aiats jee advanced 2015 test-1
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7/18/2019 Aiats Jee Advanced 2015 Test-1
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Test - 1 (Paper - II) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2015
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TEST - 1 (Paper - II)
PHYSICS
1. (B)
2. (A)
3. (B)
4. (D)
5. (C)
6. (B)
7. (A, B, C, D)
8. (A, B, C)
9. (A, B, C, D)
10. (A, D)
11. (A, C)
12. (B)
13. (C)
14. (B)
15. (A)
16. (A)
17. (C)
18. (4)
19. (5)
20. (6)
CHEMISTRY
21. (B)
22. (B)
23. (C)
24. (D)
25. (B)
26. (C)
27. (A, D)
28. (B, C, D)
29. (A, C)
30. (C)
31. (A, D)
32. (B)
33. (A)
34. (C)
35. (A)
36. (B)
37. (A)
38. (5)
39. (8)
40. (4)
MATHEMATICS
41. (B)
42. (D)
43. (B)
44. (B)
45. (A)
46. (B)
47. (A, B, C, D)
48. (A, D)
49. (A, C, D)
50. (A, B)
51. (A, B, C, D)
52. (A)
53. (C)
54. (B)
55. (B)
56. (A)
57. (A)
58. (6)
59. (3)
60. (1)
ANSWERS
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1. Answer (B)
120°mg
T
In equilibrium when right spring is cut
2T cos60° = mg T = mg ...(i)
F nett
=2 2
2 cos120mg T mg T
a = g
2. Answer (A)
N = 2 mg cos ...(i)
For wedge, N = 5mg + 2mg cos230°
35 2
4mg mg
3 135
2 2mg mg mg
N
2 c o
s
m g
2 s i n
3 0 °
m g
Also, 2mg cos30°sin30° = N
Wedge
5mg
N
N
m g
= 2
c o s
f
3 0 °
PART - I (PHYSICS)
ANSWERS & HINTS
3 1 132
2 2 2mg mg
3
13
3. Answer (B)
For block,
2
2 –2
22 – 2 0 – 4 –4 ms B
dH d H H t t a
dt dt ⇒ ⇒
aB cos60° = a
W cos30°
60°
aW
aB
60°
aW =
cos60·cos30
Ba
–21 2 44 ms
2 3 3
4. Answer (D)
N
E
i
j
– –2 2
PQ P Q
v v v i j v v
...(i)
2 2– – –
2 2
QR Q R
v v v i j v v ...(ii)
– – – – RP R P R Q P Q
v v v v v v v
= 2 2– –
2 2 2 2
v v v v i j i j
=3
2 2
v v i j
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5. Answer (C)
x = u cos × t ...(i)
21sin –
2y u t gt ...(ii)
2
2 2 2 21
2 x y gt u t
...(iii)
2 4 2
2 2 2 22
4 2
g t gt x y y u t
36 + 64 + 25t 4 + 80t 2 = 625t 2
25t 4 – 545t 2 + 100 = 0
2 2 2 2
1 2 1 2
545 100,
25 25t t t t
2 2
1 2 1 2 1 2
545– – 2 – 2 2
25t t t t t t t
4.2 s
6. Answer (B)
2
rel cos 37 – sin 37v v at v t
2
4 310 – 5 10
5 5t v
rel0
d v
dt
t = 1.6 s
7. Answer (A, B, C, D)
For vertical motion, 21–
2
h ut gt ...(i)
1 2
1
2h g t t
For projectile motion,21
sin · –2
h u t gt ...(ii)
1 2
1
2h g t t
1 2 1 2
2 sin 2 cos 1, ,
2
u u t t R g t t
g g
8. Answer (A, B, C)
(i) Time of crossing is minimum when v u and
t min
=d
v
(ii) When v > u , the swimmer can reach, directly
opposite the point of start, t crossing
= 2 2–
d
v u
u
v
(iii) For v < u , drift cannot be zero
u
v v s
9. Answer (A, B, C, D)
F = kt k a t
m ...(i)
2
2
k v t
m
...(ii)
2
2
2
k mv a
m k
...(iii)
a t , v t 2, v a2
10. Answer (A, D)
In (A), For m, 2mg – mg = ma1
...(i)
2
3 3
2 2 2 –In (B), ,
2 2
In (C), – and 2 –
m m m m g T g a
m m m m
T mg ma mg T ma
1 2 3
1 2 3
Solving , ,3 2
84 , , 3
3
g ga g a a
T mg T mg T mg
11. Answer (A, C)
For bead, 1 1
7–
8 8
mg g mg ma a ⇒ ...(i)
For mass block, 2 2
3–
4 4
mg g mg ma a ⇒
For bead w.r.t. string, arel
= a1+ 2 × a
2
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=7 3 7 12 19
8 2 8 8
g g g g g
2 8 16
19 19
l l t
g g
12. Answer (B)
1 2 1 1 2 2
1 2
sin 37 – cos 37m m g m m g a
m m
1 1 2 2
1 2
sin 37 – ·cos 37m m
gm m
3 4
10 – 0.2 2 0.1 15 5 3
3 410 – 0.5
5 5 3
3 0.410 –
5 3
= 4.67 ms–2
13. Answer (C)
m1g sin37° + f –
1m
1g cos 37° = m
1a f = 0.53 N
14. Answer (B)
A
B C a
1
a2
a1
2a32a
3
a3
Let acceleration of A, B and C are a1 , a
2 and a
3 ,
respectively.
Clearly, a2 = a
3...(i)
mg – T = ma1
...(ii)
mg – 2T = ma2
...(iii)
Pulley constraints1 3
2
2–
2
a aa
...(iv)
From (iv), a1 + 2a
3 = –2a
2 a
1 + 2a
2 = –2a
2
a1 = 4a
2...(v)
From (ii) and (iii), 1
2
2 – 2 2
2 –
mg T ma
T mg ma
1
4
9a g
1
2
4 1
4 9 4 9
a g ga
15. Answer (A)
16. Answer (A)
17. Answer (C)
18. Answer (4)
2 2
1
2
1– 10 20 –5 2– 2 s
2
2 2 s
h ut gt t t t
t
⇒ ⇒
t 1+ t
2 = 4 s
19. Answer (5)
20. Answer (6)
f 2 =
2m
2g = 0.5 × 2 × 10 = 10 N,
Tension T = 15 – 10 = 5 N
For m1, T = F
2 + f
1 f
1 = 1 N
21. Answer (B)
Equivalent weight of 2–
4SO ion
96
2 = 48
Total equivalents of sulphates of two metals
2 2 7
12 48 32 48 120
Equivalents of BaSO4
7
120
Mass of BaSO4 7 233 6.8 g
120 2
PART - II (CHEMISTRY)
22. Answer (B)
Let the normalities of HCl and H2SO
4 be N
1 and N
2
respectively.
25(N1 + N
2) = 10 × 2.0 = 20
N1 + N
2 = 0.8
H2SO
4 + BaCl
2 BaSO
4 + 2HCl
2
0.699 100010 N
233
2
; N2 = 0.60
N1 = 0.20
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23. Answer (C)
FeCl3 + 3NaOH Fe(OH)
3 + 3NaCl
Moles of Fe(OH)3
1.425
107
Molarity of FeCl3
1.425
0.1 107
Normality of FeCl3
1.425 30.40 N
0.1 107
24. Answer (D)
Kinetic energy of proton = eV
2p pm v 2 eV ; p p p
m v 2m eV
Similarly, for particle m v 4m eV
p2m eV VV
4m e 8
(∵ m = 4mp)
25. Answer (B)
Energy of photon when electron returns from 6th orbit
to 3rd orbit of Li2+ 1 1
13.6 9 – eV9 36
3
13.6 eV4
Electron in 2nd orbit of He+ ion absorbs this energyand jumps to nth orbit
2
3 1 113.6 13.6 4 –
4 4 n
n = 4
Energy of electron in 4th orbit of He+
4
–13.6 –3.4 eV16
26. Answer (C)
Fact
27. Answer (A, D)
104.5% oleum means 100 g oleum reacts with 4.5 g
H2O.
10 g oleum reacts with 0.45 g H2O
SO3 + H
2O H
2SO
4
Mass of SO3 in 10 g oleum
0.4580 2.0 g
18
10 g oleum is mixed with 0.18 g H2O
Mass of SO3 that combines with
2
0.18 80H O 0.80 g18
Mass of free SO3 left in 10.18 g solution = 1.2 g
Percentage of free SO3
1.2 10011.79%
10.18
Mass of H2SO
4 in 10.18 g solution
0.18 988 8.98 g18
28. Answer (B, C, D)
Let the volume of CO2 in the given mixture be x ml
CO2(g) + C(s) 2CO(g)
Volume of CO after the reaction = 125 – x + 2x = 205
x = 80 ml
Volume of CO in the mixture = 45 ml
Mole percent of CO45 100
36%125
Mole fraction of CO2 80
0.64125
29. Answer (A, C)
Since the number of photons absorbed and
transmitted is same, the electron moves from 1st orbit
to 2nd orbit only. The energy of the photon absorbed is
given by
E = (10.2)Z2 eV
If Z = 2, E = 40.8 eV and if Z = 3, E = 91.8 eV
30. Answer (C)
Electronic configuration of Mn is 1s22s22 p63s23 p63d 54s2
l + m = 0 if l = 0 and m = 0 i.e., all s-orbitals
l = 1 and m = –1 i.e., one of p-orbitals
l = 2 and m = –2 i.e., one of d -orbitals
Total number of electrons having l + m = 0 is equal to 13.
31. Answer (A, D)
Fact
32. Answer (B)
For particle (X),X
x x
h
m v
For particle (Y),Y
y y
h
m v
x xY X
y y
m v 2.416 Å
m v 0.3 0.5
33. Answer (A)
1 n
v Z
2 2
1 1
n
n
2
1
n 3n
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34. Answer (C)
Fact
35. Answer (A)
Fact
36. Answer (B)37. Answer (A)
38. Answer (5)
K4[Fe(CN)
6] + 6H
2SO
4 + 6H
2O 2K
2SO
4 + FeSO
4
+ 3(NH4)2SO
4 + 6CO
Millimoles of K4[Fe(CN)
6] =
1
6 × millimoles of CO
33.6x 0.05
6 22.4
x = 5 ml
39. Answer (8)
(COOH)2 + 2NaOH (COONa)
2 + 2H
2O
NaOH + HCl NaCl + H2O
Milliequivalents of oxalic acid = Milliequivalents of
NaOH – Milliequivalents of HCl
1. 89 2 100040 0.80 – 0.25 x
126
On solving, x = 8 ml
40. Answer (4)
2
2
ZKE
n
He
H
(KE) 4
(KE) 1
PART - III (MATHEMATICS)
41. Answer (B)
(4, 4)(–4, 4)
(4, –4)(–4, –4)
|z – (4 + 4i )| 4
|iz – (–4 + 4i )| 4
|–z – (–4 – 4i )| 4
|–iz – (4 – 4i )| 4
A = 8 × 8 – × 42
= 16(4 – )
42. Answer (D)
2
2 2 2
1 1 1 1 1 1
– – –( – ) ( – ) ( – ) x y y z z x x y y z z x
1 1 1– 2
( – )( – ) ( – )( – ) ( – )( – )y z z x x y z x x y y z
=
21 1 1 – – –
– 2– – – ( – )( – )( – )
x y y z z x
x y y z z x x y y z z x
=
2
1 1 1
– – – x y y z z x
Hence ture for all x R , y R , z R except the
case when x = y or y = z or z = x.
43. Answer (B)
Clearly all the terms of x 28 to x –28 are possible
So, total number of terms are = 28 + 28 + 1 = 57
44. Answer (B)
(72)10 = (70 + 2)10
10 32 2 · 8
27 7
⇒ will be remainder
45. Answer (A)
Coefficient of x 49 is =
2 2 501 2
0 1 49
2 ..... 50 C C C
C C C
50
1
– (51– ) –22100
r
r r
∑
46. Answer (B)
[ x 7 + (1 + x 5)]20
= 20C 0(1 + x 5)20 ( x 7)0 + 20C
1(1 + x 5)19( x 7) + 20C
2(1 + x 5)18
( x 7)2 + ........
Coefficient of x 17 = 20C 1 × 19C
2
20 19 18
2
= 3420
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47. Answer (A, B, C, D)
S = 20C 7 – 20C
8 + 20C
9 – 20C
10 + ....... – 20C
20
We know
(1 + x )20 = 20C 0 + 20C
1x + 20C
2 x 2 + ....... + 20C
20 x 20
(1 + x )–1 = 1 – x + x 2 – x 3 + x 4 – ......
(1 + x )19 = x 13[20C 13
– 20C 12
+ ....... – 20C 0] + .....
20C 7 – 20C
8 + ....... – 20C
20= 19C
13
19 18 17 16 15 14
6!
48. Answer (A, D)
2 2
2
1
| |
z z z z E
z z
Let z = x + iy
2 2
2 2 2
.
| |
x y x iy x iy z z z z E
z x y
2 2 2 2
2 2
x x y y x y y i E
x y
y ( x 2 + y 2) – y = 0
either y = 0 or x 2 + y 2 = 1
49. Answer (A, C, D)
S = 0.100C 02 + 100C
12 + 2.100C
22 + ..... + 100.100 2
100C
S = 100.100 2
100C + 99.100 2
99C + ................ + 100C
12
___________________________________________________________________________
2S = 100 [100C 02 + 100C
12 + 100C
22 + ..... + 100 2
100C ]
S = 50.200C 100
50. Answer (A, B)
51. Answer (A, B, C, D)
3 + 2 = 1
1– 32 1– 3
2
⇒
= 1,1
5
= 11
5and rejected
1
5
= –1
The polynomial will become
( x + 1)2( x – 1)3 = ( x 2 – 1)2( x – 1)
= ( x 4 – 2 x 2 + 1)( x – 1)
= x 5 – x 4 – 2 x 3 + 2 x 2 + x – 1
52. Answer (A)
2
0
1
nn n r
r
x C x
∑
On integrating between 0 to 1
21201 2
0
2 1.....
2 3 21 21
C C C C
... (1)
On integrating between –1 to 0
201 2
0
1.....
2 3 21 21
C C C C
... (2)
(1) + (2), we get
201931 2 1
.....2 4 20 21
C C C
53. Answer (C)
1
1
0 0 0
1 1 1
1 1
n n n
n n n
r r r
r r r
r C C C
r n
∑ ∑ ∑
112 2 1
1
n n
n
Since,
8
0
2 2 1
1 6
n
n
r
r
r C
r
∑
1 8
2 1 2 12 5
1 6
n
nn
n
⇒
54. Answer (B)
55. Answer (B)
Solution of Q. No. 54 and 55
x
y
(–10, 0) (–3, 0) (+5, 0)
(0, 4)
| +3| + | –3| = 10z z
(+3, 0)
– 4 4– 4Re Re 0
4 4 4
z z z
z z z
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put z = x + iy
x 2 + y 2 = 4
– 255
– 1
z
z
put z = x + iy
x 2 +y 2 52 , Clearly B C has no intersection points
area of ( A C ) = area ( A) = 20
area of (C D) = 25 – 20 = 5
area( A) = 20
area(C ) = 25
20 + 5 + 20 – 25
= K
K = 20
56. Answer (A)
(P)
64
64 821
3 5
r r
r r t C
For a rational term r must be a multiple of 8,
hence number of multiples of 8 in 0 to 64 = 9
(Q) By using properties of Binomial coefficient
2 2 2
69 69 69 69 70 70
3 1 3 31r r r r r r
C C C C C C ⇒
r 2 = 3r or r 2 + 3r – 70 = 0
r = 3, 7
(R) 26.3
2 3
x r
x
r = 5
(S) 262 = 4(1 + 7)20 = 4 + 7k ,
Hence remainder is 4
57. Answer (A)
58. Answer (6)
a0
( x – 2)0 + a1
( x – 2)1 +.........+a10
( x – 2)10
= b0( x – 3)0 + b
1( x – 3)1 +........+b
29( x – 3)29
comparing both the side
b5( x – 3)5 = ( x – 2)5 + ( x – 2)6+........+( x – 2)10
= ( x – 3 +1)5 + ( x – 3 + 1)6 +..........+( x – 3 + 1)10
b5 = 5C
5 + 6C
5 + 7C
5+.......+10C
5
b5 = 6C
6 + 6C
5 + 7C
5+.......+10C
5
= 11C 6
Then the value of = 6
59. Answer (3)
1 2 3
18 2
81 9r r r ... (1)
1 2 2 3 3 1
36 4
81 9
r r r r r r
... (2)
1 2 3
8
81r r r ... (3)
From (2) and (3)
1 2 3
1 1 1 4 / 9 9
8 / 81 2r r r
2
2
3 9 2
2 3r
r
⇒
1 3 1 3
4 4,
27 9r r r r
1 2
2 2,
3 9r r
2 2 2
3 9 3
1 6 4 2 1 18 36 6
8 2 / 3 2 / 9 2 / 3 8 2
= 3
60. Answer (1)
5 3 322 1z z z z z
5 3 322 1z z z z z
3 2 22 1z z z z
2 2 2 2
2 2 2 x y x iy x y xyi
3/ 2
2 2
1
x y
(where z = x + iy )
2
3/ 22 2
12 2y x
x y
and y = 0
3
12 x
x
3 1
2 x x
only possibility is z > 0 and 4 1
2 x
1
4
1
10
2z i
,
1
4
2
1– 0
2z i
= +