2019 apr jee-main q p k s - sri chaitanya · 2019-04-15 · 2019_jee-main question paper_key &...
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2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
PHYSICS 1. The magnetic field of a plane electromagnetic wave is given by:
0 1cos cosB B i kz t B j kz t
Where 50 3 10B T and 6
1 2 10B T .
The rms value of the force experienced by a stationary charge 410Q C at z = 0 is
1. 0.6 N 2. 0.1 N 3. 23 10 N 4. 0.9 N
Ans: 1
Sol: 2 2
0 0
2 2rmsCB CBF g
5 6
83 10 2 1010 3 102 2
0.63
2. A solid sphere of mass ‘M’ and radius ‘a’ is surrounded by a uniform concentric
spherical shell of thickness 2a and mass 2M. The gravitational field at distance ‘3a’
from the centre will be:
1. 2
23GMa
2. 29GM
a 3. 2
29GMa
4. 23GM
a
Ans: 4
Sol: Field at
2 2
23 3
G MGMPa a
P 3a Ma
Field = 23GM
a
3. A moving coil galvanometer has resistance 50 and it indicates full deflection at 4
mA current. A voltmeter is made using this galvanometer and a 5 k resistance.
The maximum voltage, that can be measured using this voltmeter, will be close to:
1. 10 V 2. 15 V 3. 40 V 4. 20V
Ans: 4
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
Sol: max 4I MA
50GR
In voltmeter
max s KV I R S
20V
4. For a given gas at 1 atm pressure, rms speed of the molecules is 200 m/s at 127ºC. At
2 atm pressure and at 227ºC, the rms speed of the molecules will be:
1. 100 /m s 2. 80 /m s 3. 80 5 /m s 4. 100 5 /m s
Ans: 4
Sol: rmsV T
2 500200 400V
2 100 5V
5. Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as
660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be:
1. 642.7 nm 2. 889.2 nm 3. 488.9 nm 4. 388.9 nm
Ans: 3
Sol: 1st line Balmer
N=N NH 36 6605R
2
1 1 14 16
R
2
163R
1 5 66036R
216 5 66013 36
2 488.9
6. The pressure wave, 20.01sin 1000 3P t x Nm , corresponds to the sound produced
by a vibrating blade on a day when atmospheric temperature is 0ºC. On some other
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
day when temperature is T, the speed of sound produced by the same blade and at
the same frequency is found to be 1336ms . Approximate value of T is:
1. 12ºC 2. 11ºC 3. 15ºC 4. 4ºC
Ans: 4
Sol: soundV T
2336 31000 273
T
2 277T K
2 4ºT C
7. The following bodies are made to roll up (without slipping) the same inclined plane
from a horizontal plane : (i) a ring of radius R, (ii) a solid cylinder of radius 2R and
(iii) a solid sphere of radius 4R . If, in each case, the speed of the center of mass at the
bottom of the incline is same, theratio of the maximum heights they climb is:
1. 2 : 3 : 4 2. 4 : 3 : 2 3. 10 : 15 : 7 4. 14 : 15 : 20
Ans: 3
Sol: 21 12 cmMgh mV K
1 21 1:1 :12 5
Ratio
20 :15 :14
8. An NPN transistor is used in common emitter configuration as an amplifier with
1k load resistance. Signal voltage of 10 mV is applied across the base-emitter. This
produces a 3 mA change in the collector current and 15 A change in the base
current of the amplifier. The input resistance and voltage gain are:
1. 0.33 ,300k 2. 0.67 ,300k 3. 0.33 ,1.5k 4. 0.67 ,200k
Ans: 2
Sol: 200C
B
II
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
B iI R V
23iR
Voltage gain = C C
B I
I RI R
Voltage gain = 300
9. A ball is thrown vertically up (taken as +z-axis) from the ground. The correct
momentum – height (p-h) diagram is:
1. 2.
3. 4.
Ans: 4
Sol: 2 2 2v u gh
2 2 2v v gh
2p A Bh
10. The electric field of light wave is given as 3 147
210 cos 2 6 105 10
x NE t xC
.
The light falls on a metal plate of work function 2eV. The stopping potential of the
photo-electrons is:Given, E (in eV) 0
12375
Ain
1. 0.48 V 2. 2.48 V 3. 0.72V 4. 2.0
Ans: 1
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
Sol: 0 .hv hv s potential
12375 .5000
w s potential
2.48 2 .ev s potential
0.4stop potential eV
11. A string is clamped at both the ends and it is vibrating in its 4th harmonic. The
equation of the stationary wave is 0.3sin 0.157 cos 200Y x t . The length of the
string is: (All quantities are in SI units.)
1. 40 m 2. 20 m 3. 80 m 4. 60 m
Ans: 3
Sol: ln 4th harmonic
4. 22
2 2 3.14 20 400.157
mk
2 80m
12. A signal cosA t is transmitted using 0 0sin t as carrier wave. The correct
amplitude modulated (AM) signal is:
1. 0 0sin cost A t
2. 0 0 0 0sin sin sin2 2A At t t
3. 0 0cos sinA t t
4. 0 0sin 1 0.01 sinA t t
Ans: 2
Sol: Mododulater wave = 0 0sin cosC rmsA V t t w
= 0 0cos sinV A t t
0 0 0 0sin sin sin2 2A AV t t t
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
13. In the density measurement of a cube, the mass and edge length are measured as
10.00 0.10 kg and 0.10 0.01 m , respectively. The error in the measurement of
density is:
1. 30.31 /kg m 2. 30.07 /kg m 3. 30.01 /kg m 4. 30.10 /kg m
Ans: 1
Sol: 10 0.1mass kg
Length = 0.1 0.01 m
Denisty = 3
m DL
0.1 0.01310 0.1
=0.01+0.3
= 0.31 kg/m3
14. A simple pendulum oscillating in air has period T. The bob of the pendulum is
completely immersed in a non-viscous liquid. The density of the liquid is 116
th of the
material of the bob. If the bob is inside liquid all the time, its period of oscillation in
this liquid is:
1. 1210
T 2. 1214
T 3. 1414
T 4. 1415
T
Ans: 4
Sol: 1Tg
1
1 1
1516
T gT g
15' 4
TT
4'15TT
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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15. If ‘M’ is the mass of water that rises in a capillary tube of radius ‘r’, then mass of
water which will rise in a capillary tube of radius ‘2r’ is:
1. 4 M 2. M 3. 2M 4. 2 M
Ans: 4
Sol: 2Threg
2mass r h
mass r
So final mass rise = 2M
16. Following figure shows two process A and B for a gas. If AQ and BQ are the
amount of heat absorbed by the system in two cases, and AU and BU are changes
in internal energies, respectively, then:
1. ,A B A BQ Q U U 2. ,A B A BQ Q U U
3. ,A B A BQ Q U U 4. ,A B A BQ Q U U
Ans: 1
Sol: A BU U is U is state function
A A AQ U W
B B BQ U w
A B A BQ Q w w
0
A BQ Q
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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17. The total number of turns and cross-section area in a solenoid is fixed. However, its
length L is varied by adjusting the separation between windings. The inductance of
solenoid will be proportional to:
1. 21 / L 2. L 3. 2L 4. 1/L
Ans: 4
Sol: WKT
Self inductance 2L n volume
2
2
N A KL
2N A
L
Self inductance 1L
18. A uniform cable of mass ‘M’ and length ‘L’ is placed on a horizontal surface such
that its 1 th
n
part is hanging below the edge of the surface. To lift the hanging part
of the cable upto the surface, the work done should be:
1. 2
gLn 2. n gL 3. 22
gLn
4. 2
2 gLn
Ans: 3
Sol: The resultant is the centre of mass of the part AB is shifter to top i.e. 2lhn
U mg hcm
22 2M L L MgLgL n N n
22MgLU
n
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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19. A body of mass 2 kg makes an elastic collision with a second body at rest and
continues to move in the original direction but with one fourth of its original speed.
What is the mass of the second body?
1. 1.2 kg 2. 1.8 kg 3. 1.0 kg 4. 1.5 kg
Ans: 1
Sol: Momentum conservation
22 2 14vv m
132vmv
Elatic collision
Energy conservation
2
2 21
1 1 12 22 2 4 2
vv mv
22 2 212 15 915
16 8 4mvv v v
m m
1.2m kg
20. A capacitor with capacitance 5 F is charged to5 C . If the plates are pulled apart
to reduce the capacitance to 2 F , how much work is done?
1. 63.75 10 J 2. 62.55 10 J 3. 66.25 10 J 4. 62.16 10 J
Ans: 1
Sol: work done by force F S
2
0
32 2
q dA
2
0
34
q dA
63 5 10
4x
63.75 10 J
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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21. A rectangular coil (Dimension5 2.5cm cm ) with 100 turns, carrying a current of 3A
in the clock-wise direction, is kept centered at the origin and in the X-Z plane. A
magnetic field of 1 T is applied along X-axis. If the coil is tilted through 45º about
Z-axis, then the torque on the coil is
1. 0.38 Nm 2. 0.42 Nm 3. 0.55 Nm 4. 0.27 Nm
Ans: 4
Sol: Torque M B
= sin 4Nxi A B J
= 4 1100 3 12.5 10 1 12
0.265
22. The figure shows a Young’s double slit experimental setup. It is observed that when
a thin transparent sheet of thickness t and refractive index is put in front of one
of the slits, the central maximum gets shifted by a distance equal to n fringe widths.
If the wavelength of light used is ,t will be:
1. 2
1nD
a
2.
2
1D
a
3.
1nD
a
4.
1D
a
Ans: 3
Sol: 1 2 1 0S P S P t
2 11 t S P S P
1 t nl
1
nlt
23. A wire of resistance R is bent to form a square ABCD as shown in the figure. The
effective resistance between E and C is: (E is mid-point of arm CD)
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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A B
CD E
1. R 2. 34
R 3. 764
R 4. 116
R
Ans: 3
Sol: 4 4 4 8AB AD BC CR R R RR R R RO R
Reff 4 4 4 8
4 4 4 8
R R R R
R R R R
778 864
R RR
R
24. A rigid square loop of side ‘a’ and carrying current 2I is lying on a horizontal
surface near a long current 1I carrying wire in the same plane as shown in figure.
The net force on the loop due to the wire will be:
1. Attractive and equal to 0 1 2
3I I
2. Zero
3. Repulsive and equal to 0 1 2
4I I
4. Repulsive and equal to 0 1 2
2I I
Ans: 3
Sol: 0AB DCF F
2ADF BI a
022
I I
(right wards)
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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1 2
2I I
2'BCF B I a
0 1 2
4I I
(up + wards)
0 1 2 0 1 2
2 4netI I I IF
0 1 2
4I I
(right wards ev (repulsive)
25. The stream of a river is flowing with a speed of 2 km/h. A swimmer can swim at a
speed of 4 km/h. What should be the direction of the swimmer with respect to the
flow of the river to cross the river straight?
1. 150º 2. 60º 3. 90º 4. 120º
Ans: 4
Sol: to cross the river straight, Valong the river = 0
Vsteam = Vswimmer cos
2 = 14cos cos2
60º
Direction w.r.t flow of river = 180º-60º = 120º
26. A stationary horizontal disc is free to rotate about its axis. When a torque is applied
on it, its kinetic energy as a function of , where is the angle by which it has
rotated, is given as 2k . If its moment of inertia is I then the angular acceleration of
the disc is:
1. 2kI 2.
2kI 3.
4kI 4. k
I
Ans: 1
Sol: 2KE K
212
IW K
Differentiating on both sides w.r.t to
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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2 212
d dI kd d
22 kI KI
27. A system of three charges are placed as shown in the figure:
If D > > d, the potential energy of the system is best given by:
1. 2
20
14
q qQdd D
2. 2
20
14 2
q qQdd D
3. 2
20
1 24
q qQdd D
4. 2
20
14
q qQdd D
Ans: 4
Sol: The potential energy of +q, -q system is 2
1kqUd
The system of +q, -q act as an electric dipole as d<<D
Hence, Potential energy between Q and dipole is 2 2
kpU Qr
Hence, 1 20
14
U U U
2
2q d dd D
28. An HCl molecule has rotational, translational and vibrational motions. If the rms
velocity of HCl molecules in its gaseous phase is v
, m is its mass and Bk is Boltzman
constant, then its temperature will be :
1. 2
6 B
mvk
2. 2
7 B
mvk
3. 2
3 B
mvk
4. 2
5 B
mvk
Ans: 1
Sol: 23 3rms
KT KTV Vm m
2
3mvT
k
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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29. A concave mirror for face viewing has focal length of 0.4 m. The distance at which
you hold the mirror from your face in order to see your image upright with a
magnification of 5 is:
1. 0.24 m 2. 0.16 m 3. 1.60 m 4. 0.32 m
Ans: 4
Sol: 1 1 1f u v 1 1 1
0.4u v
5vmu
54v
1 1 15 0.4u u
1.6 0.325
u
30. Determine the charge on the capacitor in the following circuit:
1. 10 C 2. 200 C 3. 60 C 4. 2 C
Ans: 2
Sol: 4 12 616TR
9
72 89TI A
1248 2
16I A
10 2I A
10 20V V
20 20 10 200C CV V Q F F
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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CHEMISTRY 31. Among the following, the set of parameters that represents path functions, is:
A) q+w B) q C) w D) H-TS
1. A, B and C 2. B, C and D 3. B and C 4. A and D
Ans: 3
Sol: Except q, w all other thermodynamic variables are state functions q, w are path
functions.
32. The correct IUPAC name of following compound is
2NO
3CHCl
1. 2-methyl – 5 - nitro 1- chlorobenzene
2. 2-+chloro – 1- methyl -4 - nitrobenzene
3. 5 – chloro – 4 – methyl – 1 – nitrobenzene
4. 3- chloro – 4 – methyl – 1 – nitrobenzene
Ans: 2
Sol:
3CH
2NO
Cl12
34
2NO - Nitro Cl - Chloro 3CH - Methyl
33. The degenerate orbitals of 3
2 6Cr H O
are
1. yzd and 2zd 2. 2and dx xzd 3. and dxz yzd 4. 2 2and dxyx yd
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
Ans: 3
Sol: ,zx yzd d are degenerate in octahedral splitting.
34. The organic compound that gives following qualitative analysis is:
Test Inference
a) Dil. HCl Insoluble
b) NaOH solution Soluble
c) Br2/ water Decolourization
1.
2NH
2.
OH
3.
2NH
4.
OH
Ans: 3
Sol:
on
insoluble in Dil. HCl, Soluble in NaOH solution , & Decolourization of
2Br / water
35. Excessive release of 2CO into the atmosphere results in
1. depletion of ozone 2. polar vertex
3. formation of smog 4. global warming
Ans: 4
Sol: Excessive release of 2CO into the atmosphere results in global warming.
36. The element having greatest difference between its first and second ionization
energies, is
1. K 2. Ca 3. Ba 4. Sc
Ans: 1
Sol: In given options k has greatest difference between its first & second ionisation
energies.
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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37. The major product of the following reaction is:
21.
2 .PBr
KOH alcOH
O
1. O 2. O
3. O
OH
4. O
Ans: 4
Sol:
BrOH
O
3PBr .alc KOH
OO
38. The number of water molecules not coordinated to copper ion directly in
24.5CuSOl H O , is
1. 2 2. 3 3. 1 4. 4
Ans: 3
Sol: In 4 2.5CnSO H O , four water molecules are coordinates to 2Cu ion and one water
molecules is outside the coordinate sphere.
Number of water molecules not coordinated to copper ion is
39. 60 ,C an allotrope of carbon contains:
1. 18 hexagons and 14 pentagons 2. 20 hexagons and 12 pentagons
3. 12 hexagons and 20 pentagons 4. 16 hexagons and 16 pentagons
Ans: 2
Sol: In 60C , 102n hexagonal, 12 pentagonal
20 hexagonal, 12 petagonal
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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40. Aniline dissolved in dilute HCl is reacted with sodium nitrite at 0ºC. this solution
was added dropwise to a solution containing equimolar mixture of aniline and
phenol in dil. HCl. The structure of the major product is
1.
N=N NH2
2.
N=N O
3.
N=N OH
4.
N=N NH
Ans: 1
Sol:
2. ,0ºNaNO
dil HCl C
2NH2N Cl
In equimolar mixture of aniline, phenol in dil.HCl, amiline in more reactive, as
slightly acidic medium supports.
41. The major product of the following reaction is
1 .3
i DCl equivii DlCh C CH
1. 3 2CH C l Cl CHD 2. 3CH CD Cl CHD l
3. 3 2CH CD CH Cl l 4. 3CH CD l CHD Cl
Ans: 1
Sol: 3
i DCl eqilateralentii DICH C C H
3CH C C H
I
Cl
D
D
42. The standard Gibbs energy for the given cell reaction in kJ mol-1 at 298 K is:
2 2 , º 2 298Zn s Cu aq Zn aq Cu s E V at K
1. -192 2. 384 3. -384 4.192
Ans: 3
Sol: 22aqZn s Cu aq Zn Cu s
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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0º cellG nFE
2 96000 2
4 96kJ 384kJ
43. The given plots represent the variation of the concentration of a reactant R with
time for two different reactions (i) and (ii). The respective orders of the reactions
are
ln R
time
(i)
R
time
(ii)
1. 0,1 2. 0,2 3. 1,0 4. 1,1
Ans: 3
Sol:
ln R
time
R
time
T is proportional [R] it is 1st order t is proportional to [R] it is 0th order
44. The increasing order of reactivity of the following compounds towards aromatic
electrophilic substitution reaction is
A B C D
Cl OMe Me CN
1. D<A<C<B 2. B<C<A<D 3. D<B<A<C 4. A<B<C<D Ans: 1
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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Sol:
Cl OMe Me CN
B > C > A > D(dueto + M, Directing)
45. The correct order of the oxidation states of nitrogen in 2 2 2 3, ,NO N O NO and N O is
1. 2 2 3 2NO NO N O N O 2. 2 2 3 2N O N O NO NO
3. 2 2 3 2N O NO N O NO 4. 2 2 3 2NO N O NO N O
Ans: 3
Sol: Compound Oridation state
No +2
2N O +1
2NO +4
2 3N O +3
46. Match the catalysts (colum I) with products (column II).
Column I Column II
Catalyst Product
A) 2 5V O i) Polyethylene
B) 4 3/TiCl Al Me ii) ethanal
C) 2PdCl iii) 2 4H SO
D) Iron oxide iv) 3NH
1. a-iii, b-i, c-ii, d-iv 2. a-ii, b-iii, c-i, d-iv
3. a-iv, b-iii, c-ii, d-i 4. a-iii, b-iv, c-i, d-ii
Ans: 1
Sol: Catalyst Product 2 5V O 2 4H SO 4 3/TiCl Al Me Polyethelene
2PdCl Ethanal Iron oxide 3NH
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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47. Which of the following statement is not true about sucrose? 1. On hydrolysis, it produces glucose and fructose
2. It is also named as invert sugar
3. It is non-reducing sugar
4. The glycosidic linkage is present between C1 of glucose and C1 of fructose
Ans: 4
Sol: coshydrolysisSucrose glu e fructose
48. The aerosol is a kind of colloid in which:
1. Solid is dispersed in gas 2. liquid is dispersed in water
3. gas is dispersed in liquid 4. gas is dispersed in solid
Ans: 1
Sol: aerosol is a kind of colloid is disersion phase(s) dispersion medium(g)
49. The major product of the following reaction is
2 3CH CH
4
3
alkaline KMnOiii H O
1.
2CH CHO
2.
COOH
3.
2CH COOH
4.
3COCH
Ans: 2
Sol:
CH2CH3
4
3
i alkaline KMnOii H O
COOH
2019_Jee-Main Question Paper_Key & Solutions
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50. Consider the van der walls constants, a and b, for the following gases
6 2
2 3 1
/ 1.3 0.2 5.1 4.1
/ 10 3.2 1.7 1.0 5.0
Gas Ar Ne Kr Xe
a atmdm mol
b dm mol
Which gas is expected to have the highest critical temperature
1. Kr 2. Ar 3. Ne 4. Xe
Ans: 1
Sol: To have highest critical temperature a should be high b should below so, it is Kr
51. Among the following, the molecule expected to be stabilized by anion formation is:
2 2 2, , ,C O NO F
1. NO 2. 2C 3. 2F 4. 2O
Ans: 2
Sol: C2 has vacant and if e—enters into this if will be stabilized
52. The ore that contains the metal in the form of fluoride is
1. cryolite 2. Malachite 3. Magnetite 4. Sphalerite
Ans: 1
Sol: Cryolife Na3AlF6
Malachite CuCO3.CU(OH)2
Magnetite Fe3O4
Sphalarite ZnS
53. The one that will show optical activity is:
(en = ethane – 1,2, -diamine)
1. 2. 3. 4.
Ans: 4
Sol: Given which will show optical activity
The Compound should not have plans of symmetry only compounds in options
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
54. For any given series of spectral lines of atomic hydrogen, let max minv v v be the
difference in maximum and minimum frequencies in 1cm . The ratio balmerv is:
1. 4:1 2. 9:4 3. 5:4 4. 27.5
Ans: 2
Sol: 2 2 2 2
2 2 2 2
1 1 1 191 1 2
1 1 1 1 42 2 3
vyman
Balmer
R RVV R R
55. The major product of the following reaction is:
1. 2.
3. 4. Ans: 3 Sol: 56. The osmotic pressure of a dilute solution of an ionic compound XY in water is four
times that of a solution of 0.01 M BaCl2 in water. Assuming complete dissociation of the given ionic compounds is water, the concentration of XY (in mol L-1) in solution is
1. 44 10 2. 416 10 3. 24 10 4. 26 10
Ans: 4
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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Sol: 2 – Chloro – 1 – methyl - 4 – nitro benzene
2 3 0.01 4C RT RT
for 2BaCli-factor for xy
C = 0.06 M 57. For a reaction, 2 2 33 2 :N g H g NH g identify dihydrogen 2H as a limiting
reagent in the following reaction mixtures.
1. 35g of N2 + 8g of H2 2. 28g of N2+6g of H2
3. 56g of N2 + 10g of H2 4. 14g of N2+ 4g of H2
Ans: 3
Sol: 2 2 33 2N H NH 56g 10 g 2 mol 5 mol Actually 6 mol of 2H is required but only 5 mol present so it is limiting reagent. 58. Magnesium powder burns in air to give:
1. 3 3 22and MgMg NO N 2. 3 2and MgMgO N
3. MgO only 4. 3 2and MgMgO NO
Ans: 2
Sol: 2 3 2Mh O MgO Mg N
59. The major product of the following reaction is
43 2 3
LiAIHCH CH CHCO CH
1. 3 2 2 2 3CH CH CH CO CH 2. 3 2 2 2CH CH CH CH OH
3. 3 2 2CH CH CH CHO 4. 3 2CH CH CHCH OH
Ans: 4
Sol:
O
O4LiAlH
OH
60. Liquid ‘M’ and liquid ‘N’ form an ideal solution. The vapour pressures of pure liquids ‘M’ and ‘N’ are 450 and 700mmHg, respectively, at the same temperature. Then correct statement is:
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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(xM = Mole fraction of ‘M’ is solution; xN = Mole fraction of ‘N’ in solution ; yM = Mole fraction of ‘M’ in vapour phase; yN = Mole fraction of ‘N’ in vapour phase)
1. M M
N N
x yx y
2. M M
N N
x yx y
3. M M
N N
x yx y
4. M M N Nx y x y
Ans: 3
Sol: 0 450PM
0 700PN 450700
M m
N n
y xy x
0
M mm
t
P XyP
. 1m N
n M
X YX Y
0
nN
PN XYPt
m m
n n
X YX Y
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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MATHS 61. Let 2S 2 ,2 : 2cos 3sin 0 . Then the sum of the elements of S is:
1. 53
2. 2 3. 13
6
4.
Ans: 2
Sol: solving the trigonometric equation, we get 1sin
2
2 ,2 so 5 7 11, , ,
6 6 6 6
so sum 2
62. If the standard deviation of the numbers 1,0,1,k is 5 where k 0 , then k is
equal to:
1. 1023
2. 2 6 3. 543
4. 6
Ans: 2
Sol:
22 2 2i i2 x x k 2 k5 k 2 6
n n 4 16
63. All the points in the set iS : Ri
i 1 lie on a:
1. circle whose radius is 2 2. circle whose radius is 1
3. straight line whose slope is -1 4. straight line whose slope is 1
Ans: 2
Sol: 2 2
2 2 2
i 1 2x iy x , y1 1 1
So 2 2x y 1 , is a circle of radius 1
64. For any two statements p and q, the negation of the expression p ~ p q is:
1. p q 2. ~ p ~ q 3. p q 4. ~ p ~ q
Ans: 2
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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Sol: p ~ p q p ~ p p q T p q p q
So ~ p ~ p q ~ p q ~ p ~ q (or) draw truth tables
65. If the function f : R 1, 1 A defined by 2
2
xf x1 x
, is surjective, then A is
equal to:
1. R 1,0 2. R 1,0 3. R 1 4. 0, Ans: 2
Sol: Let 2
2 22
xy y x y x1 x
2x 1 y y
2 yx 0 y 11 y
y y 1 0 y , 1 0,
So A R 1,0
66. If the fourth term in the Binomial expansion of g
6log x2 x x 0
x
is 720 8 , then
a value of x is:
1. 28 2. 8 3. 38 4. 28
Ans: 1
Sol: equate 7
4T 20 8
x8
3 3log6 7
32C . . x 20 8x
Take logaritham on both side to base 8, we get
2x x x8 8 8log 2 log log 2 (or) 21 x 8 (or)
18
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
67. Slope of a line passing through P 2,3 and intersecting the line, x y 7 at a
distance of 4 units from P, is:
1. 1 51 5
2. 1 71 7
3. 7 17 1
4. 5 15 1
Ans: 2
Sol:
x y 7
2
m,4
090
P 2,3
2 1 m 1 m 1tan
1 m 1 1 m14 7
m 1 11 m 7
simplify get 1 7m1 7
(or) 1 7m1 7
68. If f x is a non-zero polynomial of degree four, having local extreme points at
x 1,0,1 ; Then the set S x R : f x f 0 contains exactly:
1. four irrational numbers
2. two irrational and two rational numbers
3. two irrational and one rational number
4. four rational numbers
Ans: 3
Sol: Let 4 2
1 2 x xf x x x 1 f x c4 2
f 0 f x c 4 2x x c
4 2
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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2 2x x 1 0
2 2
x 0 (or) x 2
69. Let 3i j
and 2i j 3k
. If 21
, where 1
is parallel to
and 2
is
perpendicular to
, then 1 2
is equal to:
1. 1 3i 9 j 5k2 2. 1 3i 9 j 5k
2
3. 3i 9j 5k 4. 3i 9j 5k
Ans: 1
Sol: Let 1 1k , 2 2k i 3j k
Given 1 2 equating i, j,k coefficient we get 1 21 1k ,k2 2
find
1 21 3i 9 j 5k2
70. If the function f defined on ,6 3
by 2 cos x 1,xcot x 1 4f x
k, x4
is continuous,
then k is equal to:
1. 2 2. 12
3. 12
4. 1
Ans: 2
Sol: x
4
2 cos x 1 0k Lt formcot 1 0
by LH rule we get
1k2
71. Let and be the roots of the equation 2x x 1 0 . Then for y 0 in R,
y 1y 1
1 y
is equal to:
1. 2y y 1 2. 3y 3. 2y y 3 4. 3y 1
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
Ans: 2
Sol: 2 21, 1, 1
1 1 2 3R R R R & take common y 1
Given determinant 1 1 1
y 1 1 y 11 y
expanding simplify
2 3y y y
72. If the line y mx 7 3 is normal to the hyperbola 2 2x y 1
24 18 , then a value of m
is:
1. 52
2. 152
3. 25
4. 35
Ans: 3
Sol: use 22 22 2
2 2 2
a ba bm n
here 2 2a 24,b 18, m,m 1,n 7 3
Simplify we get 2m5
73. The value of /2 3
0
sin x dxsin x cos x
is:
1. 2
4
2. 2
8
3. 1
4
4. 1
2
Ans: 3
Sol: a a
0 0
use f x dx f a x dx
Simplify we get, given integral 1
4
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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74 If the line, x 1 y 1 z 2
2 3 4
meets the plane, x 2y 3z 15 at a point P, then
the distance of P from the origin is:
1. 52
2. 2 5 3. 92
4. 72
Ans: 3
Sol: Let P 2t 1,3t 1,4t 2 be any point on the line, P lies on plane also
1 1 9t ,P 2, ,4 ,OP2 2 2
75. Let 10
10
k 1
f a k 16 2 1
, where the function f satisfies f x y f x f y for
all natural numbers x, y and f 1 2 . Then the natural number ‘a’ is
1. 16 2. 2 3. 3 4. 4
Ans: 3
Sol: xf x 2 & simplify (using G.P formula) we get a 3
76. If one end of a focal chord of the parabola, 2y 16x is at 1,4 , then the length of
this focal chord is:
1. 24 2. 22 3. 20 4. 25
Ans: 4
Sol: use 1 2 1 21t t 1, t , t 22
So Q 16, 16 , P 1,4 given PQ 25
77. The solution of the differential equation 2dyx 2y x , x 0dx
with y 1 1 , is:
1. 32
4 1y x5 5x
2. 22
3 1y x4 4x
3. 2
2
x 3y4 4x
4. 3
2
x 1y5 5x
Ans: 3
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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Sol: LDE 2 dx 2xIF e x
4
2 2 x 3y.x x.x dx c,c4 4
2
2
x 3y4 4x
78. If the tangent to the curve, 3y x ax b at the point 1, 5 is perpendicular to
the line, x y 4 0 , then which one of the following points lies on the curve?
1. 2, 1 2. 2,2 3. 2, 2 4. 2,1 Ans: 3
Sol: f 1 5 1 a b 5 a b 6 ……(1)
1 2 1f x 3x a f 1 3 a 1 a 4
b a 6 2
3f x x 4x 2 check options 2, 2 satisfies
79. If 1 1 1 2 1 3 1 n 1 1 78
..............0 1 0 1 0 1 0 1 0 1
then the inverse of
1 n0 1
is
1. 1 120 1
2. 1 0
12 1
3. 1 0
13 1
4. 1 130 1
Ans: 4
Sol: n 1 n1 78 1 1 2 3... n 1 1
20 1 0 1 0 1
n 1 n 156 13 2 n 13 n N
So increase of 1 13 1 130 1 0 1
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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80. The value of is 2 0 0 0 2 0cos 10 cos10 cos50 cos 50 is :
1. 34
2. 03 1 cos 202
3. 03 cos 204 4.
32
Ans: 1
Sol: 2 2 3cos A cos B cos A cos B ,
4 if 0A B 60
81. Let the sum of the first n terms of a non-constant A.P., 1 2 3a ,a ,a ,... be
n n 750n A
2
, where A is a constant, If d is the common difference of this A.P.,
then the ordered pair 50d,a is equal to:
1. 50,50 45A 2. A,50 45A 3. 50,50 46A 4. A,50 46A Ans: 4
Sol: n n n n 1
n n 7S 50n A, t S S n 4 A 50
2
50d A,a 46A 50
82. A committee of 11 members is to be formed from 8 males and 5 females. If m is the
number of ways the committee is formed with at least 6 males and n is the number
of ways the committee is formed with at least 3 females, then:
1. m n 68 2. n m 8 3. m n 68 4. m n 78 Ans: 4
Sol: 8 5 8 5 8 5
6 5 7 4 8 3m C . C C . C C . C 28 40 10 78
8 5 8 5 8 56 3 7 4 6 5n C . C C . C C . C 78
83. Four persons can hit a target correctly with probabilities 1 1 1, ,2 3 4
and 18
respectively. If all hit at the target independently, then the probability that the
target would be hit, is:
1. 1
192 2.
2532
3. 25
192 4.
732
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, [email protected]
Ans: 2
Sol: 1 2 3 7 7 251 P no one hiting the t arg et 1 . . . 12 3 4 8 32 32
84. Let S be the set of all values of x for which the tangent to the curve
3 2y f x x x 2x at x, y is parallel to the line segment joining the points
1,f 1 ,and 1,f 1 , then S is equal to:
1. 1 , 13
2. 1 ,13
3. 1 ,13
4. 1 , 13
Ans: 2
Sol: 1,f 1 1, 2 , 1,f 1 1,0
2slope 1 3x 2x 2 1
23x 2x 1 0 x 1 3x 1 0
1x 1 or x3
85. The integral 2/3 4/3sec x cosec x dx is equal to (Here C is a constant of integration)
1. 1/33cot x C 2. 1/33 tan x C 3. 1/33tan x C 4. 4/33 tan x C4
Ans: 2
Sol: put 2 22 2 2 2 43 3 3 3 3
1 1dx dx put tan x t
cos x sin x cos x. tan x .cos x
1 1
3 34/3dt t . 3 c 3 tan x Ct
86. Letp,q R . If 2 3 is a root of the quadratic equation, 2x px q 0 , then :
1. 2q 4p 16 0 2. 2p 4q 12 0
3. 2p 4q 12 0 4. 2q 4p 14 0
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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Ans: 3
Sol: 2 3 also root, S.R 2 3 2 3 P p 4
P.R 4 3 q q 1
So 2p 4q 12 16 4 12 0
87. If a tangent to the circle 2 2x y 1 intersects the coordinate axes at distinct points
P and Q, then the locus of the mid-point of PQ is:
1. 2 2 2 2x y 16x y 0 2. 2 2 2 2x y 2x y 0
3. 2 2x y 2xy 0 4. 2 2 2 2x y 4x y 0 Ans: 4
Sol: Let P cos ,sin be any point on circle, tangent at P is x cos ysin 1
1 1A ,0 ,B 0,
cos sin
1 1P x , y be mid point of
1 10 0cos sinAB ,
2 2
1 11 12x ,2y
cos sin
1 1
1 1cos ,sin2x 2y
2 21 1
1 1 14x 4y
2 2 2 21 1 1x y 4x y
Locus of 1 1P x , y is 2 2 2 2x y 4x y 0
88. The area (in sq. units) of the region 2A x, y : x y x 2 is:
1. 92
2. 103
3. 136
4. 316
Ans: 1
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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Sol: 2x y x 2 2x x 2 0
Use
3 32 2 2
2
b 4ac 1 4.1. 2 27 96a 6.1 6 2
(or)
Draw diagram and find 2
2
1
9x 2 x dx2
89. Let f x 15 x 10 ;x R. Then the set of all values of x, at which the function,
g x f f x is not differentiable, is:
1. 5,10,15 2. 10 3. 10,15 4. 5,10,15,20 Ans: 1
Sol: f x 15 x 10 , x R
x 10 if x 520 x 5 x 10
g x f f xx 10 x 1530 x if x 15
90. A plane passing through the points 0, 1,0 and 0,0,1 and making an angle 4
with the plane y z 5 0 , also passes through the point:
1. 2, 1,4 2. 2, 1, 4 3. 2,1, 4 4. 2,1,4
Ans: 4
Sol: x y z 1a 1 1
is the equation of plane ……(1)
y z 0 ……(2)
2
1 .0 1.1 1 1acos
4 1 1 1 1 1a
2019_Jee-Main Question Paper_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
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2
2
21 14 2a2 12 2
a
21 12 aa 2
So plane equation is
2x y z 1 2x y z 1 , 2x y z 1
Check options 2,1,4 satisfies second equation
Prepared by
Sri Chaitanya Faculty