part test – iii

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com

PART TEST – III

Time Allotted: 3 Hours Maximum Marks: 360 Please r ead the inst ruct ions carefu l l y. You are a l lot ted 5 m inutes

speci f i ca l l y for th is purpose. You are not a l lowed to leave the Exam inat ion Hal l before the end of

the test .

INSTRUCTIONS

A. General Instructions 1. Attempt ALL the questions. Answers have to be marked on the OMR sheets. 2. This question paper contains Three Parts. 3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics. 4. Each part has only one section: Section-A. 5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be

provided for rough work. 6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic

devices, in any form, are not allowed. B. Filling of OMR Sheet 1. Ensure matching of OMR sheet with the Question paper before you start marking your answers

on OMR sheet. 2. On the OMR sheet, darken the appropriate bubble with black pen for each character of your

Enrolment No. and write your Name, Test Centre and other details at the designated places. 3. OMR sheet contains alphabets, numerals & special characters for marking answers. C. Marking Scheme For All Three Parts.

1. Section-A (01 – 30, 31 – 60, 61 – 90) contains 90 multiple choice questions which have only one correct answer. Each question carries +4 marks for correct answer and –1 mark for wrong answer.

Name of the Candidate

Enrolment No.

ALL

IND

IA T

ES

T S

ER

IES

FIITJEE JEE (Main)-2018

AITS-PT-III-PCM-JEE(Main)/18


2

Useful Data

PHYSICS

Acceleration due to gravity g = 10 m/s2

Planck constant h = 6.6 1034 J-s

Charge of electron e = 1.6 1019 C

Mass of electron me = 9.1 1031 kg

Permittivity of free space 0 = 8.85 1012 C2/N-m2

Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY

Gas Constant R = 8.314 J K1 mol1 = 0.0821 Lit atm K1 mol1 = 1.987 2 Cal K1 mol1 Avogadro's Number Na = 6.023 1023 Planck’s constant h = 6.625 1034 Js = 6.625 10–27 ergs 1 Faraday = 96500 coulomb 1 calorie = 4.2 joule 1 amu = 1.66 10–27 kg 1 eV = 1.6 10–19 J Atomic No: H=1, He = 2, Li=3, Be=4, B=5, C=6, N=7, O=8,

N=9, Na=11, Mg=12, Si=14, Al=13, P=15, S=16, Cl=17, Ar=18, K =19, Ca=20, Cr=24, Mn=25, Fe=26, Co=27, Ni=28, Cu = 29, Zn=30, As=33, Br=35, Ag=47, Sn=50, I=53, Xe=54, Ba=56, Pb=82, U=92.

Atomic masses: H=1, He=4, Li=7, Be=9, B=11, C=12, N=14, O=16, F=19, Na=23, Mg=24, Al = 27, Si=28, P=31, S=32, Cl=35.5, K=39, Ca=40, Cr=52, Mn=55, Fe=56, Co=59, Ni=58.7, Cu=63.5, Zn=65.4, As=75, Br=80, Ag=108, Sn=118.7, I=127, Xe=131, Ba=137, Pb=207, U=238.



3

PPhhyyssiiccss PART – I

SECTION – A

Straight Objective Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE option is correct. 1. A spherical ball of mass m, radius r is connected to string of length r . This string is then

connected to fixed point and this arrangement allowed to oscillate in a medium whose density is and viscosity is . If amplitude of oscillation becomes half of the initial value after n seconds. Then viscosity of liquid is given by (Assume that resistance force on ball follows stoke’s law).

(A) mrn

(B) 1/3m n 2

rn

(C) m n8rn (D) m n2

rn

2. An ideal diode is connected in a

circuit with resistance R = 5 and V = 10 volt as shown in figure maximum and minimum value of output voltage. When no load applied is (assume diode to be ideal) – 25

+25

Vinput R = 5

V = 10V

Output t

(A) 10V, – 25V (B) 10V, – 15V (C) 25V, – 25V (D) 25V, – 15V 3. A vessel is half filled with a liquid of refractive index . The other half of the vessel is filled with

an immiscible liquid of refractive index 1.5. The apparent depth of the vessel is 50% of the actual depth. Then is

(A) 1.4 (B) 1.5 (C) 1.8 (D) 1.67

Space for Rough work



4

4. A spring block system with mass of block m and spring constant K (all the surfaces of block are perfectly absorbing and smooth) is placed on a smooth horizontal plane as shown in the diagram at rest. A light beam of intensity I is switched on its right side. Find the amplitude of oscillations of the block. (face area of block is A and c is the speed of light in vacuum)

(A) I AKc

(B) 2I AKc

(C) 4I AKc

(D) Zero

m

5. What should be the approximate kinetic energy of an electron so that its de-Broglie wavelength is

equal to the wavelength of X-ray of maximum energy, produced in an X-ray tube operating at 24800 V? (given that h = 6.6 1034 joule-sec, mass of electron = 9.1 1031 kg)

(A) 600 eV (B) 365 eV (C) 120 eV (D) 300 eV 6. A 2 kg block moving with 10 m/sec strikes a spring of constant 2 N/m which is attached to 2 kg

block at rest kept on a smooth horizontal floor. The time for which rear moving block remain in contact with spring is

(A) 2 sec (B) 12

sec

(C) 1 sec (D) 1/2 sec 7. Unpolarised light of intensity 32 watt m–2 passes through three polarisers such that the

transmission axes of the first and second polarises makes an angle 30º with each others the transmission axes of the second and third polarises makes an angle 60º with each other. The intensity of final emerging light will be

(A) 32 watt m–2 (B) 3 watt m–2 (C) 8 watt m–2 (D) 4 watt m–2. 8. Choose the incorrect statements (A) An electron beam is used to obtain interference in a simple Young’s double-slit experiment. If

speed of electron is increased the fringe width decreases. (B) de-Broglie wavelength of electron is inversely proportional to the speed of electrons. (C) Two coherent point sources of light having non-zero phase difference are separated by a

small distance. On the perpendicular bisector of line segment joining both the point sources, no Constructive interference can be obtained.

(D) For two waves from coherent point sources to interference constructively at a point the magnitude of their phase difference at that point must be 2n(where n is non-negative integer)




5

9. In radio-receiver, the short wave and medium wave stations transmit signal using same capacitors but ideal coils of different inductors Ls and Lm respectively. Choose the correct option

(A) Ls > Lm (B) Ls = Lm (C) Ls < Lm (D) None. 10. A string of length 2L, obeying Hooke’s Law, is stretched so that its extension is L. The speed of

the tranverse wave travelling on the string is v. If the string is further stretched so that the extension in the string becomes 4L. The speed of transverse wave travelling on the string will be.

(A) 2V (B) V (C) 2 2V (D) 2V 11. A rigid bar of mass M is supported symmetrically by three wires each of length . Those at each

end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to

(A) copper

iron

YY

(B) iron

copper

YY

(C) 2iron2copper

Y2Y

(D) iron

copper

Y23 Y

12. A close organ pipe of diameter 10 cm has length 42 cm. The air column in pipe vibrates in its

second overtone with the maximum amplitude Po. The pressure amplitude at middle of pipe is

(A) Δ oP2

(B) Δ oP

2

(C) Δ o3 P2

(D) Po

13. For the same objective, the ratio of the least separation between two points to be distinguished by

a microscope using electrons accelerated through 100 V and light of 5500 Å used as the illuminating substance is

(A) 2 × 10–4 (B) 2 × 10–3 (C) 2 × 10–2 (D) 4 × 10–4




6

14. Mx and My denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The Q-value for a -decay is Q1 and that for a + decay is Q2. If me denotes the mass of an electron, then which of the following statements is correct?

(A) 2 21 x y 2 x y eQ M M c and Q M M 2m c

(B) 2 21 x y 2 x yQ M M c and Q M M c

(C) 2 21 x y e 2 x y eQ M M 2m c and Q M M 2m c

(D) 2 21 x y e 2 x y eQ M M 2m c and Q M M 2m c

15. Tritium is an isotope of hydrogen whose nucleus Triton contains 2 neutrons and 1 proton. Free

neutrons decay into p e If one of the neutrons in Triton decays, it would transform into a He3 nucleus. This does not happen because

(A) Triton energy is less than that of a He3 nucleus. (B) The electron created in the beta decay process cannot remain in the nucleus. (C) Both the neutrons in triton have to decay simultaneously resulting in a nucleus with 3 protons,

which is not a He3 nucleus. (D) Because free neutrons decay due to external perturbations which is absent in a triton

nucleus. 16. A tuning fork A is being tested using an accurate oscillator. It is found that they produce 2 beats

per second when the oscillator reads 514 Hz and 6 beats per second when it reads 510 Hz. The actual frequency of the fork in Hz is

(A) 512 Hz (B) 504 Hz (C) 516 Hz (D) 510 Hz 17. The objective of telescope has diameter 12 cm. The distance at which two small green object

placed 30 cm apart can be barely resolved by telescope, assuming the resolution to be limited by diffraction by objective only ( = 5.4 × 10–5 cm)

(A) 54.6 km (B) 57.0 km (C) 546 km (D) 52.4 km




7

18. Two cars A and B are moving towards each other with same speed 25 m/s. Wind is blowing with speed 5 m/s in the direction of motion of car A. The car A blows horn of frequency 300 Hz and the sound is reflected from car B. The wavelength of the reflected sound received by the driver of car A is (velocity of sound in air = 330 m/s)

(A) 25 m36

(B) 31m36

(C) 35 m36

(D) 37 m36

19. The value of displacement current at t = 1 time constant is

(A) 120 mAe

(B) 2

120 mAe

(C) 120 mA (D) None of these

100

12V

100 F

20. An insect of negligible mass is sitting on a block of mass M, tied with

a spring of force constant k. The block performs simple harmonic motion with amplitude A in front of a plane mirror placed as shown in the figure. The maximum speed of insect relative to its image will be

(A) A km

(B) A 3 k2 m

(C) kA 3m

(D) mAk

M 60

21. Mass of a sphere of radius r = 10.00 cm is 30.000 g If measured values are expressed upto the correct significant figures then maximum fractional

error in calculation of density of the sphere is (A) 53 10 (B) 33.3 10 (C) 33.033 10 (D) 53.03 10




8

22. Maximum pressure difference between inner and outer surface of a thin walled circular tube of radius r and thickness r r r , so that tube won’t break is (Breaking stress of the material is 0)

(A) 0rr

(B) 0rr

(C) 02 rr

(D) 0

23. 7 th8

of nuclei have been disintegrated in 2 minutes after fresh sample was prepared. The half life

of the sample is (A) 30 seconds (B) 1 minute (C) 40 seconds (D) 20 seconds 24. An astronomical telescope has angular magnification m = 7 for distant object. Distance between

objective and eyepiece is 40 cm. Final image is formed at infinitely. Focal length of objective and eyepiece respectively are:

(A) 35 cm, 5 cm (B) 28 cm, 14 cm (C) 5 cm, 35 cm (D) 70 cm, 10 cm 25. The length of an elastic chord is 1 when tension in it is T1 and 2 when tension is T2. The

natural length of wire is

(A) 1 1 2 2

1 2

T TT T

(B) 2 1 1 2

1 2

T TT T

(C) 2 1 1 2

1 2

T TT T

(D) 1 1 2 2

1 2

T TT T

26. The angular speed of rotation of earth about its axis when the weight of a man standing on

equator will become half of its weight at the north pole (assume earth to be a perfect sphere) R-6400 km.

(A) 275 × 10–4 rad sec–1 (B) 9.75 × 10–4 rad sec–1 (C) 8.75 × 10–4 rad sec–1 (D) 8.0 × 10–4 rad sec–1




9

27. In a certain pump whose rated power is P. Ideal fluid of density enters with speed V0 which lifts water upto height h through a pipe of uniform cross section area without discontinuity. Volumetric flow rate of the liquid is Q. Speed of fluid when it comes out of the pipe at point A is {Assume that pump works at 100% efficiency and flow is streamline}

(A) 20

2P V 2ghQ

(B) 20

2P VQ

V0

A

h

(C) 20

2P V 2ghQ

(D) 20

2P V 2ghQ

28. In above diagram × = ? (A) A + BC (B) AB + C (C) AC + B (D) None of above

A

B

C

X

29. An object is floating in water with fraction of its volume submerged in water filled in beaker which

is at rest with respect to ground. Now this beaker is accelerated with constant acceleration downwards (a < g). Then,

(A) Fraction of volume submerged will not change. (B) Buoyant force on the block will remain same. (C) Pressure at any point in the water will be more than earlier. (D) Fraction of volume submerged will change. 30. A block of mass M is kept in gravity free space and touches the two

springs as shown in the figure. Initially springs are in its natural length. Now, the block is shifted (0/2) from the given position in such a way it compresses a spring and released. The time-period of oscillation of mass will be

M k 4K

0 20

(A) M2 K (B) M2

5K

(C) 3 M2 K (D) M

2K




10

CChheemmiissttrryy PART – II

SECTION – A


This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 31. The pair of compounds having metals in their highest oxidation state (A) MnO2, FeCl3 (B) 4 2 2MnO ,CrO Cl

(C) 6 3

3Fe CN , Co CN

(D) 4 4

2NiCl , CoCl 32. The density of solid Ar (Ar = 40 g/mole) is 1.68 g/ml at 40 K. If the argon atom is assumed to be a

sphere of radius 1.50 × 10-8 cm, then % of solid Ar is apparently empty space? (A) 35.64 (B) 64.36 (C) 74 (D) None these 33. Resistance of a decimolar solution between two electrodes is 0.02 meter apart and 0.0004 m2 in

area was found to be 50 ohm. Specific conductance will be (A) 0.1 Scm2 (B) 1 Scm2 (C) 10 Scm2 (D) 4 × 10-4 Scm2 34. Poling process is used for (A) the removal of Cu2O from Cu (B) the removal of Al2O3 from Al (C) the removal of Fe2O3 from Fe (D) All of these 35. A solution of Na2S2O3 is standarised idometrically against 0.167 gram of KBrO3. This process

requires 50 mL of the Na2S2O3 solution. What is the normality of the Na2S2O3? (A) 0.2 N (B) 0.12 N (C) 0.72 N (D) 0.02 N 36. A 0.10 M solution of a mono protic acid (d = 1.01 g/cm3) is 5% dissociated. What is the freezing

point of the solution. The molar mass of the acid is 300 and Kf(H2O) = 1.86oC/m (A) - 0.189oC (B) - 0.194oC (C) - 0.199oC (D) None




11

37. In a spinel structure of a mixed oxide, oxide ion are arranged in FCC packing whereas 18

th of

tetrahedral voids are occupied by A2+ ion and 12

of octahedral voids are occupied by B3+ cations.

The general formula of the compound having structure is (A) A2B2O4 (B) AB2O4 (C) A2B4O2 (D) A4B2O2 38. In the purification of impure Ni by Mond’s process, metal is purified by (A) Electrolytic reduction (B) Vapour phase thermal decomposition (C) Thermite reduction (D) Carbon reduction 39.

4 32

Black

Dil. HNODil. H SO

GasX Y Colloidal sulphur

Identify X? (A) CuS (B) FeS (C) As2S3 (D) CdS 40. Fixed mass of an ideal gas collected in a 24.63 litre sealed rigid vessel at 1 atm is heated from

– 73oC to 27oC. Calculate change in Gibbs free energy if entropy of gas is a function of temperature as S = 2 + 10-2 T (J/K) (1 atm = 0.1 kJ)

(A) 1231.5 J (B) 1281.5 J (C) 785.1 J (D) None

41. CsCl has bcc structure with Cs+at the centre and Cl ion at each corner. If Csr is 1.69

oA and

Clr is 1.81

oA . What is the edge length of the cube?

(A) 3.5 oA (B) 3.80

oA

(C) 4.04oA (D) 4.5

oA

42. The cell 2 2 2Pt | H g, 0.1bar | H aq , pH x || Cl 1M | Hg Cl | Hg | Pt has emf of 0.5755 V at

25oC. The SOP of Calomel electrode is – 0.28 V, then pH of the solution will be (A) 11 (B) 4.5 (C) 5.5 (D) None




12

43. For a reversible adiabatic ideal gas expansion dpp

is equal to

(A) dVv

(B) dvv

(C) dV1 V

(D) dVV

44. A freshly prepared Fe(OH)3 precipitate is peptized by adding FeCl3 solution. The charge on the

colloidal particle is due to preferential adsorption of (A) Cl (B) Fe3+ (C) OH (D) None of these 45. Consider the following apparatus. Calculate the partial pressure of He after opening the valve.

The temperature is remain constant at 16oC

He Ne

1.2 litre.0.63 atm 3.4 litre.

2.8 atm (A) 0.164 atm (B) 1.64 atm (C) 0.328 atm (D) 1 atm 46. On adding KI to a metal salt solution, no precipitate was observed but the salt solution gives

yellow precipitate with K2CrO4 in the presence CH3COOH. Then the salt is (A) Sr(NO3)2 (B) Pb(CH3COO)2 (C) AgNO3 (D) BaCl2 47. Which of the following is true in respect of chemical adsorption? (A) H O S O G O (B) H O S O G O (C) H O S O G O (D) H O S O G O 48. For the reaction 2 4 3 2 2N H IO 2H Cl ICl N 3H O The eq. mass of N2H4 and KIO3 is respectively are (A) 8 and 53.5 (B) 16 and 53.5 (C) 8 and 35.6 (D) 8 and 87




13

49. Which of the following statement is false? (A) Optical isomerism is observed in nabcdM tetrahedral complexes. (B) Geometrical isomerism does not exists while optical isomerism exists in the complex

3

2 4 3Fe C O

.

(C) Both cis and trans forms are optically inactive in 2

2 3 4PtCl NH

complex ion.

(D) 2

2Pt en

shows geometrical isomerism as well as optical.

50. Consider the following reaction: 2 2 1H O H O g H 44 kJ

3 2 2 2 22CH OH 3O 4H O 2CO g H 1453 kJ What is the value of H for second reaction if water vapour instead of liquid water is formed as

product? (A) - 1409 kJ (B) - 1629 kJ (C) - 1277 kJ (D) None of these 51. Smog is essentially caused by the presence of (A) O3 and N2 (B) O2 and N2 (C) Oxides of sulphur and N2 (D) O2 and O3 52. Which of the following is not tranquilizer? (A) Luminal (B) Seconal (C) Reserpine (D) Piperazine 53. Which of the following will not be oxidised by O3? (A) KI (B) FeSO4 (C) KMnO4 (D) K2MnO4 54. Which of the following process is used in the extractive metallurgy of Mg? (A) Fused salt electrolysis (B) Self – reduction (C) Aqueous solution electrolysis (D) Thermite reduction 55. Which of the following compound is not an antacid? (A) Aluminium hydroxide (B) Cimetidine (C) Phenelzine (D) Ranitidine 56. The gas leaked from a storage tank of the union carbide plat in Bhopal gas tragedy was (A) Ammonia (B) Phosgene (C) Methyl isocynate (D) Methyl amine




14

57. 36 mL of pure water takes 100 sec to evaporate from a vessel when a heater of 806 watt is used. The vaporisationH of H2O is: (density of water = 1 g/cc)

(A) 40.3 kJ/mole (B) 43.2 kJ/mole (C) 4.03 kJ/mole (D) none 58. 108 g fairly concentrated solution of AgNO3 is electrolysed by using 0.1 F charge. The mass of

resulting solution is (A) 94 g (B) 11.6 g (C) 96.4 g (D) None 59. The geometry, hybridization and magnetic moment of [Ni(CO)4] are……..,….and …..respectively (A) tetrahedral, sp3, zero (B) square planar, sp3, 3 (C) tetrahedral, sp3, 2 2 (D) square planar, dsp2, zero 60. X = amount of gas adsorbed P = Pressure T = Temperature L = Physisorption M = Chemisorption For the following graphs I, II, III and IV

T I

Pconstant xm

T II

xm

x

P III

200 K 250 K

Pot

entia

l ene

rgy

(Distance of molecule from surface)

IV

1adsH 150 kJ mol

Chose correct combination: (A) I → L, II → M (B) I → L, III → M (C) IV → M, II → L (D) IV → M, III → M




15

MMaatthheemmaattiiccss PART – III

SECTION – A


This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

61. If ˆ ˆ ˆa xi yj 2k

, ˆ ˆ ˆb i j k

, ˆ ˆc i 2 j

, a ^b2

, and a c 4

, then

(A) 2

a b c a

(B) a b c a

(C) a b c 0

(D) 2a b c a

62. The value of m for which the straight line 2x – 3y + 4 = 0 = x – 4y + 4z + 5 is parallel to the plane 2x – y + mz – 2 = 0 is (A) 3 (B) 2

(C) 165 (D) –4

63. A line L passing through the point P(0, 1, –1), and is perpendicular to both the lines

x 2 y 4 z 22 1 4

and x 2 y 4 z 13 2 2

. If the position vector of point Q on L is (a, b, c)

such that (PQ)2 = 357, then a + 2b + 3c is equal to (A) 26 (B) 24 (C) –24 (D) 7

64. Let O be the interior point of ABC such that 3OA 4OB 6OC 0

, then Area of ABCArea of AOB

is

equal to

(A) 137

(B) 136

(C) 87

(D) 713




16

65. If ˆ ˆ â 3i 2 j 3k

and ˆ ˆ ˆb 2i 3 j 3k

then the vector ˆ ˆˆ â b a b is collinear to the vector

(A) ˆ ˆ ˆ4i j 4k (B) ˆ ˆ î 5 j 6k

(C) ˆ ˆ ˆ5i 4 j 5k (D) ˆ ˆ5i 5k 66. If a, b

and c

are three non-zero non-coplanar vectors and x a 2b c

, y a 2b 2c

and

z a 4b 2c

are three vectors such that volume of the parallelopiped formed by a, b, c

and

x, y, z

as their coterminous edges are v1 and v2 respectively. Then 2

1

vv

is

(A) 10 (B) 15 (C) 18 (D) none of these

67. If 100100

100k

k 0

x 2 yk Ck 1 z

where x, y, z N then least value of xy2z is equal to

(A) 9999 (B) 1010 (C) 9090 (D) none of these 68. A two digit positive number is selected at random then the probability that its tens digit is at least

four more than its unit digit, is

(A) 16

(B) 730

(C) 1445

(D) none of these

69. Consider the system of equations x – y + z = ; x – y + z = 1; x – y + z = 1. If L, M, and N

denotes the number of integral values of in interval [–5, 7] for which the system of equation has unique solution, no solution and infinite solution respectively, then the value of L2 + M2 + N2 is

(A) 125 (B) 13 (C) 15 (D) 121 70. If , , be the roots of the equation 3x3 – 5x2 + 6x – 2 = 0 and A(, , ), B(, , ), C(, , )

are the vertices of the triangle, then centroid of ABC lies on the line (A) x = 2y = 3z (B) x = y = z (C) x = –2y = z (D) –x = y = –z




17

71. The coefficient of x20 in the polynomial (1 + x)22 + x(1 + x)21 + x2(1 + x)20 + ..... + x20(1 + x)2 is (A) 22C19

(B) 23C3 (C) 22C2 (D) 23C19

72. Number of complex numbers z such that |z| < 13

and n

rr

r 1a z 1

(where |ar| < 2) is

(A) 1 (B) 2 (C) n (D) 0

73. Let kk kA cos isin20 20

where 1 i , then

2018

k 1 kk 0

673

2k 1 2k 2k 1

A A

A A

is equal to

(A) 3 (B) 2018673

(C) 2019674

(D) none of these

74. The value of

2 1 1

2 1 1

2 1 1

a p a p p

a q a q q

a r a r r

is

(A)

2

2

a p q

pqr a p

(B)

2

2

a p q

prq a p

(C)

2

2

b p q

pqr a p

(D)

2

2

c p q

pqr a p




18

75. If ar, br, cr (r = 1, 2, 3) are non-negative real numbers and 3

r r rr 1

a b c A

, then maximum

value of 1 2 3

1 2 3

1 2 3

a a ab b bc c c

is

(A) A3 (B) 3A

8

(C) 3A

27 (D)

2A9

76. If a point z1 satisfying |z + 8| 2 and z2 satisfying |z + 3| + |z – 3| 10 then the range of |z1 – z2| is (A) [10, 15] (B) [1, 15] (C) [2, 10] (D) none of these 77. Area of the triangle whose vertices are A(–3 – 2i), B(2 – 3i) and C(–1 –5i), is

(A) 74

(B) 142

(C) 132

(D) 2 14

78. If k1 is the number of arrangements of the letters of the word “EQUATION” in which all vowels are

together and k2 is the number of arrangements of the letters of the word “EQUATION” in which

vowels are in alphabetical order then 1

2

7kk

is

(A) 120 (B) 4! 5!

(C) 60 (D) 8!5!

79. 100 identical objects are distributed among 10 persons in which 1st person gets at least one

object, 2nd person gets at least 2 objects, 3rd person gets at least 3 objects, ....., 10th person gets at least 10 objects, then total number of ways of distribution is

(A) 5510C (B) 55C9

(C) 54C9 (D) 54C44




19

80. A palindrome is a word, number, phrase or sequence of words that reads the same backwards as forwards e.g. “SOLOS”. The number of palindromes that can be formed using the letters AABBBBCCCDDDD are

(A) 6!2!

(B) 6!2! 2!

(C) 7!2! 2!

(D) none of these

81. Two cards are missing from a pack of 52 playing cards. One card is selected at random then the

probability that it is a king, is

(A) 326

(B) 113

(C) 1013

(D) 413

82. All the words formed using the letters of the word “ALGEBRA” taken all at a time and one of the

word is selected from it, then the probability that the word contains the word “AGE”, is

(A) 7

2C5!

(B) 5!7!

(C) 72

1P

(D) 72

1C

83. If / 4

nn

0

I tan xdx

and 2 4

1I I

, 3 5

1I I

, 4 6

1I I

, 5 7

1I I

form an A.P., then the common difference of

the A.P. is (A) 1 (B) 2 (C) 3 (D) none of these

84. The sum of the series 8 16 24 325 65 325 1025 ..... upto 20 terms is

(A) 1280841

(B) 1680841

(C) 1880841

(D) none of these




20

85. If x2, A1, A2, y2 are in arithmetic progression, x2, G1, G2, y2 are in geometric progression and x2, H1, H2, y2 are in harmonic progression, then the value of

(G1G2)(H1 + H2) – H1H2(A1 + A2) is (A) 0 (B) 1 (C) 2 (D) none of these

86. Let A and B are two events such that P(A) = 0.4, P(B) = 0.6 and BP 0.5A . Then AP B

equals to

(A) 14

(B) 23

(C) 12

(D) 13

87. If f(x) = x2 + x + (where , R) and f(f(x)) = 0 has two roots 1 and 2, then f(0) is equal to (A) 1 (B) –2

(C) 32 (D) –1

88. A bag contains 5 balls of unknown colours. A ball is drawn at random from it and is found to be

white. The probability that bag contains only white balls is

(A) 35

(B) 15

(C) 23

(D) 13

89. The arithmetic mean of a set of observation is x . If each observation is divided by and then

increased by 10, then mean of the new series is

(A) x

(B) x 10

(C) x 10

(D) x 10

90. The coefficient of x203 in the expression (x – 1)(x2 – 2)(x3 – 3) ..... (x20 – 20) is (A) 11 (B) 12 (C) 13 (D) 15


AITS-PT-III-PCM (Sol.)-JEE(Main)/18


1

ANSWERS, HINTS & SOLUTIONS

PART TEST – III (Main)

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

1. B 31. B 61. D 2. A 32. B 62. C 3. D 33. B 63. B 4. A 34. A 64. B 5. A 35. B 65. B 6. C 36. C 66. C 7. B 37. B 67. A 8. C 38. B 68. B 9. C 39. B 69. A 10. C 40. C 70. B 11. B 41. C 71. B 12. C 42. C 72. D 13. A 43. B 73. A 14. A 44. B 74. B 15. A 45. A 75. C 16. C 46. D 76. B 17. A 47. B 77. C 18. B 48. A 78. C 19. A 49. D 79. C 20. C 50. C 80. B 21. C 51. C 81. B 22. B 52. D 82. D 23. C 53. C 83. A 24. A 54. A 84. B 25. B 55. C 85. A 26. C 56. C 86. C 27. A 57. A 87. C 28. A 58. C 88. D 29. A 59. A 89. C 30. C 60. A 90. C

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2


1. b/ 2mt

0A A e b = 6 r

b/2m. t00

AA e

2

6 rn2 .n2m

m n23 rn

2. Positive half cycle Voutput = 10V For negative half cycle. potential difference across output = – 25 V 3. The distance by which image appears to be raised is

1 21 2

1 1d t 1 t 1

t t 1 t 21 12 2 2 3

5 1.673

4 Leftward Force = IA Pc t

5. The wavelength min emitted by the X-ray tube operating at a voltage V is given by

eV = min

hc

Kinetic energy of the electron = 2 2

21 1 m vmv2 2 m

2

21

h2m

Now the wavelength of an electron moving with the velocity v is given by

= h/mv = h2meV

K.E. = 600 eV

6. T 2k

2T T 2sec

Required time T T T 1sec.4 4 2

7. I0 is intensity of unpolarized incident light.

0II2

cos230º cos260º = 3 watt m–2.



3

8. hmv

9. cf

fm = sm s

c c& f

m > s fm < fs

m sm s

1 1f & f2 L c 2 L c

Ls < Lm .

10. Tv

T can be calculated by using Hooke’s Law and on stretching also changes.

11. T YA

12. Here

5 42 0.3 10 36cm

4

pressure amplitude at middle of pipe is

o o

2P P sinKx P sin 2436

o3 P

2

13. min1.22d2sin

Where is the angle subtended by the objective at the object. For light of 5500 Å

7

min1.22 5.5 10d m

2sin

For electrons accelerated through 100V the deBroglie wavelength is

9h 1.227 0.13 10 mp 100

10

min1.22 1.3 10d

2sin

10

min1.22 1.3 10d

2sin

10

3min7

min

d 1.3 10 ~ 0.2 10d 5.5 10

14. E = mc2



4

16. For 1st reading of oscillator fA = (514 ± 2)Hz fA = 516 Hz or 512 Hz For 2nd reading of oscillator fA = (510 ± 6) Hz fA = 516 Hz or 504 Hz A has a frequency of 516 Hz

17. 1.22d

51.22 5.4 10 30

12 x

x 54.645km

18. Frequency received by car B is 1v w uf fv w u

Now the car B will be treated as a source of frequency f1 wavelength of reflected sound received by the driver of car A is

1

v w u v w u v w u'

f v w u f

330 5 25 330 5 25 31m

330 5 25 300 36

19. For continuity displacement current should be equal to conduction current

t /RC t /RC00

qq q 1 e i e CR t

RC

1200 F 0.0012 1.2i A mAex1 e e

20. vmax = A

21. 3

m m 3 r4 m rr3

0.001 3 0.0130 10

5 33.3 10 3 10

3 30.033 10 3 10

33.033 10

22. 2T sin P Area

0 rT . P PA r



5

23. 2 minutes = 3 half life

1 half life = 120 40 sec3

24. 0 ef f 40

00

e

f7 f 35cm

f

ef 5cm 25. 1 1 0T k

2 2 0T k

2 1 1 20

1 2

T TT T

26. 2g g R2

g2R

27. 2 20

1P Q V V Qgh2

20

2P 2gh V VQ

28. A B A B 29. Buoyancy = effective weight of liquid displaced

30. M 1 M 3 MTK 2 K 2 K



6


31. Oxidation number of Mn in 4MnO is 7 and Cr in CrO2Cl2 +6. 32. Volume of all atoms in 1.68 g of Ar

323 81.68 4 226 10 1.50 10 0.35640 3 7

% of empty space = (1 – 0.3564) × 100 = 64.36

33. * 10.02G 50 mA 0.0004

1 1R 50

Specific conductance * 21G 1ScmR

34. When impure metal has impurity of its own metal oxide, then poling process is used e.g. impure

Cu and Sn are purified by this method.

35. Equivalent weight of KBrO3 167

6 n factor n-factor = 6

2 2 3Na S O

0.167 1N 6 0.12 N167 0.05

36. Mass of 1 litre of solution = 1010 gram Mass of solvent = 1010 – 300 × 0.1 = 980 gram

0.1m 0.1020.98

f f fT iK m 1 K .m

ofT 0.199 C

o ofT 0 0.199 C 0.199 C

37. Effective number of 2O in a unit cell 1 18 6 48 2

Effective number of A2+ in a unit cell = 18 18

Effective number of B3+ in a unit cell = 14 22

General formula = AB2O4 38. o

4200 to 300 C

Thermal decompositionImpure PureVolatile compound

Ni 4CO Ni CO Ni 4CO

39.

4 32

2 2 2Black

Dil. HNODil. H SOColloidal sulphur

FeS H S S NO 2H O



7

40. At constant V 1 22

1 2

P PP 1.5 atm

T T

For single phase dG VdP SdT 2G v. P 2 10 T dT = 781.5 J 41. 3a 2 r r

2 1.69 1.81

a1.732

o

a 4.04 A 42. For normal Calomel electrode o

RP RPE E

cell0.059E 0.28 0

2

2

2

H

Hlog pH 5.5

P

43. As 1PV cons tan t p. v dv V .dp 0

1dP v. dv

P v

dp dVp V

44. Preferential adsorption of Fe3+ takes place so positive charge develop. 45.

He Ne

1.2 litre.0.63 atm 3.4 litre.

2.8 atm

He1.2 0.63n

0.082 289

Ne3.4 2.8n

0.0821 289

= 0.032 = 0.4

T0.432 0.0821 289P 2.223

4.6

He0.032P 2.223 0.164 atm0.432

46. 2BaCl KI No ppt.

22 4 4

Yellow ppt.BaCl aq CrO BaCrO

47. H, S & G should all be – ve to hold chemical adsorption.



8

48. Eq. wt. of N2H4 = Molar mass4

Eq. wt. of 3Molar massIO

4

49. n

2M AA

does not show geometrical isomerism and can not show optical.

50. 3 2 2 22CH OH 3O 4H O 2CO g 1453

2 24 44 4H O 4H O

3 2 2 22CH OH 3O g 4H O 2CO g 1277 H 1277 kJ 51. The word smog is originated from smoke and fog. Hence it is oxide of sulphur and N2. 52. Piperazine is not a tranquilizer. 53. As in higher oxidation state transition elements show acidic nature. 54. 2

2Fused anhydrous

MgCl Mg 2e

At cathode: 2Mg 2e Mg

At anode: 22Cl 2e Cl 55. Phenelzine is a tranquilizer. It is not an antacid. 57. 1 watt = 1 J/sec Total heat for 36 mL H2O = 806 × 100 = 80600 J

vaporisation80600H 18

36

= 40300 J or 40.3 kJ/mole 58. At anode, 22HgO 4H aq O g 4e At cathode Ag aq e Ag s Ag deposited = 0.1 mole = 10.8 gram

O2 expelled out = 0.1 0.8 gram4

Total wt. loss of the solution = 11.6 gram wt. of remaining solution = 108 – 11.6 = 96.4 gram 59. On applying valence bond theory Ni has electronic configuration [Ar]3d8 4s2 and in presence of CO (strong field ligand), electronic

configuration is 3d10. Hybridization is sp3 and all the electrons are paired.



9


61. x – y + 2 = 0 x + 2y = 4 x = 0, y = 2 ˆ ˆa 2j 2k

and 2a 8

20 2 2

a b c 1 1 1 8 a1 2 0

62. Vector ˆ ˆ ˆ ˆ ˆ2i 3 j i 4 j 4k is perpendicular to ˆ ˆ ˆ2i j mk

2 3 01 4 4 02 1 m

m = 165

63. Equation of line L is ˆ ˆ ˆi j k

ˆ ˆr j k 2 1 43 2 2

ˆ ˆ ˆ ˆ ˆr j k 10i 16 j k

Point Q is (–10, 16 + 1, – 1)

2 2PQ 357 357

2 = 1 = 1 Q (–10, 17, 0) or (10, –15, –2) then a + 2b + 3c = 24 or –26 64. Let OA a

; OB b

; OC c

Area of ABC = 1 a b b c c a2

Now, r 3a 4b 6c 0

a r 4a b 6a c 0

4a b 6c a

2c a a b3

b r 3b a 6b c 0

A a

B b

C c

O

3a b 6b c

1b c a b2

and Area of AOB is 1 a b2

Now,

1 a b b c c aArea of ABC 21Area of AOB a b2

=

1 1 2a b 1132 2 3

1 6a b2



10

65. ˆ ˆ ˆ ˆˆ ˆ ˆ ˆa b a b 1 a b b a

66. 1v a b c

2v a 2b c a 2b 2c a 4b 2c 18 a b c

67. 100100 100 100

100 100 kk k

k 0 k 0 k 0

Ck 1 1 C Ck 1 k 1

= 100

100 100k

k 0

1 1012 C101 k 1

= 10099 2 1

101 (x = 99, y = 1 and z = 101) xy2z = 9999

68. Tens place (A) 1, 2, 3 not possible 4 5 6 7 8 9

Unit place (B) 0 0, 1 0, 1, 2 0, 1, 2, 3 0, 1, 2, 3, 4 0, 1, 2, 3, 4, 5

21 7P90 30

69. D = –( + 2)( – 1)2 D1 = –( – 1)2( + 2) D2 = D3 = 0 From = 1, –2 infinite solution 1, –2 unique solution So, L = 11; M = 0; N = L; L2 + M2 + N2 = 125

70. 53

53 9

71. 22C2 + 21C2 + 20C2 + 19C2 + ..... + 3C2 + 2C2 = 23C3

72. n

rr

r 1a z 1

a1z + a2z2 + ..... + anzn = 1 |a1z + a2z2 + ..... + anzn| = 1 1 |a1||z| + |a2||z|2 + ..... + |an||zn| < 2[|z| + |z|2 + ..... + |z|n |ar| < 2

nz 1 z1 2

1 z

1 – |z| < 2|z| – 2|z|n + 1 2|z|n + 1 _ 3|z| + 1 < 0

n 11 2z z3 3

1z3

which is contradiction



11

73. ki20

kA e

k i2018 i20 20

k 0

k i673 i20 20

k 1

e e 1

3

e 1 e

74.

2

2

2

2

a p1 a p

p

a q1 1 a qqa p

a r1 a r

r

Apply R2 R2 – R1 and R3 R3 – R1 and then on explained, we get =

2

2

a p q

prq a p

75. A = r r r r r ra b c a b c

1/3 r r rr r r

a b ca b c

3

3

r r rAa b c27

r 2 3

r 2 3

r 2 3

a a ab b bc c c

=

32

r r3

32r r r r r r

r r

32

r r

aa1a a

bb Aa b c 1 a b cb b 27

cc1c c

76. AD is maximum distance BC is minimum distance Range is [1, 15]

A

(–8, 0)

B

(–5, 0)

(–6, 0)

C

(0, 0)

D (5, 0)

77. Area of the triangle formed by z, iz and z + iz is 21 z2

78. k1 = 4! 5!

28!k5!

1

2

7k60

k

79. 100 55 10 1 54

10 1 9C C



12

80. Letters C must be in middle of word

Required number are 6!2! 2!

81. Required probability is 113

82. 72

5! 2! 17! C

83. n n 21I I

n 1

Hence, common difference = 1

84. r 4 2 2

8r 8rt4r 1 2r 1 2r 2r 1 2r

n 21S 2 1

2n 1 2n

by method of difference

Hence, 201680S841

85. 1 2 1 2

1 2 1 2

G G A AH H H H

in this case

86. P(A B) = BP A P 0.4 0.5 0.2A

Now P(A B) = P(A) + P(B) – P(A B) = 0.4 + 0.6 – 0.2 = 0.8

Also,

1 P A BA B 0.2 1AP PB 1 P B 0.4 2P B

87. Given f(f(–1)) = 0 and f(f(1)) = 0 i.e., f(f(x)) = 0 has of two roots f(1) and f(2) 88. Let i (i = 1, 2, 3, 4, 5) denotes 1, 2, 3, 4, 5 white balls in the bag.

i

iP5

(i = 1, 2, 3, 4, 5)

i1P5

(i = 1, 2, 3, 4, 5)

Now,

55

55i

ii 1

P PP

P P

=

1 1 5 151 1 2 3 4 5 15 35 5 5 5 5 5



13

89. x1, x2 ..... xn be n observations then, n

ii 1

1x xn

Let ii

xy 10

then n n

i ii 1 i 1

1 1 1 1y y x 10nn n n

x x 10y 10

90. Expression = x·x2·x3 ..... x20· 11x

2102 3 20

2 3 201 1 ..... 1 x Ex x x

Where 2 201 2 20E 1 1 ..... 1x x x

Now, coefficient of x203 in original expression = coefficient of x–7 in E

But 2 3 6 2 5 3 4 2 41 2 3 1 6 2 5 3 4 1 2 4E 1 ..... ..... .....x x xx x x x x x x x x

Coefficient of x–7 = –7 + 6 + 10 + 12 – 8 = 13


PART TEST – III

Paper 1

Time Allotted: 3 Hours Maximum Marks: 264 Please r ead the inst ruct ions carefu l l y . You are a l lot ted 5 minutes


the test .

INSTRUCTIONS

A. General Instructions

1. Attempt ALL the questions. Answers have to be marked on the OMR sheets. 2. This question paper contains Three Parts. 3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics. 4. Each part is further divided into Three sections: Section-A Section-B & Section-C. 5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be


devices, in any form, are not allowed.

B. Filling of OMR Sheet 1. Ensure matching of OMR sheet with the Question paper before you start marking your answers


Enrolment No. and write your Name, Test Centre and other details at the designated places. 3. OMR sheet contains alphabets, numerals & special characters for marking answers.

C. Marking Scheme For All Three Parts.

1. Section–A (01 – 10, 21 – 30, 41 – 50) contains 30 multiple choice questions which have one or more than one correct answer. Each question carries +4 marks for correct answer and –2 marks for wrong answer.

2. Section–B (11 – 12, 31 – 32, 51 – 52) contains 6 Match the following Type questions. Each

question having 4 statements in Column I & 5 statements in Column II with any given statement in Column I having correct matching with 1 or more statement (s) given in Column II. Each statement carries +2 marks for correct answer and –1 mark for wrong answer.

3. Section–C (13 – 20, 33 – 40, 53 – 60) contains 24 Numerical based questions with answers as

numerical value from 0 to 9 and each question carries +4 marks for correct answer. There is no negative marking.


Enrolment No.

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AITS-PT-III (Paper-1)-PCM-JEE(Advanced)/18


2

Useful Data

PHYSICS






Density of water water = 103 kg/m3

Atmospheric pressure Pa = 105 N/m2

Gas constant R = 8.314 J K1 mol1

CHEMISTRY


N=9, Na=11, Mg=12, Si=14, Al=13, P=15, S=16, Cl=17, Ar=18, K =19, Ca=20, Cr=24, Mn=25, Fe=26, Co=27, Ni=28, Cu = 29, Zn=30, As=33, Br=35, Ag=47, Sn=50, I=53, Xe=54, Ba=56, Pb=82, U=92.

Atomic masses: H=1, He=4, Li=7, Be=9, B=11, C=12, N=14, O=16,

F=19, Na=23, Mg=24, Al = 27, Si=28, P=31, S=32, Cl=35.5, K=39, Ca=40, Cr=52, Mn=55, Fe=56, Co=59, Ni=58.7, Cu=63.5, Zn=65.4, As=75, Br=80, Ag=108, Sn=118.7, I=127, Xe=131, Ba=137, Pb=207, U=238.




3


SECTION – A

One OR More Than One Choice Type

This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which only ONE OR MORE THAN ONE is/are correct 1. First an object is slowly lifted from the bottom (point – A) of a shaft of

depth 1Rh2

to earth’s surface (point-B) and then it is slowly lifted still

higher to attain an altitude 2Rh2

above the earth’s surface (Point C).

W1 and W2 are the work performed in two cases respectively. Choose the correct option(s)

(A) 1 2W W (B) 1 2W W

(C) 1 2

1 2

W W 17W W

(D) 1 2

1 2

W W 9W W

A

B

C

R O

Earth

2. A light stick of length rests with its one end against the smooth wall

and other end against the smooth horizontal floor as shown in the figure. The bug starts at rest from point B and moves such that the stick always remains at rest. aP is the magnitude of acceleration of bug of mass m, which depends upon its distance of x from the top end of the stick. Choose the correct option(s)

(A) 1sinP

g xa

x

m

B

A

P

(B) 1cosP

g xa

(C) The time taken by the bug to reach the bottom of the stick having started at the top end from

rest is sin2 3g

(D) The time taken by the bug to reach the bottom of the stick having started at the top end from

rest is sin2 g





4

3. Monochromatic rays of intensity I are falling on a metal plate of surface area A placed on a rough horizontal surface at certain angle as shown in figure. Choose correct statement (s) based on above information:

(A) There is a value of for which plate will not move however high the intensity of radiation is

(B) Plate will not move if plate is perfectly reflecting irrespective of the value of intensity. (C) If rays are falling perpendiculars to surface plate will not move (D) None of these 4. All the surfaces are frictionless and system is

released from rest when ideal spring of stiffness k is in its relaxed state. Choose the correct statement.

(A) The magnitude of maximum acceleration of

particle B is g3

(B) The magnitude of maximum acceleration of

particle B is 2g3

B 2m

= 0

k m

g A

(C) The maximum speed of particle A is m2g3k

(D) The maximum speed of particle A is mg3k

5. We are given the following atomic masses: 238 4

92 2U 238.05079 u He 4.00260 u 234 1

90 1Th 234.04363 u H 1.00783 u 237

91 Pa 237.05121 u Here the symbol Pa is for the element protactinium (Z = 91). (A) The energy released during the -decay of 238

92 U is 4.75 MeV approximately. (B) The energy released during the -decay of 238

92 U is 4.25 MeV approximately. (C) The emission of proton from 238

92 U can be spontaneous. (D) The emission of proton from 238

92 U can not be spontaneous.





5

6. A 50–kg ball is attached to one end of a 1 m cord that has a mass of 0.10 kg. The other end of the cord is attached to a ring that can slide on a frictionless horizontal shaft AB, as shown in the diagram. A horizontal blow is delivered to the cord and excites the fundamental vibration with a maximum transverse velocity of 15 m/s. Assume that the ball remains essentially stationary as the cord vibrates. [g = 9.8 m/s2]

(A) The frequency of the fundamental vibration is 17.5 Hz. (B) The amplitude of the motion is 13.6 × 10–2 m (C) If the blow is delivered as an impulse to a stationary cord, the

wave function for the cord is y(x, t) = A sin (x/2L) sin t, where the ball is located at x = 0 and positive x-axis is along the string.

50 kg

1 m

Ring

A B

(D) The period of the pendulum motion of the hanging ball is 2 sec. approximately.

7. Mass 0

2

2

mm

v1c

as a function of velocity is shown in

the graph. m0 is the rest mass of the system. The kinetic energy of system is given as 2 2

cK mc m c Choose correct option(s)

0.5c c v O

m0

m

(A) If a system is moving with speed v = 0.5c, the mass of system is increased by 15.5% (B) Electrons in Cornell university synchrotron reach a velocity of v = 0.8c. The mass of these

electrons are approximately 53

time the rest mass of electron.

(C) The Bevatron a proton accelerator generates the accelerated protons with kinetic energy 10–9 Joules. The mass of accelerated protons is 8.68 times the actual mass of proton approximately 0

27pm 1.67 10 kg

(D) If the kinetic energy of particle is equal to its rest mass energy then speed of particle is 0.75c approximately.





6

8. A rod of length l = 100 cm is fixed at 30 cm from both end. If velocity of transverse wave in rod is v(cms–1), then choose the correct option(s)

(A) Fundamental frequency (in Hz) of transverse wave in rod is v/40 (B) Second overtone in rod will be 5th harmonic (C) Frequency of third overtone (in Hz) is 7v/40 (D) Third overtone is 5th harmonic. 9. If voltage applied to an X-ray tube increased from V = 10 kV to 20 kV. The wave length interval

between K-line and short wave cut off of continuous X-ray increases by factor of 3 Rhc 13.6 V

e , where R is Rydberg constant, h is plank’s constant, c is speed of light in vacuum

and e is charge on electron. (A) Atomic no. of target metal used is 29 (B) Cut-off wavelength when V = 10 kV is 1.2 Å (C) Cut off wavelength when V = 10 kV is 2 Å (D) Atomic no. of target metal used is 26 10. Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at

a distance 2m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å.

(A) Number of photon emitted by bulb is 5 × 1019 per second (B) Time required by atomic disk to receive energy equal to work function of metal is 22.4 sec. (C) Number of photon received by disk to receive 2 eV energy is 4 (D) Number of photons emitted by bulb is 5 × 1017 per sec.





7

SECTION – B

(Matching Type) This section contains 2 multiple choice questions. Each question has matching Column(s). The Column(s) have choices (A), (B), (C) and (D) out of which only ONE OR MORE THAN ONE is/are correct 11. In the photoelectric effect experiment, if f is the frequency of radiations incident on the metal

surface and I is the intensity of the incident radiations, then match the following columns. Column -I Column -II

(A) If f is increased keeping I and work-function constant

(p) Stopping potential increases

(B) If distance between cathode and anode is increased

(q) Saturation current increases

(C) If I is increased keeping f and work-function constant

(r) Maximum kinetic energy of photoelectron increases

(D) Work-function is decreased keeping f and I constant

(s) Stopping potential remains same

(t) Stopping potential decreases





8

12. Two thin lens made of glass 32

and water 43

kept in contact with each other. An

object is placed at point O. Assume all curved surfaces have equal radius. Match the following Columns:

Column – I Column - II (A)

O

5 cm

R = 10 cm

glass water

(p) Image of O will be Real

(B) O

5 cm

R = 5 cm

glass water

(q) Image of O will be Virtual

(C) O

50 cm

R = 10 cm

glass water

(r) Focal length = 10 cm

(D)

O

R =10 cm

glass water

(s) Focal length = 15 cm

(t) Focal length = 25 cm





9

SECTION – C

(One Integer Value Correct Type)

This section contains 8 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive). 13. A square hinged-structure (ABCD) is formed with four massless rods

which lies on the smooth horizontal table. The hinge-D of structure is fixed with table and hinge A, hinge-B and hinge-C are free to move on the table. A point mass m is attached at hinge-B, and an ideal spring of stiffness k is connected between hinge-A and hinge-C as shown in the figure. System is in equilibrium. The mass m is slightly displaced along line BD and released to perform SHM. The time period of SHM is

nm.k

Find

D C

B A

14. The mean pressure is 6K5

, which rain renders to vertical windshield of automobile, moving with

constant velocity of magnitude v = 12 m/s. Consider that raindrops fall vertically with speed u = 5 m/s. The intensity of rainfall deposits h = 2 cm of sediments in time = 1 minute. [ = 103 kg/m3 is the density of liquid] (Assume collisions are inelastic). Calculate K.

15. On the bottom of lake at a depth h = 100m, a

horizontal pipe of length 80m with piston is kept as shown in the figure. The piston is light and movable. Between the piston and pipe some air is captured (x0 = 9m). The pipe is slowly raised to vertical position and open end upward. The length of air column (x) in SI units is 5K. Find value of K. (Disregard the atmospheric pressure & piston and pipe are air tight. Temperature of gas remains constant).

h–x

x x0





10

16. In the figure shown rod AB is light and rigid while rod CD is also rigid and hinged at midpoint O Rod can rotate without friction about O. The time period of small oscillation of

rod CD is mT 2nk

. Where m is mass of

rod CD and k is spring constant of each spring. The value of n is

k

D O C

B

k k

A

k

17. Consider a cuboidal vessel (2R × 2R × R) with a

hemispherical cavity of radius R, kept at a horizontal smooth surface as shown in the figure. The vessel has very small hole of cross-sectional area “a”. Now water of density is poured into space developed due horizontal surface and vessel through hole very slowly. When the height of water level is h, the vessel lifts off the surface and liquid leaks through space generated. (R = 2m, = 103 kg/m3, m = 9 kg). If the value of h is 5k cm, find the valve of k.

(Atmospheric pressure = 105 N/m2).

h Water

Horizontal Surface

a

18. In an undershot water wheel, the cross-sectional area 2a 0.1m of the stream is striking the

series of radial flat vanes of the wheel. The velocity of stream is 6 m/s. The velocity of vanes is 3 m/s. If the power supplied by jet (in watts) is 2700K, find K.

O

a

v

Water jet

u

19. A man of height = 3/2 m, wants to see himself in plane mirror from top to bottom. The plane

mirror is inclined with vertical wall at angle = 53º. If the least size of mirror to see him is 3 mn

.

the distance of eye from mirror is d = 3m, find the value of n. 20. If 0, B, V represent permittivity of free space, magnitude of magnetic field and volume of space

respectively, then the dimension of 0B2V is a b cM L T . Find a + b + c.





11


SECTION – A


This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which only ONE OR MORE THAN ONE is/are correct 21. Select the correct statement (A) Viscosity in gases arises principally from the molecular diffusion that transport momentum

between layers of flow. The net effect will be decrease in the relative rate of movement of one layer with respect to the other.

(B) Viscosity of ideal gas increases as temperature increases. (C) Viscosity of liquids decreases with the rise in temperature. (D) At critical temperature the surface of separation between liquid and gas disappear and their

viscosity is same. 22. An ideal gas whose adiabatic exponent is expanded according to the law P V , where is a

constant. The initial volume of the gas is equal to Vo. As a result of expansion, the volume increases 4 times. Select the correct option for above information.

(A) Molar heat capacity of gas in the process is

R 12 1

(B) Work done by the gas is 2o15V

2

(C) Change in internal energy of gas is 2o15V

2

(D) Both B and C are correct 23. Which of the following complexes exists in facial and meridional forms? (A) [Co(dien)2]3+ (B) [Co(NH3)3Cl3] (C) [Co(en)3]3+ (D) [Co(gly)3] 24. Select the correct statement, for non stoichiometric cuprous oxide Cu1.8O. (A) % of Cu2+ in total copper is 11.11% (B) % of Cu1+ in total copper is 11.11% (C) It behaves like p-type semiconductor (D) Defect is metal deficiency defect 25. Which of the following are colourless? (A) Ce3+ (B) La3+ (C) Lu3+ (D) Gd3+





12

26. One mole of component P and two moles of component Q are mixed at 27oC to form an ideal binary solution. Select the correct option for given information?

(A) 1mixS 3.82 cal k (B) mixH 0

(C) mixS 0 (D) mixG 1145.7 cal 27. Select the correct statement for lead storage battery: (A) It is a reversible cell. (B) Salt bridge is not required for lead storage battery. (C) It is recharged by using DC source of current. (D) Specific gravity of sulphuric acid solution decreases during discharging. 28. Which of the following statements are true for froth floatation process (for the concentration of

sulphide ores)? (A) Pine oil used as collectors. (B) Cresol enhance non wettability of the mineral particles. (C) NaCN selectively prevents PbS from coming to the froth but allows ZnS to come with the

froth. (D) The mineral particles become wet by oils while the gangue particles by water.

29. Which of the following elements can give NO2 as a by product on reaction with conc. HNO3? (A) C (B) Cu (C) Zn (D) Sn 30. Chlorine has great affinity for hydrogen. It reacts with compounds containing hydrogen to form

HCl. Which of the following reaction/s give/s HCl? (A) 2 2H O Cl (B) 2 2H S Cl

(C) 10 16 2C H 8Cl (D)

3 2excessNH 3Cl





13

SECTION – B

(Matching Type) This section contains 2 multiple choice questions. Each question has matching Column(s). The Column(s) have choices (A), (B), (C) and (D) out of which only ONE OR MORE THAN ONE is/are correct 31. Match the following column of precipitate/mass listed in Column I with the reagent (s) listed in

Column II: Column – I (Observations) Column – II (Reagents)

(A) Mg2+ gives pale pink mass with (p) NaOH solution (B) Pb2+ gives yellow ppt. with (q) H2S gas (C) Ag+ gives black/brown ppt. with (r) K2CrO4 solution (D) 2

2Hg gives black ppt. with (s) KI (not in excess) or KI in excess (t) Cobalt nitrate in charcoal cavity test





14

32. In Column I, certain thermodynamic process are given and in Column II, the value of physical quantities are given. Match the Column I with Column II suitably. (Given : density of gas )

Column – I Column – II (A)

oT o2T T

V T 1 mole of N2

V

(p) oQ 2RT

(B)

oT o2T T

P T1 mole of He

P

(q) o

3U RT2

(C)

oT o2T T

1VT

1 mole of gaseous mixture

having adiabatic index 1.5

V

(r) oW RT

(D)

oT o2T T

1PT

1 mole of gas having

degree of freedom f = 4

P

(s) vs T graph for the process is a straight line

(t) P vs graph for the process is a parabola





15

SECTION – C


This section contains 8 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive). 33. How many types of isomerism is exhibited by the complex [Co(NH3)4(NO2)2]Cl?

34. How many millilitres of 0.05 M K4[Fe(CN)6] solution is required for titration of 60 ml of 0.01 M

ZnSO4 solution, when the product of reaction is K2Zn3[Fe(CN)6]2? 35. A molecule Ax dissolve in water and is non volatile. A solution of certain molality showed a

depression of 0.93 K in freezing point. The same solution boiled at 100.26oC. When 7.87 g of Ax was dissolved in 100 g of water, the solution boiled at 100.44oC. Given Kf for water = 1.86 K kg mol-1, atomic mass of A = 31 u. Assume no association or dissociation of solute. Calculate the value of x……….

36. What volume of air (in m3) is needed for the combustion of 1 m3 of a gas having the following

composition in percentage volume : 2% of C2H2, 8% of CO, 35% of CH4, 50% of H2 and 5% of non-combustible gas. The air contains 20.8% (by volume) of oxygen.





16

37. Find the total number of metals, which makes a thin protective layer of its oxide on treatment with conc. HNO3.

Zn, Cu, Al, Pt, Cr, Au, Pb, Fe 38. The number of salts formed by telluric acid when treated with strong base NaOH is/are…. 39. If 6.53 × 10-2 g of metallic zinc is added to 100 ml saturated solution of AgCl, it reacts with Ag+ of

solution as following reaction 2Zn s 2Ag aq Zn aq 2Ag s and approximately 10-x moles of Ag will be precipitated. Calculate the value of x……(Given

o oZn / Zn Ag / Ag

E 0.76V E 0.8 V, Ksp of AgCl = 10-10, atomic mass of Zn = 65.3u,

1052.8813 = 7.61 × 1052) 40. Initial volume of H2 gas saturated with water vapour is confined under a piston in a container is

10 litres as shown in the given figure:

H2(gas)

Water(liq.) the container also contains some liquid water. The total pressure over liquid water is 80 cm of Hg.

If now the piston is removed such that volume of container is doubled, then final total pressure over liquid water in the container is P. The vapour pressure of water is 20 cm of Hg and volume of liquid is negligible. Calculate P/10…….





17


SECTION – A


This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which only ONE OR MORE THAN ONE is/are correct 41. If D1, D2, D3, ..... D1000 are 1000 doors and P1, P2, P3, ..... P1000 are 1000 persons. Initially all doors

are closed. Changing the status of doors means closing the door if it is open or opening it if it is closed. P1 changes the status of all doors. Then P2 changes the status of D2, D4, D6, ..... D1000 (doors having numbers which are multiples of 2). Then P3 changes the status of D3, D6, D9, ..... D999 (doors having number which are multiples of 3) and this process is continued till P1000 changes the status of D1000, then the doors which are finally open is/are

(A) D961 (B) D269 (C) D413 (D) D729 42. Which of the following options are correct?

(A) [(nC0 + nC3 + nC6 +.....) – 12

(nC1 + nC2 + nC4 + nC5 +.....)]2 + 34

(nC1 – nC2 + nC4 – nC5 +.....)2 = 1

(B) If a and b are two positive numbers such that a5b2 = 4 then the maximum value of 1/5 1/ 2

2 22 2log a log b is equal to 4

(C) Constant term in ((((((x – 2)2 – 2)2 – 2)2 – 2)2 – 2)2 ..... 2)2 is equal to 2

(D) The coefficient of x24 in 2525 25 25

2 2 231 2 425 25 25 25

0 1 2 3

CC C Cx x 2 x 3 x 4

C C C C

.....

25

2 2525

24

Cx 25

C

is equal to 2925

43. x1, x2, x3 are three real numbers satisfying the system of equations x1 + 3x2 + 9x3 = 27, x1 + 5x2 + 25x3 = 125 and x1 + 7x2 + 49x3 = 343, then which of the following

options are correct

(A) number of divisors of x1 + x3 is 16 (B) 1 2x x2

is a prime number

(C) x3 – x2 is a prime number (D) x1 + x2 + x3 is square of an integer

Space for rough work




18

44. a1, a2, a3, ..... are distinct terms of an A.P. We call (p, q, r) an increasing triad if ap, aq, ar are in G.P. where p, q, r N such that p < q < r. If (5, 9, 16) is an increasing triad, then which of the following option is/are correct

(A) if a1 is a multiple of 4 then every term of the A.P. is an integer (B) (85, 149, 261) is an increasing triad

(C) if the common difference of the A.P. is 14

, then its first term is 13

(D) ratio of the (4k + 1)th term and 4kth term can be 4 45. If z1, z2, z3 z4 are complex numbers in an Argand plane satisfying z1 + z3 = z2 + z4. A complex

number ‘z’ lies on the line joining z1 and z4 such that 3 22

1 2 2

z zz zArg Argz z z z

. It is given that

|z – z4| = 5, |z – z2| = |z – z3| = 6 then (A) area of the triangle formed by z, z1, z2 is 3 7 sq. units

(B) area of the triangle formed by z, z3, z4 is 15 74

sq. units

(C) area of the quadrilateral formed by the points z1, z2, z3, z4 taken in order is 27 72

sq. units

(D) area of the quadrilateral formed by the points z1, z2, z3, z4 taken in order is 27 74

sq. units

46. Which of the following is/are true? (A) 100300 < 300! (B) 300300 300!

(C) 100300 > 300! (D) 300300 300! 47. Which of the following is/are correct? (A) If A is a n n matrix such that 2 2

ija i j 5ij j i i and j then trace (A) = 0

(B) If A is a n n matrix such that 2 2ija i j 5ij j i i and j then trace (A) 0

(C) If P is a 3 3 orthogonal matrix, , , are the angles made by a straight line with OX, OY,

OZ and

2

2

2

sin sin sin sin sin

A sin .sin sin sin sin

sin sin sin sin sin

and Q = PTAP, then PQ6PT = 32A

(D) If matrix ij 3 3A a

and matrix ij 3 3

B b

where ij jia a 0 and ij jib b 0 i and j then

A6B7 is a singular matrix





19

48. The vertices of a triangle ABC are A (2, 0, 2), B(–1, 1, 1) and C (1, –2, 4). The points D and E divide the sides AB and CA in the ratio 1 : 2 respectively. Another point F is taken in space such that the perpendicular drawn from F to the plane containing ABC, meets the plane at the point of intersection of the line segments CD and BE. If the distance of F from the plane of triangle ABC is

2 units, then

(A) the volume of the tetrahedron ABCF is 73

cubic units

(B) the volume of the tetrahedron ABCF is 76

cubic units

(C) one of the equation of the line AF is ˆ ˆ ˆ ˆr 2i 2k 2k i

( R)

(D) one of the equation of the line AF is ˆ ˆ ˆ ˆr 2i 2k i 7k

49. The direction cosines of two lines are connected by the relations m n 0 and

mn 2 m n , then

(A) 1 1 1

2 2 2

m nm n

is equal to 32

(B) 1 2 1 2 1 2m m n n is equal to 12

(C) 1 1 1 2 2 2m n m n is equal to 23 3

(D) 1 2 1 2 1 2m m n n is equal to 13 6

50. Let the equation of a straight line L in complex form be az az b 0 , where a is a complex

number and b is a real number, then

(A) the straight line i z cz c 0

a a

makes an angle of 45º with L and passes through a point

c (where c is a complex number)

(B) the straight line i z cz c

a a

makes an angle of 45º with L and passes through a point c

(where c is a complex number)

(C) the complex slope of the line L is aa

(D) the complex slope of the line L is aa





20

SECTION – B

(Matching Type) This section contains 2 multiple choice questions. Each question has matching Column(s). The Column(s) have choices (A), (B), (C) and (D) out of which only ONE OR MORE THAN ONE is/are correct 51. Match the following Column-I with Column-II

Column – I Column – II (A) If the polynomial x3 + ax2 + bx + c is divisible by (x2 + 1), where a,

b, c belong to {1, 2, 3, 4, ....., 10} then a + b + c may be equal to (p) 6

(B) A and B are 3 3 matrices of real numbers, where A is symmetric a matrix and B is a skew symmetric a matrix, and

(A + B)(A – B) = (A – B)(A + B). If (AB)T = (–1)kAB then k may be equal to

(q) 3

(C) Sum of the digits of (10050 – 43) is divisible by (r) 9

(D) Let ˆ ˆa cos i sin j

, ˆ ˆb sin i cos j

, ˆc k

,

ˆ ˆ ˆr 7i j 10k

, if r xa yb zc

then the value of 2 2x y

z is

less than

(s) 10

(t) 11 52. Match the following Column-I with Column-II

Column – I Column – II (A) If x and y are two integers such that 289 – x2 + y4 = 0 then

the possible value(s) of unit digit of x + 12y + 4 is/are (p) 1

(B) If P, Q, R, S be four points in space satisfying PQ 3

,

QR 7

, RS 11

, SP 9

, then the value of PS QR9

is

less than

(q) 2

(C) The sum of three positive integer is 20. If the probability that they form the sides of a triangle is P then 19P is equal to

(r) 3

(D) The first term of an infinite geometric series is 21. The second term and the sum of the series are both positive integers. If the value of second term is k then the possible value(s) of |k – 15| is/are

(s) 4

(t) 5





21

SECTION – C


This section contains 8 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive). 53. If 1, 2, 3, 4 are the roots of x4 + 2x3 + bx2 + cx + d = 0 such that 1 – 3 = 4 – 2, then b – c is

equal to _____ 54. A chess match between two players A and B is won by whoever first wins a total of two games.

Probability of A’s winning, drawing and losing any particular game are 16

, 13

and 12

respectively.

(The games are independent). If the probability that B wins the match in the 4th game is p, then 6p is equal to _____

55. If k1 and k2 (k1 > k2) are two non-zero integral values of k for which the cubic equation x3 + 3x2 + k = 0 has all integer roots, then the value of k1 – k2 is equal to _____

56. Let ak = nCk for 0 k n and k 1k

k

a 0A

0 a

& n 1

k k 1k 1

a 0B A A

0 b

, then a

b is equal to

_____

57. If 203

2

r 0r 2 r 1 ! 2r r 1 ! a! 2 b!

(where a, b N), then a – b is equal to _____

58. If 3 6 9 4 7 10x x x x x xa 1 ....., b x .....

3! 6! 9! 4! 7! 10! and

2 5 8 11x x x xc .....2! 5! 8! 11!

then the

value of a3 + b3 + c3 – 3abc is equal to _____

59. Let m n

A p q r1 1 1

and B = A2. If 2 2m p q 9 , (m – n)2 + (q – r)2 = 16,

2n + (r – p)2 = 25, then the value of |det(B) – 140| is _____ 60. If the number of ordered pairs (a, b) where a, b R such that (a + ib)5015 = (a – ib)3 is k, then the

unit digit of k is equal to _____



AITS-PT-III (Paper-1)-PCM (Sol.)-JEE(Advanced)/18


1


PART TEST – III PAPER-1


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




2


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.




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.




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




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




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




7


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




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




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




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




11


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




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




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




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




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



PART TEST – III

Paper 2

Time Allotted: 3 Hours Maximum Marks: 240 Please r ead the inst ruct ions carefu l l y . You are a l lot ted 5 minutes


the test .

INSTRUCTIONS

A. General Instructions

1. Attempt ALL the questions. Answers have to be marked on the OMR sheets. 2. This question paper contains Three Parts. 3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics. 4. Each part is further divided into two sections: Section-A & Section-C. 5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be


devices, in any form, are not allowed.

B. Filling of OMR Sheet 1. Ensure matching of OMR sheet with the Question paper before you start marking your answers


Enrolment No. and write your Name, Test Centre and other details at the designated places. 3. OMR sheet contains alphabets, numerals & special characters for marking answers.

C. Marking Scheme For All Three Parts.

1. Section–A (01 – 08, 21 – 28, 41 – 48) contains 24 multiple choice questions which have one or

more correct answer. Each question carries +4 marks for correct answer and –2 marks for wrong answer.

Section–A (09 – 12, 29 – 32, 49 – 52) contains 12 paragraphs with each having 2 questions with

one correct answer. Each question carries +4 marks for correct answer and –2 marks for wrong answer.

2. Section–C (13 – 20, 33 – 40, 53 – 60) contains 24 Numerical based questions with answers as

numerical value from 0 to 9 and each question carries +4 marks for correct answer. There is no negative marking.


Enrolment No.

ALL

IND

IA T

ES

T S

ER

IES

FIITJEE JEE (Advanced)-2018




2

Useful Data

PHYSICS






Density of water water = 103 kg/m3 Atmospheric pressure Pa = 105 N/m2 Gas constant R = 8.314 J K1 mol1

CHEMISTRY


N=9, Na=11, Mg=12, Si=14, Al=13, P=15, S=16,

Cl=17, Ar=18, K =19, Ca=20, Cr=24, Mn=25,

Fe=26, Co=27, Ni=28, Cu = 29, Zn=30, As=33,

Br=35, Ag=47, Sn=50, I=53, Xe=54, Ba=56,

Pb=82, U=92.

Atomic masses: H=1, He=4, Li=7, Be=9, B=11, C=12, N=14, O=16,

F=19, Na=23, Mg=24, Al = 27, Si=28, P=31, S=32,

Cl=35.5, K=39, Ca=40, Cr=52, Mn=55, Fe=56, Co=59,

Ni=58.7, Cu=63.5, Zn=65.4, As=75, Br=80, Ag=108,

Sn=118.7, I=127, Xe=131, Ba=137, Pb=207, U=238.




3


SECTION – A


This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which only ONE OR MORE THAN ONE is/are correct 1. Choose the correct statement(s). (A) The image of sun (consider point source of light) can never be produced at center of glass

sphere, (assuming the space between sphere and sun is filled with air). (B) If a lens having = 1.5 has focal length 20 cm in air, then its focal length will be 60 cm when it

is immersed in water of refractive index 43

(C) The laws of reflection and refraction are true for all surface and pairs of media at the point of incidence.

(D) The thick lenses give coloured images due to dispersion of light. 2. When a yellow light of wavelength 6000 Å from a sodium lamp falls on a certain photocell, a

negative potential of 0.30 volt is needed to stop all the electrons from reaching the collector. Then choose correct option(s).

(A) Work function of the given photocell is 2.5 eV approximately. (B) Work function of the given photocell is 1.8 eV approximately. (C) A negative potential of 0.6 Volt will be needed to stop all the electrons if light of wavelength

= 4000 Å is used (D) A negative potential of 1.3 Volt will be needed to stop all the electrons if light of wavelength

= 4000 Å is used 3. An electromagnetic radiation of wavelength ranging

between 400 nm and 1150 nm (for which the plate is penetrable) incident perpendicularly on the plate from above is reflected from both the air surfaces and interferes. In this range only two wavelength give maximum reinforcements, one of them is = 400 nm. (refractive index of air = 1 and the coefficient of linear thermal expansion of cube = 8 106 C1). The distance of bottom of cube from plate does not change during warming up. Choose the correct option(s)

cube h = 2 cm

Plate d

(A) The change in temperature of cube so that it would touch the plate is 6.2C (B) The change in temperature of cube so that it would touch the plate is 3.1C

(C) The second wavelength, which gives maximum reinforcement is = 2000 nm3

(D) The second wavelength, which gives maximum reinforcement is = 4000 nm3





4

4. A certain ideal spring stretches 20 cm when m1 = 40 gm mass is hung from it. If a mass of

22205m gm

32 gm is hung instead of 40 gm at its end pulled out 20 cm from equilibrium and

released at t = 0 sec. Choose the correct option(s). (g = 10 m/s2) (A) The magnitude of acceleration of the m2 mass when it is 2.5 cm from the equilibrium position

is 6445

m/s2

(B) The magnitude of acceleration of the m2 mass when it is 2.5 cm from the equilibrium position

is 3245

m/s2

(C) At t8

sec. m2 mass is 10 cm from equilibrium position.

(D) At t2

sec. m2 mass is 10 cm from equilibrium position.

5. I am a Vernier Calliper. I have a task to measure the length of a pencil as shown in the figure.

The least count of my Vernier scale is 0.05 times the value of one division of main-scale. Pencil

9 cm 8 cm

Portion of Main-Scale

My Vernier Scale has n-divisions and coincides with (n – 4) divisions of main scale. While

measuring the length of pencil, the n th5

division of Vernier-Scale exactly coincide the main-scale.

Now choose the correct option(s) (A) The value of n is 40 (B) The least count of Vernier scale is 0.005 cm (C) The length of Pencil is 8.44 cm (D) The length of Pencil is 8.48 cm





5

6. A triple slit experiment is performed as shown in the figure with monochromatic light of amplitude a0. S3M S2N = . The graph between resultant amplitude (A) versus is plotted. Choose the correct option(s).

D>>d

d

d

S1

S2

S3

N

M

screen

P

x

(A)

0 43

2

3

2

3

4

3 2

3

2

3

A20

A10

A

(B)

0 43

53

23

3

53 4

3

23

3

A20

A10

A

(C) 10 20

10 20

A A2

A A

, where A10 and A20 are the values shown in option A.

(D) 10 20

10 20

A A5

A A

, where A10 and A20 are the values shown in option A.





6

7. One end of thin aluminium rod (cross-sectional Area = 10–6 m2) is bult welded to the end of copper rod with same diameter. Both rods are very long and under uniform tension of 900 N. Given that

Y(Cu) = 1.6 × 1011 Nm–2 Y(Al) = 0.9 × 1011 N m–2 (Cu) = 8.1 × 103 kg m–3 (Al) = 2.5 × 103 Kg m–3

(A) The value of reflection factor r

i

AA

for longitudinal pulse approaching the junction along copper

rod is 0.15

(B) The value of reflection factor r

i

AA

for longitudinal pulse approaching the junction along copper

rod is approximately 0.3

(C) The value of reflection factor r

i

AA

for transverse pulse approaching the junction along copper

rod is 0.15

(D) The value of reflection factor r

i

AA

for transverse pulse approaching the junction along copper

rod is approximately 0.3 8. There are two cylinder-shaped wooden billets, each having a mass of

mass 45 kg, in a vertical wall sewage, which contains water in it. The two billets have the same size and the same material; they touch each other and the walls of the sewage. One of them is totally under the water, whilst only half of the other one is immersed into the water. Friction is negligible everywhere. (density of water is 1000 kg/m3 and g = 10 m/s2). Choose the correct option(s)

(A) Density of the wood is 750 kg/m3

(B) Density of the wood is 800 kg/m3 (C) The forces exerted by the billets on the vertical walls is 260 N approximately (D) The forces exerted by the billets on the vertical walls is 320 N approximately





7

Comprehension Type This section contains 2 groups of questions. Each group has 2 multiple choice question based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which only ONE is correct

Paragraph for Question Nos. 09 and 10 Consider a wave sent down a string in the positive direction whose equation is given as

0xy y sin tv

The wave is propagated along because each string segment pulls upward and downward on the segment adjacent to it at a slightly larger value of x and, as a result does work upon the string segment to which wave is travelling. For example, the portion of string at point A is going upward, and will pull the portion at point B upward as well. In fact, at any point along the string, each segment of the string is pulling on the segment just adjacent and to its right, causing the wave to propagate. It is by this process that the energy is sent along the string.

x x+x

A

B

v

Now we try to calculate how much energy is propagated down the string per second

yT Tsin T tan yyT Tx

(The negative sign appears because as shown in the figure-II, the slope is negative) the force will act through a distance

yydy v dt dtt

x

A

y

T

Therefore work done by force in time dt is

dW = Tydy = y yT dtx t

2 2

20y T xdW cos t dtv v

…(A)

Now answer the following questions. 9. The average power transmitted down the string (i.e. the average energy transfer or sent down it

per second) is

(A) 2 2

02 y Tv

(B)

2 20y T

v

(C) 2 2

0y T2v

(D)

2 20y T

4v

10. How much average power is transmitted dawn a string having density = 5 104 kg/m and T = 5 N by a 200 Hz vibration of amplitude of 0.20 cm.

(A) 0.08 Watt (B) 0.16 Watt (C) 0.32 Watt (D) 0.64 Watt





8

Paragraph for Question Nos. 11 and 12

ABC is an isosceles right angled triangular reflecting prism of average refractive index . When incident rays on face AB are parallel to face BC , then they emerges from Face AC , which are also parallel to Face BC as shown in figure-I. The prism capable to do so , known as Dove Prism. In figure-II, the Dove Prism is used for dispersion of incident light containing red colour and violet colour only. The red colour and violet colour lights are separated( displaced ) on screen by a distance . In reality , each ray of any colour has some width, which can be designated as d. It is clear that an observer can distinguish the red and violet rays that emerges from prism only if d .Otherwise the bundles of rays will overlap and mix.

[ Given for Dove Prism in figure-II: 52R , 0.02V R , 5 2.25 , EF = 1m and AB =4cm ]

Now answer the following questions.

B

A

Ray-1

C

Ray-2

900

V

B

M

A

C

900

R

E

G H

F

Screen

Figure-I Figure-II

11. Choose the INCORRECT option.

(A) As per figure-I, the average refractive index of Dove Prism may be greater than 1. (B) As per figure-I, the average refractive index of Dove Prism may be greater than 2 . (C) As per figure-II, the displacement depends upon average refractive index and length

of AB. (D) As per figure-II, the displacement depends upon average refractive index and length

of EF. 12. Find the maximum value of width d of bundle of rays that can be resolved at the outlet of Dove

Prism as shown in the figure-II. (A) 32.25 10 m (B) 42.25 10 m (C) 31.125 10 m (D) 41.125 10 m





9

SECTION – C


This section contains 8 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive)

13. If the dimension of a physical quantity EB

is a b cM L T . Find the value of b c .a

Here

E magnitude of electric field B Magnitude of magnetic field Permeability of medium 14. A 200 cm length of wire weighs 0.6 gm. If the tension in the wire caused by a 500 gm mass

hanged from it. If the approximate wavelength of a wave of frequency 400 Hz sent down by it is 2n 102 m. Find the value fo n. (g = 9.8 m/s2)

15. A glass sphere of refractive index 1.5

forms the real image of object O at point I as shown in the figure when kept in air of refractive index 1. If the value of x is kR. Find k.

R

I

x

O 0 = 1.0 0 = 1.0

= 1.5

C

R





10

16. A large cylindrical container with open top contains a liquid upto height (2H = 8m) and whose density varies as shown in the figure. Atmospheric pressure is 1 atm (= 105 N/m2)

Find the pressure at bottom of cylindrical container in atm.

(given 3 3 20 10 Kg / m , and g 10m / s ).

0

O H 2H x

20

17. The figure shows a crude type of perfume atomizer. When bulb at A is compressed, air flows

swiftly through the tiny BC with uniform speed v, there by causing a reduced pressure at the position of vertical tube DE. The liquid of density 500 kg/m3, then rises in the tube, enters tube BC and sprayed out. When bulb is in natural position the air in the bulb and tube are at atmospheric pressure P0 = 105 N/m2. When bulb A is compressed, it creates an excess pressure p = 0.001 P0 inside the bulb A. Density of air is 1.3 kg/m3. If the magnitude of minimum value of speed v, required to cause the liquid to rise to tube BC is 5k m/s. Find the value of k. (g = 10 m/s2).

1

2

h = 32 mm

D A

B

3 4

E

C





11

18. A hydrogen atom in ground state, moving with speed v collides with another hydrogen atom in

ground state at rest. If n00

H

kEv v a bc 10 m / s,

m then the collision is elastic. Here a, b and

c are whole number, less then 9. Find the value of a b c .k n

4

1802

me E 13.6eV / atom 2.18 102

J/atom ionisation energy of H-atom

27Hm 1.67 10 kg Mass of hydrogen atom.

19. A soap bubble is being blown on a tube of radius 1 cm. The surface tension of

the soap solution is 0.05 N/m and the bubble makes an angle of 60 with the tube in equilibrium state as shown. If the excess of pressure over the atmospheric pressure in the tube in Pascal (Pa) is x, find x

60 60

20. A light ray is incident on a rotating mirror, which

is rotating with angular speed 3 rad/s as shown in figure. The speed of bright spot on the wall when = 30°.

Wall

L = 1 m

Incident ray





12


SECTION – A


This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which only ONE OR MORE THAN ONE is/are correct 21. Which of the following statement is/are correct?

(A) 3 2 2 4 4 2CrI KOH Cl K CrO KCl KIO H O In the balanced chemical reaction, stoichiometric coefficient of KOH is 64. (B) 4 2 7 2 2 3 22

NH Cr O N Cr O 4H O n-Factor of (NH4)2Cr2O7 is 12. (C) 3 3

2 36Fe CN Fe CO NO

n-Factor for [Fe(CN)6]3– is 60. (D) 2

4 4 2HMnO MnO MnO

Eq. wt. of 24MnO is 178.5.

22. Which of the following on reaction with H2S will produce metal sulphide? (A) COCl2 (B) CuCl2 (C) 2

4Cd CN

(D) 2

4Zn OH

23. A dilute solution contains m mol of solute A in 1 kg of a solvent with molal elevation constant Kb.

The solute dimerises in solution as 22A A Select the correct expression of equilibrium constant ‘K’ for the dimer formation. (assume molarity

= molality)

(A)

b b b

2b b

K K m TK

K m 2 T

(B)

2K2m 1

(C)

b b b

2b b

K K m TK

2 T K m

(D)

b b

b

2 K m TK

K m

24. Select the correct statement for cloud seeding: (A) It promotes rainfall. (B) It suppress fog. (C) Most commonly used chemical for cloud seeding is silver iodide because it has a crystalline

structure to that of ice. (D) It helps in snow fall.





13

25. Which of the following combination, of compounds/elements and its packing fraction is true? Assume all cations or anions are ideally fitted in the voids without any distortion in lattice.

(A) 32Na O, 1 2 0.225

3 2

(B) 32 32Cr O , 1 0.41433 2

(C) 32 3Fe O , 1 0.225

3 2

(D) 3Diamond, 1 0.4143 2

26. In which of the following ores at least one metal is in +3 oxidation state and crystal system is

orthorhombic. (A) Haematite (B) Copper pyrite (C) Diaspore (D) Aragonite 27. Which of the following IUPAC names are correct? (A) K2[OsCl5N] : Potassium pentachloridonitridoosmate(VI) (B) [Fe(NH4C5 – C5H4N)3]Cl2 : Tris(bipyridine)iron(II) chloride (C) [Fe(C5H5)2] : Bis( 5 -cyclopentadienyl)iron(II) (D) [Pt(py)4][PtCl4] : Tetrapyridineplatinum(II) tetrachloridoplatinate(II) 28. Consider the following graph and choose the correct option

Vapour pressure of BenzeneVapour pressure

of Toluene

Mole fraction of Benzene

oTP

oBP

A

B

CD

EF

(A) A represent vapour composition and B represent liquid composition. (B) At the state E of the system the vapour phase just start forming and its vapour phase

composition will correspond to the point C. (C) At the state C of the system the vapour phase just start forming and its vapour phase

composition will correspond to the point D. (D) A and B both represent vapour composition.





14


Paragraph for Question Nos. 29 to 30

A metal crystallizes in a closed packed structure, the atoms stack together occupying maximum space and leaving minimum vacant space. Each sphere in the first layer is in contact with six neighbours. Now while arranging the second layer on the first layer, spheres are placed in depression of first layer. Spheres of third layer lie in the depression of second layer. This repeating arrangement of atoms produces a giant lattice. 29. If one of the edge length of unit cell is ‘a’ what is the shortest distance between two atoms?

(A) 3a4

(B) a or4

a3

(C) a or a2

(D) a2

30. If spheres of third layer do not lie directly over the atoms of first layer, then density of unit cell is:

(Given: atomic weight of metal is 197u and a = 4.07 oA )

(A) 86.5 g/cc (B) 19.4 g/cc (C) 56.2 g/cc (D) 40.5 g/cc





15

Paragraph for Question Nos. 31 to 32

Stereoisomerism occurs when the bond connectivity is same but the spatial arrangement of more than one type is possible. Stereoisomerism finds vast use in organic chemistry, mainly biochemistry as all the biomolecules, are generally stereospecific in their actions. Stereoisomerism is limited not only to organic compounds but also to inorganic compounds, mainly in co-ordination complexes. Stereoisomers include cis and trans isomers, chiral isomers, compound with different conformation of chelate rings and other isomers that differ only in the geometry of attachment to the metal ion. 31. Select the incorrect statement: (A) trans form of [CoCl2(trien)]+ has 3 coplanar chelate ring. (B) [Pt(py)(NH3)(NO2)(Cl)(Br)(I)] has 15 geometrical isomers. (C) Carboxylic group of salicylate ion is cis and trans to tertiary nitrogen in [Co(tren)(sal)]+. (D) [Pt(NH3)2(py)2Cl2]2+ complex has 5 non chiral isomer. 32. Select the correct option for tetraamminedichlorocobalt(III) ion, the trans isomer (A) does not have any symmetry (B) has plane of symmetry (C) trans form is meso form (D) trans form exists as a pair of enantiomer





16

SECTION – C


This section contains 8 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive) 33. The enthalpy change involved in oxidation of glucose is - 2880 kJ mol-1, 25% of this energy is

available for muscular work. if 100 kJ of muscular work is needed to walk one km, the maximum distance (km) that a person will be able to walk after taking 150 g of glucose is……

34. 1 litre solution containing 4.5 millimoles of 2 7Cr O and 15 millimoles of Cr3+ shows a pH of 2.0.

The potential of the reduction half reaction is approximately x V. Then [x] is ……., where [ ] is GIF. o 2 3

2 7E of Cr O / Cr is 1.33 V 35. Assuming 100% polymerization of an organic compound in its aqueous solution, the number of

moles of organic compound undergoing polymerization containing 9.4 g of organic compound per 100 g of the solvent is……... Freezing point depression is 0.93 K, Kf of water is 1.86 K kg mol-1 and molecular weight of organic compound is 94u.

36. A gas consists of 5 molecules with a velocity of 5 ms-1, 10 molecules with a velocity of 3 ms-1 and

6 molecules with a velocity of 6 ms-1. The ratio of average velocity and rms velocity of the molecules is 9.5 × 10-x. Where x is …..





17

37. Number of S – S linkage in Marshall acid = a Number of P – P linkage in Pyrophosphoric acid = b Number of O – O linkage in Caro acid = c Number of C = C linkage in Carbon suboxide = d Find out the value of a + b + c + d?

38. In an ionic solid o

cr 1.6A and o

ar 1.864A . Find edge length of cubic unit cell in oA

39. Percentage of silver in german silver….. 40.

KCN added slowly In excess KCN precipitate dissolves

Reddish brown precipitate Yellow solutionA B C

4 6Fe CNK NaOH

Intense blue precipitateA D Red precipitate

2 4Na HPO

Yellowish white precipitateA E

How many products are correctly match? A = FeCl3 A = FeCl2 A = CuCl2 B = Fe(CN)3

B = K3Fe(CN)6 C = K3Fe(CN)6 C = K4[Fe(CN)6] D = Fe4[Fe(CN)6]3

D = Fe[Fe(CN)6] E = FePO4





18


SECTION – A


This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which only ONE OR MORE THAN ONE is/are correct 41. A is a matrix of order 3 3 and aij is its elements of ith row and jth column. If aij + ajk + aki = 0 holds

for all 1 i, j, k 3 then (A) A is a non-singular matrix (B) A is a singular matrix (C) ij

1 i, j 3a

is equal to zero (D) A is a symmetric matrix

42. If x and y satisfy the equation xy – 2x2 – 9x + 3y – 16 = 0, then (A) number of ordered pair (x, y) is 4 where x, y Z (B) number of ordered pair (x, y) is 1 where x, y N (C) if x y, x, y N then number of ordered pair is zero (D) if x y, x, y N then number of ordered pairs are two 43. Let p, q be integers and let , be the roots of the equation x2 – 2x + 3 = 0 where , if n = 0,

1, 2, 3, ..... Let an = pn + qn where n {0, 1, 2, .....} then the value of a9 is equal to (A) 3a8 – 5a7 + 3a6 (B) a7 – 6a6 (C) a7 – 12a5 + 18a4 (D) 2a8 + 3a6

44. For any two positive integers x and y f(x, y) =

1 1 1 1.....x 1 ! x 2 ! x 3 ! x y !

, then

which of the following options is/are correct

(A) 1 1 1f x, yx x! x y !

(B)

y

1lim f x, yx!

(C) 1f x, x

x 1 !

(D) f(2, 2) is equal to 2

3





19

45. Let a > 0, b > 0, c > 0 and a + b + c = 6 then 2 2 2

2 2 2

ab 1 bc 1 ca 1b c a

may be

(A) 754

(B) 35

(C) 15 (D) 10 46. A and B play a game in which they alternately call out positive integers less than or equal to n,

according to the following rules. A goes first and always calls out an odd number, B always calls out an even number and each player must call out a number which is greater than the previous number (except for As first turn). The game ends when one player cannot call out a number, then which of the following is/are correct?

(A) For n = 6, number of possible games is 8 (B) For n = 6, number of possible games is 10 (C) For n = 10, number of possible games is 55 (D) For n = 8, number of possible games is 21 47. Which of the following is/are correct? (A) The point of intersection of AB

and CD

where A(4, 7, 8), B(–1, –2, 1), C(2, 3, 4) and

D(1, 2, 5) is 3 5 9, ,2 2 2

(B) If (, , ) be a point on the plane 2x + 6y + 15z = 7 then the least value of 2 + 2 + 252 is 7 (C) The circumcentre of the triangle formed by the points (3, 2, –5), (–3, 8, –5) and (–3, 2, 1) is

the point (–1, 4, –3)

(D) The length of intercept on the line x 1 y 12 z 71 5 2

by the surface 11x2 – 5y2 + z2 = 0 is

30 units 48. Consider all 10 digit numbers formed by using all the digits 0, 1, 2, 3, ....., 9 without repetition

such that they are divisible by 11111, then (A) the digit in tens place for smallest number is 6 (B) the digit in tens place for largest number is 3 (C) total numbers of such numbers is 3456 (D) total numbers of such numbers is 4365





20


Paragraph for Question Nos. 49 to 50 Read the following write up carefully and answer the following questions: Let A(z1), B(z2), C(z3) and D(z4) be the vertices of a trapezium in an Argand plane such that AB || CD. Let

|z1 – z2| = 4, |z3 – z4| = 10 and the diagonals AC and BD intersect at P. It is given that 4 2

3 1

z zArgz z 2

and 3 2

4 1

z zArg

z z 4

49. Which of the following option(s) is/are correct?

(A) Area of the trapezium ABCD is equal to 1403

sq. units

(B) Area of the trapezium ABCD is equal to 703

sq. units

(C) Area of the triangle BCP is equal to 10021

sq. units

(D) Area of the triangle BCP is equal to 20042

sq. units

50. Which of the following option(s) is/are incorrect?

(A) value of |CP – DP| is equal to 1021

(B) PB : PD = 2 : 5

(C) PC : AC = 5 : 7 (D) PC : AC = 2 : 3





21

Paragraph for Question Nos. 51 to 52 Read the following write up carefully and answer the following questions: In the equation A + B + C + D + E = FG, where FG is the two digit number whose value is 10F + G and letters A, B, C, D, E, F and G each represents different digits and FG is as large as possible 51. A five digit number is made using digits A, B, C, D, E, F, G (repetition not allowed), then which of

the following is incorrect

(A) probability that number made is divisible by 5 is 17

(B) probability that number made is divisible by 3 is 27

(C) probability that number made is divisible by 4 is 27

(D) probability that number made is divisible by 4 is 17

52. If a five digit number is made using the digits A, B, C, D, E without repetition then

(A) the probability that the number is divisible by 11 is 25

(B) the probability that the number is divisible by 11 is 15

(C) the probability that the number is divisible by 4 is 14

(D) the probability that the number is divisible by 4 is 25

SECTION – C


This section contains 8 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive) 53. If a, b, c are three positive real numbers then the minimum value of

a 3c 4b 8ca 2b c a b 2c a b 3c

is 2 (where a, b Z), then | + | is equal to _____





22

54. If the quadratic equation a1x2 – a2x + a3 = 0 where a1, a2, a3 N has two distinct real roots belonging to the interval (1, 2), then least value of a1 is _____

55. Let a1, a2, a3, a4, a5 be five terms of an increasing geometric progression such that a1, a2, a3, a4,

a5 N, and a5 < 100. Then the number of possible geometric progression (is/are) _____

56. If the coefficient of 2n n 18

2x

in (x – 1)(x2 – 2)(x3 – 3)(x4 – 4) ..... (xn – n), where (n 8) is k, then k is equal to _____

57. If a, b, c are three positive numbers then the minimum value of

4 6 8

3 2

a b cab c 2 2 is equal to _____

58. Let k when k = 0, 1, 2, 3, 4, ....., 253 are 254th roots of unity then the unit digit of 253 2k 2

k 0z z

is

_____ (where 2i7z e

)

59. Let A = [aij]3 3 be a matrix such that AAT = 4I and aij + 2cij = 0 where cij is the cofactor of aij i & j,

I is the unit matrix of order 3 and AT is the transpose of the matrix A.

If 11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

a 4 a a a 1 a aa a 4 a 5 a a 1 a 0a a a 4 a a a 1

, then a

b where a and b are

coprime positive integers then the value of a + b is _____ 60. If x3 – ax2 + bx + c = 0 has the roots 2 + 3 + 4, 2 + 3 + 4 and 2 + 3 + 4 where , , are the

roots of x3 – x2 – 1 = 0 then the value of a + b + c is equal to _____





1 1


PART TEST – III PAPER-2


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

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

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

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

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

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

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

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

9. C 29. C 49. A

10. B 30. B 50. D

11. D 31. D 51. D

12. B 32. B 52. C

13. 3 33. 6 53. 5

14. 5 34. 1 54. 5

15. 5 35. 2 55. 8

16. 2 36. 1 56. 0

17. 4 37. 3 57. 1

18. 6 38. 4 58. 8

19. 5 39. 0 59. 7

20. 8 40. 5 60. 6

ALL

IND

IA T

ES

T S

ER

IES

FIITJEE JEE(Advanced)-2018


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


2 2


SECTION – A

1. (A) 1(not possible)R R

The higher the refractive index of the material of sphere (), the closer one could come to fulfilling the objective

S P 1 = 1 2 =

(B) w

a aw

a w

ff

3 41 1 3 12 3 2

3 4 6 4 412 3

(C) basic (NCERT-point to ponder) (D) basic (NCERT-point to ponder) 2. As we know that,

hC K

s12400 v

, where is in electron Volt and is in Å

When light of wavelength = 6000 Å is used

12400 0.36000

= 1.77 1.8 eV (approximately)

When light with wavelength = 4000 Å is used

12400 eV4000

eV = (3.1 1.8) e volt = 1.3 eV v = 1.3 Volt 3. Condition for maximum intensity due to thin film

2 d n2

2d = 12p 12

(2p + 1)1 = 4 d

Here 1 = 400 nm and n = p An integer Let = 1150 nm. Since only two wavelength give maximum intensity, so

4 d 4 d2(p 1) 1 2(p 2) 1

1 1(2p 1) (2p 1)2(p 1) 1 2(p 2) 1

1(2p 1)2(p 1) 1

or 1(2p 1)2(p 2) 1

1

1

1 1 1150 400p 1......2 2 1150 400

, or




3 3

1

1

31 1 3 1150 400p 2......2 2 1150 400

The only integer which satisfies both inequalities is 2, so

9(2 2 1)400 10d 500

4 1

nm

9

24 d 4 1 500 10 666.7

2(p 1) 1 2 1

nm

9

6 2d 500 10T 3.1h 8 10 2 10

C

4. k = stiffness of string 3

2max max

F mg 40 10 9.8 1.96 N / mx x 20 10

When we replace the mass 40 gm by the mass 22205m gm

32 , we have

m 3T 2k 8

sec

2 k 16T m 3

rad/sec

x = x0 cos t = 0.2 cos 16 t3

v = x0sin t = 16 16sin t15 3

a = x0 2cos t = 256 16cos t45 3

If x = 10 cm 16 1cos t cos3 2 3

16 t (2n )3 3

3nt sec8 8

A

B

2

5. L.C. = 1 0.1 0.005 cm20

Let 1 V.S.D. = x

1 1 x x 0.95 mm20

n 4 1 n 0.95 n 80

Length of pencil = 808.4 0.005 8.48 cm5

6. 2

2 0y a sin( t kx) 1 0y a sin( t kx ) and 3 0y a sin t kx )




4 4

0 0 0A | a 2a cos | a | (1 2cos ) |

02 xdA a | (1 2cos |

D

02a | (1 2cos |

a0

a0

a0 x

y

Second Method : Suppose and represent the phase difference and path difference between

two connective waves so Phasor representation Resultant amplitude 0 0 A

a0 a0 a0

A = 3a0

60 6

A

a0

a0

a0

60

60

A = 2a0

120 23 6

120

a0

a0

a0

A = 0

180 32 6

A

a0

a0

a0

A = a0

240 2 43 6

240

a0

a0

a0

A = 0




5 5

300 56

A

a0

300

300

A = 2a0

360 66

A

a0 a0 a0

3a0

A = 3a0

7. For transverse wave, TF Fv

S

Al 3 6 3

900 30V 600 m / s2.5 10 10 50 10

Cu 3 6 3

900 30 1000V m / s38.1 10 10 90 10

Al Cur

i Al Cu

800V Va 23 0.3

2800a V V 73

For longitudinal wave YV

10 5

Al 3

9 10 3 10V 6000 m / s502.5 10

10 5

Cu 3

16 10 4 10 40000V m / s9 10 98.1 10

Al Cur

i Al Cu

1400V Va 79 0.15

94000a V V 479

8. B1 B22mg F F

3wood water

3 750kg / m4

Taking torque about O1 B2N R mg F 2Rcos30

145 10 1.732 259.8 260N3

FB1

FB2

N

mg

mg

30° O2 N

O1




6 6

9. During one time period (T0)

0T2 2

0

0

T y xW 1 cos 2 t dt2v v

Average Power = 2 2

0

0

y TWT 2v

10. Since the work done will be same at any place along the string in one cycle so choose x = 0, for

convenience. Then work done in a cycle = 2 /2 2 2

20 0

0

y T y Tcos tdt

v v

Average power = work done per unit time = work done in a cycleTime period

=

20 2 2

0

y Ty Tv

2 2v

Average power = 2 2 2 2 2 20 02 f y T 2 f y T

T

4 42 10 200 200 0.2 0.2 10 5 10 5 = 160 103 Watt = 0.16 Watt.

11.-12. In figure-I, total internal reflection takes place at face AB. EMD MDE BEM

90 45

1sin 45

D B

M

A

N

E

Ray-1

C

450 Ray-2

900

2sin cos

……………….( 1)

According to law of refraction at point M, we can write

1 sin 45 sin 1sin2

2

1cos 12

Putting these value in equation ( 1), we have

2

1 1 2122

22 1 1 1

In figure-II: Replace the Dove Prism, by glass cube of side as AB.

D B

M

A

E

+

V

C

450

900

R

-

B

M

A

V

C

900

R

+

L




7 7

2L

……………..( 2)

From geometry, we can write

1sin2R

and

1sin2R R

tan tanL a a = 2 22 1 2 1R R

a a

2

22

2 11

2 12 1R

R R

aL

2

22

2 11

2 12 1R

R R

a

2

222

2 11

2 4 2 12 1R

RR R

aL

Neglecting the value 2 , we have

12

22 2

2

411 1 14 2 12 1 2 112 1

RR R

R

R

R

a aL

Expand binomially, we have

32 2

2

2 1R

RaL

Putting this value in equation ( 2), we have

32 2

2

2 1R

Ra

Second method:

dL a tand

2L asec

Also R

1sin2

2R

1cos2

2

2R

asecL2 cos

R

R3

2 2

2aμ ΔμΔL=2μ -1

Third method:

r v

L 1 1, sin , sin2 2 2

L a tan atan




8 8

2 2r v

1 1a2 1 2 1

r

1 1a2 2 1

1/2r

a 1 12

ra 11 12 2

ra4

rL a 4cm 5 2

1002 4 2 4 2 2

45 cm 5 10 m100

42.25 10 m Since: d 4d 2.25 10 m

22v r2 1 2 1

22r r2 2 4 1

2r4 4 0

r2 1

a

SECTION – C

13. 2EB B C energy

density speed

= Energy per unit volume speed

= Energy speedvolume

2 2

1 0 33

ML T LT ML TL




9 9

Here a = 1, b = 0 c = 3 b c 0 ( 3) 3a 1

14. = mass per unit length = 3m 0.6 10

2

= 3 104 kg/m

T = mg = 0.5 9.8 = 4.9 N Tension in the string

v = speed of wave in string = T

= 44.9 128

3 10

m/s

wavelength of the wave = 2v 128 32 10f 400

m = 25 102 m

n = 5 15. We can reverse the question. Let I behaves as object and O behaves as image, because

retracing the path will not alter the distances. Consider first refraction at point N

1.5 1.0 1.5 1.0 1.5 1 1 v 3Rv R R v 2R R

I1 3R P2 P1 R I x O

0 = 1.0

0 = 1.0

= 1.5

N M

C

R

Consider second refraction at point M, for this I, will behaves as normal object so

1.0 1.5 1.0 1.5v 5R R

1 3 1 5v 10R 2R 10R

1 5 3 1v 10R 10R 5R

v 5R x 5R k 5 16. We have to calculate the partial pressure, generated due the liquid of variable density

00y

H

W = weight of liquid column above the point A

H H

0y y

yW S gdy gS 1 dyH

2 20

0

S H y 3H y gH ygs H y2H 2H




10 10

P1 = Partial pressure at interface (y = 0) due to liquid with

variable density = y 0

WS

03 gH2

P Pressure at bottom = 00 1 0 0

5 gHP P gH

2

3

5 55 10 10 410 2 10 2 atm2

x H

2H

y

0

17. Use Bernoulli’s equation at point-3 and point-4, we can write

20 BC

air air

P P P 1 v2

PBC = P0 + P 0.65 v2

Using Bernoulli’s equation at point -1 and point -2, we can write

21 2liquid

liquid liquid

P P1 v2

BC iquid 2 0

iquid

P gh P1 v2

2

20 0P P gh 0.65v P1 v2

2 21 v 0.65v P gh 02

2 P ghv0.65

minP gh 100v 200.65 5

m/s

18. According to Bohr’s model

00 min2

3E1E E 1 E4n

During inelastic collision, a part of kinetic energy of colliding particles is converted into internal energy. The internal energy of the system of two hydrogen atoms, considered in the problem cannot be changed less than Emin. It means if the change in kinetic energy of system in ground frame is less than Emin (or if the kinetic energy of colliding atoms with respect to their centre of mass is less than Emin), then collision must be an elastic one. Hence considering the critical case

2

0 0 0H 0

H

v 3E 3E1 m 2 v2 2 4 m

18

4 427

3 2.18 10 39.1617 10 6.257 10 m / s1.67 10

46.26 10 m / s

a b c 6 2 6 6k n 3 4




11 11

19. r 1cosR 2

24 rTcos r P

20. Ytan 2 90L

y Lcot 2

2dy dL cosec 2 2dt dt

2 d2L cosec 2dt

Wall

L = 1 m

P

vsin2

v




12 12


SECTION – A

21. (A) 3 2 2 4 4 22CrI 27Cl 64KOH 2K CrO 54KCl 6KIO 32H O (B) n-Factor of (NH4)2Cr2O7 is 6.

(C) 3 2 3 4 5

32 3

6

Fe CN Fe CO NO

n-Factor = 60.

(D) 6 7 4

24 4 2

2n Factor 3

MnO H MnO MnO

Hence, eq. wt. of 24

119 3MnO 178.52

22. B. 2 2CuCl H S CuS 2HCl

C. 2

24Cd CN H S CdS 2H 4CN

D. 2

2 24Zn OH H S ZnS 2OH 2H O

23. 2

eq

2A Amoles at t 0 m 0

mmoles at t m m2

Total moles = mm m 12 2

b b bT iK m K m 12

b

b

T1

2 K m

b b

b

2 K m TK m

Equilibrium constant K for the dimer formation is

22 2

A mKA 2 m 1

2K

2m 1

2

b b b b

b b

2 K m T 2 K m TK 2m 1

K m K m

b b b

2b b b

K K m TK

K m 2K m 2 T

b b b

2b b

K K m TK

2 T K m




13 13

24. Cloud seeding is a form of weather modification, a way changing the amount or type of precipitation that falls from clouds by dispersing chemicals into the air.

25. 32 32Fe O , 1 0.41433 2

3

3

4 rDiamond, 83 2 2r

26. Diaspore is an aluminium oxide hydroxide mineral crystallizing in the orthorhombic system. 28. A represent liquid composition and B represent vapour composition. 29. Above packing represent two possible arrangements, hexagonal close packing or cubic close

packing, hence minimum distance, a or a2

.

30. Above information represent cubic close packing

3 24 23

197 44.07 10 6.02 10

19.41 g / cc 31. (A)

Co

NN

NN

Cl

Cl

trans form has 3 coplanar ring

(C)

COO trans to tertiary N

OC

O

CoN N

ON

N

COO cis to tertiary NCoN N

ON

N

C OO

[Pt(NH3)2(py)2Cl2]2+ has 4 non chiral isomer.




14 14

32.

Co

H3N NH3

NH3H3N

Cl

Cltrans achiral

It has axis of symmetry and plane of symmetry

Co

H3N Cl

NH3H3N

Cl

NH3cis achiral

SECTION – C

33. The energy involved during oxidation of glucose is 2880 150 2400 kJ180

Energy available for muscular work 2400 600kJ4

Energy required for 1 km walk = 100 kJ Therefore the distance that a person will be able to walk taking 150 g glucose = 6 km. 34. 3

2 7 214H Cr O 6e 2Cr 7H O

3 32 7 2 7

23o

10 14Cr O /Cr Cr O /Cr2 7

Cr0.059E E log6 Cr O H

32 7

142

10 2Cr O /Cr

4.5 100.059 1000E 1.33 log

6 151000

31

10 4

0.059 4.5 101.33 log6 2.25 10

2710

0.0591.33 log 2 106

= 1.06 = 1 V 35. Assume organic compound is A and n molecules of A polymerise f fT iK m

f

f

T 0.93 94 100iK m 1.86 9.4 1000

1i 0.5n

n = 2

36. Average velocity 5 5 10 3 6 6 4.3321

rms velocity 5 25 10 9 6 36 4.5321




15 15

1av

rms

V0.955 9.5 10

V

x = 1 37.

OH S

O

O

O O S

O

O

OH Marshall acid a = 0

Pyrophosphoric acid b = 0P

O

OH

OHOH P

OH

O

O

Caro acid c = 1O HOH S

O

O

O

O C C C O Carbon suboxide d = 2

38. c

a

r0.858

r

From above ratio it is clear that cation occupy cubic void in simple cubic unit cell, hence c a3a 2 r r

a = 4 40. 3 3A B

FeCl 3KCN Fe CN 3KCl

33 6C

Fe CN 3KCN K Fe CN

3 4 46 6 3D

4FeCl 3K Fe CN Fe Fe CN 12KCl

4 46 3 63Fe Fe CN 12NaOH 4Fe OH 3Na Fe CN

3 2 4 4E

FeCl Na HPO FePO 2NaCl HCl




16 16


SECTION – A

41. Put i = j = k aij = 0 and put k = i aij = –aji So, matrix is skew symmetric of odd order

42. 7y 2x 3

x 3

x, y I; x + 3 = 1, 7 (x, y) (–2, 6), (–4, –12), (4, 12), (–10, –18) 43. n 2 n 1 na 2a 3a

44.

x x i 1 1 1x i ! x i ! x i 1 ! x i !

y

i 1

1 1 1xx i ! x! x y !

..... (1)

y

1 1lim f x, yxx! x!

..... (2)

1 1 1 x 1f x, x .....

x 1 ! x 2 ! x x ! x! x 1 !

45. A.M. H.M. for 1a

, 1b

, 1c

, we get

1 1 13a b c

3 6

1 1 1 3a b c 2

Now,

2 2 2 2 21 1 1 1 1 1 3a b c a b c 6b c a b c a 23 3 3

2 2 21 1 1 75a b c

b c a 4

46. Pn = Pn – 1 + Pn – 2, P1 = 1, P2 = 1 P3 = 1 + 1 = 2 P4 = 2 + 1 = 3 P5 = 5 P6 = 8 P7 = 13 P8 = 21 P9 = 34 P10 = 55 47. (A) Intersection point is mid-point of AB and CD (B) 2 + 6 + 15 = 7




17 17

2 2 2ˆ ˆ ˆ ˆ ˆ ˆ2i 6 j 3k i j 5 k 7 25

2 2 22 6 15 7 25 2 + 2 + 252 1 (C) Triangle is equilateral (D) Intersection points are (1, 2, 3) and (2, –3, 1) 48. Let N = x1x2x3 ..... x10 be one of such number ix 45 , hence N is divisible by 9. Hence N should by divisible by 11111 9 = 99999 Now x1x2x3 ..... x10 = x1x2x3x4x5 105 + x6x7x8x9x10 = x1x2x3x4x5 99999 + (x1x2x3x4x5) + (x6x7x8x9x10) Now x1x2x3x4x5 < 99999 x6x7x8x9x10 < 99999 x1x2x3x4x5 + x6x7x8x9x10 < 2 99999 x1x2x3x4x5 + x6x7x8x9x10 = 99999 x1 + x6 = x2 + x7 = x3 + x8 = x4 + x9 = x5 + x10 = 9 So, smallest number 1023489765 Largest number 9876501234 Total number 9 8 6 4 2 1 1 1 1 1 = 3456 49.-50. ABP and CDP are similar if AP = 2x, BP = 2y then CP = 5x, DP = 5y

Area of trapezium 49ABCD xy2

2x 2ytan , tan5y 5x

also + = 45º

2 210xy x y21

(z1)A B(z2)

Q

4 2x 2y

5x 5y

10 (z4)D C(z3)

Also AB2 = AP2 + BP2 x2 + y2 = 4 40xy21

Ar(PCD) = 5xy 51. 9 + 8 + 6 + 5 + 4 = 32 Set P(A , B, C, D, E, F, G) = (2, 3, 4, 5, 6, 8, 9) n(S) = 7C5 5! Total numbers divisible by 5 = 6C4 4! Total numbers divisible by 3 = 3C1 5! + 3C2 5! = 6! Total numbers divisible by 4 = 720 52. 9, 8, 6, 5, 4 (A and B) ..... 9 + 8 + 6 + 5 + 4 = 32 |(9 + 8) – (6 + 5 + 4)| = 2 not divisible by 11 |(9 + 6) – (8 + 5 + 4)| = 3 not divisible by 11 |(9 + 5) – (8 + 6 + 4)| = 4 not divisible by 11 |(9 + 4) – (8 + 6 + 5)| = 6 not divisible by 11 |(8 + 6) – (9 + 5 + 4)| = 4 not divisible by 11 |(8 + 5) – (9 + 6 + 4)| = 6 not divisible by 11 |(8 + 4) – (9 + 6 + 5)| = 8 not divisible by 11 |(6 + 5) – (9 + 8 + 4)| = 10 not divisible by 11 |(6 + 4) – (9 + 8 + 5)| = 12 not divisible by 11 |(5 + 4) – (9 + 8 + 6)| = 14 not divisible by 11 (C) ..... 4 8




18 18

5 6 6 4 8 4 9 6

Probability = 5 3! 15! 4

SECTION – C

53. Let a + 2b + c = x, a + b + 2c = y, a + b + 3c = z a = –x + 5y – 3z, b = z + x – 2y, c = z – y Apply A.M. G.M.

a 3c 4b 8ca 2b c a b 2c a b 3c

17 12 2

54. 2

1

aa

, 3

1

aa

, , (1, 2)

– 1, – 1, 2 – , 2 – (0, 1)

A.M. G.M.

1 21 2

2

11 24

similarly ( – 1)(2 – ) 14

10 1 2 1 216

(Both can’t equal to 14

simultaneously ( )

3 32 2

1 1 1 1

a aa 2a 10 1 4a a a a 16

1 2 3 1 2 3

21

a a a 4a 2a a 1016a

21

1 2 3 1 2 3a

a a a 4a 2a a16

21a

f 1 f 216

2a 1

16

Least integral value of a is 5

55. Let nm

be the common ratio of G.P., where n and m are coprime integers n > m

4

5 1 4na am

so let a1 = km4 (k I+)

G.P. km4, km3n, km2n2, kmn3, kn4 kn4 < 100 n can’t be greater than or equal to 4 Case-I: n = 3 and m = 2 a5 = kn4 < 100

100k81

k = 1

G.P. (16, 24, 36, 54, 81) Case-II: n = 3 and m = 1 a5 = 81k < 100 k = 1 G.P. (1, 3, 9, 27, 81)




19 19

Case-III: n = 2, m = 1

100k16

k = 1, 2, 3, ..... 6 G.P.’s (1, 2, 4, 8, 16), (2, 4, 8, 16, 32), (3, 6, 12, 24, 48), (4, 8, 16, 32, 64), (6, 12, 24, 48, 96), (5, 10, 20, 40, 80)

56. Highest power of n n 1

x m2

(let) have to find coefficient of xm – 9

–9 + (–8 –1) + (–7 –2) + (–6 –3) + (–5 –4) + (–6 –2 –1) + (–5 –3 –1) + (–4 –3 –2) = 0

57.

4 4 6 6 6 6 8 8 18 24 16 8

12

a a b b b b c ca b c2 2 4 4 4 4 2 2

8 2

Equality hold for a = 1, b = 21/6, c = 1 58. |z + k z2|2 = |z|2 + |k| |z2|2 + 0 59. AAT = 4I |A| = 2

T 1 4Adj A

A 4AA

11 21 31 11 21 31

12 22 32 12 22 32

13 23 33 13 23 33

a a a c c c4a a a c c cA

a a a c c c

Now, aij = 4A

cij = –2cij |A| = –2

Now, |A + 4I| = |A + AAT| = |A| |I + AT| = –2|(I + A)T| = –2|I + A| so, A 4I

2 5A I

25

60. x3 – x2 – 1 = 0 , , 2 + 2 + 2 = 1 2 + 3 + 4 = 2 + 2 + 1 + 3 + = 3 + In given equation x x – 3 x3 – 10x2 + 33x – 37 = 0


part test – iii

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