Matrix€¦ · Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999,...
Transcript of Matrix€¦ · Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999,...
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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
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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
Vµ
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
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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 %
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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
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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
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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
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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
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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.
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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
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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
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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
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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
t½
A
B
1
4
B
A
t½ 1
t½ 4
By option (1)
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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