Aits-1920-Ft-Iii-Jeem - Sol PDF

FIITJEE
ALL INDIA TEST SERIES

FULL TEST – III
JEE (Main)-2020
TEST DATE: 29-12-2019
ANSWERS, HINTS & SOLUTIONS
Physics PART – I
SECTION – A
1. B
u2
Sol. Maximum range up the inclined plane =
g(1  sin )
u2
Maximum range down the inclined plane =
g(1  sin )
The maximum possible distance between the two bullets after they hit the inclined plane
u2 u2
 dmax  
g(1  sin ) g(1  sin )
2u2
dmax 
gcos2 
2. B
Sol. Use the basic concept of interference of light waves.
3. C
Sol. Acceleration of block = 3 m/s2
Acceleration of plank = 7 m/s2
1
After 3 sec, the plank will come to rest. In 3 sec, the block will move  4  3  3  18 m relative to
2
the plank.
Now, let the block will cover remaining 18 m relative to plank in next t seconds.
1
12t   3t 2  18
2
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AITS-FT-III-PCM(Sol.)-JEE(Main)/20 2
8t  t 2  12  t = 2 sec
Total time after which the block will get off the plank, t0 = 3 + 2 = 5 sec.
4. B
Sol. At min
gR(1  sin )  gR(1  cos )  gR(sin   cos )  0
sin  cos 
 cos   sin    0
3 3
  = tan1(1/2)
5. B
Sol. f T
T0 = nwL3g
Tnew  n wL3 g  2w L2hg = w L2 g(nL  2h)
L h
f w L2 g(nL  2h)
 new 
f0 nw L3 g
1/2
 nL  2h 
fnew   f0
 nL 
6. C

Sol. 
2
t /2

T 2
 m
 t = T/4 =
2 k
7. D
Sol. 3L(1   eff )  L(1  2)  2L(1  )
3L eff   4L
4
  eff  
3
8. A
Sol. F.B.D. vg
(6/5)v 6v
(vg)/5
9. A
Sol. I = neAvd
I = kV2
3 AITS-FT-III-PCM(Sol.)-JEE(Main)/20
10. D
Sol. VCC = IBRB + VBE IB IC
V  VBE 7.5  1
IB  CC  3
 50  10 6  50A RL
RB 130  10 RB C
IC  IB  100  5  10 5  5  103 A  5 mA B
VCE VCC = 7.5 V
Now, ICRL  VCE  VCC VBE E
V  VCE 7.5  3.5 4000
RL  CC    800
IC 5  103 5
11. A
Sol. d(K.E.) = mv v
Change in velocity v has fixed magnitude
Greatest change in kinetic energy is achieved when impulse is parallel to the velocity and when
the speed is as large as possible.
12. C
2
I0 I 4
Sol. Then intensity of the emerging light, I  cos2 (37)  0   
2 2 5
8I0 2
I  8 W/m
25
13. C
ka2
Sol. V0  = potential at the centre of a cube of side ‘a’
0
4a2
k   4V0 = potential at the centre of a cube of side ‘2a’
0
Six such pyramids make a cube
4V0 2V0
Potential at the tip of pyramid = 
6 3
14. B
dQ1 dQ2
Sol.  0
T1 T2
dT1 dT
  2  T1T2 = constant
T1 T2
When their temperatures become equal
T 1T 2 = T 2
 Final temperature, T = T1T2
15. A
Sol. t = 1.5 h
Total charge transferred = 4.5 Ah
Qmax = 60 Ah
4.5
Percentage charged =  100  7.5 %
60
16. C
Sol. Use Faraday’s Law
17. A
100  60
Sol. nh  = 60 J
100
60 60  3.9  10 7
n   34 8
= 1.18  1020
hc 6.63  10  3  10
18. A
Sol. Using conservation of momentum
mu = mv1 + Mv2
 2v 2 + v1 = u …(i)
v2  v1 = u …(ii)
from (i) and (ii)
v2 = 2u/3
2
1  2u 
 2m   
 Fraction of kinetic energy lost by neutron=
2  3  8
1 9
mu2
2
19. C
2
1  p sin   2p sin  p cos 
Sol. 2    
2g  m  mg m
1 2
sin   sin  cos   tan  = 2
2
1
 cos  
5
(pcos )2 p2 1 p2
Minimum kinetic energy =   
2m 2m 5 10m
20. D
bt

Sol. A  A0e 2m
n2 20
  A0
A  A0e 102 2  A 0 e  n 2

2
SECTION – B
21. 5
Sol. H = B/0
ni = 2  104
4
40  100 i = 2  10
20
i A  5A
4
22. 7
Sol. m1g  T1  m1a …(i)
T2  m2g = m2a …(ii)
m1g
T1  T2   m1a …(iii)
2
mg
(m1  m2 )g  1  (2m1  m2 )a
2
3gm2
2m2 g 
a 2
7m2
g
amax 
2
g
amin 
14
amax
7
amin
SECTION – C
23. 00004.80
d y
Sol. (S  1)t 0 
D
D
y  (S  1)t 0
d
 velocity of central maxima,
dy 2.4  10 4  2
v  (2t  12) 
dt 2  10 3
At t = 4 sec
v = 20  2.4  101 = 4.80 m/s
24. 00086.60
Sol. N  mgcos   F sin  F
F cos  = N + mg sin 
F(cos  + sin ) = mg (sin  +  cos ) N 
mg(sin    cos  )
F 10 kg
(cos    sin )
mg(sin    cos  ) N = fsmax
 Fmin  [when  = tan1() = 30]
2  =30
1 
 Fmin = 86.60 N
25. 00000.25
5
Sol.  1  e …(i)
4
9
 2  e …(ii)
4
From (i) and (ii)
 = 2  1 = 200  111 = 89 cm
From (i), end correction
5 5
e  1   89  111
4 4
 e = 0.25 cm
Chemistry PART – II
SECTION – A
(One Options Correct Type)
This section contains 20 multiple choice questions. Each question has four choices (A), (B), (C) and
(D), out of which ONLY ONE option is correct.
26. A
Sol. H3C CH CH3 will respond the test.
Cl
27. A
O O
|| ||
Sol. Aqueous solution of CH3  C OH and CH3  C ONa is a buffer solution.
28. D
Sol. Sketch for 2s orbital is
O r
29. A
Sol. Ecell is an intensive property, where as mass, internal energy, heat capacity are extensive
properties.
30. C
Sol. dE  dq  dW
dE dV dq
dE  PdV  dq  P 
dT dT dT
PdV dq
 nCV, m  
dT dT
fR PdV dq
 1  
2 dT dT
5R PdV dq
   …. (1)
2 dT dT
PV  RT  V 3  RT
dV PdV R
 3V 2 R 
dT dT 3
5R R dq
 
2 3 dT
17R dq

6 dT
31. A
Sol. Forsterite [Mg2(SiO4)] is orthosilicate.
32. B
Sol. Given complex have four possible coordination isomers.
en CN CN en
3 0]3 6 0]3 
2 2]1 4 1]1
1 4]1 2 2]1
0 6]3 0 3]3 
Co  en 3  Cr  CN6 
Co  en 2  CN2   Cr  CN4  en  
Co  en 1  CN 4  Cr  CN2  en 2 
Co  CN6  Cr  en 3 
33. A
Sol. Number of streo isomers possible for the given complex is 8.
34. A
Sol. P4O10 is not an oxidizing agent.
P4O10  NaCl 
 POCl3  Na3PO 4
35. A
Sol. Minimum 4 hydrogen atom should be present in sample to get 6 different spectral lines.
36. B
Sol. Barytes is a sulphate ore.
37. C
Sol. Two optical isomers are possible for the given organic molecule
HO
H3C CH CH CH2
Cl Cl
38. A
Sol. Total 5 dihalogen derivatives are formed including stereoisomers in which two are diastereomers.
x + y = 10.
39. C
Sol. CH3
O
CH3
CH3 1 CH MgX

3 OH

 2 H O

2
Cl CH3
40. C
O OH
Sol. will tautomerise to form .
41. B
Sol.
O O O
C CH2 X C CH2 X


NaBH4 
H
42. D
Sol. CH3
i BH
3 / THF
H OH CH3 CH3

ii H O /OH
 
2 2
CH3
CH3 CH3 H OH
Number of products = 2.
43. C
PVm Vm 11b 11
Sol. Z   
RT Vm  b 11b  b 10
P  V  nb  RT V  nb
  RT   = Vm – b
n P n
44. B
Sol. Cl
dry ether
 Na  
Br
45. B
Sol. B  3D  4A  4C
d B  1 d D 1 d  A  1 d C
Rate of reaction    
dt 3 dt 4 dt 4 dt
SECTION – B
46. 1
Sol. Be2 : 12 * 1s2 2s 2  * 2s1
1
B. O. =
2
 4  3  0.5
 2x = 1
47. 4
Sol. Following are dibasic acid
H2SO4, H3PO3, H2SO5 and H2S2O7.
SECTION – C
48. 00200.00
Sol. Let the final composition, liquid (10 mole)
A = x mole, B = (10 – x) mole
Vapour (10 moles), A = 10 – x mole; B = x mole
YA X A PAo 10  x x 400
Now,  o
  
YB XB PB x 10  x 100
10
x=
3
Now, P  X APAo  XBPBo
x 10  x
  400   100
10 10
= 200.00
49. 00085.40
Sol. K 2Cr2O7  14HCl  2KCl  2CrCl3  3Cl2  7H2O
n-factor of K2Cr2O7 = 6 and
3
n-factor of HCl 
7
M 7M
 EHCl    85.40
3 3
7
= 85.40
50. 00006.50
pKa1  pKa2 4  9
Sol. pH at isoelectric point =   6.5
2 2
Mathematics PART – III
SECTION – A
51. B
3  9
Sol. Area is minimum when c  3
2
52. B
2
25!  1
Sol.  6900
22! 2! 1!
53. A
Sol. b + 3 = 4, b = –7, 1
54. A
Sol. Coefficient of x in (x–3 + x–2 + ….. + x4)3 – 1 = 11
C9  3  3 C1  1  46  1  45
55. B
Sol. Number of even divisor of 80 is 8
56. B
Sol. k = 2, a > 3
57. A
Sol. x + y = 47, number of matrix = 46C1 = 46
58. D
17 17
Sol.  ,  = 0,  
2 2
59. A
Sol. Using graph clearly see
f(x) is continuous everywhere but not differentiable at exactly 3 points
60. A
Sol. m = 10 and n = 12
61. A
Sol. Let A(a, 0), B(0, b)
za 3 4 C(z)
 ei   i B(0, b)
z  ib 5 5
5(z – a) = (z – ib)(3 + 4i)
5(x – a + iy) = (x + iy – bi)(3 + 4i)
Comparing, we get 5(x – a) = 3x – 4(y – b) A(a, 0)
2x + 4y = 5a + 4b ….. (1)
5y = 4x + 3(y – b)
4x – 2y = 3b ….. (2)
4  4x  2y 
2x + 4y = 5a   2x – 4y + 3a = 0
3
62. D
1 1 1
Sol.  
 1  2 4a
63. A
5 5
Sol. a2  , b2 
2 12
64. D
Sol. P(x) = (x + 3)(x – 3)3 + x2 + 2
65. B
Sol. Radius of circle = 34
66. C
9110  8210
Sol. Required probability =
9110
67. C
Sol. It is neither tautology nor contradiction
68. D
Sol. Data are multiplied by 7, then variance should be 49 times
69. A
x x
dy 1
Sol. tx = y  dt    f  y  dy  n f  x    f  y  dy  x  n  f  x 
x x0 0
f x  1 n
Differentiating f(x) = n(f(x) + xf(x))  
f x  nx
 1 n   1n 
 1 n       
Integrating, ln f  x     ln cx  f  x    cx  n   f  x   c   x  n 
 n 
70. D
 
Sol. =0x+y=  Equation of mirror image is x – y =
4 4
SECTION – B
71. 1
 
Sol. A = –1, B = 1, f     1
 4
72. 1
Sol. Use A.M.  G.M.
SECTION – C
73. 00001.75
 2  1  2  a 7
Sol.   tan1    sin    2b  4  1.75
 3   7 
74. 00000.44
x3 y 2 z0  3  4  1  8 1 10 8  10 8
Sol.    2   ; x ; y ; z ;   3    7  0
1 2 1  1 4  1  3 3 3 3 3 3 3
11
25 = –11,  
25
75. 00001.65
12!
14 
5!  2!
Sol.  1.65
7! 8
 P5
2!

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FIITJEE

ALL INDIA TEST SERIES

ANSWERS, HINTS & SOLUTIONS

CH3 1 CH MgX

Sol. will tautomerise to form .

Mathematics PART – III

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