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AITS-FT-III (Paper-2)-PCM (Sol.)-JEE(Advanced)/18

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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

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T S

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FIITJEE JEE(Advanced)-2018

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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




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




4

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




5

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




6

20. Ag x (1 + sin 30) = 2

2d xA R cosec30

2 dt

2

2d x gx 3 / 2

Rdt 22

4R( 4)T3g




7

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




8

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




9

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




10

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




11

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




12

|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


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