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FIITJEE Solutions to JEE(Main)-2020

PHYSICS, CHEMISTRY & MATHEMATICS

Time Allotted: 3 Hours

Maximum Marks: 300

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

Important Instructions:

1. The test is of 3 hours duration. 2. This Test Paper consists of 75 questions. The maximum marks are 300. 3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics

having 25 questions in each part of equal weightage out of which 20 questions are MCQs and 5 questions are numerical value based. Each question is allotted 4 (four) marks for correct response.

4. (Q. No. 01 – 20, 26 – 45, 51 – 70) contains 60 multiple choice questions which have only one correct

answer. Each question carries +4 marks for correct answer and –1 mark for wrong answer. 5. (Q. No. 21 – 25, 46 – 50, 71 – 75) contains 15 Numerical based questions with answer as numerical

value. Each question carries +4 marks for correct answer. There is no negative marking. 6. Candidates will be awarded marks as stated above in instruction No.3 for correct response of each

question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.

7. There is only one correct response for each question. Marked up more than one response in any

question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.

Paper - 1

Test Date: 5th September 2020 (First Shift)

http://www.fiitjee.com.

JEE-MAIN-2020 (5th September-First Shift)-PCM-2


PART –A (PHYSICS) 1. A hollow spherical shell at outer radius R floats just submerged under the water surface.

The inner radius of the shell is r. If the specific gravity of the shell material is 278

w.r.t

water, the value of r is.

(A) 2R3

(B) 4 R9

(C) 1R3

(D) 8 R9

2. A balloon is moving up in air vertically above a

point A on the ground. When it is at a height h1, a girl standing at a distance d (point B) from A (see figure) sees it at the angle 450 with respect to the vertical. When the balloon climbs up a future height h2, it is seen at the angle 600 with respect to the vertical if the girl moves further by a distance 2.464 d (point C). Then the height h2 is (given tan 300 = 0.5774):

(A) 0.464 d (B) 0.732 d (C) 1.464 d (D) d

3. A physical quantity z depends on four observables a, c, c and d, as z = 2

2 3

3

a bc d

. The

percentages of error in the measurement of a, b, c and d are 2%, 1.5%, 4% and 2.5% respectively. The percentage of error in z is:

(A) 13.5% (B) 16.5% (C) 12.25% (D) 14.5% 4. Three different processes that can occur in an ideal

monoatomic gas are shown in the P vs V diagram. The paths are labeled as A B, A C and A D. The change in internal energies during these process are taken as EAB, EAC and EAD and the work done by WAB, WAC and WAD. The correct relation between these parameters are:

(A) AB AC AD AB AC ADE E E ,W 0,W W (B) AB AC AD AB AC ADE E E ,W 0,W 0,W 0 (C) AB AC AD AB AC ADE E E ,W W W (D) AB AC AD AB AC ADE E E ,W 0,W 0,W 0 5. A wheel is rotating freely with an angular speed on a shaft. The moment of inertia of

the wheel is I and the moment of inertia of the shaft is negligible. Another wheel of moment of inertia 3I initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is:

(A) 34

(B) 56

(C) 14

(D) 0




6. In a resonance tube experiment when the tube is filled with water up to a height of 17.0 cm from bottom, it resonates with a given tuning fork. When the water level is raised the next resonance with the same tuning fork occurs at a height of 24.5 cm. If the velocity of sound in air is 330 m/s, the tuning fork frequency is:

(A) 1100 Hz (B) 2200 Hz (C) 3300 Hz (D) 550 Hz 7. An electron is constrained to move along the y – axis with a speed of 0.1 c (c is the

speed of light) in the presence of electromagnetic wave, whose electric field is 7 2ˆE 30 jsin 1.5 10 t 5 10 x V / m.

The maximum magnetic force experienced by

the electron will be: (given 8 1c 3 10 ms and electron charge = 191.6 10 C ) (A) 183.2 10 N (B) 191.5 10 N (C) 182.4 10 N (D) 194.8 10 N 8. With increasing biasing voltage of a photodiode, the photocurrent magnitude: (A) remains constant (B) increases initially and after attaining (C) increases linearly (D) increases initially and saturates finally 9. A bullet of mass 5 g, traveling with a speed of 210 m/s, strikes a fixed wooden target.

One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0 70.030 cal / g C 1cal 4.2 10 ergs close to:

(A) 83.30 C (B) 38.40 C (C) 87.50 C (D) 119.20 C

10. The value of the acceleration due to gravity is g1 at a height Rh2

(R= radius of the

earth) from the surface of the earth. It is again equal to g1 at a depth d below the surface

of the earth. The ratio dR

equals:

(A) 79

(B) 13

(C) 59

(D) 49




11. For a concave lens of focal length f, the relation between object and image distances u and is , respectively, from its pole can best be represented by (u - is the reference line):

(A)

(B)

(C)

(D)

12. A solid sphere of radius R carries a charge Q + q

distributed uniformly over its volume. A very small point like piece of it of mass m gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge q. If it acquires a speed when it has fallen through a vertical height h (see figure), then: (assume the remaining portion to be spherical)

(A) 22

0

qQv y g4 R ym

(B)

2

0

qQv 2y g4 R R y m

(C)

2

0

qQv y g4 R R y m

(D)

23

0

QqRv 2y g4 R y m

13. A galvanometer of resistance G is converted into a voltmeter of range 0 – IV by

connecting a resistance R1 in series with it. The additional resistance that should be connected in series with R1 to increase the range of the voltmeter to 0 – 2V will be:

(A) R1 + G (B) R1 (C) G (D) R1 – G 14. Number of molecules in a volume of 4 cm3 of a perfect monoatomic gas at some

temperature T and at a pressure of 2 cm of mercury is close to? (Given, mean kinetic energy of a molecule (at T) is 4 x 10–14 erg, g = 980 cm / s2, density of mercury = 13.6 g / cm3)

(A) 185.8 10 (B) 164.0 10 (C) 165.8 10 (D) 184.0 10




15. A square loop of side 2a, and carrying current I, is kept in XZ plane with its centre at origin. A long wire carrying the same current I is placed parallel to the z – axis and passing through the point (0, b, 0), (b > > a). The magnitude of the torque on the loop about z – axis is given by:

(A) 2 3

02

2 I ab

(B) 2 2

0I a2 b

(C) 2 3

02

I a2 b

(D) 2 2

02 I ab

16. Assume that the displacement (s) of air is proportional to the pressure difference p

created by a sound wave. Displacement (s) further depends on the speed of sound (v), density of air () and the frequency (f). If p 10Pa, 300 m/ s, 1 lg / m3 and f 1000 Hz , then s will be of the order of (take the multiplicative constant to be 1)

(A) 1 mm

10 (B) 10 mm

(C) 3 mm

100 (D) 1 mm

17. A helicopter rises from rest on the ground vertically upwards with a constant acceleration

g. A food packet is dropped from the helicopter when it is at a height h. The time taken by the packet to reach the ground is close to [g is the acceleration due to gravity]:

(A) ht 1.8g

(B) 2ht3g

(C) ht 3.4g

(D) 2 ht

3 g

18. Activities of three radioactive substances A, B and C are represented by the curves A, B and C in the figure. Then their half lives 1 1 1

2 2 2

T A : T B : T C are in the radio:

(A) 3:2:1 (B) 2:1:3 (C) 4:3:1 (D) 2:1:1

19. An electrical power line, having a total resistance of 2, delivers 1 kW at 220 V. The

efficiency of the transmission line is approximately: (A) 96% (B) 72% (C) 91% (D) 85% 20. Two capacitors of capacitances C and 2C are charged to potential differences V and 2V,

respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is:

(A) 23 CV2

(B) 29 CV2

(C) 225 CV6

(D) zero




21. A force ˆ ˆ ˆF i 2 j 3k

N acts at a point ˆ ˆ ˆ4i 3 j k m. Then the magnitude of torque

about the point ˆ ˆ ˆi 2 j k m will be x N-m. The value of x is __________. 22. A particle of mass 200 MeV/c2 collides with a hydrogen atom at rest. Soon after the

collision the particle comes to rest, and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in eV) is N

4. The value of N is: (Given the

mass of the hydrogen atom to be 1 GeV / c2) __________.

23. Two concentric circular coils, C1 and C2 are placed in the XY plane. C1 has 500 turns, and a radius of 1 cm. C2 has 200 turns and radius of 20 cm. C2 carries a time dependent current 2I t 5t 2t 3 A where t is in s. The emf induced in C1 (in mV), at the

instant t = 1 s is 4x

. The value of x is __________.

24. A beam of electrons of energy E scatters from a target having atomic spacing of 0

IA . The first maximum intensity occurs at 060 . Then E (in eV) is __________.

(Planck constant h = electron mass m = 319.1 10 kg )

25. A compound microscope consists of an objective lens of focal length 1 cm and an eye piece of focal length 5 cm with a separation of 10 cm. The distance between an object and the objective lens, at which the strain on the eye is minimum is n

40 cm. The value of

n is __________.




PART –B (CHEMISTRY) 26. The structure of PCI5 in the solid state is: (A) square planar [PCI4]+ and octahedral [PCI6]– (B) trigonal biphyramidal (C) tetrahedral [PCI4]+ and octahedral [PCI6] (D) square pyramidal 27. In the sixty period, the orbitals that are filled are: (A) 6s, 5f, 6d, 6p (B) 6s, 6p, 6d, 6f (C) 6s, 4f, 5d, 6p (D) 6s, 5d, 5f, 6p 28. The increasing order of basicity of the following compounds is:

(A) (A) < (B) < (C) < (D) (B) (D) > (A) > (B) < (D) (C) (B) < (A) < (C) < (D) (D) (B) < (A) < (D) < (C) 29. Identify the correct molecular picture showing what happens at the critical micellar

concentration (CMC) of an aqueous solution of a surfactant

(A) (B) (B) (A) (C) (C) (D) (D) 30. Which of the following is not an essential amino acid? (A) Valine (B) Tyrosine (C) Leucine (D) Lysine 31. A flask contains a mixture of compounds A and B. Both compounds decompose by first

– order kinetics. The half lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the line required for the concentration of A to be four times that of B (in s) is : (Use In 2 = 0.693)

(A) 120 (B) 300 (C) 900 (D) 180




32. The equation that represents the water – gas shift reactions is: (A)

(B)

(C)

(D)

33. In the following reaction sequence the major products A and B are:

(A)

(B)

(C)

(D)

34. A diatomic molecule X2 has a body centered cubic (bcc). The density of the molecule is

6.17 g cm–3. The number of molecules present in 200 g of X2 is: (Avogadro constant (NA) = 6 x 1023 mol–1) (A) 2 NA (B) 8 NA (C) 14 NA (D) 4 NA 35. If a person is suffering from the deficiency of nor – adrenaline, what kind of drug can be

suggested? (A) Antidepressant (B) Antishistamine (C) Analgesic (D) Anti – inflammatory 36. The potential energy curve for the H2 molecule as a function of internuclear distance is :

(A)

(B)

(C)

(D)




37. Consider the following reaction:

For each of the following cases (a, b) the direction in which the equilibrium shifts is:

(a) Temperature is decreased (b) Pressure is increased by adding N2 at constant T (A) (a) towards reactant, (b) towards product (B) (a) towards reactant, (b) no change (C) (a) towards product, (b) no change (D) (a) towards reactant 38. The most appropriate reagent for conversion of C2H5CN into CH3CH2CH2NH2 is: (A) CaH2 (B) Na(CN)BH3 (C) LiAIH4 (D) NaBH4 39. Which of the following derivates of alcohol is unstable is an aqueous base?

(A)

(B) RO – CMe3

(C)

(D)

40. The condition that indicates a polluted environment is: (A) 0.03% of CO2 in the atmosphere (B) pH of rain water to be 5.6 (C) eutrophication (D) BOD value of 5 ppm 41. The values of the crystal field stabilization energies for a high spin d6 metal ion in

octahedral and tetrahedral fields, respectively, are:

(A) (B)

(C) (D)

42. The difference between the radii of 3rd and 4th orbits of Li2+ in R1.. The difference between the radii of 3rd and 4th orbits of He+ is R2. Ratio R1 : R2 is:

(A) 8:3 (B) 2:3 (C) 3:8 (D) 3:2 43. The increasing order of the acidity of the -hydrogen of the following compounds is:

(A) (C) < (A) < (B) < (D) (B) (B) < (C) < (A) < (D) (C) (D) < (C) < (A) < (B) (D) (A) < (C) < (D) < (B) 44. The correct electronic configuration and spin – only magnetic moment (BM) of Gd3+ (Z =

64), respectively, are: (A) 7Xe 4f and 7.9 (B) 7Xe 4f and 8.9

(C) 7Xe 5f and 8.9 (D) 7Xe 5f and 7.9




45. An Ellingham diagram provides information about: 1. the condition of pH and potential under which a species is thermodynamically stable. 2. the standard Gibbs energies of formation of some metal oxides 3. the kinetics of the reduction process 4. standard electrode potentials of reduction reactions involved in the extraction of

metals 46. The total number of coordination sites in ethylenediaminetetraacetate (EDTA4-) is ____. 47. The minimum number of moles of O2 required for complete combustion of 1 mole of

propane and 2 moles of butane is ___ 48. An oxidation – reduction in which 3 electrons are transferred has a G0 of 17.37 kJ mol–1

at 250 C. The value of 0cellE (in V is ____ 10–2)

(1 F = 96,500 C mol–1) 49. The number of chiral carbon(s) present in peptide, Iie-Arg-Pro, is ___ 50. A soft drink was bottled with a partial pressure of CO2 of 3 bar over the liquid at room

temperature. The partial pressure of CO2 over the solution approaches a value of 30 bar when 44 g of CO2 is dissolved in 1 kg of water at room temperature. The approximate pH of the soft drink is ___ x 10–1/

(First dissociation constant of 72 3H CO 4.0 10 ;log 2 0.3; density of the soft drink =

1 g mL–1)




PART–C (MATHEMATICS) 51. If the four complex numbers, z,z,z 2Re z and z – 2Re(z) represent the vertices of a

square of side 4 units in the Argand plane, then z is equal to:

(A) 2 2 (B) 2 (C) 4 2 (D) 4 52. The negation of the Boolean expression x y is equivalent to (A) x y v x y (B) x y v x y

(C) x y xv y (D) x y v x y 53. If 10 9 1 8 2 9 10 112 2 .3 2 .3 ... 2.3 3 S 2 , then S is equal to

(A) 11

103 22

(B) 11 123 2

(C) 112.3 (D) 113 54. Let R. The system of linear equations. 1 2 32x 4x x 1 1 2 3x 6x x 2 1 2 3x 10x 4x 3 Is inconsistent for: (A) exactly one positive value of (B) exactly one negative value of (C) every value of (D) exactly two values of

55. If x x x xe e e e2x x xe 2e e 1 e dx g x e +c where c is a constant of integration,

then g(0) is equal to: (A) e (B) 2 (C) e2 (D) 1 56. If 2sin2 13 , 14 and 4 2sin23 are the first three terms of an AP for some , then the sixth

term of this AP is: (A) 65 (B) 78 (C) 66 (D) 81 57. A survey shows that 73% of the persons working in an office like coffee, whereas 65%

like tea. If x denotes the percentage of them, who like both coffee and tea, then x cannot be:

(A) 63 (B) 54 (C) 38 (D) 36




58. If the common tangent to the parabolas 2 2y 4x and x 4y also touches the circle, 2 2 2x y c , then c is equal to

(A) 1

2 2 (B)

12

(C) 12

(D) 14

59. The product of the roots of the equation 29x 18 x 5 0 , is:

(A) 59

(B) 259

(C) 527

(D) 2581

60. The value of /2

sin x/2

1 dx1 e

is:

(A) 32

(B) 2

(C) (D) 4

61. If S is the sum of the first 10 terms of the series

1 1 1 11 1 1 1tan tan tan tan ....,3 7 13 21

then tan(S) is equal to:

(A) 56

(B) 1011

(C) 65

(D) 511

62. If (a, b, c) is the image of the point (1, 2, –3) in the line, x 1 y 3 z

2 2 1

, then a + b+

c is equal to (A) 3 (B) 1 (C) 2 (D) – 1

63. If y = y (x) is the solution of the differential equation x

x5 e dy. e2 y dx

= 0 satisfying y(0) =

1, then a value of ey log 13 is: (A) 0 (B) 1 (C) – 1 (D) 2




64. If is the positive root of the equation, 2p x x x 2 0,then

x

1 cos p xlim

x 4

is equal to:

(A) 12

(B) 12

(C) 32

(D) 32

65. If the co-ordinates of two points A and B are 7,0 and 7,0 respectively and P is

ny point on the conic 29x 16y 144 , then PA + PB is equal to: (A) 8 (B) 16 (C) 9 (D) 6

66. If the function 2

1

2

k x 1 xf x

k cos x x

is twice differentiable, then the ordered pair

1 2k ,k is equal to: (A) (1, 0) (B) (1, 1)

(C) 1, 12

(D) 1,12

67. If the minimum and the maximum values of the function f : , R2 2

, defined by

2 2

2 2

sin 1 sin 1f cos 1 cos 1

12 10 2

are m and M respectively, then the ordered pair (m,

M) is equal to: (A) (0, 4) (B) 0,2 2

(C) (–4, 4) (D) (–4, 0) 68. The mean and variance of 7 observations are 8 and 16, respectively. If five observations

are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is: (A) 2 (B) 4 (C) 1 (D) 3 69. If the point P on the curve, 2 24x 5y 20 is farthest from the point Q(0, –4) then PQ2

is equal to: (A) 21 (B) 48 (C) 36 (D) 29




70. If the volume of a parallelepiped, whose coterminus edges are given by the vectors ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆa i j nk, b 2i 4 j nkand c i nj 3k n 0

, is 158 cu.units. then:

(A) a.c 14

(B) n=7 (C) b.c 10

(D) n=9

71. The natural number m, for which the coefficient of x in the binomial expansion of

22m

21xx

is 1540, is ____

72. Let xf x x.2

, for – 10 < x < 10, where [t] denotes the greatest integer function.

Then the number of points of discontinuity of f is equal to ___

73. If the line, 2x y 3 0 is at a distance 15

and 25

from the lines 4x 2y 0

and 6x 3y 0, respectively, then the sum of all possible values of and is ____. 74. Four fair dice are thrown independently 27 times. Then the expected number of times, at

least two dice show up a three or a five, a ___. 75. The number of words, with or without meaning, that can be formed by taking 4 letters at

a time from the letters of the word ‘SYLLABUS’ such that two letters are distinct and two letters are alike, is ___.




FIITJEE

Solutions to JEE (Main)-2020

PART –A (PHYSICS) 1. D Sol. As it just floats B = mg

3 3 3

s4 R 4 R 4 r( ) (g) ( ) (g)

3 3 3

R3 = 3 3 27(R r )8

On solving we get,

8r R (approx)9

2. D

Sol. 1h 1d

1 2h h 13.464d 3

23.464dd h

3

h2 = 2d – d = d

A B C 45° 30°

h1

h2

d 2.464 d

3. D 4. No option. 5. A Sol. As Torquenet = 0 Hence, L = constant I = (3I + I)

4




Loss in K.E. =

221 1I (I 3I)

2 2 16

=

21 1I 12 4

Fractional loss = 2

2

1 3I 32 41 4I2

6. B

Sol.

V(n)f174 1

100

(as closed from end)

(n 2) (v)f24.54 1100

nV(100) (n 2) (v) (100)4(83) (4) (75.5)

75.5 n = 83 (n – 2) 75.5 n = 83 (M – 2) 7.5 n = 166 n = 22 (approx)

(330) (22) (100)f4(83)

= 2200 Hz

7. D

Sol. 0

0

E CB

00

EBC

Force, Fmax = QVB 8. D Sol. Photodiode operate in reverse bias. The photocurrent increases initially and saturates

finally. 9. C

Sol. 21 m(210) m(0.03) (4.2) 1000 T4

; Q = mSt




T = (210) (210)(4) (4.2) (0.03) (1000)

= 87.5°C

10. C

Sol. 2Rg g

R h

above

22g g3

Rh2

dg g 1R

above

d 4IR 9

d 5R 9

11. B Sol. V is always less than u.

1 1 1v u F

1 1 1v u F

V =

uF uuu F 1F

12. B

Sol.

d

dwef = Eqdx

et 2

kQdu dxqx

=

R y

R

1kQq2

ef

kQq(y)W(R) (R y)

Wall = k

Wmg + Wef = 21 mv2




2 2 kQqyV mg ym (R) (R y)

2

0

kQq 1V 2y g ; km(R) (R y) 4

13. A

Sol.

R1 G

V = IG (G + R1)

1 = IG (G + R1) …(1)

2 = IG (G + R1 + R2)

2 = 1 + IGR2

IGR2 = 1

R2 = G + R1

14. D Sol. Use the equation to find ‘N’:

22 1( )( )3 2

NP muV

15. D

Sol. o1

IB2 ( )

r F

= 2aF

=

oI2(a) (I) (2a)

2 b

2 2o2 I ab

2a

b

i i

a

B on the also equal to B1 as

02

2 0

IB

2 b4

b >>> a




16. C

Sol.

3

2

kg s mdisp (s)m kgms

disp= (1) (16) 1300

11000

(1)

disp in mm = 1000 0.03333300 100

3100

17. C

Sol.

h

2gh

V2 = 2gh

V = 2gh

h = 2gt2gh

2

gt2 – 2 2gh t 2h 0

2 2gh 8gh 8ght

2g

2 2gh 4 ght

2g

as cannot be negative.

ht (2 2)g

h3.4g

18. B Sol. 0ln R ln R t

and 1/21T

where ‘N’ is slope

1/2 1/2 1/210 5 5T (A) : T (B) : T (C) : :6 6 2

= 2 : 1 : 3

19. A Sol. Power delivered = 1000 W Voltage = 220 V

Transmission ‘I’ = 1000220




Power loss = I2R

Efficiency = 2

1000 1001000 I R

20. A

Sol.

4CV q CV q

2 c (as same point)

4cV – q = –2cV + 2q

q = 2cV

Energy = 2 2 2(2cv) (cV) 3cV

4c 2c 2

–CV +CV

+4CV –4CV

q

21. 195.00

Sol. ˆ ˆ ˆF (i 2 j 3k)N

ˆ ˆ ˆr (4 1)i (3 2) j ( 1 1)k

r F

= ˆ ˆ ˆ ˆ ˆ ˆ3i j 2k i 2 j 3k

ˆ ˆ ˆi j k 3 1 – 2 ˆ ˆ ˆi(3 4) j(11) (6 1)k 1 2 3 ˆ ˆ ˆ7i 11j 5k 121 25 49

195 ; x = 195

22. 51.00 Sol. Mo = 200 MeV/C2, m = 1 Gev/C2 Initial velocity of particle is ‘Vo’. Final velocity of hydrogen atom is ‘V’ MoVo = mV

o oM VVm

…(1)

Also, 2 20 0

1 1 3M V mV 13.62 2 4

Put (1) and get answer

20 0

1 51M V eV2 4

Hence, N = 51.




23. 5.00

Sol. o

2

I (200)B2R

R2 = 20 cm ; R1 = 1 cm

2

1d dBe (R ) (500)dt dt

= 2 o1

2

I(R ) (500) (200) (10t 2)2R

= o(500)(200) ( ) (8) (5)110000 (2)

= 162 × 10 × 5 × 10–7 = 8002 × 10–7 e = 0.7887 mv

4 0.8x ; x = 5.

24. 50.47

Sol. We know, P 2 Em and h2 Em

…(1)

For 1st maxima, 2nd sin = Put (1) and get the answer E = 50.47 eV. 25. 50.00 Sol. Image by objective must be at focus of eye piece.

o

1 1 1v u f

1 1 15 u 1

; 5u cm4

Hence, N = 50.




PART –B (CHEMISTRY) 26. A Sol. In solid state PCl5 exists as an ionic solid with constituent ions 4PCl (Tetrahedral) and

6PCI (Octahedral) 27. C Sol. According to Aufbau’s principal, the increasing energy of atomic orbitals for sixth period

element follows the order 6s < 4f < 5d < 6p 28. C Sol. B is least basic as lone pair of electron is present in resonance so as to make the

system aromatic in nature D is most basic as it results in formation of equivalent resonating structures upon attack

of H+ Among A and C, the former is less basic as sp2 hybridistion of nitrogen decrease its

basic strength. Hence option 3 follows Sol.

N

N

H

H

N

N

H

H

N

N

H

H

29. D Sol. At CMC, the particles cluster together through lyophobic end to form associated colloid

called micelle. Further all lyophilic ends(polar head) get projected towards water. 30. B Sol. Tyrosine is a non essential amino acid. 31. C Sol. For A

0

t

A0.693 2.303 log300 t A

For B

0

t

B0.693 2.303 log180 t B

Given A0 = B0 & At = 4Bt Substituting & solving we get t = 900 s 32. A Sol. In water gas shift reaction carbon monoxide is oxidized into carbon dioxide by treating it

with steam in presence of catalyst.




33. B

Sol.

34. D

Sol. MV

2X310

A

M26.17N 300 10

Solving 2XM = 50 g

No. of molecules in 200 g

A A200 N 4N50

35. A Sol. Among its different uses noradrenaline is used as an anti-depressant. 36. D Sol. With decrease in inter-nuclear distance, the potential energy of the system decreases,

reaches a minimum value and then sharply increases due to rise in inter-electronic as well as inter-nuclear repulsions

37. B Sol. In endothermic reaction formation of reactants is favoured upon decrease in

temperature. Addition of inert gas at constant volume and temperature has no effect on equilibrium.

38. C Sol. LiAlH4 is a versatile reducing agent which can be used for reduction of cyanide into 1o

amine. 39. C Sol. Esters are susceptible to reaction in basic medium. 40. C Sol. Excess of nitrogen and phosphorus is primarily responsible for eutrophication and hence

an indicator of polluted environment. 41. C Sol. Octahedral high spin

d6




CFSE = 4(-0.40) + 2(0.60) = -0.440 Tetrahedral

d6

CFSE = 3(-0.6t) + 3(0.4t) = -0.60 42. B

Sol. 01

aR 16 93

0

2

1

2

aR 16 92

R 2R 3

43. C Sol. B is the most acidic as it is active methylene group. D is least acidic due to cross-

conjugation in conjugate base. Option 3 follows 44. A Sol. Gd : [Xe]4f75d16s2 Gd3+ is [Xe]4f7

Also n n 2 B.M

7 9 B.M 7.9 B.M

45. B Sol. Ellingham diagram provides information on Gibb’s free energy for formation of oxides as

a function of temperature. 46. 6 Sol. EDTA as shown is hexadentate

47. 18 Sol. 3 8 2 2 2C H 5O 3CO 4H O

4 10 2 2 213C H O 4CO 5H O2

2 mol C4H10 is given 13 mol O2 is required for combustion of Butane Total mol of O2 = 5 + 13 = 18




48. -6 Sol. G0 = -nFEo 17.37 103 = -3 96500 Eo Eo = -6 10–2 V

49. 4 Sol. The structure of the given compound is as show below

50. 37 Sol. Mass of CO2 dissolve at 3 bar pressure in 1 kg water(which is same as 1 litre since

density is 1 gm/mL) = 4.4 gm = 0.1 mol. For weak electrolytes of Ostwald’s dilution law we have pH = ½(pKa – log C) pH = ½ [6.4 + 1] = 37 = 37 10–1




PART–C (MATHEMATICS) 51. A Sol. length of side = 4 then z z 4

2iy 4

y 2

also z z 2Re z 4

2x 4 x 2

2 2z x y 2 2

52. D Sol. Negation of x ~ y ~ x ~ y

x ~ ~ y x y x y ~ x ~ y 53. D Sol. 10 9 8 2 9 10S' 2 2 .3 2 .3 ........ 2.3 3

G.P. 10 3a 2 ,r , n 112

11

1110 11

11

3 12 3S' 2 . 2 13 212

11 113 2 54. B

Sol. Here 2 4

D 1 6 1 3 3 210 4

D 0 23,3




1

1 4D 2 6 1 2 3

3 10 4

For 12 , D 03

55. B

Sol. x xe e2x x xI e 2e e 1 e dx

x x x xe e2x x x x e eI e e 1 e dx e e e dx

x x e xx x e e x eI e 1 e e dx e

x xe e x du x xe e 1 dx du

x x x x x xe e e e e e xI e e e e 1 then xg x e 1

g 0 2 56. C Sol. a, b, c are in AP then 2b a c 2sin2 1 2sin228 3 3 Put 2sin23 x

2x 8128 x 84x 243 03 x

x 3 x 81 0

2sin23 3 or 43 2sin2 1 or 4

1sin22

terms are 1, 14, 27,…………….then 6T 1 5 13 57. D Sol. n C 73, n T 65, n C T x

n C T 100

n C n T n C T 100 x 38 n C T min n C ,n T x 65

38 x 65




58. B Sol. 2y 4x and 2x 4y

any tangent of 2y 4x is 1y mxm

it also tangent for 2x 4y

21 mm

m 1

common tangent is y x 1, it also touches 2 2 2x y c

2 2 11 c . 1 1 c2

59. D

Sol. 22x x t let

29t 18t 5 0 3t 1 3t 5 0

1 5x ,3 3

Product of roots 1 1 5 5 253 3 3 3 81

60. B

Sol. /2 /2

sinx sinx/2 /2

1 1I dx.I dx1 e 1 e

/2 sinx

sinx/2

ReplaceeI dxx a b xe

b b

a c

f x dx f a b x dx

/2 /2

/2 /2

12I 1dx I dx2

/2/2

1I x I2 2

61. A

Sol. 1 1 11 1 1S tan tan tan .....3 7 13

upto 10 term

1 1 1 12 1 3 2 4 3 11 10S tan tan tan ..... tan1 1.2 1 2.3 1 3.4 1 11.10




1 1 1 1 1 1S tan 2 tan 1 tan 3 tan 2 ...... tan 11 tan 10

1 1S tan 11 tan 1

1S tan 114

5tan S6

62. C Sol. R 1 2r, 3 2r, r

dr's of PR are 2 2r, 1 2r, 3 r

Then 2 2 2r 2 1 2r 1 3 r 0 9 9r 0 r = 1 R (1, 1, –1) then a 1 2 b 2 2 c 3 2 a 1 b 0 c 1 a b c 2

63. C

Sol. Given x

xdy e dx

2 y 5 e

xn 2 y n 5 e nC

xCy 2

5 e

y 0 1 C = 18

x18y 2

5 e

ey log 13 1 64. C Sol. P x 0

2x x 2 0 x 2 x 1 0 x 2, 1 2

Now 2

x 2

1 cos x x 2lim

x 2

22

x 2

x x 22sin2

limx 2




2

x 2

x x 2sin2

limx 2

for

2x x 2x 2 , 02

2 2

2x 2

x x 2 x x 22 sin2 2lim .

x 2x x 22

x 2

x 2 x 11 3lim .x 22 2

65. A

Sol. For ellipse 2 2x y 9 71, a 4, b 3, e 1

16 9 16 4

A and B are foci then PA PB 2a 2 4 8 66. D Sol. f(x) is differentiable then will also continuous then 1

2f 1, f k

2k 1

Now 1

2

2k x xf ' x

k sinx x

then f ' f ' 0

1

2

2k xf " x

k cos x x

then 1 22k k

11k2

67. D Sol. 3 2 1C C C

2

2 2 2

sin 1 1

f cos 1 1 4 cos sin 4 cos2 , ,4 2

12 2 2

maxf M 0

minf m 4 68. A

Sol. 2 4 10 12 14 x yx 8

7

42 x y 56 x y 14




22

i i2 x xn n

2 2

24 16 100 144 196 x y16 87

2 2460 x y16 64

7

2 2560 460 x y 2 2x y 100 ………(2) xy 48

2 2x y x y 4xy 4

x y 2 69. C

Sol. Equation 2 2x y 1

5 4 then P 5 cos ,2sin

2 22 2PQ 5cos 4 sin 2 cos 16sin 20

2sin 16sin 21

285 sin 8

2maxPQ 85 49 36,

2sin 8 49, 81 70. C Sol. Volume of parallelepiped v a b c

1 1 n

v 2 4 n 1581 n 3

21 12 n 1 6 n n 2n 4 158

23n 5n 152 0 or 23n 5n 164 0 D 0 (no real roots)

19n 8,3

n = 8

then b.c 2 4n 3n 10

a.c 1 n 3n 33

71. 13.00

Sol. 22 r22 m 2rr 1 rT C x x




m 22 r 2r22r 1 rT C x

22m mr 2r 1 22m 1 r m 2

22m 1rm 2

22m 44 45r

m 2

3.3.5r 22m 2

So possible value of m = 1, 3, 7, 13, 43 but 20

rC 1540 only possible condition is m = 13 72. 08.00

Sol. x5 52

l x 5, 4, 3, 2, 1,0,1,2,3,42

Hence, function is discontinues at = –4, –3, –2, –1, 1, 2, 3, 4 Number of values is 8. 73. 30.00 Sol. 2x y 3 0 ……….(i)

4x 2y 0 2x y 02

………….(ii)

6x 3y 2x y 03

…………(iii)

1 2 2

312d52 1

6 2 6 2, 2 8, 4

2 2 2

323d52 1

9 6 9 6, 6 15,3

Sum of all value of and 30 . 74. 11.00

Sol. P (at least 2 show 3 or 5) 2 2 3 4

4 4 42 3 4

2 4 2 4 2C . C C6 6 6 6 6

4384 128 16 11

276




n = 27 expectation of number of times = np

1127. 1127

75. 240.00 Sol. SYLLABUS S – 2, L – 2, A, B, Y, U

Required 2 51 2

4! 24C . C . 2.10. 2402! 2


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