s.
shrink the metal ligand bond.
Solution :
For l.s., 0 increase as r decreases.
Curve ‘1’ will be correct.
28. Answer (D)
Hint : Field strength of H2O, that does not cause
pairing in Mn+2.
Solution :
H2O is WFL for Mn2+ therefore, Mn2+ in H
2O form h.s.
complex and CFSE is zero for this system.
29. Answer (B)
Hint : Factual.
Solution :
FeSO + H O Fe(OH)SO 4 2 2 4
yellow
Test - 2A (Paper - 2) (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
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30. Answer (C)
Hint : Factual.
Solution :
FeSO + NaOH Fe(OH) 4 2
Fe(OH)
Brown 3
White green
31. Answer (C)
Hint : Factual.
Solution :
Mn(NO ) + NaOH Mn(OH) 3 2 2
MnMnO
(Black)
3
White
32. Answer (C)
Hint : Stability of isotope of P.
Solution :
White P is less stable.
33. Answer (C)
Hint : Properties of fullerene.
Solution :
All C atoms are sp2 in fullerene.
34. Answer (B)
Hint : Properties of group-13.
Solution : B > Al > Tl > In > Ga
35. Answer (A)
Hint : Properties of sodium.
Solution: Na forms peroxide.
CsH is the most ionic.
36. Answer A(P, S); B(P, S); C(P, Q, T); D(P, S)
Hint : Reaction of metals with HNO3.
Solution :
R = NH4NO
3.
P1 = N
2O
P2 = N
2O
P3 = NO
2
P4 = N
2O
4
P5 = NO
37. Answer A(P, Q); B(R, S); C(T); D(P)
Hint :
Factual
38. Answer (04)
Hint : Kdesorption
a–E /RT
= A e
Solution :
Life time (y) = a
E /RT
1
Ae
= 10–13
15 1000 3exp
25 300
⎛ ⎞⎜ ⎟⎝ ⎠
= 10 –13 × 400
y = 4 × 10–11 s
39. Answer (09)
Hint : Packing type of lattice.
Solution :
Cu, Ag, Au all have CCP structure, CN = 12
x + y + z = 12 + 12 + 12 = 36
40. Answer (07)
Hint : Electronic configuration of lanthanoids.
Solution :
x = 1
y = 1
z = 7
All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper - 2) (Code-E) (Hints & Solutions)
6/9
PART - III (MATHEMATICS)
41. Answer (A, C)
Hint : Find domain
Solution : Domain of f(x) is x = 1 f(x) = 0
Domain of g(x) is x = –1 g(x) = 3
2
42. Answer (C, D)
Hint : Concept of director circle.
Solution :
O
y
(4, 3)x
y
(3, 4)
xO
y
( , )h k
C
Ox
Locus of C is circle x2 + y2 = a2 + b2 = 25
Hence C moves in a circle of radius 5
Distance covered by C in the motion is equal to
1 14 35 tan tan
3 4
⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠
1 75tan
24
⎛ ⎞ ⎜ ⎟⎝ ⎠
43. Answer (B, C, D)
Hint : Parallel lines are neither concurrent nor forms
a triangle.
Solution : Lines are concurrent if = 0
2
1 1 1
0( 1) 7 5
2 2 5 0
n n
n n
2 3
2
10 25 5 10 2 2 5 5
7 2 14 0
⇒
n n n n n n
n n
n3 – 4n
2 + 5n – 6 = 0
(n – 3)(n2 – n + 2) = 0 n = 3
but for n = 3, lines become parallel
so these are concurrent for no real value of n.
44. Answer (B, C, D)
Hint : Equation of normal y = mx – 2am – am3
Solution : Equation of normal to y2 = 4x
is y = mx – 2m – m3 ...(1)
Equation of normal to y2 = (x – k) is
3
2 4
m my mx km ...(2)
Equations (1) & (2) represent same normal
3
3
1 2
1
2 4
m m m
m m mkm
3m2 = 4k – 6 4k – 6 0
3
2k
45. Answer (A, C)
Hint : 1 1
sin cos2
x x
Solution : Domain of f(x) is [–3, 3]
maximum value of f(x) is
3 397
at 38 4
x
and minimum value of f(x) is
3 311
at 38 4
x
46. Answer (A)
Hint : 1
2r (distance between parallel lines)
Test - 2A (Paper - 2) (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
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47. Answer (C)
Hint : L4 are concurrent at (12, 12)
48. Answer (A)
Hint : For cyclic quadrilateral L3 and L
4 are parallel
Solution of Q. Nos. 46 to 48
A(2, 7) 4 – 3 + 13 = 0x y
4 – 3 – 37 = 0x yB(10, 1)
(6, 4)
(3, 0
)
(9, 8)
3 + 4 – 34 = 0x y
Centre of circle is either (3, 0) or (9, 8)
+ = 3 or + = 17
(x + y – 24) + (x – y) = 0 are concurrent at (12, 12)
y
x
3 + 4 – 34 = 0x y
4 –
3 +
13 = 0
x
y
4 –
3 – 3
7 =
0
x
y(12, 12)
(2, 7)
(10, 1)
O
1
1m
should not belong to 1 11,
2 2
⎡ ⎤⎢ ⎥⎣ ⎦
= –2, –1, 0, 1
For cyclic quadrilateral
1 3 1
1 4 7
⇒
C (9, 8), r = 5
(x – 9)2 + (y – 8)2 = 52
49. Answer (C)
Hint : Take parametric point on hyperbola.
50. Answer (D)
Hint : Use parametric equation of line.
51. Answer (D)
Hint : 2
( 2) and 4y m x xym
intersect at
exactly one point.
Solution of Q. Nos. 49 to 51
Let a variable point on xy = 4 is 2
2 ,P tt
⎛ ⎞⎜ ⎟⎝ ⎠
Equation of tangent is 2
2 8x y t
t
4x
tyt
4(4 , 0), 0,A t B
t
⎛ ⎞⎜ ⎟⎝ ⎠
Circumcentre is midpoint of AB
2
2 ,h t kt
hk = 4 locus of (h, k) is xy = 4
2e R(2, 0)
r(2, 0)R
x
T
y
S(2
+ co
s,
sin
)
r
r
(2 + rcos)(rsin) = 4
r2sincos + (2sin)r – 4 = 0
1 2
4
sin cosr r
RSRT = 8|cosec2|
(RSRT)min
= 8
y2 = 8(x + 2), xy = 4
y = m(x + 2) + 2
m
is also tangent to xy = 4
242 xmx m
m
⎛ ⎞ ⎜ ⎟⎝ ⎠
has equal roots
D = 0
22
4 ( 4) 12 m mm
m
⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠
Equation of common tangent is y = –x – 4
2m – 3c = –2 + 12 = 10
All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper - 2) (Code-E) (Hints & Solutions)
8/9
52. Answer (A)
Hint : Find mirror image of (1 + t2, 2t) about line
Solution :
Let variable point on the parabola is P(1+ t2, 2t) and
its reflection in the given line is (h, k)
2 2(1 ) 2 2(1 2 2)
1 1 2
h t k t t t
( 1)( 1) 2
12(1 )
h t t kt
k t h
⇒
2
2 22
hh k
Locus of (h, k) is 4x – 4y = x2 A + B = 0
53. Answer (A)
Hint : Image of C2 in the line, C
2, C
3 are collinear
Solution :
P(7, 5)
(–7, 3)
(1, –3) 3 + 4 – 16 = 0x y
Image of C2(1, –3) in the line 3x + 4y – 16 = 0 is
P(7, 5)
C1C
2 + C
2C
3 + C
3C
1 is minimum
C1, C
3 and P lie on same line
C3(0, 4)
3 150 5 2C C
Radius of 1
3 2C
Radius of 3
2 2C
Equation of C3 is x2 + (y – 4)2 = 8
x2 + y2 – 8y + 8 = 0 a + b + c = 0
54. Answer (B)
Hint : a = A(t 2 + 1) and b = A
2
2
(1 )t
t
Solution :
4A = 1
2
1
8A
a = A(t2+1), b =
2
2
(1 )A t
t
1 1 1
8 a b A
55. Answer (C)
Hint : f(x) = 1 – xn
Solution :
f(x) = xn + 1 or 1 – xn
f(x) = 1 – x4
f(3) = –80
56. Answer A(Q, S); B(Q, S, T); C(P, R); D(Q, R, T)
Hint : For increasing fnf (x) > 0
Solution :
f1(x) is many-one and into
f2(x) is many-one, into and even
f3(x) is one- one and onto
f4(x) is many-one, onto and even
57. Answer A(P); B(Q); C(P, T); D(P, Q, R)
Hint : Let f(x) = ax2 + bx + c
Solution :
Let f(x) = ax2 + bx + c
f (–1) = 0
f(1) = 4 and f(–1) = 2
a + b + c = 4
a – b + c = 2
–2a + b = 0
Solving, we get
1 51, ,
2 2b a c
21 5( )
2 2f x x x
21
[( 1) 4]2
x
1 1cos 1
( )f x
⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
[1 + sgn(f(x))] = 2
1 1tan 0, 1
2f x
⎡ ⎤⎛ ⎞ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
1 22cot 1, 2, 3
( )f x
⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
Test - 2A (Paper - 2) (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
9/9
58. Answer (04)
Hint : Let circle (x – r)2 + y2 = r2
Solution :
( , 0 )r
y
xr
Let the equation of circle be
(x – r)2 + y2 = r2
solving with y2 = 8x
x = 0 and x = 2r – 8
2r – 8 0
rmax
= 4
59. Answer (08)
Hint : Let 10 cos , 2sinP
Solution :
Let 10 cos , 2sinP
10 4 3
10 5e
2
2
1 2
320 (given)2 10 cos
5PF PF
⎛ ⎞ ⎜ ⎟⎝ ⎠
2 25 1cos , sin
6 6
2 2
2 10 2 10
1 510sin 4cos10 4
6 6
d
2 2
60. Answer (08)
Hint : Tangent to circle 2
2(1 )y mx m
Solution :
Tangent to the circle is 2
2( 1)y mx m ...(1)
It passes through focus (2, 2) of xy = 2
22 2 2( 1)m m
m2 – 4m + 1 = 0
Line (1) intersects hyperbola xy = 2 at the points
given by 22
2( 1)mx m
x
or2 2
2( 1) 2 0mx m x
2
1 2 1 2
2( 1) 2,
mx x x x
m m
2
1 2
16( )x x
m
Also (y1 – y
2)2 = 16m
Length of focal chord 8
2 8mm
⎛ ⎞ ⎜ ⎟⎝ ⎠
28(1 )
2m
m
8 42 8
m
m
� � �
Test - 2A (Paper - 2) (Code-F) (Answers) All India Aakash Test Series for JEE (Advanced)-2019
PHYSICS
1. (A, C)
2. (A, B, D)
3. (B, C)
4. (A, B, C)
5. (A, C)
6. (A)
7. (C)
8. (D)
9. (C)
10. (B)
11. (A)
12. (C)
13. (A)
14. (C)
15. (C)
16. A(R, S)
B (T)
C(P, S)
D(Q, S)
17. A (R, T)
B(R, T)
C(Q, S)
D(P, R)
18. (02)
19. (09)
20. (05)
CHEMISTRY
21. (A, C)
22. (B, D)
23. (A, B, C)
24. (A, B, C, D)
25. (A, C)
26. (D)
27. (A)
28. (D)
29. (B)
30. (C)
31. (C)
32. (A)
33. (B)
34. (C)
35. (C)
36. A(P, Q)
B(R, S)
C(T)
D(P)
37. A(P, S)
B(P, S)
C(P, Q, T)
D(P, S)
38. (07)
39. (09)
40. (04)
MATHEMATICS
41. (A, C)
42. (B, C, D)
43. (B, C, D)
44. (C, D)
45. (A, C)
46. (A)
47. (C)
48. (A)
49. (C)
50. (D)
51. (D)
52. (C)
53. (B)
54. (A)
55. (A)
56. A(P)
B(Q)
C(P, T)
D(P, Q, R)
57. A(Q, S)
B(Q, S, T)
C(P, R)
D(Q, R, T)
58. (08)
59. (08)
60. (04)
Test Date : 25/11/2018
ANSWERS
TEST - 2A (Paper-2) - Code-F
All India Aakash Test Series for JEE (Advanced)-2019
1/9
All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper - 2) (Code-F) (Hints & Solutions)
2/9
PART - I (PHYSICS)
HINTS & SOLUTIONS
1. Answer (A, C)
Hint: 1 2
1 2
r
rE E E
r r r
Solution:
If current in ammeter is i2 and current in voltmeter is i
1
then
10 = 11i1 + i
2and 80 = 10i
2 + i
2
it gives i1 =
20
109 A and i
2 =
870
109 A
Voltage across voltmeter V = 20 10 200
volt109 109
V = 200
volt109
and current in ammeter = 870
109 A
2. Answer (A, B, D)
Hint: Q = CV
0⎛ ⎞ ⎜ ⎟
⎝ ⎠
KAC
d
Solution:
1
11–⎛ ⎞ ⎜ ⎟⎝ ⎠
Q QK
0
4
KbL
d
0 0
0
11–
4
⎛ ⎞ ⎜ ⎟⎝ ⎠
Lb VV
K d
0
2
2
3
bL K
Qd
0 0
0
11–
4
V b LV
d
⎛ ⎞ ⎜ ⎟⎝ ⎠
0
1 2
VE E
d
3. Answer (B, C)
Hint: Req
= 20
402 A
20 I
Solution:
1 2
1A 1 A
4 2 2 I I
I I=
Radius of V3 = 20 volt
4. Answer (A, B, C)
Hint: PV2 = C
Solution:
3 3– –
2 2 –1 2 2
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
pr
R R R RC R
TC
V
0
22
TT
⎛ ⎞ ⎜ ⎟⎝ ⎠
dU = 0
3(decrement)
4RT
01
2 2
TRQ
⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
.... rejected by gas
Q = U + W
W = Q + U 01
1 2
TR ⎛ ⎞ ⎜ ⎟⎝ ⎠
0
2
RTW
⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠
5. Answer (A, C)
Hint: x
y f tv
⎛ ⎞ ⎜ ⎟⎝ ⎠
v is wave speed
Solution:
2
0.8
4 – 42
yx
t
⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
v = 2 m/s
m
0.80.2 m
4 y
6. Answer (A)
7. Answer (C)
8. Answer (D)
Hint & Solution of Question Nos. 6 to 8
Hint: (VA – V
B) = 4(V
P – V
Q)
2
12
⎛ ⎞⎜ ⎟⎝ ⎠
RR
2
4 4
Test - 2A (Paper - 2) (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
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Solution:
Rr = 2 + 2 = 4
9. Answer (C)
10. Answer (B)
11. Answer (A)
Solution of Question Nos. 9 to 11
P = 5
0
0
10 (2 )Kx
P xA
dw = 5 3
010 8 10 2PA dx x dx
∫ ∫
w =
0.12
0
2 1800 2 800 164 J
2 10 200
xx
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
Tf =
4 40
4 40 0
(20 1) 10 (32 10 ) 200
20 10 24 10
f f
PV T
PV
Tf = 280 K
T = 80 K
And U = NCv T =
0 0
0
380
2
⎛ ⎞ ⎜ ⎟
⎝ ⎠
PVR
RT
U =
4 420 10 24 10 3
80 288J200 2
RR
12. Answer (C)
Hint: E inside shell is zero
Solution:
Vp =
2 2
kQ kQ kQ
R R R
13. Answer (A)
Hint: ·V E r
��
Solution:
0 0
0 0
ˆ ˆ16 123 3
E OO i j⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
������
0
0
ˆ ˆ ˆ ˆ– 16 12 · – – 53
P QV V i j i j
⎛ ⎞ ⎜ ⎟⎝ ⎠
0 0
0 0
7616 60
3 3
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
14. Answer (C)
Hint: For maximum intensity path difference (x) = nSolution:
x = 20 – 12 = 8
m
= 8 m
15. Answer (C)
Hint: T2 > T
1
U = nCv T
Solution:
TB > T
A
Q = U + W
Q1 = U
1 + W
1
Q2 = U
2 + W
2
W2 >W
1
16. Answer A(R, S); B(T), C(P, S); D(Q, S)
Hint : VA – V
B =
B
AE dl∫
����
Solution :
(VA – V
B) =
2
10
4
∫
r
r
drr
= 2
0 1
ln4
r
r
⎛ ⎞⎜ ⎟ ⎝ ⎠
17. Answer A(R, T); B(R, T); C(Q, S), D(P, R)
Hint : Principle of superposition
Solution :
14 2 sin 100 – 2y t x
24 2 cos 100 – 2y t x
(y1 + y
2) = y
R
8sin 100 – 24
t x⎛ ⎞ ⎜ ⎟
⎝ ⎠
yR
= (y1 + y
2 + y
3)
18. Answer (02)
Hint: B.F. =1 2
| |f f
Solution:
T2 = T
1 + 4
Hence, f2 = 404 Hz
19. Answer (09)
Hint: If PVn = C CPr
= CV +
1
R
nSolution:
PT = constant PV1/2 = C
All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper - 2) (Code-F) (Hints & Solutions)
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PART - II (CHEMISTRY)
21. Answer (A, C)
Hint : Surface tension first decreases and then
increases with the increase in concentration of the
surfactant.
Solution :
Correct curve for surface tension is
C
22. Answer (B, D)
Hint : M–L bond is stronger if L is a -acid ligand.
Solution :
Due to Pi-acid behaviour of –NO2, Co–Cl* bond would
be weaker than Co–Cl and the attack would take place
from the back side of the Co–Cl* bond, hence
pentagonal bipyramidal intermediate is formed.
23. Answer (A, B, C)
Hint : Factual.
Cu2S
Roasting
n limited i supply airof
Cu2S + Cu
2O
Heating in
absence of air
Cu + SO2
24. Answer (A, B, C, D)
Hint : Dut to inert pair effect, Bi+5 is not stable.
Solution :
Melting point of Bi is low due to metallic behaviour.
25. Answer (A, C)
Hint : To find the total spin, we can add spin of each
atom.
Solution :
Ortho hydrogen � para hydrogen
Total spin of ortho hydrogen is 1 and para hydrogen is 0.
At lower temperature para hydrogen is present in
greater concentration.
26. Answer (D)
Hint : Fact
Solution :
Observations from the curve.
27. Answer (A)
Hint : As given in paragraph, conversion of h.s. to l.s.
shrink the metal ligand bond.
Solution :
For l.s., 0 increase as r decreases.
Curve ‘1’ will be correct.
28. Answer (D)
Hint : Field strength of H2O, that does not cause
pairing in Mn+2.
Solution :
H2O is WFL for Mn2+ therefore, Mn2+ in H
2O form h.s.
complex and CFSE is zero for this system.
29. Answer (B)
Hint : Factual.
5 9–12 2– 1
2
⎛ ⎞ ⎜ ⎟⎝ ⎠
pr
R R RC
n = 9
20. Answer (05)
Hint: 1 2
1 2
r
eq
E E E
r r r in parallel combination
Solution:
1 2
1 2
r
r
E E E
r r r
Er = 5 unit and r
eq =
2
3
1
5 152 A
2 175
3
I
Test - 2A (Paper - 2) (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
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Solution :
FeSO + H O Fe(OH)SO 4 2 2 4
yellow
30. Answer (C)
Hint : Factual.
Solution :
FeSO + NaOH Fe(OH) 4 2
Fe(OH)
Brown 3
White green
31. Answer (C)
Hint : Factual.
Solution :
Mn(NO ) + NaOH Mn(OH) 3 2 2
MnMnO
(Black)
3
White
32. Answer (A)
Hint : Properties of sodium.
Solution: Na forms peroxide.
CsH is the most ionic.
33. Answer (B)
Hint : Properties of group-13.
Solution : B > Al > Tl > In > Ga
34. Answer (C)
Hint : Properties of fullerene.
Solution :
All C atoms are sp2 in fullerene.
35. Answer (C)
Hint : Stability of isotope of P.
Solution :
White P is less stable.
36. Answer A(P, Q); B(R, S); C(T); D(P)
Hint :
Factual
37. Answer A(P, S); B(P, S); C(P, Q, T); D(P, S)
Hint : Reaction of metals with HNO3.
Solution :
R = NH4NO
3.
P1 = N
2O
P2 = N
2O
P3 = NO
2
P4 = N
2O
4
P5 = NO
38. Answer (07)
Hint : Electronic configuration of lanthanoids.
Solution :
x = 1
y = 1
z = 7
39. Answer (09)
Hint : Packing type of lattice.
Solution :
Cu, Ag, Au all have CCP structure, CN = 12
x + y + z = 12 + 12 + 12 = 36
40. Answer (04)
Hint : Kdesorption
a–E /RT
= A e
Solution :
Life time (y) = a
E /RT
1
Ae
= 10–13
15 1000 3exp
25 300
⎛ ⎞⎜ ⎟⎝ ⎠
= 10 –13 × 400
y = 4 × 10–11 s
All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper - 2) (Code-F) (Hints & Solutions)
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PART - III (MATHEMATICS)
41. Answer (A, C)
Hint : 1 1
sin cos2
x x
Solution : Domain of f(x) is [–3, 3]
maximum value of f(x) is
3 397
at 38 4
x
and minimum value of f(x) is
3 311
at 38 4
x
42. Answer (B, C, D)
Hint : Equation of normal y = mx – 2am – am3
Solution : Equation of normal to y2 = 4x
is y = mx – 2m – m3 ...(1)
Equation of normal to y2 = (x – k) is
3
2 4
m my mx km ...(2)
Equations (1) & (2) represent same normal
3
3
1 2
1
2 4
m m m
m m mkm
3m2 = 4k – 6 4k – 6 0
3
2k
43. Answer (B, C, D)
Hint : Parallel lines are neither concurrent nor forms
a triangle.
Solution : Lines are concurrent if = 0
2
1 1 1
0( 1) 7 5
2 2 5 0
n n
n n
2 3
2
10 25 5 10 2 2 5 5
7 2 14 0
⇒
n n n n n n
n n
n3 – 4n2 + 5n – 6 = 0
(n – 3)(n2 – n + 2) = 0 n = 3
but for n = 3, lines become parallel
so these are concurrent for no real value of n.
44. Answer (C, D)
Hint : Concept of director circle.
Solution :
O
y
(4, 3)x
y
(3, 4)
xO
y
( , )h k
C
Ox
Locus of C is circle x2 + y2 = a2 + b2 = 25
Hence C moves in a circle of radius 5
Distance covered by C in the motion is equal to
1 14 35 tan tan
3 4
⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠
1 75tan
24
⎛ ⎞ ⎜ ⎟⎝ ⎠
45. Answer (A, C)
Hint : Find domain
Solution : Domain of f(x) is x = 1 f(x) = 0
Domain of g(x) is x = –1 g(x) = 3
2
46. Answer (A)
Hint : 1
2r (distance between parallel lines)
Test - 2A (Paper - 2) (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
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47. Answer (C)
Hint : L4 are concurrent at (12, 12)
48. Answer (A)
Hint : For cyclic quadrilateral L3 and L
4 are parallel
Solution of Q. Nos. 46 to 48
A(2, 7) 4 – 3 + 13 = 0x y
4 – 3 – 37 = 0x yB(10, 1)
(6, 4)
(3, 0
)
(9, 8)
3 + 4 – 34 = 0x y
Centre of circle is either (3, 0) or (9, 8)
+ = 3 or + = 17
(x + y – 24) + (x – y) = 0 are concurrent at (12, 12)
y
x
3 + 4 – 34 = 0x y
4 –
3 +
13 = 0
x
y
4 –
3 – 3
7 =
0
x
y(12, 12)
(2, 7)
(10, 1)
O
1
1m
should not belong to 1 11,
2 2
⎡ ⎤⎢ ⎥⎣ ⎦
= –2, –1, 0, 1
For cyclic quadrilateral
1 3 1
1 4 7
⇒
C (9, 8), r = 5
(x – 9)2 + (y – 8)2 = 52
49. Answer (C)
Hint : Take parametric point on hyperbola.
50. Answer (D)
Hint : Use parametric equation of line.
51. Answer (D)
Hint : 2
( 2) and 4y m x xym
intersect at
exactly one point.
Solution of Q. Nos. 49 to 51
Let a variable point on xy = 4 is 2
2 ,P tt
⎛ ⎞⎜ ⎟⎝ ⎠
Equation of tangent is 2
2 8x y t
t
4x
tyt
4(4 , 0), 0,A t B
t
⎛ ⎞⎜ ⎟⎝ ⎠
Circumcentre is midpoint of AB
2
2 ,h t kt
hk = 4 locus of (h, k) is xy = 4
2e R(2, 0)
r(2, 0)R
x
T
y
S(2
+ co
s,
sin
)
r
r
(2 + rcos)(rsin) = 4
r2sincos + (2sin)r – 4 = 0
1 2
4
sin cosr r
RSRT = 8|cosec2|
(RSRT)min
= 8
y2 = 8(x + 2), xy = 4
y = m(x + 2) + 2
m
is also tangent to xy = 4
242 xmx m
m
⎛ ⎞ ⎜ ⎟⎝ ⎠
has equal roots
D = 0
22
4 ( 4) 12 m mm
m
⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠
Equation of common tangent is y = –x – 4
2m – 3c = –2 + 12 = 10
All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper - 2) (Code-F) (Hints & Solutions)
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52. Answer (C)
Hint : f(x) = 1 – xn
Solution :
f(x) = xn + 1 or 1 – xn
f(x) = 1 – x4
f(3) = –80
53. Answer (B)
Hint : a = A(t 2 + 1) and b = A
2
2
(1 )t
t
Solution :
4A = 1
2
1
8A
a = A(t2+1), b =
2
2
(1 )A t
t
1 1 1
8 a b A
54. Answer (A)
Hint : Image of C2 in the line, C
2, C
3 are collinear
Solution :
P(7, 5)
(–7, 3)
(1, –3) 3 + 4 – 16 = 0x y
Image of C2(1, –3) in the line 3x + 4y – 16 = 0 is
P(7, 5)
C1C
2 + C
2C
3 + C
3C
1 is minimum
C1, C
3 and P lie on same line
C3(0, 4)
3 150 5 2C C
Radius of 1
3 2C
Radius of 3
2 2C
Equation of C3 is x2 + (y – 4)2 = 8
x2 + y2 – 8y + 8 = 0 a + b + c = 0
55. Answer (A)
Hint : Find mirror image of (1 + t2, 2t) about line
Solution :
Let variable point on the parabola is P(1+ t2, 2t) and
its reflection in the given line is (h, k)
2 2(1 ) 2 2(1 2 2)
1 1 2
h t k t t t
( 1)( 1) 2
12(1 )
h t t kt
k t h
⇒
2
2 22
hh k
Locus of (h, k) is 4x – 4y = x2 A + B = 0
56. Answer A(P); B(Q); C(P, T); D(P, Q, R)
Hint : Let f(x) = ax2 + bx + c
Solution :
Let f(x) = ax2 + bx + c
f (–1) = 0
f(1) = 4 and f(–1) = 2
a + b + c = 4
a – b + c = 2
–2a + b = 0
Solving, we get
1 51, ,
2 2b a c
21 5( )
2 2f x x x
21
[( 1) 4]2
x
1 1cos 1
( )f x
⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
[1 + sgn(f(x))] = 2
1 1tan 0, 1
2f x
⎡ ⎤⎛ ⎞ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
1 22cot 1, 2, 3
( )f x
⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
57. Answer A(Q, S); B(Q, S, T); C(P, R); D(Q, R, T)
Hint : For increasing fnf (x) > 0
Solution :
f1(x) is many-one and into
f2(x) is many-one, into and even
f3(x) is one- one and onto
f4(x) is many-one, onto and even
Test - 2A (Paper - 2) (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
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58. Answer (08)
Hint : Tangent to circle 2
2(1 )y mx m
Solution :
Tangent to the circle is 2
2( 1)y mx m ...(1)
It passes through focus (2, 2) of xy = 2
22 2 2( 1)m m
m2 – 4m + 1 = 0
Line (1) intersects hyperbola xy = 2 at the points
given by 22
2( 1)mx m
x
or2 2
2( 1) 2 0mx m x
2
1 2 1 2
2( 1) 2,
mx x x x
m m
2
1 2
16( )x x
m
Also (y1 – y
2)2 = 16m
Length of focal chord 8
2 8mm
⎛ ⎞ ⎜ ⎟⎝ ⎠
28(1 )
2m
m
8 42 8
m
m
59. Answer (08)
Hint : Let 10 cos , 2sinP
Solution :
Let 10 cos , 2sinP
� � �
10 4 3
10 5e
2
2
1 2
320 (given)2 10 cos
5PF PF
⎛ ⎞ ⎜ ⎟⎝ ⎠
2 25 1cos , sin
6 6
2 2
2 10 2 10
1 510sin 4cos10 4
6 6
d
2 2
60. Answer (04)
Hint : Let circle (x – r)2 + y2 = r2
Solution :
( , 0 )r
y
xr
Let the equation of circle be
(x – r)2 + y2 = r2
solving with y2 = 8x
x = 0 and x = 2r – 8
2r – 8 0
rmax
= 4