Post on 11-May-2018
AITS-FT-III-PCM (Sol.)-JEE(Main)/18
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ANSWERS, HINTS & SOLUTIONS
FULL TEST – III (Main)
Q. No. PHYSICS Q. No. CHEMISTRY Q. No. MATHEMATICS 1. D 31. B 61. B 2. B 32. C 62. C 3. A 33. C 63. A 4. B 34. B 64. B 5. C 35. B 65. A 6. D 36. C 66. A 7. C 37. D 67. D 8. D 38. C 68. A 9. D 39. A 69. D 10. D 40. B 70. D 11. C 41. C 71. B 12. C 42. C 72. A 13. D 43. D 73. C 14. C 44. C 74. A 15. B 45. B 75. D 16. B 46. A 76. D 17. A 47. B 77. A 18. D 48. C 78. A 19. D 49. C 79. C 20. D 50. D 80. B 21. B 51. B 81. C 22. A 52. B 82. B 23. D 53. C 83. C 24. B 54. C 84. C 25. B 55. D 85. C 26. B 56. A 86. D 27. C 57. B 87. C 28. D 58. D 88. B 29. D 59. D 89. D 30. A 60. D 90. B
ALL
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FIITJEE JEE(Main)-2018
AITS-FT-III-PCM (Sol.)-JEE(Main)/18
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PPhhyyssiiccss PART – I
SECTION – A
1. δR/R = 2δu/u + 2cot2θ δθ 2. If s be their separation at time t then, s² = (0.4-80t)² + (0.6-60t)² where t is in hour, s in km. Minimise s using calculus. 3. The horizontal acceleration of Block 1 is always greater than that of block 2. 4. The normal reaction vanishes when the person loses contact with the surface: mv²/r = mg cos θ; Conservation of energy gives: 1/2 mv² + mgr cos θ = 1/2 mu² + mgr; where u² = 0.5gr
5. The droplets fall a distance 21 gt2
in time t, and the number of the droplets and hence their mass
is proportional to dt. Computing the CM of the droplets using the definition, we get the result.
6. For an equilateral triangle of side a, moment of inertia is 2ma
12about an axis passing through its
CM and perpendicular to plane - the result of the integration is similar to that of a solid cone along its axis: a factor of 3/5.
23 maI
5 12
7. The maximum loss occurs when the
collision is "head-on" and the minimum is when it collides at an end. Assume that the particle strikes the rod (of length 2L) perpendicularly at a distance x from its centre. Apply conservation of momentum and angular momentum to calculate the final velocity of the rod and also its angular velocity. The fractional loss in KE = 1/[3(x/L)² + 2]
0.1 0.5 0.9
0.2
0.4
0.5
0.3
8. We take θ = (10) cos (t); calculate t when = 5. This gives t = 2/3 (s). The period is 2 times
this. 9. Use Kepler's Law of Periods.
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10. If m be the mass of water displaced, the buoyant force is mg, while the mass of the extra water that has to be pushed "out of the way" is effectively m/2.
11. Taking torques about the hinge, we get: In the 1st case, mgL/2 =p AL/2 In the second case, mgL/4 = p' AL/2 12. When the wheel gets just lifted τ = mg(3r/5). But when a torque of 2τ is applied, we can write:
mg(3r/5) = Iα, where I = 2mr². The horizontal acceleration is rα(4/5).
13. The final pressure is 101p kPa2
; kx = pr2 where x = 0.2, the compression in the spring and
r = 0.05 the radius of cylinder. 14. With reservoirs at 273 K, 173 K:
W 273 173 0.37Q 273
With reservoirs at 373 K, 173 K:
W 373 173 0.54Q 373
W W = (0.54 0.37)Q = 0.17 Q
15. 2wTv
; 1fT
16. t /3 volt e(3 5)volt
; = RC = 6 s
t 6 s.
17. Tension, 1 2
1 2
3m mT g
2(m m )
; Tv
; 2
2
mv 1 T 1v 2 T 3 m
(where, m1 2m2)
18. The change in flux = 2r B2 , where r = 0.2 m
Average emf = / 2
Current = average emf / resistance
19. Using dimensional analysis or otherwise V mtA RT
some numerical factor
Substituting the values we get t 1 1.7 0.6t 2 1.2
20. ˆ ˆ ˆ ˆForce 25(0.4i 0.4k) 2i 20j 21. The correct answer can be determined by dimensional analysis.
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22. D2d = AB = 20 m
where = 2 m D = 150 m d = 7.5 m 23. From geometry = (180 2)
So, d 2ddt dt
24. When the switch closes at VC = 2V/3 the capacitor discharges through R2 = 3R T(discharge) = 3RC n 2 (since the voltage halves). The charging occurs through R1 + R2 = 9R T (charge) = 9RCn2
T = 12 RC n 2 8.4 RC 25. When the left end is positive, upper diode conducts and lower diode is cut off. Req = R When right end is positive the lower diode conducts.
eq3R 7RR R4 4
Power = 2( V) 4 11
R 7 2
2( V) 0.8
R
26. The object must be located at the centre of the curvature of the mirror for this to happen. For the inverted image formed directly by the lens we can write
1 1 1v u 10 , v 3
u 2
27. This occurs when the angle of incidence is equal to the Brewster angle: tan = .
28. If the sphere has a uniform mass density (total mass m), then qL 2m , where L = 22 mR
5 .
29. 0n 1 n 0 2 2 4
E 2n1 1h E E En (n 1) n
, for large n.
30. 1 1 2 2n
2 1
m v m vm
v v
6 7
7 614 (4.7 10 ) 1 (3.3 10 )
3.3 10 4.7 10
1.16
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CChheemmiissttrryy PART – II
SECTION – A
31. 2
2 s 2sCaF Ca 2F
42s 2 10 4s 10 3 12
spK 4s 4 10 32. d-orbital is 5-fold degenerate in absence of magnetic field. 33. NaCl glucosei CRT 1 0.1 RT 2 0.1RT 0.1RT = 0.2 RT NaCl glucose fb NaClT 2 0.1K fb ureaT 0.1K
Hence b urea b NaClT T
f ff glucose f KClT 0.1 K T 2 0.1 K
f fT 0.2 K f glucose f KClT T
34. Ea/R 600k Ae
Ea 20
c R 500k Ae
ck k
Ea 20Ea
R 600 R 500
Ea Ea 206 5
5 Ea Ea 206
520 Ea Ea6
Ea 20 6 Ea 120 kJ / mol 35. Melting point of ice decreases hence more amount of 2H O will form on increasing pressure. 36. 2H aq. OH aq. H O
2r f f fH O H aq. OHH H H H
f OH aq
57.32 285.84 0 H
f OH aq
H 228.52 kJ / mol
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37. 3 3 4 3 224Fe 10HNO 4Fe NO NH NO 3H O
38. 22 31 61 1KE m v 9 10 2.2 102 2
231 6
19
9 10 2.2 101KE2 1.6 10
KE 13.61 eV PE 2 KE 2 13.61 27.22 eV 40. o o
M N n Factor o
M 2BaCl 160 2 320
oM 3 4K PO 140 3 420
oM KCl 100 1 100
2 3 43 4 2
o o o oM M M MBaCl K PO HClBa PO
3 2 6
3 320 2 420 6 100 = 1200
oM
K 1000s
51.2 10 1000s
1200
51 10 3 4 2
5sp Ba PO
K 108 s
55108 10
231.08 10 42. [Co(NH3)4(ONO)2]Cl [Co(NH3)4(NO2)2]Cl linkage
Co
NH3
NH3
NO2
NO2
H3N
H3N
Cl Co
NO2
NH3
NH3
NO2
H3N
H3N
Cl geometrical
[Co(NH3)4(ONO)Cl]ONO [Co(NH3)4(ONO)2]Cl ionization. 43. O
S
O O
p d
p p p d
O
S
O
p d
p p S C Sp p p p
Cl
S
O O
Cl
p d p d
45. Pb2+ and Hg2+ both will precipitate out by H2S in dil. HCl.
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46. Oxidising power increases with increase of oxidation number of central atom. Mn (O.S.) in 4MnO 7 Cr (O.S.) in 2
2 7Cr O 6 V (O.S.) in 2VO 5 47. 2 2 3 2 2 4Na S O Cl H O NaHSO HCl 48. Ozone oxidizes alakaline solution of KI. 51.
Teflon
C C
F
F F
F
n CF2 CF2 n
52. Millon’s test, Biuret test, Ninhydrin test are used for protein. 54.
HO
18 O+
H
18
H
OH
18O H
18O O
O
H 55.
1CHO
CH2 CH2 CHO
CH2 CHO
1
2
2 3
1 H attackon 1
1 H attackon 3
2 H attackon 2
CH2CHO
CHO
CHOCHO
CHO
CHO 56.
OH
CH3
OH
OH
NO2
acidic strength
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57. 3 2CH NHFO2N NO2N
CH3
CH3
NNH2
CH3
CH3
58.
3POClC
O
NH2 C N 2H O
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MMaatthheemmaattiiccss PART – III
SECTION – A
61. 1 1 1 15 9 15 23cot cot cot cot .....3 3 3 3
= 1 1 1
2 1 3 2 n 1 n3 3 3 3 3 3tan tan ..... ..... tan .....2 1 3 2 n 1 n1 1 13 3 3 3 3 3
1 1n
n 1 1S tan tan3 3
nnlim S
2 6 3
62. Clearly tangents at P and Q intersect at right angle. Let S is point of intersection P, Q, R, S are cyclic S lies on the director circle of hyperbola
2 2 2 2S a b cos , a b sin
The chord with midpoint (h, k) i.e. circumcentre will be same as equation of the chord of contact w.r.t. S
2 2
2 2 2 2xh yk h ka b a b
and 2 2 2 2
2 2x a b cos y a b sin 1
a b
are identical comparing and
removing Q locus i 22 2 2 2
2 2 2 2x y x ya b a b
63. Let O (origin) is circumcentre of ABC and position vectors of A, B, C
and P are a
, b
, c
and p
Hence OH a b c
, b cOD2
, a b c pOE2
a pDE OE OD2
, AP p a
2 2p a
DE AP2
= 0 since p a
H P
A
E
D B C
O
64. It is equal to 21 3
0
x xx f x dx4 2
which is less than
1 3
0
x 1dx4 16
65. f(x) = anxn + an – 1xn – 1 + an – 2xn – 2 + ..... + a0, [x] = m then f(m) = anmn + an–1mn – 1 + ..... + a0 f(f(m) + 1) = an(f(m) + 1)n + an – 1(f(m) + 1)n – 1 + ..... + a0 kf(m) + an + an – 1 + ..... + a0 = kf(m) + f(1)
Since
f f m 1 f 1k
f m f m
= integer
m = 1 hence x [1, 2)
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66. Total such numbers are (3 2) (3!) = 36 Sum of all such numbers 66666 18 = 1199988 67. (z + )2 = z z2 + 2 + z = 0
z 1 3 i2
arg(z) – arg() = 2
3 or 4
3
68. 2a 3p q q r r p q p r p q p r p sin60º2
69. (x + y + z)6 general term 6 x y z
+ + = 6, 0 then total number of solution = 8C2 – 7 = 21
70. 2
a a a b a c
a b c b a b b b cc a c b c c
71. 1 + + 2 = 0 and
2
2
2
x zy
z xy
z x y
= x3 + y3 + z3 – 3xyz
= (x + y + z)(x + y + z2)(x + y2 + z) = 0 = (x + y + z)((x – y)2 + (y – z)2 + (z – x)2) = 0 Either x = y = z or x + y + z2 = 0 72. Denominator of all terms equal to n 12a
Numerator = (2nd + (2n – 2)d + ..... + 2d) = 2 n n 1
d2
73. 1 1 1AL L MAB BC
1 1L M1n 1 a
L1M1 = an 1
2 2 2AL L MAB BC
2 2L M2n 1 a
2 22aL M
n 1
The required sum is a 2a 3a na an.....n 1 n 1 n 1 n 1 2
L1 L2 L3
Ln
M1 M2
M3
Mn
A
B C
74. nn n n n0 1 2 n
1 1 1 1 1 1 1 1C C C ..... C 12 3 3 4 4 5 n 2 n 3
1 1
n nn n 2 n 3 n n 10 1 2 n
0 0
1 x x dx C x C x C x ..... 1 C x dx
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1 1
n n2 n 2 n 3 n 4 n n 20 1 2 n
0 0
1 x x dx C x C x C x ..... 1 C x dx
=
1 2 1 21n 1 n 2 n 2 n 1 n 3 n 1 n 2 n 3
=
1
n 2 n 3
75. Coefficient of xn in (x + x2 .....)(x2 + x4 + x6 .....)(x3 + x6 + x9+ .....) = x6(1 – x)–1(1 – x2)–1(1 – x3)–1 None of the option can satisfy 76. Clearly (5, 15) is the mid point of the chord hence slope of normal to the
chord is 20 15 110 5
, slope of chord = –1
Equation of chord (y – 15) = –(x – 5) x + y = 20 2x + 2y = 40 p2013 + q2013 = (2)2013 + (2)2013 = 22014
27 (10, 20)
(5, 15)
77. Let origin is shifted to (1, 0) after this the equation will be
max{|x|, |y|} = k Area = 4k2 = 100 k = 5
78. 21
1 h2 1
x h 0
tan hlim x n xcot x lim nh
=
3 5
1
2 3h 0 h 0
h hh ..... h3 51 tan h 1lim 1 lim
h 3h h
79. 2 2
cos x 3 cos x 3I dx dx1 sin x 3 cos x1 2sin x
3
Dividing by sin2 x, we get
2
2cosec xcot x 3 cosec x 1I dx c
cosec x 1 3 cot xcosec x 1 3 cot x
= sin x c1 2sin x
3
80. 1 1f x 1 x x 1
2 x 2 x
1f x 2 12
(A.M. G.M.)
fmin = 2 1 which occurs when 1x2
1x n2
1n , n I2
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81.
3 2 4 4
5 55
x x x 1 x 1 x 1dx dx dx dxx 1x 1 x 1x 1 x 1
= 51 n x 1 n x 1 c5
82. Odd power of a skew symmetric matrix is always a skew symmetric matrix 83. F G = {3, 6}, E {F G} = {1, 3, 6, 5, 7} P{E {F G)} = 0.13 + 0.12 + 0.13 + 0.08 + 0.06 = 0.52 84. Plane containing the line will be given as (4x – 2y – 10) + (5y – 4z + 3) = 0 4x + (5 – 2)y + 4z + 3 – 10 = 0 The plane passing through (4, 3, 7) will give 16 + (5 – 2)y + 4z + 3 – 10 = 0 –10 = 0 = 0 i.e., 4x – 2y – 10 = 0
85. Let f(x) = sin4 x – sin x cos x + cos4 x = 21 11 sin2x sin 2x2 2
Let sin 2x = t, f(x) = g(t) g(t) = 21 1t t 12 2
Since t [–1, 1] 9f x 0,8
86. f x 3 x x 1 is well defined for x [–1, 3]
1 1f x 02 3 x 2 x 1
x R
f(–1) = 2 and 12 4x 1 4x 4 gives 31x 18
Hence, 31x 1,18
87. n = 1, n + 1 = ..... etc.
n/PP 1 is P is divisor of n 88. ABC is isosceles and circumcentre of PQR is orthocentre of ABC and lies on a line to x – 3y – 31 = 0 passing through (0, 3)
89. Roots of the equation x2 + kx + 2 are irrational since k2 – 8 is not a perfect square as k > 12
So a c sinC sin A 31 2 sinC sinA
90. At point of maxima f(x) = 0 and f(x) < 0 f(x) = x2 – f2(x) 0 Since the curve x2 – y2 = a2 and x2 – f2(x) 0 2 2 2
1 1x y a Point lies outside hyperbola Hence, 2 tangents