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FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. 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: 8 th January 2020 (Second Shift)
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  • FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.

    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: 8th January 2020 (Second Shift)

    http://www.fiitjee.com.

  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-2

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

    PART – A (PHYSICS) 1. A simple pendulum is being used to determine the vale of gravitational acceleration g at

    a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is:

    (A) 4.40% (B) 2.40% (C) 3.40% (D) 5.40%

    2. A Carnot engine having an efficiency of 110

    is being used as a refrigerator. If the work

    done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:

    (A) 99 J (B) 100 J (C) 1 J (D) 90 J 3. Consider two charged metallic spheres S1 and S2 of radii R1 and R2, respectively. The

    electric fields E1 (on S1) and E2 (on S2) on their surfaces are such that E1/E2 = R1/R2. Then the ratio of V1 (on S1) / V2 (on S2) of the electrostatic potentials on each sphere is:

    (A) (R1/R2)2 (B) 3

    1

    2

    RR

    (C) (R2/R1) (D) R1/R2 4. As shown in the figure, a battery of emf is

    connected to an inductor L and resistance R in series. The switch is closed at t = 0. The total charge that flows from the battery, between t = 0 and t = tc (tc is the time constant of the circuit) is

    (A) 2

    LeR (B) 2

    L 11eR

    (C) 2R

    eL (D) 2

    LR

    5. A very long wire ABADMNDC is shown in figure carrying

    current I. AB and BC parts are straight, long and at right angle. At D wire forms a circular turn DMND of radius R. AB, BC parts are tangential to circular turn at N and D. Magnetic field at the centre of circle is:

    (A) 0I2R (B) 0I 1

    2 R 2

    (C) 0I ( 1)2 R

    (D) 0I 12 R 2

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-3

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

    6. A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is:

    (A) 8.75 × 10–4 J (B) 8.75 × 10–3 J (C) 6.25 × 10–4 J (D) 1.13 × 10–3 J 7. In a double-slit experiment, at a certain point on the screen the path difference between

    the two interfering waves is 1 th8

    of a wavelength. The ratio of the intensity of light at that

    point to that at the centre of a bright fringe is: (A) 0.672 (B) 0.568 (C) 0.760 (D) 0.853 8. A particle of mass m is dropped from a height h above the ground. At the same time

    another particle of same mass is thrown vertically upwards from the ground with a speed of 2gh. If they collide head-on completely inelastically, the time taken for the combined mass to

    reach the ground, in units of hg

    is:

    (A) 12

    (B) 12

    (C) 34

    (D) 32

    9. A particle moves such that its position vector ˆ ˆr (t) cos i sin tj

    where is a constant

    and t is time. Then which of the following statements is true for the velocity v(t) and acceleration a (t)

    of the particle:

    (A) v

    is perpendicular to r

    and a

    is directed towards the origin. (B) v

    and a

    both are parallel to r

    (C) v

    is perpendicular to r

    and a

    is directed away from the origin. (D) v

    and a

    both are perpendicular to r

    10. Consider a mixture of n moles of helium gas and 2n moles of oxygen gas (molecules

    taken to be rigid) as an ideal gas. It CP/CV value will be (A) 19/13 (B) 40/27 (C) 67/45 (D) 23/15 11. A capacitor is made of two square plats

    each of side ‘a’ making a very small angle a between them, as shown in figure. The capacitance will be close to:

    (A) 2

    0 a a1d d

    (B) 2

    0 a a1d 4d

    (C) 2

    0 a 3 a1d 2d

    (D) 2

    0 a a1d 2d

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-4

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

    12. Two liquids of densities 1 and 2(2 = 21) are filled up behind a square wall of inside 10 m as shown in figure. Each liquid has a height of 5 m. The ratio of the forces due to these liquids exerted on upper part MN to that at the lower part NO is (Assume that the liquids are not mixing):

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

    13. A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the

    z-direction. At a particular point in space and time, the magnetic field is given by 8 ˆB 5 10 j T.

    The corresponding electric field E

    is (speed of light c = 3 × 108 ms–1)

    (A) –1.66 × 10–16 î V / m (B) –15 î V / m (C) 15 î V / m (D) 1.66 × 10–16 î V / m 14. An electron (mass m) with initial velocity 0 0ˆ ˆv v i v j

    is in an electric field 0 ˆE E k.

    If 0 is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time t is given by

    (A) 02 2 2

    02 2

    0

    e E t1m v

    (B) 02 2 2

    2 20

    e E t2m v

    (C) 02 2 2

    2 20

    2e E t1m v

    (D) 02 2 2

    2 20

    e E t12m v

    15. A transverse wave travels on a taut steel wire wit a velocity of v when tension in it is

    2.06 × 104 N. When the tension is changed to T, the velocity changed to v/2. The value of T is close to:

    (A) 10.2 × 102 N (B) 30.5 × 104 N (C) 5.15 × 103 N (D) 2.50 × 104 N 16. An object is gradually moving away from the focal point of a concave mirror along the

    axis of the mirror. The graphical representation of the magnitude of linear magnification (m) versus distance of the object from the mirror (x) is correctly given by (Graphs are drawn schematically and are not to scale)

    (A)

    (B)

    (C)

    (D)

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-5

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

    17. A particle of mass m and charge q is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed v on the distance x travelled by it is correctly given by (graphs are schematic and not drawn to scale)

    (A)

    (B)

    (C)

    (D)

    18. As shown in figure when a spherical cavity (centered

    at O) of radius 1 is cut out of a uniform sphere of radius R (centred at C), the centre of mass of remaining (shaded) part of sphere is at G, i.e. on the surface of the cavity. R can be determined by the equation:

    (A) (R2 + R + 1) (2 – R) = 1 (B) (R2 + R – 1) (2 – R) = 1 (C) (R2 – R + 1) (2 – R) = 1 (D) (R2 – R – 1) (2 – R) = 1 19. In the given circuit, value of Y is (A) 0 (B) 1 (C) toggles between 0 and 1 (D) will not execute

    20. A galvanometer having a coil resistance 100 gives a full scale deflection when a

    current of 1 mA is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of 10 V.

    (A) 9.9 k (B) 7.9 k (C) 10 k (D) 8.9 k

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-6

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

    21. A ball is dropped form the top of a 100 m high tower on a planet. In the last 1 s2

    before

    hitting the ground, it covers a distance of 19 m. Acceleration due to gravity (in ms–2) near the surface on that planet is _________.

    22. The first member of the Balmer series of hydrogen atom has a wavelength of 6561 Å.

    The wavelength of the second member of the Balmer series (in nm) is _________. 23. The series combination of two batteries, both of the same emf 10 V, but different internal

    resistance of 20 and 5 , is connected to the parallel combination of two resistors 30 and R . The voltage difference across the battery of internal resistance 20 is zero, the value of R (in ) is _________.

    24. An asteroid is moving directly towards the centre of the earth. When at a distance of 10

    R (R is the radius of the earth) from he earths centre, it has a speed of 12 km/s. Neglecting the effect of earths atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity form the earth is 11.2 km/s)? Given your answer to the nearest integer in kilometer/s _________.

    25. Three containers C1, C2 and C3 have water at different temperatures. The table below

    shows the final temperature T when different amounts of water (given in liters) are taken from each container and mixed (assume no loss of heat during the process)

    C1 C2 C3 T

    1 2 __ 60°C

    __ 1 2 30°C

    2 __ 1 60°C

    1 1 1

    The value of (in °C to the nearest integer) is _________.

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-7

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

    PART –B (CHEMISTRY) 26. The increasing order of the atomic radii of the following elements is: (a) C (b) O (c) F (d) Cl (e) Br (A) (a) < (b) < (c) < (d) < (e) (B) (d) < (c) < (b) < (a) < (e) (C) (c) < (b) < (a) < (d) < (e) (D) (b) < (c) < (d) < (a) < (e)

    27. Kjeldahl's method cannot be used to estimate nitrogen for which of the following

    compounds?

    (A) (B) C6H5NH2 (C) C6H5NO2 (D) CH3CH2 – C N

    28. Consider the following plots of rate constant versus 1T

    for four different reactions. Which

    of the following order is correct for the activation energies of these reactions?

    (A) Ec > Ea > Ed > Eb (B) Eb > Ed > Ec > Ea (C) Ea > Ec > Ed > Eb (D) Eb > Ea > Ed > Ec 29. For the following Assertion and Reason, the correct option is :

    Assertion : The pH of water increase with increase in temperature. Reason : The dissociation of water into H+ and OH– is an exothermic reaction (A) assertion is not true, but reason is true (B) both assertion and reason are false (C) but the reason is not the correct explanation for the assertion (D) both assertion and reason are true, and the reason is the correct explanation for the

    assertion 30. White phosphorus on reaction with concentrated NaOH solution in an inert atmosphere

    of CO2 gives phosphine and compound (X). (X) on acidification with HCl gives compound (Y). The basicity of compound (Y) is :

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

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-8

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

    31. The major product [B] in the following sequence is :

    32. Among (a) – (d), the complexes that can display geometrical isomerism are: (a) [Pt(NH3)3]Cl+ (b) [Pt(NH3)Cl5]– (c) [Pt(NH3)2Cl(NO2)] (d) [Pt(NH3)4ClBr]2+ (A) (d) and (a) (B) (c) and (d) (C) (b) and (c) (D) (a) and (b) 33. For the following Assertion and Reason, the correct option is : Assertion: For hydrogenation reactions, the catalytic activity increases from Group 5 to

    Group 11 metals with maximum activity shown by Group 7-9 elements. Reason : The reactants are most strongly adsorbed on group 7-9 elements. Both

    assertion and reason are true and the reason is the correct

    (A) Assertion is not true, but reason is true (B) Both assertion and reason are false (C) but the reason is not the correct explanation for the assertion (D) Both assertion and reason are true, and the reason is the correct explanation for the

    assertion

    34. The major product in the following reaction is

    (A)

    (B)

    (C)

    (D)

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-9

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

    35. Among the reactions (a)-(d), the reaction(s) that does/do not occur in the blast furnace during the extraction of iron is/are:

    (a) 2 3CaO SiO CaSiO (b) 2 3 3 4 23Fe O CO 2Fe O CO

    (c) 2 3FeO SiO FeSiO (d) 21FeO Fe O2

    (A) (c) and (d) (B) (a) (C) (a) and (d) (D) (d) 36. A metal (A) on heating in nitrogen gas gives compound B. B on treatment with H2O gives

    a colourless gas which when passed through CuSO4 solution gives a dark blue-violet coloured solution. A and B respectively, are:

    (A) Mg and Mg(NO3)2 (B) Na and NaNO3 (C) Mg and Mg3N2 (D) Na and Na3N 37. Two monomers in maltose are: (A) -D-glucose and -D-galactose (B) -D-glucose and -D-glucose (C) -D-glucose and -D-glucose (D) -D-glucose and -D-Fructose 38. The correct order of the calculated spin-only magnetic moments of complexes (A) to (D)

    is : (a) Ni(CO)4 (b) [Ni(H2O)6]Cl2 (c) Na2[Ni(CN)4] (d) PdCl2(PPh3)2 (A) (c) (d) < (b) < (a) (B) (a) (c) < (b) (d) (C) (a) (c) (d) < (b) (D) (c) < (d) < (b) < (a) 39. An unsaturated hydrocarbon X absorbs two hydrogen molecules on catalytic

    hydrogenation, and also gives following reaction:

    (A)

    O

    (B)

    (C)

    (D)

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-10

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

    40. Hydrogen has three isotopes (a), (b) and (c). If the number of neutron(s) in (a), (b) and (c) respectively, are (x), (y) and (z), the sum of (x), (y) and (z) is:

    (A) 4 (B) 3 (C) 2 (D) 1 41. Which of the following compound is likely to show both Frenkel and Schottky defects in

    its crystalline form? (A) CsCl (B) AgBr (C) ZnS (D) KBr 42. Arrange the following bonds according to their average bond energies in descending

    order: C–Cl, C–Br, C–F, C–I (A) C–F > C–Cl > C–Br > C–I (B) C–I > C–Br > C–Cl > C–F (C) C–Cl > C–Br > C–I > C–F (D) C–Br > C–I > C–Cl > C–F 43. Preparation of Bakelite proceeds via reactions: (A) Electrophilic substitution and dehydration (B) Condensation and elimination (C) Electrophilic addition and dehydration (D) Nucleophilic addition and dehydration 44. The radius of the second Bohr orbit, in terms of the Bohr radius, a0, in Li2+ is

    (A) 04a3

    (B) 02a9

    (C) 02a3

    (D) 04a9

    45. Among the compounds A and B with molecular formula C9H18O3, A is having higher

    boiling point the B. The possible structures of A and B are:

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-11

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

    46. Complexes (ML5) of metals Ni and Fe have ideal square pyramidal and trigonal bipyramidal geometries, respectively. The sum of the 90°, 120° and 180° L-M-L angles in the two complexes is _______

    47. At constant volume, 4 mol of an ideal gas when heated from 300 K to 500 K changes its internal energy by 5000 J. The molar heat capacity at constant volume is ____

    48. In the following sequence of reactions the maximum number of atoms present in molecule 'C' in one plane is

    49.

    50. NaClO3 is used, even in spacecraft, to produce O2. The daily consumption of pure O2 by

    a person is 492L at 1 atm, 300 K. How much amount of NaClO3, in grams, is required to produce O2 for the daily consumption of a person at 1 atm, 300 K? .

    NaClO3(s) + Fe(s) O2(g) + NaCl(s) + FeO(s) R = 0.082 L atm mol–1 K–1

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-12

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

    PART–C (MATHEMATICS)

    51. If the 10th term of an A.P. is 120

    and its 20th term is 1 ,10

    then the sum of its first 200

    terms is:

    (A) 100 (B) 1504

    (C) 11002

    (D) 50

    52. Which of the following statement is a tautology? (A) ~ p ~ q p q (B) ~ p ~ q p q (C) p ~ q p q (D) ~ p ~ q p q 53. The mean and variance of 20 observations are found to be 10 and 4, respectively. On

    rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is:

    (A) 4.02 (B) 3.98 (C) 4.01 (D) 3.99

    54. The mirror image of the point (1, 2, 3) in a plane is 7 4 1, ,3 3 3

    . Which of the

    following points lies on this plane? (A) (1, –1, 1) (B) (1, 1, 1) (C) (–1, –1, –1) (D) (–1, –1, 1)

    55. If a hyperbola passes through the point P (10, 16) and it has vertices at 6, 0 then the equation of the normal to it at P is:

    (A) 2x 5y 100 (B) x 3y 58 (C) 3x 4y 94 (D) x 2y 42

    56. If a line, y mx c is a tangent to the circle, 2 2x 3 y 1 and it is perpendicular to

    a line 1L , where 1L is the tangent to the circle, 2 2x y 1 at the point 1 1,

    2 2

    ; then:

    (A) 2c 7c 6 0 (B) 2c 6c 7 0 (C) 2c 6c 7 0 (D) 2c 7c 6 0 57. The area (in sq. units) of the region 2x,y R y 3 2x , is (A) 32

    3 (B) 29

    3

    (C) 313

    (D) 343

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-13

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

    58. Let S be the set of all functions : 0,1 R, which are continuous on 0,1 and differentiable on (0, 1). Then for every in S, there exists ac 0,1 , depending on f, such that:

    (A) f c f 1 1 c f ' c (B) f c f 1 1 c f ' c

    (C) f c f 1 f ' c (D) f 1 f c f ' c1 c

    59. Let ˆ ˆ ˆa i 2 j k

    and ˆ ˆ ˆb i j k

    be two vectors. If c

    is a vector such that b c b a

    and c.a 0,

    then c.b

    is equal to:

    (A) 12

    (B) 32

    (C) –1 (D) 12

    60. Let f : 1,3 R be a function defined by 2x x

    f x ,1 x

    where [x] denotes the greatest

    integer x . Then the range of f is:

    (A) 2 4,5 5

    (B) 2 1 3 4, ,5 2 5 5

    (C) 3 4,5 5

    (D) 2 3 3 4, ,5 5 4 5

    61. Let 1 i 32

    . If

    1002k

    k 0a 1

    and

    1003k

    k 0b ,

    then a and b are the roots of

    the quadratic equation: (A) 2x 101x 100 0 (B) 2x 102x 101 0 (C) 2x 102x 101 0 (D) 2x 101x 100 0

    62. x

    0x 0

    t sin 10 t dtlim

    x is equal to:

    (A) 0 (B) 110

    (C) 15

    (D) 110

    63. The system of linear equations x 2y 2z 5 2 x 3y 5z 8 4x y 6z 10 has: (A) no solution when 8 (B) infinitely many solutions when 2 (C) no solution when 2 (D) a unique solution when 8

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-14

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

    64. If and be the coefficients of 4x and 2x respectively in the expansion of

    6 62x x 1 x x 1 , then: (A) 30 (B) 60 (C) 132 (D) 60 65. The differential equation of the family of curves, 2x 4b y b ,b R , is: (A) xy" y ' (B) 2x y ' x 2yy ' (C) 2x y ' x 2yy ' (D) 2x y ' 2yy ' x 66. The length of the perpendicular from the origin, 0n the normal to the curve,

    2 2x 2xy 3y 0 at the point (2, 2) is: (A) 2 2 (B) 4 2 (C) 2 (D) 2

    67. If 2 2

    A9 4

    and 1 0

    I ,0 1

    then 110A is equal to:

    (A) 4I A (B) A 6I (C) A 4I (D) 6I A

    68. Let A and B be two events such that the probability that exactly one of them occurs is 25

    and the probability that A or B occurs is 1,2

    then the probability of both of them occur

    together is: (A) 0.01 (B) 0.10 (C) 0.20 (D) 0.02

    69. If 2

    3 21

    dxI2x 9x 12x 4

    , then:

    (A) 21 1I16 9

    (B) 21 1I9 8

    (C) 21 1I8 4 (D) 21 1I

    6 2

    70. Let S be the set of all real roots of the equation, x x x x3 3 1 2 3 1 3 2 . Then

    S: (A) contains at least four elements (B) is a singleton (C) is an empty set (D) contains exactly two elements

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-15

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    71. Let f x be a polynomial of degree 3 such that f 1 10, f 1 6 , f x has a critical point at x 1 and f ' x has a critical point at x = 1. Then f x has a local minima at x =________

    72. Let aline y mx m 0 intersect the parabola, 2y x at a point P, other than the

    origin. Let the tangent to it at P meet the x – axis at the point Q. If area OPQ 4 sq. units, then m is equal to ____________.

    73. The sum 7

    n 1

    n n 1 2n 14

    is equal to _____________.

    74. The number of 4 letter words (with or without meaning) that can be formed from the

    eleven letters of the word ‘EXAMINATION’ is ____________________.

    75. If 2 sin 1

    71 cos2

    and 1 cos2 1 ,

    2 10

    , 0,2

    , then tan 2 is equal

    to __________.

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-16

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    JEE (Main) – 2020 ANSWERS

    PART –A (PHYSICS)

    1. A 2. D 3. A 4. A 5. D 6. A 7. D 8. D 9. A 10. A 11. D 12. B 13. C 14. A 15. C 16. D 17. D 18. B 19. A 20. A 21. 8.00 or 2888.00 22. 486 23. 30 24. 16 25. 50

    PART –B (CHEMISTRY)

    26. C 27. C 28. A 29. B 30. D 31. D 32. B 33. A 34. A 35. A 36. C 37. C 38. C 39. D 40. B 41. B 42. A 43. A 44. A 45. D 46. 20.00 47. 6.25 48. 13.00 49. 2.15 50. 2130.00

    PART–C (MATHEMATICS)

    51. C 52. B 53. D 54. B 55. A 56. C 57. A 58. NA 59. A 60. B 61. B 62. A 63. C 64. C 65. C 66. A 67. B 68. B 69. B 70. B 71. 3 72. 0.5 73. 504 74. 2454 75. 1

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-17

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    HINTS AND SOLUTIONS PART A – PHYSICS

    1. A

    Sol. T 1 g LT 2 g L

    g 2 T L 1 0.12g T L 50 25.0

    = 4.4% 2. D Sol. For Carnot engine using as refrigerator

    122

    TW Q 1T

    It is given 110

    21

    T1T

    21

    T 9T 10

    So, Q2 = 90 J (as W = 10 J) 3. A

    Sol. 1 12 2

    E rE r

    2

    1 1 1 1 1 1

    2 2 2 2 2 2

    V E r r r rV E r r r r

    4. A

    Sol. cT

    0

    q idt

    c

    C

    T

    t /T1

    c c c

    C 0

    et ; T T e T1R RT

    21 L L;

    R e R R e

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-18

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    5. D Sol. 0 0 0 0 01 2 3 4B B B B B

    o o o o0 0 0i i isin 90 sin 45 (sin 45 sin 90 )4 R 2R 4 R

    = 0 0 0i i i1 11 14 R 2R 4 R2 2

    = 0i 1 11 2 14 R 2 2

    = 0 0i i 12 24 R 2 R 2

    6. A Sol. K.E. of the sphere = Translational K.E + Rotational

    K.E.

    = 2

    22

    1 Kmv 12 R

    K = Radius of gyration

    21 1 5 21

    2 2 100 5

    435 10 J4

    7. D

    Sol. 20I I cos 2

    2 20

    2 xI cos cosI 2 8

    0

    II

    = 0.853

    8. D

    Sol. Time for collision 1ht2gh

    After t1

    A 1ghV 0 gt2

    and B 11V 2gh gt gh 22

    at the time f collision i fP P

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-19

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    A B fmV mV 2mV

    fgh 1gh 2 2V2 2

    Vf = 0

    and height from ground = 211 h 3hh gt h2 4 4

    So time =

    3h3h42

    g 2g

    9. A Sol. ˆ ˆr cos t i sin tj

    dr ˆ ˆv ( sin t i cos tj)dt

    2dv ˆ ˆa (cos ti sin tj)dt

    2a r

    a

    is antiparallel to r

    v r ( sin t cos t cos t sin t) 0

    So, V r

    10. A

    Sol. 1 21 2

    1 p 2 pmix

    1 v 2 v

    n c n cn c n c

    5 7n R 2n R2 23 5n R 2n R2 2

    5 14 193 10 13

    11. D

    Sol. 0adxdcd x

    a0 0ac ln(d x)

    = 2

    0 0a aa aln 1 1d d 2d

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-20

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    12. B

    Sol. 12

    F 1F 4

    13. C

    Sol. E cB

    E = B × c = 15 N/c 14. A

    Sol. Initially 00

    hm 2 v

    Velocity as a function of time = 00 0eEˆ ˆ ˆv i v j tkm

    So wavelength 2 2

    2 200 2

    he Em 2v tm

    02 2

    202 2

    0

    e E1 t2m v

    15. C Sol. v T

    4

    1 1

    2 2

    v T v 2.06 10v T (v / 2) T

    42.06 10T N

    4

    = 0.515 × 104 N

    16. D Sol. At focus, magnification is . 17. D

    Sol. 2 2qEV xm

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-21

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    18. B

    Sol. 314M R3

    ; 324M (1) ( )3

    1 1 2 2com1 2

    M X M XXM M

    3 3

    3 3

    4 4R 0 (1) ( ) [R 1]3 3 (2 R)4 4R (1) ( )

    3 3

    3(R 1) (2 R)(R 1)

    (R 1)

    2(R 1) 2 R

    (R 1) (R R 1)

    (R2 + R + 1) (2 – R) = 1 Alternative: Mremaining (2 – R) = Mcavity (1 – R) (R3 – 13) (2 – R) = 13 [R – 1] (R2 + R + 1) (2 – R) = 1 19. A Sol. Y AB A

    AB A AB A = 0 + 0 = 0 20. A Sol. Vg = ig Rg = 0.1 V V = 10 V

    gg

    VR R 1V

    = 100 × 99 = 9.9 K 21. 8.00 or 2888.00 Sol. Area of shaded trapezium

    =

    1g t t12 19

    2 2

    …(1)

    21 gt 1002

    …(2)

    200tg

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-22

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    1g 2t 762

    2004 1g76

    g 2

    g = 8 m/s2 22. 486

    Sol. 2 2 21 2

    1 1 1RZn n

    2 2 21

    1 1 1 5RR(1)362 3

    2 2 22

    1 1 1 3RR(1)162 4

    21

    2027

    220 656127

    Å = 4860 Å

    = 486 nm 23. 30 Sol. V1 = 1 – I, r1 0 = 10 – I × 20 i = 0.5 A V2 = 2 – ir2 = 10 – 0.5 × 5 V2 = 7.5 V

    7.5 7.50.530 x

    0.5 = 0.25 + 7.5x

    7.5 7.50.25 ; x 30x 0.25

    24. 16 Sol. KEi + PEi = KEf + PEf

    2 201 GMm 1 GMmmu mv2 10R 2 R

    2 202GM 1v u 1

    R 10

    209 GMv u5 R

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-23

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    2

    2 9 (11.2)125 2

    2144 0.9(11.2) 256.896

    = 16.028 km/s 16 25. 50 Sol. 11 + 22 = (1 + 2) 60 1 + 22 = 180 …(1) 0 × 1 + 1 × 2 + 2 × 3 = (1 + 2) 30 2 + 23 = 90 …(2) 2 × 1 + 0 × 2 + 1 × 3 = (2 + 1) 60 21 + 3 = 180 …(3) and 1 + 2 + 3 = (1 + 1 + 1) …(4) from (1) + (2) + (3) 31 + 32 + 33 = 450 1 + 2 + 3 = 150 From (4) equation 150 = 3 = 50°C

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-24

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    PART –B (CHEMISTRY) 26. C

    Sol. Generally in a period L R Atomic radius decreases and in a Group T B Atomic radius increases.

    27. C Sol. Kjedahl’s method cannot be used to Test nitrogen in nitro and diazo present in ring

    because nitrogen in nitro cannot convert into Ammonium sulphate. 28. A

    Sol. EaRTk Ae

    EaLog LogA2.303RT

    EaSlope2.303RT

    Slope c > a > d > b Ec > Ea > Ed > Eb 29. B Sol. kw = [H+][OH–] Tkw For pure water [H+] = [OH–] kw = [H–]2 [H+] = kw on increasing T kw [H+] pH Dissociation of H2O is endothermic 2H O H OH H ve

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-25

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    30. D Sol.

    31. D Sol.

    32. B Sol. [Pt(NH3)2Cl(NO2)] & [Pt(NH3)4ClBr]2+ [M A2 BC] type [M A4 BC] type 33. A

    Sol. For hydrogenation reaction catalytic activity increase because reactants are more strongly adsorbed on group 7-9 element, So Assertion & Reason both are correct.

    34. A Sol.

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-26

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    35. A Sol.

    36. C Sol.

    37. C Sol. Maltose on hydrolysis give 2 mole of -D-glucose, because in maltose glucosidic linkage

    is present in between C1 & C4 of -D-glucose. 38. C Sol.

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-27

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    39. D Sol.

    40. B Sol.

    41. B Sol. AgBr show both Frenkel and Schottky defect. 42. A Sol.

    43. A

    Sol. Formation of Bakelite follows electrophillic substitution of phenol and formaldehyde followed by dehydration.

    44. A Sol.

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-28

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    45. D Sol. A is having higher boiling point than B. in case of A inter molecular H-bonding is possible

    while in case of B. intermolecular H-bonding is not possible hence have lower boiling point.

    46. 20.00 Sol.

    47. 6.25 Sol. U = ncv T 5000 = 4 Cv(500 - 300) Cv = 6.25 J k–1 mol–1

    48. 13.00 Sol.

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-29

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    49. 2.15 Sol.

    50. 2130.00 Sol. Moles of NaCl3 = mole of O2

    Mole of O2 = PV 1 492 20 moleRT 0.082 300

    Mass of NaClO3 = 20 106.5 = 2130 g

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-30

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    PART–C (MATHEMATICS) 51. C

    Sol. 101T a 9d

    20 ………..(i)

    201T a 19d

    10 …………..(ii)

    1 1a ,d200 200

    200200 2 199 201 1S 100

    2 200 200 2 2

    52. B Sol. p q ~ p ~ q p ~ q ~ p ~ q p q ~ p ~ q p q T T F F T F T T T F F T T F T T F T T F F T T T F F T T T F F T 53. D

    Sol. ix

    1020

    ………….(i)

    2ix 100 4

    20 …………(ii)

    2ix 104 20 2080 Actual mean 200 9 11 202

    20 20

    Variance 22080 81 121 202

    20 20

    22120 10.1 106 102.01 3.9920

    54. B

    Sol. d.r of normal to the plane 10 10 10, ,3 3 3

    Midpoint of P and Q is 2 1 4, ,3 3 3

    equation of plane x y z 1

    55. A Sol. Vertex is at 6, 0 a 6

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-31

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    Let the hyperbola is 2 2

    2 2x y 1a b

    Putting point P (10, 16) on the hyperbola

    2100 256 136 b

    2b 144

    hyperbola is 2 2x y 1

    36 144

    equation of normal is 2 2

    2 2

    1 1

    a x b y a bx y

    putting we get 2x 5y 100 56. C

    Sol. Slope of tangent to 2 2x y 1 at 1 1,2 2

    2 2x y 1 1 1,2 2

    2 2x y 1 2x 2yy ' 0 2

    xy ' 1y

    y mx c is a tangent of 2 2x y 1 So y x c

    now distance of (3, 0) from y x c is c 3 12

    2c 6c 9 2 2c 6c 7 0 57. A Sol. Point of intersection of 2y x and y 2x 3 is obtained by 2x 2x 3 0 x 3,1

    So, area 1

    2

    3

    3 2x x dx

    2 2 3 31 3 1 33 4 2

    2 3

    28 3212 83 3

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-32

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    58. NA Sol. NA 59. A Sol. a b c a b a

    a.b c a.a b a.b a

    ˆ ˆ ˆ ˆ ˆ ˆ4c 6 i j k 4 i 2 j k

    ˆ ˆ ˆ4c 2i 2 j 2k

    1 ˆ ˆ ˆc i j k2

    1b.c2

    60. B

    Sol.

    2

    2

    x ; x 1,2x 1f x

    2x ; x 2, 3x 1

    f x is a decreasing function

    2 1 6 4y , ,5 2 10 5

    61. B Sol. 3 6,b 1 ....... 101 2 4 198 00a 1 1 ........

    1012

    2 2

    1 1 11 1

    1 1

    Equation 2x 101 1 x 101 1 0 2x 102x 101 0 62. A Sol. Using L’ Hospital

    x 0

    x sin 10xlim 0

    1

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-33

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    63. C

    Sol. 2 2

    D 2 3 54 6

    D 8 2 for 2

    1

    5 2 2D 8 3 5

    10 2 6

    5 18 10 2 48 50 2 16 30 40 4 28 0 No solution for 2 64. C

    Sol. 2 36 6 6 4 2 6 2 2 6 20 2 4 62 C x C x x 1 C x x 1 C x 1 6 6 4 2 4 2 2 4 62 x 15 x x 15x x 2x 1 1 3x 3x x 6 4 22 32x 48x 18x 1 96 and 36 132 65. C

    Sol. x2x 4by ' b

    2y '

    So, differential equation is 2

    2 2x xx .yy y

    2 2

    2 2x x dy dyx .y x 2y xy y dx dx

    66. A Sol. 2 2x 2xy 3y 0 2 2x 3xy xy 3y 0 x y x 3y 0 x y 0 x 3y 0 (2, 2) satisfy x y 0 Normal x y 4

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-34

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    Hence x y 4

    Perpendicular distance from origin 0 0 4 2 22

    0 0 4 2 22

    67. B

    Sol. Characteristics equation of matrix ‘A’ is 22 x 2

    0 x 6x 10 09 4 x

    2A 6A 10I 0 110A A 6I 68. B

    Sol. P (exactly one) 25

    2P A P B 2P A B5

    1P A B2

    1P A P B P A B2

    1 2 5 4 1P A B2 5 10 10

    69. B

    Sol. 3 2

    1f x2x 9x 12x 4

    2

    33 2 2

    6x 18x 121f ' x2

    2x 9x 12x 4

    3

    3 2 2

    6 x 1 x 2

    2 2x 9x 12x 4

    1 1f 1 , f 23 8

    1 1I3 8

    70. B Sol. Let x3 t

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  • JEE-MAIN-2019 (8th Jan-Second Shift)-PCM-35

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    t t 1 2 t 1 t 2

    t a x3 a 3x log a so singleton set 71. 3 Sol. Let 3 2f x ax bx cx d

    1a4

    35d4

    3b4

    9c4

    3 2f x a x 3x 9x d 23f ' x x 2x 34 f ' x 0 x 3, 1

    local minima exist at x = 3 72. 0.5 Sol. 22ty x t 2Q t ,0

    2

    2

    0 0 11 t t 1 42

    t 0 1

    3t 8 t 2 t 0

    1m2

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  • JEE-MAIN-2020 (8th Jan-Second Shift)-PCM-36

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    73. 504

    Sol. 7

    3 2

    n 1

    1 2n 3n n4

    21 7.8 7.8.15 7.82 3

    4 2 6 2

    1 2 49 16 28 15 284

    1 1568 420 28 5044

    74. 2454 Sol. EXAMINATION 2N, 2A, 2I, E, X, M, T, O

    Case I All the different so 8 48!P 8.7.6.5 16804!

    Case II 2 same and 2 different so 3 71 24!C . C . 3.21.12 7562!

    Case III 2 same and 2 same so 3 24!C . 3.6 18

    2!.2!

    Total = 1680 + 756 + 18 = 2454 75. 1

    Sol. 2 sin 1

    72 cos

    and 2 sin 1

    2 10

    1tan7

    1sin10

    1tan3

    1 22. 33 3tan2 1 8 419 9

    4 211 3

    tan tan2 287 4tan 2 11 3 251 tan tan2 1 .7 4 26

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