FIITJEE JEE(Advanced)-2018

AITS-PT-III (Paper-1)-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

PART TEST – III PAPER-1

Q. No. PHYSICS Q. No. CHEMISTRY Q. No. MATHEMATICS

1. A, C 21. A, B, C, D 41. A, D

2. A, D 22. A, B 42. A, B, D

3. A, B, C 23. A, B, D 43. A, B, D

4. B, C 24. A, C, D 44. A, B, C

5. B, D 25. A, B, C, D 45. A, B, C

6. A, B, C, D 26. A, B, D 46. A, B

7. A, B 27. A, B, C, D 47. A, C, D

8. A, B, C 28. A, D 48. A, C

9. A 29. A, B, C, D 49. A, B, C, D

10. A 30. A, B, C 50. A, B, C

11.

(A) (p, r), (B) (s), (C) (q, s), (D) (p, r)

31.

(A) (t), (B) (r, s), (C) (p, q), (D) (p, q, s)

51.

(A) (q, r, t), (B) (q, r, t), (C) (p, q), (D) (p, r, s, t)

12.

(A) (q, s), (B) (q, s), (C) (p, r), (D) (q, s)

32.

(A) (p, q, r), (B) (q, s), (C) (q, r, s, t), (D) (p, q, r, t)

52.

(A) (p, r, t), (B) (s, t), (C) (s), (D) (p, r, t)

13. 4 33. 3 53. 1

14. 8 34. 8 54. 1

15. 2 35. 3 55. 8

16. 5 36. 5 m3 56. 1

17. 6 37. 3 57. 1

18. 2 38. 2 58. 1

19. 8 39. 6 59. 4

20. 1 40. 5 60. 9

ALL

IND

IA T

ES

T S

ER

IES

FIITJEE JEE(Advanced)-2018

www.puucho.com

http://www.fiitjee.com



2

PPhhyyssiiccss PART – I

SECTION – A

1. 3

ex

3

GMm r if 0 r RRF

GMm r if R rr

1 2W W

R

1R/2

4 3GMmW G mxdx3 8R

3R/ 2

2 2R

GMm GMmW dx3Rx

1 1 2

2 1 2

W W W9 17W 8 W W

Fex

R2

W1 W2

R 3R2

2. Consider the bug at point P at any time t, moving with speed v along the stick, so sin cosL mvh mv Angular momentum of bug

about point O

sin cosdL dvmdt dt

cos sin cos dvmg x mdt

1sin

dv g xdt

x

mg

B

A

v

O

P h

sin sin

dv g gxdt

Equation of SHM with mean position at point A

sin2Tg

Time required = sin4 2T

g

3. Basic concept of Intensity of light 4. Consider the system at any time t when each block is moving with speed v and spring has an

extension x.

www.puucho.com




3

2mg T 2ma …(1) T – kx = ma …(2) From equation (1) & (2), we can write

2

2

2g k d x kx 2gx a 03 3m 3m 3dt

2

2

d x k 2mg kx 03m k 3mdt

For equilibrium 2

02

d x 2mg0 xkdt

Amplitude of motion = 02mgx 0

k

2max

k 2mg 2ga A3m k 3

max2mg k mv A 2g

k 3m 3k

B

2 mg

T

A

mg

N

kx

a

T

Second Method: Using COME, we can write

2 23mv kx2mgx

2 2

For extreme position, v = 0 x = 0, 4mgk

A Amplitude of motion = 2mgk

For maximum velocity, 2

2

d x 2mg0 x Equilibrium positionkdt

2 2 2max3mv2mg 4m g2mg k

k 2 2k

22 2

2 2maxmax

3mv4m g mv 4g2k 2 3k

maxmv 2g3k

max2g k 4mg 2ga3 3m k 3

5. The energy released in -decay is given by 2

U Th HeQ M M M c Substituting the atomic masses as given in the data, we find 2Q 238.05079 234.04363 4.00260 u c 20.00456 u c 20.00456u c

0.00456u 931.5 MeV / u = 4.25 MeV.

If 23892 U spontaneously emits a proton, the decay process would be 238 237 1

92 91 1U Pa H The Q for this process to happen is 2

U Pa HM M M c 2238.05079 237.05121 1.00783 u c 20.00825u c

0.00825 u 931.5 MeV / u = – 7.68 MeV Thus, the Q of the process is negative and therefore it cannot proceed spontaneously. We will

have to supply an energy of 7.68 MeV to a 23892 U nucleus to make it emit a proton.

www.puucho.com




4

6. T 50 1 9.8v 70m / s0.1

.

v 701 4m f 17.5Hz4 4

.

pmax 2pmax

v 15v A A 0.136m 13.6 10 m2 f 2 3.14 17.5

7. (C) 22 9

02 2 27 20

0 0 0

K M cM Mc 101 7.68M M c M c 1.67 10 9 10

(D) 2 2 20 0Mc M c K 2m c

M = 2 M0

002

2

M2M

v1c

2

2

4v1 4c

2

2 3c 3v v c 0.866c4 2

8. m 402

1n 302 2

2n 1 6

m 4

2 2n 1

m3

n = 1 m = 2 = 40 cm n = 4 m = 6 v = f f/40 = f0 n = 7 m = 10 = 8 second overtone f = v/8 = 5f0 n = 10 m = 14

340 7vf7 40

.

9. Let x be the wave length for K-line

2

1

hcxeV 3hcxeV

2

4x3R Z 1

Z = 29

www.puucho.com




5

10.

19dN P 5 10 / sdt hc /

0hc 2.49eV W :Yes.

d

r

SECTION – B

11. Use maxK h 12. Use concept of combination of lenses.

SECTION – C 13. OE = x

FG 2a 2y From the property of rhombus, we can write

2 2

2a a xy a22 2

Neglecting x2 and y2, we have xy2

Using COME, we can write

2

2mv k x E cons tan t2 2

4mTk

y

O B A

D C

G

E

F

x

14. Let in time t, m mass of liquid collides with windshield. So

f ip p 0 mv mvFt t t

density of liquid molecules in air so m Svt Sv t Where S surface area of the windshield

2F Sv Mean pressure = 2F vS …(1)

Now as we assumed that is the density of liquid molecules of drops in air so in time uS0 volume of liquid will strikes on the surface of earth so mass of liquid strikes on the earth in time interval will be

0 0m S u 0 0S h S u

hu

Putting this value in equation (1), we have

Mean pressure = 2hv 48

u 5

K = 8

www.puucho.com




6

15. 2 1P g h x , P gh 2 1 0V Ax V Ax Using Boyle’s low we can write 0g h x Ax gh Ax

2

20 0

h hx hx hx 0 x hx2 4

x 10m, 90m x length of pipe

h–x

x x0

So answer will be x = 10 m

16. 2

2m 2k k12 12 4

5km

17. h

2 2 2B

0

mg F R h R z dz g

3hm

3

3

3

3m 3 9 3h h 0.3m 30cm1010

18. Mass of water striking vanes/sec av Initial velocity = v Final velocity of water = u Change in velocity = v – u F = force on the vane av(v u) Work done f u avu(v u) = 5400 watts K = 2

19. h tand

using triangle EAB

dHcos

sin2sin 90

d H

cos 2sin

2d = H cos cot sin

dHcos Hsinh

900

d

H

h Wall

A

E B

C

h = Hdcos2d Hsin

3 33 27 32 53 4 36 2 82 32 5

.

20. 2

2 00 2

EVB V

v

EBv

e2 2

u EnergyV mass, so2v speed

a = 1, b = 0, c = 0

www.puucho.com




7

CChheemmiissttrryy PART – II

SECTION – A

21. At critical temperature various physical properties of gas and liquid are identical. 22. A. P V 1PV We know for polytropic process xPV constant x 1

We know vRC C

x 1

vRC C2

R RC1 2

R 1C

2 1

B. dW PdV dW VdV On integrating above equation.

2 2

o o4V VW

2 2

2 2o oW 16V V

2

2oW 15V

2

23. [M(AAA)2], [Ma3b3] and [M(AB)3] can show fac and mer forms. 24. Say, number of Cu2+ ions = x Number of Cu+ ions = 1.8 – x x × 2 + (1.8 – x) × 1 = 1 × 2 2x + 1.8 – x = 2 x = 2 – 1.8 x = 0.2

% of Cu2+ 0.2 100 11.111.8

% of Cu1+ = 88.89 26. mix A A B BG RT n n n n

1 22 300 1 n 2 n3 3

mixG 600 1.0986 0.8109 = 1145.7 cal

1mixmix

GS 3.82 cal k

T

mixH 0

www.puucho.com




8

28. In froth floatation process pine oil used as collectors which enhance non-wettability of mineral particles. Cresol is froth stabilizers. NaCN used as depressant

29. 3 2 2 2C 4HNO CO 4NO 2H O

3 3 2 22Cu 4HNO Cu NO 2NO 2H O

3 3 2 22Zn 4HNO Zn NO 2NO 2H O

3 2 3 2 2Sn 4HNO H SnO 4NO H O 30. (A) 2 2H O Cl 2HCl O (B) 2 2H S Cl 2HCl S

(C) 10 16 2C H 8Cl 16HCl 10C (D) 3 2 4 28NH 3Cl 6NH Cl N

SECTION-B

31. Pb2+ : 2

2Yellow

Pb 2KI PbI 2K

22 4 4

YellowPb K CrO PbCrO 2K

Ag+ : 2 2Brown

2Ag 2NaOH Ag O H O 2Na

2 2Black

2Ag H S Ag S 2H

22Hg : 2

2 2 2Black

Hg 2NaOH Hg O H O 2Na

22 2

Black

Hg H S Hg HgS 2H

22 2 2Hg 2KI Hg I 2K

2 2 2 4Black

Hg I 2KI K HgI Hg

32. (A). Process is isobaric

V P5R 7RC C2 2

op P

7RTq nC T

2

oV

5RTU nC T

2

oW RT

(B). Process is isochoric, V3RC2

oV

3RTq U

2

W 0

(C). VR RC 2R

1 1.5 1

VT constant PV2 = constant 2

P constant

www.puucho.com




9

dV dV V P dV PVTV 0 RdT dT T n dT nT

o o oW RT U 2RT q RT

(D). 2 314 2

VRC 2R

1

PT = constant P2V = constant

2P constant

oU 2RT

SECTION – C 33. Ionisation isomerism

Linkage isomerism Geometrical isomerism

34.

f f

4 4 2 3 2 46 6 2n 2 n 33ZnSO 2K Fe CN K Zn Fe CN 3K SO

Milliequivalents of ZnSO4 = milliequivalents of K4[Fe(CN)6] N1V1 = N2V2

Volume of K4[Fe(CN)6] (V2) 60 0.01 2 8 ml

0.05 3

35. f fT K m

0.93m 0.51.86

b bT K m

1b

0.26K 0.52K kgmol0.5

Similarly b bT K m

bb

K w 1000T

W m.wt

m.wt. of solute b

b

K w 1000

W T

0.52 7.87 1000100 0.44

31x 93 x = 3 36. 2 2 22H O 2H O

4 2 2 2CH 2O CO 2H O

2 22CO O 2CO

2 2 2 2 25C H O 2CO H O2

www.puucho.com




10

Required oxygen for the combustion of given gas = 1.04 m3

Hence, volume of air required 31.04 100 5 m20.8

37. Al, Cr, Fe 38. Telluric acid H6TeO6 is quite different from sulphuric and selenic acid. It exists as octahedral

Te(OH)6 molecule in the solid. It is a weak dibasic acid and forms two series of salt, NaTeO(OH)5 and Na2TeO2(OH)4.

39. 2Zn s 2Ag aq Zn aq 2Ag s

ocell 10 eq

0.059E log K2

10 eq0.0590.8 0.76 log K

2

Keq = 7.61 × 1052 As Keq is very large the reaction will tend towards completion

Moles of zinc added 26.53 10

65.3

= 10-3

Moles of Ag+ in solution 5

610 100 101000

As Ag+ is limiting reagent, the moles of Ag precipitated 610 Hence x = 6 40. 1 1 2 2P V P V

P2 = pressure of H2 gas 60 1020

P2 = 30 cm of Hg Final total pressure over liquid water (P) = 30 + 20 = 50 cm of Hg

P 50 510 10

www.puucho.com




11

MMaatthheemmaattiiccss PART – III

SECTION – A

41. Dm will be open finally if m is the perfect square 42. (A) (1 + x)n = nC0 + nC1x + nC2x2 + nC3x3 + ..... + nCnxn

Put x = where 1 i 32 2

nn n n n n n n n n 20 3 6 1 4 7 2 5 8C C C ..... ..... C C C ..... C C C ..... 1

n n n n n n0 3 6 1 4 7

1 i 3C C C ..... C C C .....2 2

nn n n 22 5 8

1 i 3C C C .....2 2

n n n n n n n0 3 6 1 2 4 5

1C C C ..... C C C C .....2

nn n n n 21 2 4 5

i 3 C C C C .....2

Take modulus both sides (B) a5b2 = 4 5 log2(a) + 2 log2(b) = 2 y = 10 log2a·4 log2b = 40 log2a·log2b

A.M. G.M. 2 2

2 25log a 2log b

10log a·log b2

2 21log a log b

10 y 4, y = 4 when 5 log2a = 2 log2 b a5 = b2 = 2

(C) For constant term put x = 0 and get constant term = 4 (D) Coefficient of x24 = 1.25 + 2.24 + 3.23 + ..... + 25.1 = 2925 43. Roots of the equation y3 – x3y2 – x2y – x1 = 0 are 3, 5 and 7 x3 = 15, x2 = –71, x1 = 105 44. Let R be the common ratio of the G.P. and D be the common difference of A.P. a5 = a5, a9 = Ra5, a16 = a5R2 a9 – a5 = 4D (R – 1)a5 = 4D ..... (1) a16 – a9 = 7D R(R – 1)a5 = 7D ..... (2)

From equation (1)/(2), we get 1 4R 7

7R4

From equation (2) – (1), we get (R – 1)2a5 = 3D 59a3D

16

13 a 4D D

16 1

3 3a 1 D16 4

14Da3

www.puucho.com




12

45. a = b + 5

and b 66 a

ab = 36 b2 + 5b – 36 = 0 b = 4

z2 z3 a

b 6 6 b

z1 z4 z 5 b

46. nn n!

3

put n = 300 100300 < 300! ..... (1)

3 < 100 3150 < 100150 (300)150 < 100300 300 300300 100 ..... (2)

From equation (1) and (2), we get 300300 300! 47. (A) ij jia a ij jia a 0 A is skew symmetric matrix

(C) A2 = 2A A3 = 22A A6 = 25A (D) A6B7 is a skew symmetric matrix of odd order

48. CD : ˆ ˆ ˆ ˆ ˆr i 2 j 4k 7 j 7k3

BE : ˆ ˆ ˆ ˆ ˆ ˆr i j k 7i 7 j 7k3

ˆ ˆ ˆP i j 3k

Area of tetrahedron ABCF

= 1 Area of base triangle height3

= 7 cubic units3

F

ˆ ˆ ˆ4i 4 j 10k E3

ˆ ˆ ˆC i 2 j 4k ˆ ˆ ˆB i j k

P

ˆ ˆ ˆ3i j 5kD3

ˆ Â 2 i 2k

ˆ ÂB AC 7j 7k

, PF PF 2

units

ˆ ˆ7 j 7k ˆ ˆPF 2 j k

49 49

= P.V. of F – P.V. of P

P.V. of F = ˆ î 4k Vector equation of AF is ˆ ˆ ˆ ˆr 2 i k i 2k

49. By eliminating n 2

2 1 0m m

1

11

m

, 2

2

1m 2

On solving, we get 1 1 11 1 2, m , n , ,

36 6

and 2 2 21 2 1, m , n , ,

36 6

www.puucho.com




13

50. Let P(z) be any point on the required line.

Then, CPCP

i.e. z c

z c

is a unit vector parallel to it

Let A(z1) and B(z2) be two points on

az az b 0 then 2 1

2 1

z zz z

is a unit vector parallel to the line

az az b 0

45º

45º

C(c) P(z)

P(z) A(z1)

B(z2)

i

42 1

2 1

z zz c ez c z z

2 2i2 1 2

2 1 2 1

z c z ze

z c z c z z z z

2 1

2 1

z zz c iz c z z

..... (1)

A(z1) and B(z2) are on the line az az b 0 therefore 1 1az az b 0 2az az b 0

2 1

2 1

z zaa z z

..... (2)

From equation (1) and (2), we get i z cz c 0

a a

SECTION – B

51. (A) f(x) = x3 + ax2 + bx + c = (x2 + 1)(x + a) + (b – 1)x + (c – a) f(x) is divisible by x2 + 1 b – 1 = 0 c = a (B) (A + B)(A – B) = (A – B)(A + B) AB = BA, AT = A, BT = –B (AB)T = (–1)k – 1AB k is an odd number (C) 10100 – 43 = 999 ..... 9957 Sum of digits = 98(9) + 5 + 7 = 894 (D) x = 7 cos – sin , y = 7 sin + cos , z = 10

2 2x y 5

z

52. (A) (x + y2)(x – y2) = 172 x + y2 = 172 and x – y2 = 1 or x + y2 = 17 and x – y2 = 17 or x + y2 = –1 and x – y2 = –172 or x + y2 = –17 and x – y2 = –17 Possible cases: (x, y) (145, 12), (17, 0) x + 12y + 4 = 293, 5, 21, –13, 3, –285 (B) PQ2 + RS2 = QR2 + SP2 = 130

39 7PS QR 7 39 9

R

P Q S

9 3

7 11

(C) x + y + z = 20, number of the solution is 19C2 they form triangle if 1 x, y, z 9 Number of solutions = coefficient of x20 in (x1 + x2 + ..... + x9)3 = 19C2 – 310C2

Required probability 19 10

2 219

2

C 3 CPC

4P

19

www.puucho.com




14

(D) Let the series be 21, 21r, 21r2, 21r3, .....

Sum = 21 N1 r

and 21r N

21 21 N21 21r

21 – 21r N and 21r < 21

21 – 21r may be equal to 1, 3, 7, 9 21r = 20, 18, 14, 12 |k – 15| = 5, 3, 1

SECTION – C

53. (x4 + 2x3 + bx2 + cx + d) = (x – 1)(x – 2)(x – 3)(x – 4) = (x2 + px + q)(x2 + px + r) = x4 + 2px3 + (q + r + p2)x2 + (pr + pq)x + qr (2p)3 – 4(2p)(q + r + p2) + 8(pq + pr) = 0 23 – 4 2 b + 8c = 0 b – c = 1 54. Probability that B wins the match in the 4th game

= 31

1 1 1 1 1 1 1 16 C2 6 3 2 2 3 3 2 = 6

72 + 3 6 6 1

36 72 6

55. x3 + 3x2 + k = 0 , , + + = –3, + + = 0, = –k 2 + 2 + 2 = 9 , , I 2 = 9, 2 = 0, 2 = 0 or 2 = 4, 2 = 4, 2 = 1 2 = 4, 2 = 4, 2 = 1 Possible roots: 3, 0, 0, 2, 2, 1 But + + = 0 So, possible roots are 3, 0, 0; –3, 0, 0; 2, 2, –1; –2, –2, 1 Possible non-zero values of k are –4 and 4

56. k 1 kk k 1

k k 1

a 0 a 0A A

0 a 0 a

k 1 kk k 1

k k 1

a a 0A A

0 a a

n n

k 1 kk k 1 n n

k k 1

C C 0A A

0 C C

n 1n n

k 1 kk 1

n 1n n

k k 1k 1

C C 0a 0

B0 b

0 C C

a = b = 2nCn – 1 – n 57. (r2 + 2)(r + 1)! + 2r(r + 1)! = (r + 1)(r + 2)! – r(r + 1)! 58. a3 + b3 + c3 – 3abc = (a + b + c)(a + b + 2c)(a + 2b + c)

a + b + c = ex, a + b + c2 = ex, a + b2 + c = 2xe

www.puucho.com




15

59. det(A) is twice of area of triangle with vertices , p , (m, q), (n, r) with sides 3, 4, 5 2 = s(s – a)(s – b)(s – c) 2 = 6 3 2 1 = 6 Now det(A) = 12 det(B) = 144 60. a + ib = z |z|5015 – |z|3 = 0 |z|3 (|z|5012 – 1) = 0 |z| = 0 or |z|5012 = 1 |z| = 1 zz 1 z = 0

z5015 = 3z 501531zz

z5018 = 1

Number of complex numbers = 5018 + 1 = 5019

www.puucho.com


FIITJEE JEE(Advanced)-2018

Documents

Transcript of FIITJEE JEE(Advanced)-2018