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AITS-FT-III-PCM (Sol.)-JEE(Main)/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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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

IND

IA T

ES

T S

ER

IES

FIITJEE JEE(Main)-2018

http://www.fiitjee.com



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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.




3

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.




4

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




6

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.




7

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




8

57. 3 2CH NHFO2N NO2N

CH3

CH3

NNH2

CH3

CH3

58.

3POClC

O

NH2 C N 2H O




9

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




11

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




12

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


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