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
18
Embed
Test - 2A (Paper - 2) (Code-E) (Answers) All India Aakash Test … · 2019-06-04 · All India Aakash Test Series for JEE (Advanced)-2019 1/9 All India Aakash Test Series for JEE
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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)
4/9
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
5/9
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
7/9
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