Matrix€¦ · Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999,...

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1 Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999 Matrix JEE Academy JEE (MAIN ONLINE) 2019 PHYSICS 09 Jan. 2019 [Session : 09.30 AM to 12.00 PM] JEE MAIN PAPER ONLINE RED COLOUR IS ANSWER IN JEE-MAIN 1. A mixture of 2 moles of helium gas (atomic mass = 4 u), and 1 mole of argon gas (atomic mass = 40 u) is kept at 300 K in a container The ratio of their rms speeds rms rms V (helium) V (argon) , is close to: ,d ik=k es a 2 eksy ghfy;e (ijek.kq nzO;eku = 4 u), rFkk 1 eks y vkxZ u (ijek.kq nzO;eku = 40 u) xSlksa dk feJ.k 300K ij j[kk x;k gSA ijek.kqvksa ds oxZ ek/; ew y osxksa ds vuqikr] rms rms V (helium) V (argon) , dk fudV eku gksxk % (1) 224 (2) 3.16 (3) 0.45 (4) 0.32 A. 2 Question ID : 41652910056 Option 1 ID : 41652939684 Option 2 ID : 41652939685 Option 3 ID : 41652939683 Option 4 ID : 41652939682 sol. rms 3KT v m m = atomic mass m helium = 4µ rms 1 v m m argon = 40µ arg on rms rms helium m v (helium) 40µ 10 3.16 v (argon) m

Transcript of Matrix€¦ · Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999,...

  • 1Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    PHYSICS09 Jan. 2019 [Session : 09.30 AM to 12.00 PM]

    JEE MAIN PAPER ONLINERED COLOUR IS ANSWER IN JEE-MAIN

    1. A mixture of 2 moles of helium gas (atomic mass = 4 u), and 1 mole of argon gas (atomic mass = 40 u) is

    kept at 300 K in a container The ratio of their rms speeds rms

    rms

    V (helium)

    V (argon) , is close to:

    ,d ik=k esa 2 eksy ghfy;e (ijek.kq nzO;eku = 4 u), rFkk 1 eksy vkxZu (ijek.kq nzO;eku = 40 u) xSlksa dk feJ.k 300K ij

    j[kk x;k gSA ijek.kqvksa ds oxZ ek/; ewy osxksa ds vuqikr] rms

    rms

    V (helium)

    V (argon) , dk fudV eku gksxk %

    (1) 224 (2) 3.16 (3) 0.45 (4) 0.32

    A. 2

    Question ID : 41652910056

    Option 1 ID : 41652939684

    Option 2 ID : 41652939685

    Option 3 ID : 41652939683

    Option 4 ID : 41652939682

    sol. rms3KT

    vm

    m = atomic mass

    mhelium

    = 4µ

    rms

    1v

    m m

    argon = 40µ

    argonrms

    rms helium

    mv (helium) 40µ10 3.16

    v (argon) m 4µ

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    2. A rod, of length L at room temperature and uniform area of cross section A, is made of a metal havingcoeficient of linear expansion /ºC. It is observed that an external compressive force F, is applied on each ofits ends, prevents any change in the length of the rod, when its temperature rises by T K. Young's modulus, Y,for this metal is:

    js[kh; izlkj xq.kkad /°C okyh /kkrq ls cuh yEckbZ L, rFkk ,d leku vuqizLFk dkV ds {ks=kQy A dh ,d NM+ dks d{k rkiekuij j[kk x;k gSA tc ,d cká lankch cy F dks blds izR;sd fljksa ij yxkrs gSa] rks TK dh rkieku o`f) gksus ij] NM+ dhyEckbZ esa dksbZ ifjorZu ugha ik;k tkrk gSA bl /kkrq dk ;ax izR;kLFkrk xq.kkad Y gksxk %

    (1) 2F

    A T (2) F

    A T (3) F

    2A T (4) F

    A ( T–273)

    A. 2

    Question ID : 41652910053

    Option 1 ID : 41652939673

    Option 2 ID : 41652939671

    Option 3 ID : 41652939672

    Option 4 ID : 41652939670

    sol.

    f iL L (1 T)

    L L T

    stress F / A FY

    strain L / L A T

    3. Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of

    the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio:

    nks dyklEc) rjax L=kksrksa ls mRiUu fofHkUu rhozrkvksa dh rjaxksa dk O;frdj.k gksrk gSA O;frdj.k ds ckn vf/kdre rFkk U;wure

    rhozrkvksa dk vuqikr 16 gS] rks rjaxksa dh rhozrkvksa dk vuqikr gksxk %

    (1) 5 : 3 (2) 4 : 1 (3) 25 : 9 (4) 16 : 9

    A. 3

    Question ID : 41652910070

    Option 1 ID : 41652939741

    Option 2 ID : 41652939738

    Option 3 ID : 41652939740

    Option 4 ID : 41652939739

    sol.max

    min

    I16,

    I

    2

    1 2

    2

    1 2

    I I16

    I I

    1 2

    1 2

    I I4

    I I

    ,

    1 2 1 2

    1 2 1 2

    I I I I 4 1

    4 1I I I I

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    1 1

    22

    I I5 25,

    3 I 9I

    4. A convex lens is put 10 cm from a light source and it makes a sharp image on a screen kept 10 cm from thelens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source.To get the sharp image again, the screen is shifted by a distance d. Then d is:

    (1) 0.55 cm away from the lens (2) 0.55 cm towards the lens

    (3) 1.1 cm away form the lens (4) 0

    ,d mÙky ysal dks ,d izdk'k lzksr ls 10 cm nwjh ij j[kus ls mldk Li"V izfrfcac ysal ls 10 cm nwjh ij j[kh LØhu ij curk

    gSA tc ,d dk¡p (viorZukad 1.5) ds 1.5 cm eksVs xqVds dks izdk'k lzksr ds fcydqy lVkdj j[krs gSa rks] iqu% Li"V izfrfcEc dksikus ds fy;s LØhu dks d nwjh ls f[kldkuk iM+rk gSA rks d dk eku gksxk %

    (1) 0.55 cm ysal dh rjQ (2) 0.55 cm ysal ls nwj

    (3) 1.1 cm ysal ls nwj (4) 0

    A. 1

    Question ID : 41652910069

    Option 1 ID : 41652939736

    Option 2 ID : 41652939735

    Option 3 ID : 41652939737

    Option 4 ID : 41652939734

    sol.1 1 1

    v µ f

    1 1 1

    10 10 f

    f= 5 cmif glass block of 1.5 cm thickness is placed

    shift of object = 3 1 1

    1 0.52 1.5 2

    µ' = 10–0.5 = 9.5

    1 1 1

    v 9.5 5

    1 1 10 9

    v 5 95 95

    95v 10.55

    9

    Screen is shifted by = 10.55 – 10 = .55

    5. Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equalmasses, m while C has mass M. Block A is given an inital speed towards B due to which it collides with B

  • 4Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    perfectly inelastically. The combined mass collides with C, also perfectly inelastically 5

    6th of the initial kinetic

    energy is lost in whole process. What is value of M/m?

    fp=kkuqlkj ,d fpdus {kSfrt lery ij rhu xqVds A, B rFkk C j[ks gSaA A ,oa B dk nzO;eku cjkcj rFkk m gS] tcfd C dknzO;eku M gSA xqVds A dks ,d vkjfEHkd xfr , B dh vksj nh tkrh ftlls ;g B ls ,d iw.kZr;k vizR;kLFk VDdj djrk gSA

    ;g la;qDr nzO;eku xqVds C ls Hkh ,d iw.kZr;k vizR;kLFk VDdj djrk gSA bu VDdjksa esa vkjfEHkd xfrt ÅtkZ dk 5

    6 Hkkx

    {kf;r gks tkrk gSA M/m dk eku gksxk %

    (1) 2 (2) 3 (3) 5 (4) 4

    A. 4

    Question ID : 41652910049

    Option 1 ID : 41652939656

    Option 2 ID : 41652939654

    Option 3 ID : 41652939657

    Option 4 ID : 41652939655

    sol.

    1st collision : mv = 2mv'

    Perfectly inelastic collision v

    v '2

    2nd collision : Perfectly inelastic2mv' = (2m + M) v''

    2mv ' (2 m M) v''

    2mv(2m M) v''

    2

    mvv''

    (2m M)

    In whole process 5

    6th of initial kinetic energy is lost so remaining K.E.

    K.E. = 2 21 5 1mv . mv

    2 6 2

    2mv

    12

    So 2

    2mv 1 (2m M) v''12 2

    22mv 1 mv(2m M)

    12 2 2m M

    M/m = 4

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    6. A heavy ball of mass M is suspended from the ceiling of a car by a light string of mass m (m < < M). When thecar is at rest, the speed of transverse waves in the string is 60ms–1. When the car has accelearation a, the wave-speed increases to 60.5 ms –1. The value of a, in terms of gravitational acceleration g, is closet to:

    nzO;eku M dh ,d Hkkjh xsan dks ,d dkj dh Nr ls ,d nzO;eku m dh gYdh Mksjh (m < < M) ls yVdk;k x;k gSA tc dkjfLFkjkoLFkk esa gS rks Mksjh esa vuqizLFk rjaxksa dh xfr 60ms–1 gSA tc dkj dk Roj.k a gS] rjax xfr 60.5 ms –1 gks tkrh gSA a dk]xq:Roh; Roj.k g ds :i esa] lfUudV eku gksxk %

    (1) 20

    g(2)

    10

    g(3)

    30

    g(4)

    5

    g

    A. 4

    Question ID : 41652910058

    Option 1 ID : 41652939692

    Option 2 ID : 41652939690

    Option 3 ID : 41652939693

    Option 4 ID : 41652939691

    sol. Speed of transverse wave T

    V T

    212 2 22

    V Mg

    V M a g

    2

    2 2

    60 g

    60.5 a g

    (Solve by using binomial approximation)

    a = 5

    g

    7. If the angular momentum of a planet of mass m, moving around the Sun in a circular orbit is L, about the centerof the Sun, its areal velocity is:

    ;fn lw;Z ds ifjr% o`Ùkh; d{k esa ?kwers gq, nzO;eku m ds ,d xzg dk] lw;Z ds dsUnz ds lkis{k] dks.kh; laosx L gS rks] bldh {ks=kh;xfr gksxh %

    (1) 2

    L

    m(2)

    2L

    m(3)

    4L

    m(4)

    L

    m

    A. 1

    Question ID : 41652910052

    Option 1 ID : 41652939668

    Option 2 ID : 41652939667

    Option 3 ID : 41652939669

    Option 4 ID : 41652939666

    sol. Area velocity of circular orbit = 2Area r mvr

    Time 2 r / v 2m

    Angular momentum L = mvr

  • 6Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    Area velocity = L

    2m

    8. A copper wire is stretched to make is 0.5% longer. The percentage change in its electrical resistance if itsvolume remains unchanged is:

    ,d rk¡cs ds rkj dks [khapdj 0.5% ls yEck dj fn;k tkrk gSA ;fn bldk vk;ru ugha cnyrk gS rks] blds fo|qr&izfrjks/k esaizfr'kr ifjorZu dk eku gksxk %

    (1) 2.5% (2) 0.5% (3) 1.0% (4) 2.0%

    A. 3

    Question ID : 41652910046

    Option 1 ID : 41652939644

    Option 2 ID : 41652939642

    Option 3 ID : 41652939643

    Option 4 ID : 41652939645

    sol. RA

    l

    If % change in length = 0.5% 100 0.5%

    l

    l

    then to remain volume unchange % change in Area will be A

    100 0.5%A

    % change in resistance R

    100 100 100R

    l l

    l l

    = |0.5| + |–0.5| = 1%

    9. A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a currentof 5.2 A. The coercivity of the bar magnet is:

    ,d NM+ pqEcd dks 0.2 eh- yEch rFkk 100 Qsjksa okyh ,d ifjukfydk ds vUnj j[kdj fopqEcfdr djrs gSaA ifjukfydk esa 5.2A /kkjk izokfgr gks jgh gSA NM+ pqEcd dh fuxzkfgrk gS %

    (1) 285 A / m (2) 2600 A /m (3) 520 A/m (4) 1200 A/m

    A. 2

    Question ID : 41652910066

    Option 1 ID : 41652939724

    Option 2 ID : 41652939725

    Option 3 ID : 41652939723

    Option 4 ID : 41652939722

    sol. Magnetic field due to solenoid is B = 0Ni

    µl

    , B = µ0H,

    coercivity (H) = 0

    B N 100 5.22600A / m

    µ 0.2

    i

    l

    10. A block of mass m, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of springconstant k. The othe end of the spring is fixed, as shown in the figure. The block is initally at rest inits equilibriumposition. If now the block is pulled with a constant force F, the maximum speed of the block is:

    fpduh lrg ij j[ks m nzO;eku ds ,d xqVds dks fLizax fu;rkad k dh ,d dekuh (ftldk nzO;eku ux.; gS) ls tksM+k x;k gSAdekuh dk nwljk fljk fp=kkuqlkj] vpy gSA vkjEHk esa xqVdk viuh lkE;koLFkk esa LFkk;h gSA ;fn xqVds dks ,d fu;r cy F ls

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    [khapk tk;s rks xqVds dh vf/kdre pky gksxh %

    (1) F

    mk(2)

    F

    mk(3)

    F

    mk

    (4)

    2F

    mk

    A. 1

    Question ID : 41652910051

    Option 1 ID : 41652939664

    Option 2 ID : 41652939663

    Option 3 ID : 41652939665

    Option 4 ID : 41652939662

    sol.

    at vmax

    acceleration = 0 So F = kx

    WET – Wall

    = K

    2 2 2i f max1 1

    k x x F.x mv 02 2

    2 2max

    1 1kx Fx mv

    2 2

    2 22max2

    kF F 1mv

    2k k 2

    22max

    F 1mv

    2k 2

    max

    Fv

    mk

    11. A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on theblock. The coefiicient of static friction between the plane and the block is 0.6 What should be the minimumvalue of force P, such that the block does not move downward? (take g = 10 ms –2)

    10 kg nzO;eku dk ,d xqVdk] ,d [kqjnqjs vkur lery ij] fp=kkuqlkj j[kk gSA xqVds ij 3N dk cy yxkrs gSaA xqVds rFkkvkur&lery ds chp LFkSfrd ?k"kZ.kkad 0.6 gSA cy P dk U;wure eku D;k gksxk ftlls fd xqVdk uhps dh vksj xfr ugha djsxk\(g = 10 ms –2 yhft;s)

    (1) 32 N (2) 25 N (3) 18 N (4) 23 N

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    A. 1

    Question ID : 41652910048

    Option 1 ID : 41652939651

    Option 2 ID : 41652939650

    Option 3 ID : 41652939653

    Option 4 ID : 41652939652

    sol.

    at equilibrium

    3 + mg sin45º = P + µmg sin 45º

    10g 6g3 P

    2 2

    3 4gP 32

    2

    12. A plane electromagnetic wave of frequencyy 50 MHz travels in free space along the positive x-direction. At a

    particular point in space and time, ˆ6.3E j

    V/m. The corresponding magnetic field B

    , at that point will be:

    vko`fÙk 50 MHz dh lery fo|qr pqEcdh; rjax /kukRed x fn'kk dh fn'kk esa] eqDr vkdk'k esa tk jgh gSA vkdk'k esa ,d

    fuf'pr le; rFkk fcUnq ij ˆ6.3E j

    V/m gSA rks blds laxr pqEcdh; {ks=k B

    gksxk %

    (1) –8 ˆ2.1×10 kT (2) 8 ˆ18.9×10 kT (3) –8 ˆ18.9×10 kT (4) –8 ˆ6.3×10 kT

    A. 1

    Question ID : 41652910068

    Option 1 ID : 41652939731

    Option 2 ID : 41652939732

    Option 3 ID : 41652939733

    Option 4 ID : 41652939730

    sol. E = CB

    8

    8

    E 6.3B 2.1 10 T

    C 3 10

    Direction of magnetic field is perpendicular to electric field and propogation of wave

    B

    = –8 ˆ2.1×10 kT

    13. A resistance is shown in the figure. Its value and tolerance are given respectively by:

    ,d izfrjks/k dks fp=k esa n'kkZ;k x;k gSA bldk eku rFkk lárk Øe'k%] gksaxs %

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    (1) 27 k, 10 % (2) 27 k, 20 % (3) 270 , 5 % (4) 270 , 10 %

    A. 1

    Question ID : 41652910075

    Option 1 ID : 41652939759

    Option 2 ID : 41652939760

    Option 3 ID : 41652939758

    Option 4 ID : 41652939761

    sol.

    2 Digit × multiplier ± tolerence

    27×103 ± 10%

    digit multiplier

    B– Black – 0 100

    B– Brown – 1 101

    R– Red– 2 102

    O– Orange – 3 103

    Y– Yellow – 4 104

    G– Green – 5 105

    B– Blue – 6 106

    V– Violet – 7 107

    G– Grey – 8 108

    W– White – 9 109

    Gold – 5%

    Silver –10%

    14. Surface of certain metal is first illuminated with light of wavelength 1 = 350 nm and then, by light of wavelenght

    2 = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2.

    The work function of the metal (in eV) is close to: (Energy of photon = 1240

    eV(in nm) )

    ,d /kkrq ds i`"B dks] igys 2 = 350 nm rjaxnS/;Z ds izdk'k vkSj fQj

    2 = 540 nm rjaxnS/;Z ds izdk'k ls] izdkf'kr djrs gSaA

    blls mRlftZr QksVksbysDVªkWuksa dh vf/kdre pkyksa esa 2 dk vuqikr ik;k tkrk gSA /kkrq ds dk;ZQyu dk] eV esa] eku gksxk %

    (QksVkWu dh ÅtkZ = 1240

    eV(in nm) )

    (1) 5.6 (2) 1.4 (3) 2.5 (4) 1.8

    A. 4

    Question ID : 41652910072

    Option 1 ID : 41652939747

    Option 2 ID : 41652939748

    Option 3 ID : 41652939749

    Option 4 ID : 41652939746

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    Sol. Given 21

    1

    hc 1mv

    2

    ......(1)

    22

    2

    hc 1mv

    2

    .....(2)

    1 2v 2v .......(3)

    From eq. (1), (2) & (3)

    21

    1

    22

    2

    hc1mv

    21 hc

    mv2

    1

    2

    hc

    4hc

    2 1

    4hc hc4

    2 1

    4hc hc3

    1240 12403 4

    540 350

    1240 4 11.88 eV

    3 10 54 35

    15. A parallel plate capacitor is made of two square plates of side 'a', separated by a distance d (d < < a). Thelower triangular portion is filled with a dieletric of dielectric constant K, as shown in the figure Capacitance ofthis capacitor is:

    Hkqtk a okyh nks oxkZdkj IysVksa dks nwjh d ij j[kdj ,d lekurj IysV la/kkfj=k cuk;k tkrk gSA fn;k gS (d < < a) A blesaijkoS|qrkad K ds ijkoS|qr dks fp=kkuqlkj yxkrs gSa ftlls blds fudys f=kHkqtkdkj Hkkx esa ijkoS|qr inkFkZ jgrk gSA blla/kkfj=k dh /kkfjrk gksxh %

    (1)

    20K a In K

    d(K-1)

    (2)

    20K a1

    2 d

    (3)

    20K a

    2d(K+1)

    (4)

    20K a In K

    d

    A. 1

    Question ID : 41652910061

    Option 1 ID : 41652939705

    Option 2 ID : 41652939702

    Option 3 ID : 41652939704

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    Option 4 ID : 41652939703

    sol.a x a

    y d

    ;

    d(a x)y

    a

    01

    adC

    d y

    0

    2

    k (adx)C

    y

    01 2eq

    1 2

    a kdxC CdC

    C C kd y(k 1)

    0eq

    a kdxdC

    xkd d(k 1) 1

    a

    a

    eq eq0C dC

    20K a In K

    d(K-1)

    16. A current loop, having two circular arcs joined by two radial lines is shown in the figure. It carries a current of10 A. the magnetic field at point O will be close to:

    nks o`Ùkkdkj pkiksa rFkk f=kT;d js[kkvksa ls cuk ,d /kkjk ik'k] fp=k esa fn[kk;k gSA ik'kk esa 10 A dh /kkjk izokfgr gks jgh gSA fcUnqO ij pqEcdh; {ks=k dk lfUudV eku gksxk %

    (1) 1.0 × 10–5 T (2) 1.0 × 10–7 T (3) 1.0 × 10–7 T (4) 1.5 × 10–5 T

    A. 1

    Question ID : 41652910064

    Option 1 ID : 41652939716

    Option 2 ID : 41652939717

    Option 3 ID : 41652939715

    Option 4 ID : 41652939714

    sol.

    0 0µ i µ iB2R 2 16R

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    0 0net 2 2

    µ i 2µ i1 1B

    16 10 5 3 15 16 10

    netB = 1.0 × 10–5 T

    17. An infinitely long current carrying wire and a small current carrying loop are in the plane of the paper as shown.The radius of the loop is a and distance of its centre from the wire is d (d >> a). if the loop applies a force F onwire then:

    ,d vuUr yackbZ dk /kkjkokgd rkj rFkk ,d NksVk lk /kkjkokgd ik'k dkxt ds lery esa fp=kkuqlkj j[ks gSaA ik'k dh f=kT;ka rFkk rkj ls blds dsUnz dh nwjh d gS (d >> a)A ;fn ik'k }kjk rkj ij cy F gS rks %

    (1) a

    Fd

    (2)

    2

    3

    aF

    d

    (3) F = 0 (4)

    2a

    Fd

    A. 4

    Question ID : 41652910065

    Option 1 ID : 41652939719

    Option 2 ID : 41652939721

    Option 3 ID : 41652939718

    Option 4 ID : 41652939720

    sol.

    F is force on wire due to loop

    F= ilBdue

    to loop at d distance.

    Magnetic field due to loop at d distance 0

    2

    µ MB

    4 d

    (M – magnetic moment)

    2 20 0

    2 2

    µ a µ aB

    4 d 4d

    i i, M = iA = ia2

    F B

    F

    2a

    d

    18. An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure.If AB = BC, and angle made by AB with downward vertical is , then:

    ,dleku nzO;eku ?kuRo dh NM+ksa ls cuk;h gqbZ L-dh vkd`fr ds ,d oLrq dks fp=kkuqlkj] ,d Nksjh ls yVdk;k x;k gSA ;fnAB = BC, rFkk AB }kjk Å/okZ/kj fuEu fn'kk ls cuk;k dks.k gS] rks %

  • 13Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    (1) tan = 1

    3(2) tan =

    1

    2 3(3) tan =

    1

    3(4) tan =

    1

    2

    A. 1

    Question ID : 41652910050

    Option 1 ID : 41652939659

    Option 2 ID : 41652939661

    Option 3 ID : 41652939660

    Option 4 ID : 41652939658

    sol.

    L-shaped object at equilibrium

    A = 0

    mg sin mg cos sin2 2

    l ll

    sin cossin

    2 2

    3 cossin

    2 2

    1tan

    3

    19. A conducting circular loop made of a thin wire, has area 3.5 × 10–3 m2 and resistance 10 . It is palcedperpendicular to a time dependent magnetic field B(t) = (0.4T) sin (50t). The field is uniform in space. Thenthe net charge flowing through the loop during t = 0 s and t = 10 ms is close to:

    ,d irys pkyd rkj ls cus gq, o`Ùkkdkj ik'k dk {ks=kQy 3.5 × 10–3 m2 rFkk izfrjks/k 10 gSA bls ,d yEcor~ pqEcdh;{ks=k] tks fd le; ij fuHkZj fdarq ,dleku gS] le; B(t) = (0.4T) sin (50t) esa j[kk x;k gSA le; t = 0 s ls t = 10 msrd ik'k esa cgus okys usV vkos'k dk eku gksxk %

    (1) 6 mC (2) 7 mC (3) 14 mC (4) 21 mC

    A. 3

  • 14Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    Question ID : 41652910067

    Option 1 ID : 41652939726

    Option 2 ID : 41652939727

    Option 3 ID : 41652939728

    Option 4 ID : 41652939729

    -----5-----

    sol.d AdB

    emfdt dt

    B(t) = 0.4 sin(50t), R = 10A = 3.5 × 10–3

    emf = 3.5 × 10–3 × 0.4 cos (50t) × 50emf = 7 × 10–2 cos (50t)

    3emfdi 7 10 cos(50 t)R

    t

    0dq didt

    310 103

    0q 7 10 cos 50 t

    q = .14 mCq = 14 × 10–5 C

    20. When the switch S, in the circuit shown, is closed then the value of current i will be:

    fn;s x;s ifjiFk esa tc fLop S dks cUn djrs gSa] rks /kkjk i dk eku gksxk %

    (1) 2 A (2) 4 A (3) 5 A (4) 3 A

    A. 3

    Question ID : 41652910063

    Option 1 ID : 41652939710

    Option 2 ID : 41652939712

    Option 3 ID : 41652939713

    Option 4 ID : 41652939711

    sol.

    i = i1 + i

    2

    0 0 0V 0 20 V 10 V

    2 2 4

    0V 5054 4

    V0 = 10

  • 15Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    0V 10i 5A2 2

    21. Consider a tank made a glass(refractive index 1.5) with a thick bottom. It is filled with a liquid of refractiveindex . A student finds that, irrespective of what the incident angle i (see figure) is for a beam of light enteringthe liquid, the light reflected from the liquid glass interface is never completely polarized. For this to happen theminimum value of is:

    dk¡p (viorZukad =1.5) ls cus ,d VSad dh ryh eksVh gSA blesa viorZukad dk ,d nzo Hkjk gSA ,d Nk=k ikrk gS fd fdlhHkh vkiru dks.k i (fp=k nsf[k;s) ij nzo esa vkifrr izdk'k dh fdj.k ds fy;s nzo&dk¡p vUri`Z"B ls ijkofrZr fdj.k] dHkh Hkhiw.kZr;k /kzqfor ugha gksrh gSA ,slk gksus ds fy;s] dk U;wure eku gksxk %

    (1) 5

    3(2)

    4

    3(3)

    5

    3(4)

    3

    5

    A. 4

    Question ID : 41652910071

    Option 1 ID : 41652939745

    Option 2 ID : 41652939742

    Option 3 ID : 41652939744

    Option 4 ID : 41652939743

    sol. For all beam of light should entering the liquid and light reflected from liquid glass interface is never completely

    polarized.

    1. sin90º = µ sin

    1sin

    µ .....(1)

    If Brewster angle –

    2

    1

    n 1.5tan

    n µ

    2

    3sin

    9 4µ

    .......(2)

    By (1) and (2)

    2

    1 3

    µ 9 4µ

    9µ2 = 6+4µ2

  • 16Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    µ =3

    5

    22. For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance hfrom its centre. Then value of h is:

    f=kT;k R ds ,d ,dleku vkosf'kr oy; ds fo|qr {ks=k dk eku mlds v{k ij dsUnz ls h nwjh ij vf/kdre gSA h dk eku gksxk %

    (1) R 2 (2) R

    2(3)

    R

    5(4) R

    A. 2

    Question ID : 41652910060

    Option 1 ID : 41652939699

    Option 2 ID : 41652939700

    Option 3 ID : 41652939701

    Option 4 ID : 41652939698

    sol. Electric field at a distance X on the axis from the centre of ring.

    3/22 2

    RQxE

    R x

    dE

    0dx

    3/2 1/22 2 2 2

    32 2

    3R x x. R x 2x

    dE 2RQ 0dx R x

    2

    3/2 1/22 2 2 23xR x .2 R x2

    2 2 2R x 3x R

    X2

    23. A particle is moving with a velocity ˆ ˆυ = K(yi + j)x , where K is a constant. The general equation for its path

    is:

    (1) xy = constant (2) y2 = x + constant (3) y2 = x2 + constant (4) y = x2 + constant

    ,d d.k osx ˆ ˆυ = K(yi + j)x nj ls py jgk gS] tgk¡ K ,d fu;rkad gSA bl d.k ds iFk dk O;kid lehdj.k gksxk %

    (1) xy = fu;rkad (2) y2 = x + fu;rkad (3) y2 = x2 + fu;rkad (4) y = x2 + fu;rkad

    A. 3

    Question ID : 41652910047

    Option 1 ID : 41652939649

    Option 2 ID : 41652939648

    Option 3 ID : 41652939646

    Option 4 ID : 41652939647

  • 17Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    sol. ˆ ˆυ = K(yi + j)x

    vx = ky v

    y = kx

    dxky

    dt

    dykx

    dt

    dx / dt y

    dy / dt x

    dx y

    dy x

    2 2x yc

    2 2

    2 2x yc

    2 2

    y2 = x2 + c

    24. Mobility of electrons in a semiconductor is defined as the ratio of their driftvelocity to the applied electric field.If, for an n-type semiconductor, the density of electrons is 1019 m–3 and their mobility is 1.6 m2 / (V.s) then theresistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:

    bysDVªkWuksa dh xfr'khyrk muds viokg osx rFkk yxk, gq;s fo|qr {ks=k ds vuqikr ls ifjHkkf"kr gksrh gSA ;fn ,d n-Vkbi ds v/kZpkyd esa bysDVªkWuksa dk la[;k ?kuRo 1019m–3 rFkk mudh xfr'khyrk 1.6 m2 / (V.s) gS] rks bldh izfrjks/kdrk dk lfUudVeku gksxk] (n-Vkbi v/kZpkyd esa gksyksa dk ;ksxnku mis{k.kh; gS) :

    (1) 2 m (2) 0.4 m (3) 0.2 m (4) 4 m

    A. 2

    Question ID : 41652910074

    Option 1 ID : 41652939755

    Option 2 ID : 41652939757

    Option 3 ID : 41652939754

    Option 4 ID : 41652939756

    sol. i = neAvd, mobility

    d dV V .µE E

    l

    Vµi neA

    l V = iR

    A

    ineµ

    V

    l

    neµRA

    l

    Reistivity 19 191 1 1

    .390 0.4 mneµ 10 1.6 10 1.6 1.6 1.6

    1neµ

    25. A gas can be taken from A to B via two different processes ACB and ADB.

  • 18Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    When path ACB is used 60 J of heat flows into the system and 30 J of work is done by the system. If path ADB

    is used work done by sytem is 10 J. The heat Flow into the system in path ADB is:

    ,d xSl dks voLFkk A Lks B esa nks fHkUu izØeksa ACB rFkk ADB }kjk ys tk ldrs gSaA izØe ACB esa 60 J Å"ek fudk; esa tkrh

    gS rFkk fudk; }kjk 30 J dk;Z fd;k tkrk gSA ;fn izØe ADB esa fudk; }kjk 10 J dk;Z fd;k tkrk gS rks blesa] fudk; esa Å"ek

    izokg dk eku gksxk %

    (1) 40 J (2) 100 J (3) 20 J (4) 80 J

    A. 1

    Question ID : 41652910055

    Option 1 ID : 41652939681

    Option 2 ID : 41652939678

    Option 3 ID : 41652939679

    Option 4 ID : 41652939680

    sol.

    Q = U + W U will be same because initial & final state is same as in path ACBIn path ACB – In path ADBQ = U + PV Q = U + W60 = U + 30 Q = 30 + 10U = 30 Q = 40

    26. Drfit speed of electrons, when 1.5 A of current flows in a coppe wire of cross section 5 mm2, is . If the

    electron density in copper is 9 × 1028 / m3 the value of in mm / s is close to

    (Take charge of electron to be = 1.6 × 10–19 C).

    rk¡cs ds 5mm2 vuqizLFk dkV ds {ks=kQy ds ,d rkj ls tc 1.5 A dh /kkjk cgrh gS rks bysDVªkWuksa dk viokg osx (drift velocity)

    gSA ;fn rk¡cs esa bysDVªkWuksa dh la[;k dk ?kuRo 9 × 1028 / m3 gS] rks dk mm / s esa] lfUudV eku gksxk]

    (fn;k gS % bysDVªkWu dk vkos'k = 1.6 × 10–19 C).

    (1) 0.02 (2) 2 (3) 0.2 (4) 3

    A. 1

    Question ID : 41652910062

    Option 1 ID : 41652939709

    Option 2 ID : 41652939707

    Option 3 ID : 41652939708

    Option 4 ID : 41652939706

    sol. i = neAVd

  • 19Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    d 28 19 6

    i 1.5V

    neA 9 10 1.6 10 5 10

    33

    d

    1.5 10V .02 10 m / sec

    9 1.6 5

    Vd = .02 mm/sec.

    27. Temperature difference of 120ºC is maintained between two ends of a uniform rod AB of length 2L. Another

    bent rod PQ, of same cross-section as AB and length 3L

    2 is is connected across AB (See figure). In steady

    state, temperature difference betwen P and Q will be close to:

    2L yEckbZ dh ,d NM+ AB ds nks fljksa ds chp rkikUrj 120ºC j[kk x;k gSA ,d vkSj blh vuqizLFk dkV dh 3L

    2 yEckbZ dh

    eqM+h gq;h NM+ PQ dks fp=kkuqlkj AB ls tksM+k x;k gSA fLFkjkoLFkk esa P rFkk Q ds chp rkieku ds vUrj dk lfUudV eku gksxk %

    (1) 35 ºC (2) 75 ºC (3) 45 ºC (4) 60 ºC

    A. 3

    Question ID : 41652910054

    Option 1 ID : 41652939676

    Option 2 ID : 41652939675

    Option 3 ID : 41652939677

    Option 4 ID : 41652939674

    sol.

    = = Req

    = 8R/5

    ABP Q

    eq

    T 3RT T

    R 5

    ABP Q

    120 T 3R 120 5 3RT T 45º C

    8R / 5 5 8R 5

    28. Two masses m and m

    2 are connected at the two ends of a massless rigid rod of length l. The rod is suspended

    by a thin wire of torsional constant k at the centre of mass of the rod-mass system(see figure). Because oftorsional constant k, the restoring torque is = k for angular displacement . If the rod is rotated by

    0 and

  • 20Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    released, the tension in it when it passes through its mean position will be:

    nzO;eku m rFkk m

    2 ds nks fi.Mksa dks ,d yEckbZ l dh nzO;ekujfgr NM+ ds fljksa ij tksM+k x;k gSA bl NM+ dks ,d ejksM+kad k

    ds rkj ls] NM+&nzO;eku la;kstu ds nzO;eku dsUnz ls] fp=kkuqlkj] yVdk;k x;k gSA ejksM+kad k ds dkj.k NM+ ds dks.kh; foLFkkiu ls] ml ij cy vk?kw.kZ = kyxrk gSA ;fn NM+ dks

    dks.k ls ?kqek dj NksM+ nsrs gSa rks] blesa ruko dk eku tc NM+ viuh

    ek/; voLFkk ls xqtjrh gS] gksxk %

    (1) 2

    0kθ

    l(2)

    202kθ

    l(3)

    203kθ

    l(4)

    20kθ

    2l

    A. 1

    Question ID : 41652910057

    Option 1 ID : 41652939687

    Option 2 ID : 41652939688

    Option 3 ID : 41652939689

    Option 4 ID : 41652939686

    Sol.

    work done to rotated by 0

    0

    0d d

    0

    0d k d

    20kW

    2

    Work done = 21 I

    2

    220k 1 I

    2 2

    22 220k 1 m 2 m.

    2 2 2 3 9

    l l

    2 22 20

    2m mk

    9 9

    l l

    22 20

    mk

    3

    l

  • 21Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    22 0

    2

    3k

    m

    l

    Tension force rotated the mass so T = 2 2m 2. m

    2 3 3

    l l =

    20kθ

    l

    29. Three charges + Q, q, + Q are placed rspectively, at distance , 0 d / 2 and d from the origin, on the x-axis. Ifthe net force experienced by + Q, placed at x = 0, is zero, then value of q is:

    + Q, q rFkk + Q ds rhu vkos'kksa dks x-v{k ij ewyfcUnq ls Øe'k% nwjh 0, d /2 rFkk d ij j[kk x;k gSA ;fn x = 0 ij j[ks +Qvkos'k ij dqy cy 'kwU; gS] rks q dk eku gksxk %

    (1) – Q/2 (2) – Q/4 (3) + Q/2 (4) + Q/4

    A. 2

    Question ID : 41652910059

    Option 1 ID : 41652939697

    Option 2 ID : 41652939695

    Option 3 ID : 41652939696

    Option 4 ID : 41652939694

    sol.

    2

    2 2

    kQq kQ0

    dd

    2

    2kQ

    4q Q 0d

    q Q / 4

    30. A sample of radioactive mateial A, that has an activity of 10 mCi(Ci = 3.7 × 1010 decays/s), has twice thenumbe of nuclei as anothe sample of a different radioactive mateial B which has an activity of 20 mCi. Thecorrect choices for half-lives of A and B would then be respectively:

    (1) 20 day and 5 days (2) 10 days and 40 days

    (3) 20 days and 10 days (4) 5 days and 10 days

    jsfM;ks/kehZ inkFkZ A ds ,d uewus dh ,fDVork 10 mCi(Ci = 3.7 × 1010 decays/s) gSA bl uewus esa ukfHkdksa dh la[;k nwljsjsfM;ks/kehZ inkFkZ B ds uewus ds ukfHkdksa dh la[;k dh nqxquh gSA nwljs uewus dh ,fDVork 20 mCi gSA A vkSj B dh] Øe'k%]v/kZvk;q ds ckjs esa dkSu&lk dFku lR; gS\

    (1) 20 fnu ,oa 5 fnu (2) 10 fnu ,oa 40 fnu

    (3) 20 fnu ,oa 10 fnu (4) 5 fnu ,oa 10 fnu

    A. 1

    Question ID : 41652910073

    Option 1 ID : 41652939751

    Option 2 ID : 41652939753

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    Option 3 ID : 41652939750

    Option 4 ID : 41652939752

    sol. Rate of radioactive decay A = n

    AN

    A = 10

    BN

    B = 20 Given

    A A

    B B

    N 1

    N 2

    A

    B

    N2

    N

    A

    B

    2 1

    1 2

    1

    A

    B

    1

    4

    B

    A

    t½ 1

    t½ 4

    By option (1)

  • 23Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    CHEMISTRY09 Jan. 2019 [Session : 09.30 AM to 12.00 PM]

    JEE MAIN PAPER ONLINERED COLOUR CONSIDER OFFICIAL ANSWER

    1. Correct statements among a to d regarding silicones are:

    (a) They are polymes with hydrophobic character.

    (b) They are biocompatible.

    (c) In general, they have high thermal stability and low dielectric strength.

    (d) Usually, they are resistant to oxidation and used as greases.

    (1) (a), (b) and (c) only (2) (a), (b) and (d) only

    (3) (a) and (b) only (4) (a) , (b), (c) and (d)

    a ls d esa ls flfydkWu ds laca/k esa lgh dFku gSa%

    (a) ;s cgqyd ty&fojkxh izd`fr ds gksrs gSaA

    (b) budh tSolaxfrrk gksrh gSA

    (c) lk/kkj.kr;k] budk mPp Å"ek LFkkf;Ro rFkk fuEu ijkoS|qr lkeF;Z gksrk gSA

    (d) lkekU;r;k] ;s vkWDlhdj.k izfrjks/kh gksrs gSa rFkk xzht dh rjg mi;ksx esa yk;s tkrs gSaA

    (1) dsoy (a), (b) rFkk (c) (2) dsoy (a), (b) rFkk (d)

    (3) dsoy (a) rFkk (b) (4) (a), (b), (c) rFkk (d)

    A. 2

    Question ID : 41652910092

    Option 1 ID : 41652939826

    Option 2 ID : 41652939829

    Option 3 ID : 41652939828

    Option 4 ID : 41652939827

    sol. Silicones are polymer with Si–O–Si linkages and are strongly hydrophobic. They are highly thermally stable

    with high dielectric strength. Now a days silicone greases are commony used.

    2. For emission line fo atomic hydrogen from ni = 8 to n

    f = n, the plot of wave number ( ) against 2

    1

    n

    will

    be (The Rydbeg constant RH is in wave numbe unit)

    (1) Linear with slope – RH

    (2) Linear with slope RH

    (3) Linear with intecept – RH

    (4) Non linear

    ijek.kq gkbMªkstu ds ni = 8 ls n

    f = n rd dh mRltZu ykbu ds fy, 2

    1

    n

    ds fo:) rjax la[;k ( ) dk IykV gksxk]

    (fjMcxZ fLFkjkad] RH rjax la[;k ds ek=kd esa)

    (1) – RH

    Lyksi ds lkFk jSf[kd (2) RH

    Lyksi ds lkFk jSf[kd

    (3) – RH

    vUr% [k.M ds lkFk jSf[kd (4) vjSf[kd

    A. 1

    Question ID : 41652910098

    Option 1 ID : 41652939851

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    Option 2 ID : 41652939853

    Option 3 ID : 41652939850

    Option 4 ID : 41652939852

    sol.2

    H 2 22 1

    1 1v R z (z 1)

    n n

    H 22

    1 1v R

    n 8

    H H2

    R Rv

    n 64

    y mx c

    H2

    1x ,m R (slope)

    n

    3. Which one of the following statements regarding Henry's law is not correct?

    (1) The partical pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution

    (2) The value of KH increases with increase of temperature and K

    H is function of the nature of the gas

    (3) Different gases have different KH (Henry's law constant) values at the same temperature

    (4) Higher the value of KH at a given pressure, higher is the solubility of the gas in the liquids.

    gsujh fu;e ds laca/k esa fuEufyf[kr dFkuksa esa ls dkSu lk ,d lgh ugha gS\

    (1) ok"i izkoLFkk esa xSl dk vkaf'kd nkc foy;u esa xSl ds eksyka'k ds lekuqikrh gksrk gSA

    (2) KH

    dk eku rki ds c

  • 25Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    Sol. Copper pyrites CuFeS2

    Dolomite MgCO3.CaCO

    3

    Malachite CuCO3.Cu(OH)

    2

    Azurite 2CuCO3.Cu(OH)

    2

    Copper pyrites contains both copper and iron

    5. The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4

    electrolyzed in g during the process is: (Molar mass of PbSO4 = 303 g mol–1)

    ,d ysM&vEy cSVjh ds ,uksMh v)Z&lsy dks 0.05 QSjkMs fo|qr dk mi;ksx djds iqu% vkosf'kr fd;k tkrk gSA bl izØe esafo|qr vi?kfVr PbSO

    4 dh ek=kk (g esa) gS % (PbSO

    4 dk eksyj nzO;eku = 303 g mol–1)

    (1) 22.8 (2) 11.4 (3) 7.6 (4) 15.2

    A. 3

    Question ID : 41652910103

    Option 1 ID : 41652939872

    Option 2 ID : 41652939873

    Option 3 ID : 41652939871

    Option 4 ID : 41652939870

    sol.

    6. In general , the properties that decrease and increase down a group in the periodic table, respectively, are:

    (1) Electronegativity and electron gain enthalpy

    (2) Electron gain enthalpy and electronegativity

    (3) Electronegativity and atomic radius.

    (4) Atomic radius and electronegativity

    lekU;r%] vkoÙkZ lkj.kh ds oxZ esa uhps tkus ij ?kVus rFkk c

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    Option 3 ID : 41652939802

    Option 4 ID : 41652939804

    sol. Down the group

    Electronegativity decrease as size increases

    1EN

    size

    7. According to molecular orbital theory which of the following is true with respect to 2Li and 2Li

    ?

    (1) Both are stable (2) 2Li is stable and 2Li

    is unstable

    (3) 2Li is unstable and 2Li

    is stable (4) Both are unstable

    vkf.od d{kd fl)kUr ds vuqlkj 2Li rFkk 2Li

    ds laca/k esa fuEufyf[kr esa ls dkSu lR; gS?

    (1) nksuksa LFkk;h gSa (2) 2Li LFkk;h gS rFkk 2Li

    vLFkk;h gS

    (3) 2Li vLFkk;h gS rFkk 2Li

    LFkk;h gS (4) nksuksa vLFkk;h gSa

    A. 1

    Question ID : 41652910099

    Option 1 ID : 41652939857

    Option 2 ID : 41652939855

    Option 3 ID : 41652939856

    Option 4 ID : 41652939854

    sol. Electronic configurations of Li2+ and Li

    2– are

    Li+2: 1s2 *1s2 2s1

    Li–2: 1s2 *1s2 2s2 *2s1

    Bond order of Li+2 = ½ (3 – 2) = ½

    Bond order of Li–2 = ½ (4 – 3) = ½

    Since both Li+2 and Li–

    2 have +ve bond order, both are stable (reference : NCERT)

    8. The major product of following reaction is:

    fuEufyf[kr vfHkfØ;k dk eq[; mRikn gSA

    2

    2

    (1)AlH(i Bu)

    (2)H OR – C N ?

    (1) RCH2NH

    2(2) RCOOH (3) RCONH

    2(4) RCHO

    A. 4

    Question ID : 41652910079

    Option 1 ID : 41652939774

    Option 2 ID : 41652939776

    Option 3 ID : 41652939775

    Option 4 ID : 41652939777

    sol. 22

    (1)AlH(i Bu)

    (2)H OR – C N R—CHO

    AlH (i – Bu)2 is DIBALH which reduces nitrites to aldehydes.

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    9. The correct match between Item-I and Item-II is: Item-I Item-II (drug) (test)A Chloroxylenol P Carbylamine testB Norethindrone Q Sodium hydrogen carbonate testC Sulphapyridine R Ferric chloride testd Peniclillin P Bayer's test

    enksa-I rFkk II ds e/; lgh lqesy gS % en-I en-II (vkS"kf/k) (ijh{k.k)A Dyksjkstbfyuky P dkfcZy,sehu ijh{k.kB ukj,fFkuMªku Q lksfM;e gkbMªkstu dkcksZusV ijh{k.kC lYQkfifjMhu R Qsfjd DyksjkbM ijh{k.kd isfuflfyu P csvj ijh{k.k

    (1) A Q; B P ; C S ; D R (2) A Q; B S ; C P ; D R

    (3) A R; B P ; C S ; D Q (4) A R; B S ; C P ; D Q

    A. 4

    Question ID : 41652910084

    Option 1 ID : 41652939795

    Option 2 ID : 41652939796

    Option 3 ID : 41652939794

    Option 4 ID : 41652939797

    sol. * Chloroxylenol is dettol contain phenolic group so give FeCl3 test

    * Norethindrone has double bond so will give Baeyer's reagent test* Sulphapyridine has – NH

    2 group it give carbyl amine test

    * Penicillin has – COOH group so will respond to NaHCO3 test

    10. The compounds A and B in the following reaction are, respectively:

    fuEufyf[kr vfHkfØ;k esa ;kSfxd A rFkk B Øe'k% gS %

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    (1) A = Benzyl alcohol, B = Benzyl isocyanide (2) A = Benzyl chloride, B = Benzyl isocyanide

    (3) A = Benzyl alcohol, B = Benzyl cyanide (4) A = Benzyl chloride, B = Benzyl cyanide

    (1) A = csfUty ,sYdksgky, B = csfUty vkblkslk;ukbM

    (2) A = csfUty DyksjkbM, B = csfUty vkblkslk;ukbM

    (3) A = csfUty ,sYdksgky, B = csfUty lk;ukbM

    (4) A = csfUty DyksjkbM, B = csfUty lk;ukbM

    A. 2

    Question ID : 41652910081

    Option 1 ID : 41652939782

    Option 2 ID : 41652939783

    Option 3 ID : 41652939785

    Option 4 ID : 41652939784

    sol.

    11. The highest value of the calculated spin only magnetic moment (in BM) among all the transition metal complexesis:

    lHkh laØe.k /kkrq ladqyksa esa lokZf/kd ifjdfyr izpØ.k ek=k pqacdh; vk?kw.kZ (BM esa) gS %

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    (1) 5.92 (2) 6.93 (3) 4.90 (4) 3.87

    A. 1

    Question ID : 41652910093

    Option 1 ID : 41652939830

    Option 2 ID : 41652939832

    Option 3 ID : 41652939831

    Option 4 ID : 41652939833

    sol. The transition metal atom/ion in a complex may have unpaired electrons ranging from zero to 5. So, maximum

    number of unpaired electrons that may be present in a complex is 5. Magnetic moment is given as

    µ n(n 2) BM [no. of unpaired electrons = n)

    Maximum value of magnetic moment

    5(5 2) 35 = 5.92 BM

    12. The increasing order of pKa of the following amino acids in aqueous solution is:

    tyh; foy;u esa fuEufyf[kr ,sehuksa vEyksa ds pKa dk c Lys > Gly > Asp.

    13. Aluminium is usually found in + 3 oxidation state. In contrast, thallium exists in + 1 and + 3 oxidation states.This is due to:

    (1) Diagonal relationship (2) Inert pair effect

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    (3) Lanthanoid contraction (4) Lattice effect

    ,syqehfu;e lkekU;r;k + 3 vkWDlhdj.k voLFkk esa ik;k tkrk gSA blds foijhr] FkSfy;e + 1 rFkk + 3 vkWDlhdj.k voLFkkvksa

    esa jgrk gSA bldk dkj.k gS %

    (1) fod.kZ laca/k (2) vfØ; ;qXe izHkko

    (3) ySUFksukW;M vkdqapu (4) ySfVl izHkko

    A. 2

    Question ID : 41652910091

    Option 1 ID : 41652939823

    Option 2 ID : 41652939825

    Option 3 ID : 41652939822

    Option 4 ID : 41652939824

    sol. +1 is more stable form of Thallium due to inert pair effect. For Tl +1 > +3 oxidation state.

    14. Major product of the following reaction is:

    fuEufyf[kr vfHkfØ;k dk eq[; mRikn gS %

    (1) (2)

    (3) (4)

    A. 2

    Question ID : 41652910076

    Option 1 ID : 41652939765

    Option 2 ID : 41652939763

    Option 3 ID : 41652939764

    Option 4 ID : 41652939762

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    sol.

    15. Which amongst the following is the strongest acid?

    fuEu esa ls dkSu izcyre vEy gS\

    (1) CHBr3

    (2) CHl3

    (3) CH(CN)3

    (4) CHCl3

    A. 3

    Question ID : 41652910083

    Option 1 ID : 41652939791

    Option 2 ID : 41652939792

    Option 3 ID : 41652939793

    Option 4 ID : 41652939790

    sol. Of the given compounds CH(CN)3 is strongest acid because its conjugate base is stabilised by resonance.

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    CHBr3 and CHI

    3 are less stable as their conjugate bases are stabilised by inductive effect of halogens.

    Conjugate base of CHCl3 involves back bonding between 2p and 3p orbitals.16. A solution of sodium sulfate contains 92 g of Na + ions per kilogram of water. The molality of Na + ions in that

    solution in mol kg –1 is:

    lksfM;e lYQsV ds ,d foy;u esa izfr fdyksxzke ty esa 92 g Na + vk;u gSaA Na+ vk;u dh ml foy;u esa eksykfyVh (molkg –1 esa) gksxh %

    (1) 16 (2) 12 (3) 8 (4) 4

    A. 4

    Question ID : 41652910096

    Option 1 ID : 41652939845

    Option 2 ID : 41652939844

    Option 3 ID : 41652939843

    Option 4 ID : 41652939842

    sol.

    17. The correct decreasing order for acid strength is:

    vEy lkeF;Z ds fy, lgh ?kVuk Øe gS %

    (1) FCH2COOH > NCCH

    2COOH > NO

    2CH

    2COOH > ClCH

    2COOH

    (2) NO2CH

    2COOH > NCCH

    2COOH > FCH

    2COOH > ClCH

    2COOH

    (3) NO2CH

    2COOH > FCH

    2COOH > CNCH

    2COOH > ClCH

    2COOH

    (4) CNCH2COOH > O

    2NCH

    2COOH > FCH

    2COOH > ClCH

    2COOH

    A. 2

    Question ID : 41652910080

    Option 1 ID : 41652939781

    Option 2 ID : 41652939778

    Option 3 ID : 41652939780

    Option 4 ID : 41652939779

    sol. The acidic strength of the given compounds is decided on the basis of (–I) effect of the substituents of carboxylic

    acids. Higher the (–I) effect of substituent, higher will be the acidic strength. The decreasing order of (–I) effect

    of the given substituents is NO2 > CN > F > Cl.

    Therefore, correct decreasing order of acidic strength

    O2NCH

    2COOH > NCCH

    2COOH > FCH

    2COOH > ClCH

    2COOH

    18. A water sample has ppm level concentration of the following metals Fe = 0.2 ; Mn = 5.0 ; Cu = 3.0 ; Zn = 5.0.The metal that makes the water sample unsuitable for drinking is:

    ,d ty ds izfrn'kZ esa fuEufyf[kr /kkrqvksa ds ppm lkUnzrk dk Lrj gS %

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    Fe = 0.2 ; Mn = 5.0 ; Cu = 3.0 ; Zn = 5.0.

    /kkrq ftlds dkj.k ty izfrn'kZ ihus ;ksX; ugha gS og gS %

    (1) Fe (2) Mn (3) Cu (4) Zn

    A. 2

    Question ID : 41652910095

    Option 1 ID : 41652939838

    Option 2 ID : 41652939839

    Option 3 ID : 41652939840

    Option 4 ID : 41652939841

    Sol. Prescribed level of Mn is 0.5 ppm. So water sample containing Mn = 5ppm is water unsuitable for drinking.

    19. 20 mL of 0.1 M H2SO

    4 solution is added to 30 mL of 0.2 M NH

    4OH solution. The pH of the resultant mixture

    is : [pkb of NH

    4OH = 4.7].

    20 mL of 0.1 M H2SO

    4 ds foy;u dks 30 mL 0.2 M NH

    4OH ds foy; esa feykus ij izkIr feJ.k ds pH dk eku gS %

    [pkb of NH

    4OH = 4.7].

    (1) 5.2 (2) 5.0 (3) 9.0 (4) 9.4

    A. 3

    Question ID : 41652910102

    Option 1 ID : 41652939867

    Option 2 ID : 41652939866

    Option 3 ID : 41652939868

    Option 4 ID : 41652939869

    sol.

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    20. The major product of the following reaction is:

    fuEufyf[kr vfHkfØ;k dk eq[; mRikn gS %

    (1) (2)

    (3) (4)

    A. 3

    Question ID : 41652910082

    Option 1 ID : 41652939788

    Option 2 ID : 41652939787

    Option 3 ID : 41652939786

    Option 4 ID : 41652939789

    sol.

    21. The major product of the following reaction is:

    fuEufyf[kr vfHkfØ;k dk eq[; mRikn gS %

    (1) (2) (3) (4)

    A. 1

    Question ID : 41652910085

    Option 1 ID : 41652939798

    Option 2 ID : 41652939800

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    Option 3 ID : 41652939801

    Option 4 ID : 41652939799

    sol.

    22. 0.5 moles of gas A and x moles of gas B exert a presture of 200 Pa in a container of volume 10 m3 at 1000 K.Given R is the gas constant in JK – 1 mol–1, x is:

    1000 K ij 10m3 vk;ru ds ,d ik=k esa 0.5 mol xSl A rFkk x mol xSl B, 200 Pa dk nkc cukrs gSaA ;fn R xSl fLFkjkad(JK – 1mol–1 esa) gks rks x gS %

    (1) 2R

    4 + R(2)

    4 – R

    2R(3)

    4 + R

    2R(4)

    2R

    4 – R

    A. 3

    Question ID : 41652910097

    Option 1 ID : 41652939849

    Option 2 ID : 41652939846

    Option 3 ID : 41652939847

    Option 4 ID : 41652939848

    sol. PV = nRT (ideal gas equation)

    200 × 10 × 1000 = (0.5 + x) × R × 1000 × 1000

    x = 4 + R

    2R

    23. Consider the revesible isothermal expansion of an ideal gas in closed system at two different temperatures T1

    and T2 (T

    1 < T

    2). The correct graphical depiction of the dependence of work done (w) on the final volume (V)

    is:

    nks fHkUu rkiksa T1 rFkk T

    2 (T

    1 < T

    2) ij ,d can fudk; esa ,d vkn'kZ xSl ds mRØe.kh; lerkih izlkj ij fopkj dhft,A fd;s

    x;s dk;Z (w) dh vafre vk;ru (V) ij fuHkZjrk dk lgh vkysf[kd fp=k.k gS %

    (1) (2) (3) (4)

    A. 3

    Question ID : 41652910100

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    Option 1 ID : 41652939859

    Option 2 ID : 41652939860

    Option 3 ID : 41652939861

    Option 4 ID : 41652939858

    24. The isotopes of hydrogen are:

    (1) Protium, deuterium and tritium (2) Protium and deuterium only

    (3) Tritium and protium only (4) Deuterium and tritium only

    gkbMªkstu ds leLFkkfud gSa %

    (1) izksfV;e] M~;wVhfj;e rFkk VªkbfV;e (2) izksfV;e rFkk M~;wVhfj;e ek=k

    (3) VªkbfV;e rFkk izksfV;e ek=k (4) M~;wVhfj;e rFkk VªkbfV;e ek=k

    A. 1

    Question ID : 41652910088

    Option 1 ID : 41652939812

    Option 2 ID : 41652939810

    Option 3 ID : 41652939811

    Option 4 ID : 41652939813

    sol. Hydrogen has three isotopes :

    Protium1H1

    Deuterium1H2

    Tritium1H3

    Their natural abundance is in order H > D > T.

    25. Two complexes [Cr(H2O)

    6]Cl

    3 (A) and [Cr(NH

    3)

    6]Cl

    3 (B) are violet and yellow coloured, respectively. The

    incorrect statement regarding them is:

    (1) Both are paramagnetic with three unpaired electrons.

    (2) 0 values of (A) and (B) are calculated from the energies of violet and yellow light, respectively.

    (3) Both absorb energies corresponding to their complementary colors.

    (4) 0 value for (A) is less than that of of (B).

    nks ladqy [Cr(H2O)

    6]Cl

    3 (A) rFkk [Cr(NH

    3)

    6]Cl

    3 (B) Øe'k% cSxuh rFkk ihys jax ds gSaA buds laca/k esa xyr dFku gS %

    (1) nksuksa rhu v;qfXer bysDVªkWuksa ds lkFk vuqpqacdh; gSA

    (2) (A) rFkk (B) ds 0 ekuksa dk ifjdyu Øe'k% cSaxuh rFkk ihys izdk'k dh ÅtkZvksa ds }kjk fd;k tkrk gSA

    (3) nksuksa vius iwjd jaxksa ds vuqdwy ÅtkZ dk vo'kks"k.k djrs gSaA

    (4) (A) ds fy, 0 dk eku (B) dh rqyuk esa de gSA

    A. 2

    Question ID : 41652910094

    Option 1 ID : 41652939834

    Option 2 ID : 41652939837

    Option 3 ID : 41652939836

    Option 4 ID : 41652939835

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    sol.

    Both (A) and (B) are paramagnetic with 3 unpaired electrons each. The splitting energy (0) values of (A) and

    (B) are calculated from the wavelengths of light absorbed and not from the wavelengths of light emitted.

    H2O is a weak field ligand causing lesser splitting than NH

    3 which is relatively stronger field ligand.

    26. Arrange the following amines in the decreasing order of basicity :

    {kkjdrk ds ?kVrs Øe esa fuEu ,sehuksa dks O;ofLFkr dhft, %

    (1) I > III > II (2) III > II > I (3) III > I > II (4) I > II > III

    A. 3

    Question ID : 41652910078

    Option 1 ID : 41652939772

    Option 2 ID : 41652939770

    Option 3 ID : 41652939771

    Option 4 ID : 41652939773

    sol. most basic

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    lone pair is not involved in resonance but N atom is sp2 hybridsed

    lone pair of nitrogen is involved in aromaticity..

    27. The following results were obtained during kinetic studies of the reaction :2A + B Products

    fuEufyf[kr vfHkfØ;k ds xfrd v/;;uksa ds nkSjku fuEufyf[kr ifj.kke izkIr gq, %2A + B mRikn

    1 1 1 1

    3

    3

    2

    [A] [B] Initial Rate of reactionExperiment

    in(mol L ) in(mol L ) (in mol L min )

    I 0.10 0.20 6.93 10

    II 0.10 0.25 6.93 10

    III 0.20 0.30 1.386 10

    The time (in minutes) required to consume half of A is:

    A ds vk/ks Hkkx dks lekIr djus ds fy, vko';d le; (feuV esa) gksxk %

    (1) 5 (2) 10 (3) 1 (4) 100

    A. 1

    Question ID : 41652910104

    Option 1 ID : 41652939875

    Option 2 ID : 41652939877

    Option 3 ID : 41652939874

    Option 4 ID : 41652939876

    Sol. From experiment I and II, it is observed that order of reaction w.r.t. 3 is zero.

    From experiment II and III,

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    28. The one that is extensively used as a piezoelectric material is:

    (1) Tridymite (2) Quartz (3) Amorphous silica (4) Mica

    nkc&fo|qr inkFkZ dh rjg foLrh.kZ mi;ksx esa vkus okyk v;Ld gS %

    (1) VªkbMkbekbV (2) DokV~tZ (3) vfØLVyh; flfydk (4) ekbdk

    A. 2

    Question ID : 41652910090

    Option 1 ID : 41652939819

    Option 2 ID : 41652939818

    Option 3 ID : 41652939821

    Option 4 ID : 41652939820

    Sol. Quartz exhibits piezoelectricity. It is fact based.

    29. The alkaline earth metal nitrate that does not crystallise with water molecules, is:

    {kkjh; e`nk /kkrq ukbVªsV ftldk ty ds v.kqvksa ds lkFk fØLVyhdj.k ugha gksrk gS] og gS %

    (1) Ca(NO3)

    2(2) Mg(NO

    3)

    2(3) Sr(NO

    3)

    2(4) Ba(NO

    3)

    2

    A. 4

    Question ID : 41652910089

    Option 1 ID : 41652939815

    Option 2 ID : 41652939814

    Option 3 ID : 41652939816

    Option 4 ID : 41652939817

    sol. Down the group as the charge density decreases so chances of formation of hydrate decreases.So, Ba(NO

    3)

    2 does not crystallise with water molecules.

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    30. Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the mass of the adsorbed on

    mass m of the adsorbent at prssure P. m

    x is proportional to :

    ,d xSl dk vf/k'kks"k.k ÝkW;UMfyd vf/k'kks"k.k lerki oØ dk vuqlj.k djrk gSA fn;s x;s IykV esa] p nkc ij vf/k'kks"k.k ds m

    nzO;eku ij vo'kksf"kr xSl dk nzO;eku m gSA m

    x lekuqikfrd gS %

    (1) p½ (2) p¼ (3) p (4) p2

    A. 1

    Question ID : 41652910105

    Option 1 ID : 41652939878

    Option 2 ID : 41652939881

    Option 3 ID : 41652939880

    Option 4 ID : 41652939879

    sol. In Freundlich adsorption of a gas on the surface of solid, the extent of adsorption(x/m) is related to pressure of

    gas (P) as

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    MATHS09 Jan. 2019 [Session : 09.30 AM to 12.00 PM]

    JEE MAIN PAPER ONLINERED COLOUR CONSIDER OFFICIAL ANSWER

    1. For 2 1,x n n N (the set of natural numbers), the integral

    2 2

    2 2

    2sin 1 sin 2( 1)

    2sin 1 sin 2( 1)

    x x

    x dxx x

    equal to:

    (where c is a constant of integration)

    2 1,x n n N (izkd`r la[;kvksa dk leqPp;), ds fy,] lekdy

    2 2

    2 2

    2sin 1 sin 2( 1)

    2sin 1 sin 2( 1)

    x x

    x dxx x

    cjkcj gS %

    (tgk¡ c ,d lekdyu vpj gS)

    (1)

    221 1log sec

    2 2e

    xc

    (2)

    2 21log sec ( 1)2

    e x c

    (3) 21 log sec( 1)

    2 e x c (4)

    2 1log sec

    2

    e

    xc

    A. 1,4

    Sol.

    22

    2 2

    22 22

    x 12sin

    2sin x 1 1 cos x 1 2x x

    x 12sin x 1 1 cos x 12cos

    2

    2 2x 1 x 1x tan dx t

    2 2

    2x 1tan t dt ln sec C

    2

    Indefinite Integration

    Question ID : 41652910119

    Option 1 ID : 41652939937

    Option 2 ID : 41652939936

    Option 3 ID : 41652939934

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    Option 4 ID : 41652939935

    2. Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that canbe formed from this class, if there are two specific boys A and B, who refuse to be the members of the sameteam, is:

    5 yM+fd;ksa rFkk 7 yM+dksa dh ,d d{kk dk fopkj dhft,A bl d{kk dh 2 yM+fd;ksa rFkk 3 yM+dksa dks ysdj cu ldus okyhfHkUu Vheksa (teams) ;fn nks fo'ks"k yM+ds A rFkk B ,d gh Vhe ds lnL; cuus ls euk djrs gSa] dh la[;k gS %

    (1) 300 (2) 200 (3) 350 (4) 500

    A. 1

    Sol. AB AB AB 5 5 5 5 5 5

    2 2 2 2 2 3C C C C C C 300

    P & C

    Question ID : 41652910111

    Option 1 ID : 41652939903

    Option 2 ID : 41652939902

    Option 3 ID : 41652939904

    Option 4 ID : 41652939905

    3. Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the origin,

    on the positive x-axis then which of the following points does not lie on it?

    ,d ijoy; dk v{k] x-v{k ds vuqfn'k gSA ;fn blds 'kh"kZ rFkk ukfHk] x-v{k dh /kukRed fn'kk esa ewyfcanq ls Øe'k% 2 rFkk 4dh nwjh ij gSa] rks buesa ls dkSu&lk fcnqa bl ijoy; ij fLFkr ugha gS\

    (1) (4, – 4) (2) (5, 2 6) (3) (8, 6) (4) (6, 4 2)

    A. 3

    Sol. 2

    4

    0(0, 0)

    x

    y

    2y 8 x 2 a = 2

    PARABOLA

    Question ID : 41652910126

    Option 1 ID : 41652939962

    Option 2 ID : 41652939964

  • 43Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    Option 3 ID : 41652939965

    Option 4 ID : 41652939963

    4. Let 3 2 sin

    , :2 1 2 sin

    iA

    i

    is purely imaginary

    . Then the sum of the elements in A is:

    eku 3 2 sin

    , :2 1 2 sin

    iA

    i

    iw.kZr% dkYifud gS

    , rks A ds vo;oksa dk ;ksx gS %

    (1) (2) 3

    4

    (3)

    5

    6

    (4)

    2

    3

    A. 4

    Sol.

    2

    2

    3 2i sin 1 2isin 3 4sin 8i sinz

    1 2isin 1 2isin 1 4sin

    , purely imaginary

    So 2

    2

    3 4sin 3 20 sin , ,

    1 4sin 2 3 3 3

    Complex No.

    Question ID : 41652910107

    Option 1 ID : 41652939886

    Option 2 ID : 41652939888

    Option 3 ID : 41652939889

    Option 4 ID : 41652939887

    5. The plane through the intersection of the planes x + y + z = 1 and 2x + 3y – z + 4 = 0 and parallel to y - axispasses through the point :

    y-v{k ds lekarj rFkk leryksa x + y + z = 1 vkSj 2x + 3y – z + 4 = 0 ds izfrPNsnu ls gksdj tkus okyk lery fuEu esa lsfdl fcanq ls Hkh gks dj tkrk gS\

    (1) (– 3, 0, – 1) (2) (– 3, 1, 1) (3) (3, 2, 1) (4) (3, 3, – 1)

    A. 3

    Sol. 1 2P P 0

    (x + y + z –1) + (2x + 3y – z + 4) is parallel to y-axis (0, 1, 0)

    ˆ ˆ ˆ ˆ1 2 i 1 3 j 1 k .j 0

    1

    3

    1

    x y z 1 2x 3y z 4 03

    x + 4z – 7 = 0

    VEctor 3D

    Question ID : 41652910128

    Option 1 ID : 41652939970

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    Option 2 ID : 41652939971

    Option 3 ID : 41652939972

    Option 4 ID : 41652939973

    6. If the fractional part of the number 4032

    15 is

    15

    k, then k is equal to :

    ;fn la[;k 4032

    15 dk fHkUukRed Hkkx (fractional part)

    15

    k gS] rks k cjkcj gS %

    (1) 4 (2) 8 (3) 6 (4) 14

    A. 2

    Sol.

    100 100403 8. 16 8 15 12

    15 15 15

    8.15k 1 8 8

    15 15

    is fractional part.

    Bin. TH.

    Question ID : 41652910112

    Option 1 ID : 41652939906

    Option 2 ID : 41652939907

    Option 3 ID : 41652939909

    Option 4 ID : 41652939908

    7. Three circles of radii a, b, c (a < b < c) touch each other externally. If they have x-axis as a common tangent,then :

    (1) a, b, c are in A.P. (2) 1 1 1

    a b c (3) , ,a b c are in A.P.. (4)

    1 1 1

    b a c

    a, b, c (a < b < c) f=kT;kvksa okys rhu o`Ùk ijLij cká Li'kZ djrs gSaA ;fn x-v{k mudh ,d mHk;fu"B Li'kZ js[kk gS] rks %

    (1) a, b, c ,d lekarj Js

  • 45Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    similarly 2 bc

    2 bc ...(2)

    and – = 2 ac ...(3)

    (1) + (2) + (3) bc ab ac

    1 1 1

    a b c

    Question ID : 41652910125

    Option 1 ID : 41652939958

    Option 2 ID : 41652939960

    Option 3 ID : 41652939959

    Option 4 ID : 41652939961

    8. If 1 12 3 3cos cos

    3 4 2 4x

    x x

    , then x is equal to:

    ;fn 1 12 3 3cos cos

    3 4 2 4x

    x x

    gS] rks x cjkcj gS %

    (1) 146

    12(2)

    145

    12(3)

    145

    10(4)

    145

    11

    A. 2

    Sol. 1 1 1 13 2 3 2cos cos cos cos cos sin cos .sin cos 0

    4x 3x 4x 3x

    2 2

    3 2 9 41 . 1 0

    4x 3x 16x 9x

    (9x2 – 4) (16x2 – 9) = 36

    144x4 – 145x2 + 36 = 36

    2 145x144

    Question ID : 41652910134

    Option 1 ID : 41652939995

    Option 2 ID : 41652939994

    Option 3 ID : 41652939997

    Option 4 ID : 41652939996

    9.4

    lim0 4

    1 1 2y

    y

    y

    (1) Exists and equals 1

    2 2( 2 1)(2) Exists and equals

    1

    4 2

  • 46Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    (3) Exists and equals 1

    2 2(4) Does not exist

    (1) vfLrRo gS rFkk 1

    2 2( 2 1) ds cjkcj gSA (2) vfLrRo gS rFkk

    1

    4 2 ds cjkcj gSA

    (3) vfLrRo gS rFkk 1

    2 2 ds cjkcj gSA (4) vfLrRo ugha gSA

    A. 2

    Sol.

    4 4

    y 0 4 4

    1 1 y 2 1 1 y 2

    Lim

    y 1 1 y 2

    4 4

    y 0 4 4

    1 y 1 1 y 1Lim

    y 2 2 1 y 1

    4

    4y 0

    y 1Lim

    4 2y . 2 2 .2

    Question ID : 41652910115

    Option 1 ID : 41652939921

    Option 2 ID : 41652939920

    Option 3 ID : 41652939919

    Option 4 ID : 41652939918

    10. The area (in sq. units) bounded by the parabola y = x2 – 1, the tangent at the point (2, 3) to it and the y - axisis:

    ijoy; y = x2 – 1, bl ijoy; ij fLFkr ,d fcanq (2, 3) ij [khaph xbZ Li'kZ js[kk rFkk y-v{k ls f?kjs {ks=k dk {ks=kQy(oxZ bdkb;ksa esa) gS %

    (1) 14

    3(2)

    56

    3(3)

    32

    3(4)

    8

    3

    A. 4

    Sol.y 3

    PT 2x 12

    4x – y – 5 = 0

    3 3

    5 1

    y 5A y 1 dy

    4

    y 1 0

    T(0, –5)

    (1, –1)

    (2, 3)P

  • 47Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    3 32 3

    2

    15

    1 y 25y y 1

    4 2 3

    1 9 25 1615 25

    4 2 2 3

    1 16

    17 154 3

    16 88

    3 3

    Question ID : 41652910121

    Option 1 ID : 41652939943

    Option 2 ID : 41652939944

    Option 3 ID : 41652939945

    Option 4 ID : 41652939942

    11. Let f : R R be a function defined as:

    5, if 1

    , 1 3( )

    5 , 3 5

    30, 5

    x

    a bx if xf x

    b x if x

    if x

    Then, f is :

    (1) Continuous if a = – 5 and b = 10 (2) Continuous if a = 5 and b = 5

    (3) Continuous if a = 0 and b = 5 (4) Not continuous for any values of a and b

    ekuk Qyu f : R R be a function defined as:

    5, if 1

    , 1 3( )

    5 , 3 5

    30, 5

    x

    a bx if xf x

    b x if x

    if x

    }kjk ifjHkkf"kr gS] rks f :

    (1) larr gS ;fn a = – 5 rFkk b = 10 (2) larr gS ;fn a = 5 rFkk b = 5

    (3) larr gS ;fn a = 0 rFkk b = 5 (4) a rFkk b ds fdlh Hkh eku ds fy, larr ugha gSA

    A. 4

    Sol. a + b = 5 ...(1)

    b + 25 = 30 ...(2)

    a + 3b = b + 15

    a + 2b = 15 ...(3)

  • 48Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    for continous (1), (2) and (3) should be satisfy for same value of a, b

    So no solution & not continous.

    Question ID : 41652910116

    Option 1 ID : 41652939922

    Option 2 ID : 41652939923

    Option 3 ID : 41652939924

    Option 4 ID : 41652939925

    12. The system of linear equations :x + y + z = 22x + 3y + 2z = 52x + 3y + (a2 – 1 )z = a + 1

    (1) Is inconsistent when |a| = 3 (2) Has a unique solution for |a| = 3

    (3) Has infinitely many solution for a = 4 (4) is inconsistent when a = 4

    jSf[kd lehdj.k fudk;x + y + z = 22x + 3y + 2z = 52x + 3y + (a2 – 1 )z = a + 1

    (1) vlaxr gS tc |a| = 3 (2) dk |a| = 3 ds fy, vf}rh; gy gSA

    (3) ds a = 4 ds fy, vuUr gy gSaA (4) vlaxr gS tc a = 4

    A. 1

    Sol. 2

    1 1 1

    2 3 2

    2 3 a 1

    a 3

    x 0 for a = 3 so system of equations are inconsistent

    Question ID : 41652910110

    Option 1 ID : 41652939898

    Option 2 ID : 41652939901

    Option 3 ID : 41652939899

    Option 4 ID : 41652939900

    13. Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denote therandom variable of number of aces obtained in the two drawn cards. Then P(X = 1) + P(X = 2) equals:

    52 iÙkksa dh ,d vPNh izdkj ls QsaVh xbZ rk'k dh xM~Mh esa ls] ,d ds ckn ,d] nks iÙks izfrLFkkiuk lfgr fudkys x,A eku X,nksuksa ckj esa izkIr bDdksa dh la[;k dks n'kkZus okyk ;knf̀PNd pj gS] rks P(X = 1) + P(X = 2) cjkcj gS %

    (1) 24/169 (2) 25/169 (3) 52/169 (4) 49/169

    A. 2

    Sol. Ace Ace Ace Ace Ace . Ace

  • 49Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    4 48 4 4 252

    52 52 52 52 169

    Question ID : 41652910132

    Option 1 ID : 41652939986

    Option 2 ID : 41652939987

    Option 3 ID : 41652939989

    Option 4 ID : 41652939988

    14. If a, b and c be three distinct real numbes in G.P. and a + b + c = xb, then x cannot be:

    ;fn rhu fHkUu okLrfod la[;k;sa a, b rFkk c ,d xq.kksÙkj Js

  • 50Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    J(x) = 1

    x 1

    =

    1

    1 x = f

    3(x)

    Question ID : 41652910106

    Option 1 ID : 41652939884

    Option 2 ID : 41652939882

    Option 3 ID : 41652939883

    Option 4 ID : 41652939885

    16. The maximum volume (in cu.m) of the right circular cone having slant height 3 m is

    3 eh- fr;Zd (slant) Å¡pkbZ okys yacòÙkh; 'kadq dk vf/kdre vk;ru (?ku eh. esa) gS %

    (1) 2 3 (2) 6 (3) 3 3 (4) 4

    3

    A. 1

    Sol.

    DrB C

    h

    A3

    V = 21 r h

    3

    3 21V . 3 sin cos

    3

    2 3dv 27 2sin cos sin 0d 3

    1cos

    3

    1 2 1V 27

    3 3 3

    V 2 3

    Question ID : 41652910118

    Option 1 ID : 41652939932

    Option 2 ID : 41652939930

    Option 3 ID : 41652939931

    Option 4 ID : 41652939933

    17. Consider the set of all lines px + qy + r = 0 such that 3p + 2q + 4r = 0. Which one of the following statementsis true?

    (1) The lines are concurrent at the point3 1

    ,4 2

    (2) Each line passes through the origin

    (3) The lines are not concurrent (4) The lines are all parallel

    ,slh lHkh js[kkvksa px + qy + r = 0 ds leqPp; ij fopkj dhft, ftuds fy, 3p + 2q + 4r = 0 gS] rks fuEu esa ls dksu&lk

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    ,d dFku lR; gS\

    (1) js[kk;sa fcUnq 3 1

    ,4 2

    ij laxkeh gSA

    (2) izR;sd js[kk ewy fcanq ls gks dj tkrh gSA

    (3) js[kk,¡ laxkeh ugha gSaA (4) lHkh js[kk,¡ lekarj gSaA

    A. 1

    Sol.

    px qy r 03 1

    x , y3 24 2p q r 0

    4 4

    Question ID : 41652910123

    Option 1 ID : 41652939951

    Option 2 ID : 41652939953

    Option 3 ID : 41652939952

    Option 4 ID : 41652939950

    18. The equation of the line passing through (– 4, 3, 1) parallel to the plane x + 2y – z – 5 = 0 and intersecting the

    line 1 3 2

    3 2 1

    x y z

    is:

    fcanq (– 4, 3, 1) ls gks dj tkus okyh js[kk] tks lery x + 2y – z – 5 = 0 ds lekarj gS rFkk js[kk 1 3 2

    3 2 1

    x y z

    dks dkVrh gS] dk lehdj.k gS %

    (1) 4 3 1

    1 1 1

    x y z(2)

    4 3 1

    3 1 1

    x y z

    (3) 4 3 1

    2 1 4

    x y z(4)

    4 3 1

    1 1 3

    x y z

    A. 2

    Sol.x 1 y 3 z 2

    x 3 1, y 2 3, z 23 2 1

    m n

    3 1 4 .1 2 3 3 .2 2 1 . 1 0

    –3 + 3 + 4 + –1 = 0

    2 + 2 = 0 = –1

    x 4 y 3 z 1

    6 2 2

    Question ID : 41652910129

    Option 1 ID : 41652939974

    Option 2 ID : 41652939976

    Option 3 ID : 41652939977

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    Option 4 ID : 41652939975

    19. Equation of a common tangent to the circle, x2 + y2 – 6x = 0 and the parabola, y2 = 4x, is :

    o`Ùk x2 + y2 – 6x = 0 rFkk ijoy; y2 = 4x, dh ,d mHk;fu"B Li'kZ js[kk dk lehdj.k gS %

    (1) 3 3y x (2) 3 3 1y x (3) 2 3 12 1y x (4) 2 3 12y x

    A. 1

    Sol. (x – 3)2 + y2 = 9

    y2 = 4x

    Tangent y = Mx + 1

    M

    2

    13M

    m

    1 M

    = 3 9M2 + 6 + 2

    1

    M = 9 + 9M2

    1M

    3 for

    1M

    3

    T 3y 3x 3

    Question ID : 41652910124

    Option 1 ID : 41652939957

    Option 2 ID : 41652939956

    Option 3 ID : 41652939955

    Option 4 ID : 41652939954

    20. Let and be two roots of the equation x2 + 2x + 2 = 0, then 15 + 15 is equal to:

    ekuk rFkk lehdj.k x2 + 2x + 2 = 0 ds nks ewy gSa] rks 15 + 15 cjkcj gS %

    (1) 256 (2) – 256 (3) – 512 (4) 512

    A. 2

    Sol. 1 i, 1 i,

    i3

    42e

    , i5

    42e

    i45 i75

    15 1515 15 4 4

    1 1 1 i2 e e 2

    2 2 2 2

    15

    2 2

    15 15 256

    Question ID : 41652910108

    Option 1 ID : 41652939890

    Option 2 ID : 41652939891

    Option 3 ID : 41652939893

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    Option 4 ID : 41652939892

    21. If the Boolean expression

    (p q) (~ p q) is equivalent to

    p q , where {, }, then the ordered pair ( ) is:

    ;fn cwyh; O;atd (p q) (~ p q)

    p q ds rqY; gS] tgk¡ {, } gS] rks Øfer ;qXe ( ) gS %

    (1) (, ) (2) (, ) (3) (, ) (4) (, )

    A. 4

    Sol. (i) p q n ~ p q

    p q ~ p q p q (i) is correct

    (ii) p q ~ p q p ~ p q

    (iii) p q ~ p q q

    (iv) p q ~ p q q p ~ p q q

    Question ID : 41652910135

    Option 1 ID : 41652940000

    Option 2 ID : 41652939998

    Option 3 ID : 41652940001

    Option 4 ID : 41652939999

    22. If A = cos sin

    sin cos

    , then the matrix, AA–50 when = 12

    , is equal to:

    ;fn A = cos sin

    sin cos

    , rks vkO;wg AA–50 tc = 12

    , cjkcj gS %

    (1)

    3 1

    2 2

    1 3

    2 2

    (2)

    3 1

    2 2

    1 3

    2 2

    (3)

    1 3

    2 2

    3 1

    2 2

    (4)

    1 3

    2 2

    3 1

    2 2

    A. 1

    Sol.2 cos sin cos sin cos 2 sin 2A

    sin cos sin cos sin 2 cos 2

    50 50

    3 1cos50 sin 50 cos50 sin 50 2 2A , Asin 50 cos50 sin 50 cos50 1 3

    2 2

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    Question ID : 41652910109

    Option 1 ID : 41652939894

    Option 2 ID : 41652939895

    Option 3 ID : 41652939896

    Option 4 ID : 41652939897

    23. For any ,4 2

    the exprssion 3(sin – cos)4 + 6(sin + cos)2 + 4sin6 equals :

    fdlh ,4 2

    ds fy, O;atd 3(sin – cos)4 + 6(sin + cos)2 + 4sin6 cjkcj gS %

    (1) 13 – 4 cos2 + 6sin2cos2 (2) 13 – 4 cos4 + 2sin2cos2

    (3) 13 – 4 cos2 + 6cos4 (4) 13 – 4 cos6

    A. 4

    Sol. ,4 2

    3(1–2sin cos)2 + 6(1 + 2sincos) + 4 sin6

    3(1 + 4sin2cos2– 4 sincos) + 6 + 12sincos+ 4sin6

    3 + 12sin2cos2+ 6 + 4sin6

    9 + 12 cos2(1 – cos2) + 4(1 – cos2)3

    9 + 12 cos2– 12cos4+ 4(1 – cos6 + 3cos4– 3cos2)

    13 – 4cos6

    Question ID : 41652910133

    Option 1 ID : 41652939990

    Option 2 ID : 41652939992

    Option 3 ID : 41652939991

    Option 4 ID : 41652939993

    24. 5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is156 cm, joined them. The variance (in cm2) of the height of these six student is:

    ,d d{kk ds 5 fo|kfFkZ;ksa dh Å¡pkb;ksa dk ek/; 150 cm rFkk izlj.k 18 cm2 gSA 156 cm Å¡pkbZ okyk ,d u, fo|kFkhZ mulsvk feykA bu N% fo|kfFkZ;ksa dh Å¡pkb;ksa dk izlj.k (oxZ ls-eh- esa) gS %

    (1) 18 (2) 16 (3) 20 (4) 22

    A. 3

    Sol. x1, x

    2, x

    3, x

    4, x

    5 students

    i

    i

    xx 150 x 750

    5

    2 2

    2 2i i2x x

    x 18 15 1505 5

    2ix 18 11500 5 112590

    1 2 3 4 5new

    x x x x x 156 750 156x 151

    6 6

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    22

    2 2i2

    new

    x 112590 156x 151

    6 6

    2new

    20

    Question ID : 41652910131

    Option 1 ID : 41652939984

    Option 2 ID : 41652939983

    Option 3 ID : 41652939982

    Option 4 ID : 41652939985

    25. If denotes the acute angle betwen the curves , y = 10 – x2 and y = 2 + x2 at a point of their intersection, then|tan | is equal to :

    ;fn oØksa y = 10 – x2 rFkk y = 2 + x2 ds chp ,d izfrPNsn fcUnq ij U;wu dks.k gS] rks |tan | cjkcj gS %

    (1) 8

    17(2)

    7

    17(3)

    4

    9(4)

    8

    15

    A. 4

    Sol.

    21

    22

    C y 10 x

    C y 2 x

    POI is (+2, 6)

    for 1dy

    C 2 2 4dx

    for 2dy

    C 2 2 4dx

    1 2

    1 2

    M M 8tan

    1 M M 15

    Question ID : 41652910117

    Option 1 ID : 41652939926

    Option 2 ID : 41652939929

    Option 3 ID : 41652939928

    Option 4 ID : 41652939927

    26. If y = y(x) is the solution of the differential equation, dy

    xdx

    + 2y = x2 satisfying y(1) = 1, then 1

    2y

    is equal

    to :

    ;fn y = y(x), vody lehdj.k dy

    xdx

    + 2y = x2 dk gy gS tks y(1) = 1 dks larq"V djrk gS] rks 1

    2y

    cjkcj gS %

    (1) 49

    16(2)

    7

    64(3)

    1

    4(4)

    13

    16

    A. 1

  • 56Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    Sol.2dyx 2y x

    dx

    dy 2tx

    dx x I.F.

    2dx

    2 ln x 2xe e x

    y.x2 = 4

    2 xx .xdx C4

    1 3C C

    4 4

    42 x 3y.x

    4 4

    y 1 3

    4 16 4 4

    49y

    16

    Question ID : 41652910122

    Option 1 ID : 41652939949

    Option 2 ID : 41652939947

    Option 3 ID : 41652939946

    Option 4 ID : 41652939948

    27. Let 0 < < 2

    . If the eccentricity of the hyperbola

    2 2

    2 21

    cos sin

    x y

    is greater than 2, then the length of its

    latus rectum lies in the inteval :

    ekuk 0 < < 2

    gSA ;fn vfrijoy;

    2 2

    2 21

    cos sin

    x y

    dh mRdsanzrk 2 ls vf/kd gS] rks blds ukfHkyac dh yackbZ ftl

    vUrjky esa gS] og gS %

    (1) (3, ) (2) (2, 3] (3) (1, 3/2] (4) (3/2, 2]

    A. 1

    Sol.2 2

    2 2

    x y1

    cos sin

    e = sec > 2 ,

    3 2

    22sin 3LL ' 2 tan sin 2 3

    cos 2

    min

    LL ' 3

    min

    LL '

    Question ID : 41652910127

    Option 1 ID : 41652939969D

    Option 2 ID : 41652939968

    Option 3 ID : 41652939966

    Option 4 ID : 41652939967

  • 57Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    28. Let ˆˆ ˆ ˆ ˆ,a i j b i j k

    and c

    be a vector such that 0a c b and 4a c

    , then

    2

    c

    is equal to :

    ekuk ˆˆ ˆ ˆ ˆ,a i j b i j k

    rFkk c

    ,sls lafn'k gSa fd 0a c b rFkk 4a c

    gS] rks

    2

    c

    cjkcj gS %

    (1) 9 (2) 8 (3) 19

    2(4)

    17

    2

    A. 3

    Sol. ˆ ˆ ˆC pi qj rk

    a.c 4

    ˆ ˆ ˆi j k

    ˆ ˆ ˆ1 1 0 i j k 0

    p q r

    ˆ ˆ ˆr 1 i r 1 j p q 1 k 0

    r = 1

    p + q = –1

    3 5P ,q

    2 2

    2

    2 2 2 19C p q r2

    Question ID : 41652910130

    Option 1 ID : 41652939980

    Option 2 ID : 41652939981

    Option 3 ID : 41652939978

    Option 4 ID : 41652939979

    29. The value of 3

    0

    | cos |x

    dx is :

    3

    0

    | cos |x

    dx dk eku gS %

    (1) 4

    3 (2)

    2

    3(3)

    4

    3(4) 0

    A. 3

    Sol.

    /23 3

    0 0

    2 4cos x dx 2 cos xdx 2 .1

    3 3

    walli's formulae

    Question ID : 41652910120

  • 58Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

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    Option 1 ID : 41652939941 Option 2 ID : 41652939939

    Option 3 ID : 41652939940 Option 4 ID : 41652939938

    30. Let a1, a

    2, ........., a

    30 be an A.P., S =

    30

    1 ii

    a and T = 15

    1i

    a

    (2i – 1). If a

    5 = 27 and S – 2T = 75, then a

    10 is equal to

    :

    ekuk a1, a

    2, ........., a

    30 ,d lekUrj Js