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AITS-FT-III (Paper-2)-PCM (Sol.)-JEE(Advanced)/18
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
1
ANSWERS, HINTS & SOLUTIONS
FULL TEST – III PAPER-2
Q. No. PHYSICS Q. No. CHEMISTRY Q. No. MATHEMATICS
1. A, C 21. B 41. A, B, C, D
2. C, D 22. A 42. A, B, C, D
3. B, D 23. B, C 43. A, B, C
4. A, D 24. A, B, C 44. A, B, C
5. B, C 25. B, C, D 45. A, B, C
6. A, C 26. B, C, D 46. A, B, C, D
7. A, C 27. C 47. A
8. A, B, D 28. A, B, D 48. A, B, C
9. C 29. B 49. C
10. C 30. D 50. C
11. B 31. A 51. B
12. A 32. C 52. C
13. 5 33. 1 53. 4
14. 1 34. 8 54. 1
15. 2 35. 5 55. 1
16. 8 36. 9 56. 2
17. 2 37. 3 57. 3
18. 5 38. 9 58. 3
19. 3 39. 6 59. 7
20. 4 40. 7 60. 4
ALL
IND
IA T
ES
T S
ER
IES
FIITJEE JEE(Advanced)-2018
AITS-FT-III (Paper-2)-PCM (Sol.)-JEE(Advanced)/18
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
2
PPhhyyssiiccss PART – I
SECTION – A 1.
2ay a(1 cos )2
and ab2
2 2
2a b by2 2a 8a
2
2by ( t)8a
acc. of rod’s centre = 2
2b4a
2
2mbF4a
y
a
2. y = a0 sin (kx) sin (t + )
kL kL 1n and n3 2 2
n = 1, 3kL
Also; 0 03 5La a sin aL 6
3. 3ah4
0v 2gh
0v 1.5ag and 0v v cos(90 26.5 ) 0v sin(26.5 )
ov 1.5ag sin(26.5 )
37°
a
53°
26.5°
5a/4
3a/4
v
4. Let be the angle that the rod rotates about its central axis and be the
angle made by string with the vertical.
2 2
2
L mL d(2T )2 12 dt
2
2
d 6gdt L
or 2
2
d 3gdt
2T 23g
AITS-FT-III (Paper-2)-PCM (Sol.)-JEE(Advanced)/18
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
3
5. 20
GMm GMm 1 mv2R R 2
0GMvR
While in the tunnel;
2
2 3
d x GM xdt R
2 2 2 20
GMv (a R ) a 2RR
3
A B
1 RT 24 GM
45°
45°
a
a
6. 22
2tg[cos sin cos ]
For min. time; 2d (cos sin cos ) 0d
1tan2
K + U = W friction 0 0v 2g (tan )
7. 2 2
2
Z n Zv , R and En Z n
vvR n and nE
8. 0mv 4m m v
0vv
5
2
2 00
v1 1mv . 5m . E2 2 25
2
20
1 1 vmv . m E2 2 25
20
1 4mv E2 5
20
1 5mv E2 4
E 13.6
1 3E 13.6 1 . 13.64 4
20
1 5 3mv 13.6, 13.62 4 5
= 17 eV, 8.16 eV
AITS-FT-III (Paper-2)-PCM (Sol.)-JEE(Advanced)/18
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
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9. mg T = ma1 T cos 30 = ma2 and a2 cos 30 = a1
1 21mg ma 1
cos 30
13ga7
10. Height dropped by A
a a 3ahcos60 cos37 4
2 2A B
1 1mgh mv mv2 2
Also, vB sin 37 = vA
22
Av3ag 514 2 2
A3 3agv2 17
11. Plower Pupper = P0
0 00
3P (8) 2P 12P
(8 x) (28 x)
x = 4 cm 12. Wtotal = Wupper + Wlower
= 24 12300R n n 300R n(3)12 8
SECTION – C
13. 2A
2m 2F3 3
, 2
Cm 3F2 4
A
C
F 32F 27
n = 5 14. 1htan = 2R
1
2
R / h 1tan
B
C
h
h sec sin = h tan
AITS-FT-III (Paper-2)-PCM (Sol.)-JEE(Advanced)/18
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
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15. 00
k A dTIdxcos x
6L
T = 00
0
6I LT sin x
k A 6L
T = 2T0 at x = L
16. 1j E2
qI jA2
dq qdt 2
t
2q Qe
Or Qq8
at t = 6 n(2)
17. 0v ˆ ˆv 3i j2
For crossing origin,
02(v / 2)2 mkqB qE / m
; k is an integer
0
2 EB kv
For minimum value; k = 1
min0
2 EBv
18. For leaving the surface, cos = 2/3 y = (v sin )t + ½ gt2 = R(1 + cos) … (1) and x = (v cos )t … (2)
and 2 2v gR3
… (3)
On solving;
AP = R sin + x = 5 ( 5 4 2)R27
A P
B
D
y
x
R sin
19. 3atan2h
2 4a T cos = 4ag
cos = g2T
2 2 2
3 gah2 4T g
h
3a/2
AITS-FT-III (Paper-2)-PCM (Sol.)-JEE(Advanced)/18
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
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20. Ag x (1 + sin 30) = 2
2d xA R cosec30
2 dt
2
2d x gx 3 / 2
Rdt 22
4R( 4)T3g
AITS-FT-III (Paper-2)-PCM (Sol.)-JEE(Advanced)/18
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
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CChheemmiissttrryy PART – II
SECTION – A 21.
2 2NH NHH
Ph
O
Ph
H
NNH2
OH
Ph
H
NNHPh
H
N NH
2H O
Ph
H
N NHH
OH Ph
HH
H
22. Initial m. mole of Cu2+ = 5 m. mole of H+ produced = 1
m. mole of Cu2+ converted to Cu = 1 0.52
m. mole of Cu2+ remaining in solution = 5 – 0.5 = 4.5 2
2 2 22Cu 4I Cu I I
2 2 2 3 2 4 6I 2Na S O 2NaI Na S O 4.5 = 0.04 × V V = 112.5 mol
23. B 2R,3S
Both are meso.C 2R,3S
24. Only (D) option is correct. electron cloud of C2 – C3 is present in same plane. 25. O
Claisenrearrangement
3 3 2O / CH S
O
X
OCHO
MeKOH/EtOH
O
AITS-FT-III (Paper-2)-PCM (Sol.)-JEE(Advanced)/18
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26. 4 2 3 2 32FeSO Fe O SO SO
2 2Br 2OH BrO Br H O
2 22 3 2 4Cl SeO H O SeO 2Cl 2H
2 3 4 2Cl IO 2OH 2Cl IO H O 27. 2 22 YX Z
COOH CO CO H O
CO2 → non polar and acidic → sp CO → polar and neutral → sp H2O → polar, pH = 7 → sp3
28. 4 3 2 2NH NO N O 2H O
4 2 2 2NH NO N 2H O
4 2 7 2 2 3 22NH Cr O N Cr O s 4H O
4 3NH Cl NH HCl Solution for the Q. No. 29 to 30. X2 = O2, X3 = O3, Y2 = I2 2 3 2 22I H O O 2OH I O
2 22 2 3 4 6I 2S O 2I S O
2I Starch solution Blue colouration
2 3I I excess Brown coloured I ion O2 is thermodynamically more stable as compared to O3 O3 is pale blue gas which has characteristic smell. 32. P
V 20 40 V1
A B
C
A → P1, 300 K B → P1, T2 C → P2, 300 K
For state A 1
nRTP 2.46 atmV
For isobaric process AB 1 2
1 2
V VT T
T2 = 600 K For adiabatic process BC 1TV Constant 1 1
1600 40 300 V V1 = 113.13 lit.
SECTION – C 33.
22 4
Hexadentate ligand Octahedral complexMg EDTA Mg EDTA CN 6
34. 5 2 3 4PCl 4H O H PO 5HCl
AITS-FT-III (Paper-2)-PCM (Sol.)-JEE(Advanced)/18
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35. 2 2 2 2 2H ,Li ,B ,N ,F 36. Atomic number of metal = 29 Cu = 3d104s1 Cu2+ = 3d9 37. 22 2 4 42KI HgI K HgI 2K HgI i = 3
38. |C
*Cl C C C C C → one chiral carbon pair of enantiomers.
C C
Cl
C
C
C C 2 chiral carbon2 pair of enantiomers4 pair of diastereoisomers
C C C
CCl
C C achiral molecule
C C C
C
C C
Cl
achiral molecule
x = 3, y = 4, z = 2 x + y + z = 9 39.
O
I I
IMg MgI
2 3CO / H O
H2C CH2
H2C CH2
COOH
COOH 40. sp spK CuCl K AgCl
Thus assume Cl in solution comes from CuCl only.
3spCl K CuCl 10 M
For AgCl 10Ag Cl 1.6 10
7Ag 1.6 10 x = 7
AITS-FT-III (Paper-2)-PCM (Sol.)-JEE(Advanced)/18
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MMaatthheemmaattiiccss PART – III
SECTION – A
41. 2 2d f x f x 2f x f x 2f x f xdx
= –2x g(x)(f(x))2 0
42. |PR|2 + |RQ|2 2|PR| |RQ| |PR|·|RQ| 2Ar(PQR) 8Ar(PQR) |PR|2 + |RQ|2 + 4Ar(PQR) |PR|2 + |RQ|2 + 2·|PR|·|RQ| < 8Ar(PQR) + 1 8Ar(PQR) = |PQ|2 + |QR|2 + 4Ar(PQR) and |PQ|2 + |QR|2 = 2|PQ|·|QR| = 4Ar(PQR)
B C
A
R Q
P
R = 90º and RP = RQ 43. p q r 0
So, p, q, r
can form a triangle
p c b c a c c a b c 0
p c
q a
and r b
44. Take, = 1 |1 + |2 = |(1 + cos ) + i sin |2 = 2 + 2 cos 2 3
3cos2
so, we get 332 angles
332p1996
45. 2x = CBD 2y = ABD In CBD
sin 2y x sin 2x yBD BD
sin x BA BC siny
sin(2y + x) sin y = sin(2x + y) sin x
x y
A B
D C
1 cos y x cos 3y x2 = 1 cos x y cos 3x y
2
0 < x + y = 1 ABC2 2
0 < (3y + x) + (3x + y) < 2 3y + x = 3x + y x = y ABD = CBD AD = CD
46. ag x f x fx
2a a ag x f x f f x fx x x
= 2a ax 0x x
g(x) = b
AITS-FT-III (Paper-2)-PCM (Sol.)-JEE(Advanced)/18
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af x f bx
abxf x b
a
47. Let, OA = x1, OB = y1, OC = z1 be points on X, Y, Z axis then equation of plane is
1 1 1
x y z 1x y z
1 1 1
1 2 3 1x y z
Locus is 1 2 3 1x y z
48. Total ways =
6 48 8 4 8 8 53 2
3 4 2 2 3 1 4
C C 8C C C C C C2 2 2 4
49.-50. f(x2 + y f(z)) = x·f(x) + z f(y) x = y = z = 0 f(0) = 0 x = 0 f(y·f(z)) = z·f(y) y = z = t f(t·f(t)) = t·f(t) Using this and substitution, we get f(x) = x and f(x) = 0
51.-52. Ar.(B0, B1, ....., Bn – 1) = n cot4 n
Ar.(A0 A1 ..... An – 1) = n 2sin2 n
0 1 n 2
0 1 n
nsin cosAr. A A ..... A n n 4sinnAr. B B ..... B ncot4 n
SECTION – C
53. x(f(x))2 – x2f(x) = 3/2 3 32 x x xxf x 2 x f x2 4 4
23/2 3x xx f x
2 4
B – A = 21 13/2 3
0 0
x xx f x dx dx2 4
A – B 116
54. Say, < 0
AITS-FT-III (Paper-2)-PCM (Sol.)-JEE(Advanced)/18
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|I + A2| = |I – 2A2| = |I – A||I + A| As, A = –A I – A = I + A = (I + A) |I + A2| = |I + A||I + A| = |I + A|2 0 Same, can be seen for 0 55. If x > 2 then x3 – 3x > 4x – x = x x 2 |x| 2 take x = 2 cos for some [0, ] 2cos3 2 1 cos
7 52sin sin 04 4
= 0, 4 4,7 5
x = 2, 4 12cos , 1 57 2
56. B2 – tr(B)·B + I = 0 AB – (tr(B)A + AB–1 = 0 tr(AB) – tr(A)tr(B) + tr(AB–1) = 0 57. Using property 58. Using property of parabola
59.
x
f x f xlim 0
x
For any > 0 there is > 0 such that |x| < and |f(x) – f(x)| < |x| using triangle inequality |f(x) – f(n x) |f(x) – f( x)| + |f( x) – f(2 x)| + ..... + |f(n – 1 x) – f(n x)|
< |x|(1 + + 2 + ..... + n – 1) = n1 x
1
x
1
As, n
f x x1
x
f xlim 0
x
60. Let g(x) = f(x)e–f(x) g(a) = g(b) so using Rolle’s theorem in C (a, b) g(c) = 0 g(x) = e–f(x) (f(x) – (f(x)2) f(c) – (f(c))2 = 0