2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
PHYSICS 1. Two plane mirrors are inclined to each other such that a ray of light incident on the first
mirror 1( )M and parallel to the second mirror 2( )M is finally reflected from the second
mirror 2( )M parallel to the first mirror 1( )M . The angle between the two mirrors will be:
1) 090 2) 045 3) 075 4) 060
Ans. 4
Sol.
060
2. A particle having the same charge as of electron moves in a circular path of radius 0.5 cm
under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it
to move in a straight path, then the mass of the particle is (Given charge of electron
–191.6 10 C )
1) –242.0 10 kg 2) –191.6 10 kg 3) –271.6 10 kg 4) –319.1 10 kg
Ans. 4
Sol. 2mvqvB
R
mv qBR
Path to be straight line, qE qvR
E v B
So 2
242 10qB R
m kgE
3. In a Young's double slit experiment, the slits are placed 0.320 mm apart. Light of
wavelength 500 nm is incident on the slits. The total number of bright fringes that are
observed in the angular range 0 030 30 is:
1) 320 2) 641 3) 321 4) 640
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
Ans. 2
Sol. 030d Sin n
320n
So, Number of maxima (Including central maxima) 2 1 641n
4. The pitch and the number of divisions, on the circular scale, for a given screw gauge
are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any
object, the zero of its circular scale lies
3 divisions below the mean line.
The readings of the main scale and the circular scale, for a thin sheet, are
5.5 mm and 48 respectively, the thickness of this sheet is
1) 5.755 mm 2) 5.725 mm 3) 5.740 mm 4) 5.950 mm
Ans. 2
Sol. 20.5 10
PitchLC mn
Noof Divisions
Positive zero error 2 23 0.5 10 1.5 10mm mm
Reading MSR CSR- ve zeroerror =5.725mm
5. At a given instant, say 0t , two radioactive substances A and B have equal activities. The
ratio B
A
R
R of their activities after time t itself decays with time t as –3t
e , If the half-life of
A is 2l n , the half-life of B is
1) 2
2
l n 2) 2 2l n 3)
2
4
l n 4) 4 2l n
Ans. 3
Sol. 30
0
B
B A
A
tt tB
t
A
R R ee e
R R e
So, 3B A
Hence n 2
n 2 1A A
A
lT l
n 2 n 2
4B
B
l lT
6. The top of a water tank is open to air and its water level is maintained. It is giving out
30.74 m water per minute through a circular opening of 2cm radius in its wall. The depth
of the centre of the opening from the level of water in the tank is close to:
1) 9.6m 2) 4.8m 3) 2.9m 4) 6.0m
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
Ans. 2
Sol. 40.744 10 2
60gh
4.8h m
7. Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure
if Ge diode connection are reversed, the value of 0V changes by: (assume that the Ge
diode has large breakdown voltage)
1) 0.6 V 2) 0.8 V 3) 0.4 V 4) 0.2 V
Ans. 3
Sol. Initiall current will flow through Ge diode drop will be 0.3v.
In the second case, Si diode wll conduct and drop across the combination is 0.7v. so,
change in Voltage =0.4V
8. A power transmission line feeds input power at 2300 V to a step down transformer with
its primary windings having 4000 turns. The output power is delivered at 230 V by the
transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%
, the output current would be:
1) 25 A 2) 50 A 3) 35 A 4) 45 A
Ans. 4
Sol. 0.9out s s
in p p
P V I
P V I
45sI A
9. A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits
torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its
centre on both sides, it reduces the oscillation frequency by 20% . The value of ratio m/M
is close to:
1) 0.17 2) 0.37 3) 0.57 4) 0.77
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
Ans. 2
Sol. 1
2
Kf
I
and 1 1
2
Kf
I
where K is torsigional constant
so, 1
1
20 11
100 5
f f I
f I
1
4
5
I
I
1
16
25
I
I
24
12
M LI
and
21
2
mLI I So, 0.37
m
M
10. In a car race on straight road, car A takes a time t less than car B at the finish and passes
finishing point with a speed 'v' more than that of car B. Both the cars start from rest and
travel with constant acceleration 1a and 2a respectively. Then 'v' is equal to:
1) 1 2
2
a at
2) 1 22a a t 3) 1 2
1 2
2a at
a a 4) 1 2a a t
Ans. 4
Sol. Given 1 2 1 2and t t t
Also, 1 1 1 2 2 2a t and v a t
So, 1 1 2 2v a t a t 1 2a a t 1 2
2 1
t a
t a
11. A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 027 C . Amount of heat
transferred to the gas, so that rms velocity of molecules is doubled, is about:
[Take 8.3 / R J K mole ]
1) 10 kJ 2) 0.9 kJ 3) 6 kJ 4) 14 kJ
Ans. 1
Sol. 15 54
28 2v
RnC T T T
10kJ
12. A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force
is applied on the rope at some point, the rope deviated at an angle of 045 at the roof point.
If the suspended mass is at equilibrium, the magnitude of the force applied is –210 g ms
1) 200 N 2) 100 N 3) 140 N 4) 70 N
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
Ans. 2
Sol.
F
T
mg
2
Tmg and
2
TF So, 100f mg N
13. A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis,
about 0x . When its potential Energy (PE) equals kinetic energy (KE), the position of
the particle will be:
1) 2
A 2)
2 2
A 3)
2
A 4) A
Ans. 3
Sol. 2 2 2 2 21 1
2 2m A x mx
2
Ax
14. In the given circuit the internal resistance of the 18 V cell is negligible. If 1 400R ,
3 100R and 4 500R and the reading of an ideal voltmeter across 4R is 5V, then the
value 2R will be:
1) 300 2) 230 3) 450 4)550
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
Ans. 1
Sol. As, 3R and 4R are in series, voltage across 3 1R V
So, voltage across 3 4,R R and 2R combination is 6V
Therefore , 2 3 4
1 1 1
200R R R
2 300R
15. A musician using an open flute of length 50 cm produces second harmonic sound waves.
A person runs towards the musician from another end of a hall at a speed of 10 km/h. If
the wave speed is 330 m/s, the frequency heard by the running person shall be close to:
1) 753 Hz 2) 500 Hz 3) 333 Hz 4) 666 Hz
Ans. 4
Sol. 2
nf
l
For 2nd harmonic, 2n and 660f Hz velocity of observers Vo=10Km/h=2.78m/s
Hence apparent frequency 0v vf f
v
=666Hz
16. The position co-ordinates of a particle moving in a 3 D coordinate system is given by
cosx a t
siny a t
and z a t
The speed of the particle is:
1) a 2) 3a 3) 2a 4) 2a
Ans. 3
Sol. 2 2 2 2x y z
v v v v a
17. In a communication system operating at wavelength 800 nm, only one percent of source
frequency is available as signal bandwidth. The number of channels accommodated for
transmitting TV signals of band width 6 MHz are (Take velocity of light
8 –343 10 / , 6.6 10c m s h J s )
1) 63.75 10 2) 54.87 10 3) 63.86 10
4) 56.25 10
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
Ans. 4
Sol. 143.75 10C
f hz
1% of 17 63.75 10 3.75 10f hz Mhz
Number of channels 63.75 10
6
56.25 10
18. A force acts on a 2 kg object so that its position is given as a function of time as 23 5x t .
What is the work done by this force in first 5 seconds?
1) 850J 2) 900J 3) 950J 4) 875J
Ans. 2
Sol. 6 /dz
v t m sdt
212 36 900
2w k t J
19. Two point charges 1 10q C and 2 (–25 )q C are placed on the x-axis at 1x m and 4x
m respectively. The electric field (in V/m) at a point 3y m on y-axis is,
9 2 2
0
19 10
4take Nm C
1) 263 27 10i j 2) 281 81 10i j 3) 263 27 10i j 4) 281 81 10i j
Ans. 3
Sol. 21 9 27 10
vE i j
m
and 22 72 54 10
vE i j
m
So, 21 2 63 27 10
vE E E i j
m
20. A carbon resistance has a following colour code. What is the value of the resistance ?
1) 1.64 5%M 2) 530 5%k 3) 64 10%k 4) 5.3 5%M
Ans. 2
Sol. 453 10 5%R 530 5%k
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
21. A parallel plate capacitor with square plates is filled with four dielectrics of dielectric
constants 1 2 3 4, , ,K K K K arranged as shown in the figure. The effective dielectric constant
K will be:
1)
1 2 3 4
1 2 3 42
K K K KK
K K K K
2)
1 2 3 4
1 2 3 4
K K K KK
K K K K
3)
1 4 2 3
1 2 3 42
K K K KK
K K K K
4)
1 3 2 4
1 2 3 4
K K K KK
K K K K
Ans. (Bonus)
Sol.
1K C 2K C
3K C 4K C
Kc
Hence
1 2 3 4
1 2 3 4
K K K KK
K K K K
22. Two Carrnot engines A and B are operated in series. The first one, A, receives heat at
1 600T K and rejects to a reservoir at temperature 2T . The second engine B receives heat
rejected by the first engine and, in turn, rejects to a heat reservoir at 3 400 T K . Calculate
the temperature 2T if the work outputs of the two engines are equal:
1) 400 K 2) 600 K 3) 500 K 4) 300 K
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
Ans. 3
Sol. 1 2 2 3 1 2 1 22 2 2
T TT
23. A rod of length 50cm is pivoted at one end. It is raised such that if makes an angle of 030
from the horizontal as shown and released from rest. Its angular speed when it passes
through the horizontal –1 in rad s will be –210g ms
1) 30 2) 30
2 3)
302
4) 20
3
Ans. 1
Sol. Applying energy conservation
2
21
4 2 3
l mLmg
36 /rad s
24. Charge is distributed within a sphere of radius R with a volume charge density
2 /
2,r aA
r er
where A and a are constants. If Q is the total charge of this charge
distribution, the radius R is:
1) log 12 2
a Q
aA
2) log 12
Qa
aA
3) 1
log1
2
aQ
aA
4) 1
log2 1
2
a
Q
aA
Ans. 4
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
Sol. 2
2
24
R R r
a
O O
tsdv e r dr
r
2
4 12
R
aa
A e
1log
2 12
aR
Aa
25. One of the two identical conducting wires of length L is bent in the form of a circular
loop and the other one into a circular coil of N identical turns. If the same current is
passed in both, the ratio of the magnetic field at the central of the loop LB to that at
the centre of the coil CB , i.e. L
C
B
Bwill be:
1) 1
N 2) 2
N 3) 2
1
N 4) N
Ans. 3
Sol. For loop 22
LL R R
0 0
2L
i iB
R L
2for coil L N r 2
Lr
20 0
2C
Ni iB N
r L
So,
2
1L
C
B
B N
26. The magnetic field associated with a light wave is given, at the origin, by
0B B 7 7sin 3.14 10 sin 6.28 10ct ct . If this light falls on a silver plate having a work
function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons?
8 –1 –343 10 , 6.6 10c ms h J s
1)7.72eV 2)8.52eV 3)12.5eV 4)6.82eV
Ans. 1
Sol. To calculate maxU we must take the highest frequency
So 34 7 8
max 19
6.6 10 6.28 10 3 104.7
2 3.14 1.6 10U
7.675 7.7ev
27. The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth
36.4 10 km ) is 1E and kinetic energy required for the satellite to be in a circular orbit at
this height is 2E . The value of h for which 1E and 2E are equal, is:
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
1) 41.28 10 km 2) 36.4 10 km 3) 33.2 10 km 4) 31.6 10 km
Ans. 3
Sol. 1
1 1E GMm
R R h
and
2 2
GMmE
R h
If 1 2E E , 3200
2
Rh km
28. Expression for time in terms of G (universal gravitational constant), h (Planck constant)
and c (speed of light) is proportional to:
1) 3
Gh
c 2)
5hc
G 3)
3c
Gh 4)
5
Gh
c
Ans. 4
Sol. 1 3 2G M LT
2 1h ML T
1C LT
x y zt G H C
1 1,x yz z
and 5z so 5
Ght
c
29. The energy associated with electric field is EU and with magnetic field is BU for an
electromagnetic wave in free space. Then:
1) 2B
E
UU 2) E BU U 3) E BU U 4) E BU U
Ans. 3
Sol. A BU U
30. A series AC circuit containing an inductor 20 mH , a capacitor (120 )F and a resistor
(60 ) is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in
60 s is:
1) 32.26 10 J 2) 33.39 10 J 3) 25.65 10 J 4) 25.17 10 J
Ans. 4
Sol. 6.28LX L
1
26.53cXc
So, 22 2 4010.5L Cz R X X
Energy dissipated 2
2517
vR t J
z
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
CHEMISTRY 31. Good reducing nature of H3PO2 attributed to the presence of
1) One P-OH bond 2) One P-H bond
3) Two P-H bonds 4) Two P-OH bonds
Ans. 3
Sol: Then is two P-H bonds in the structure.
32. The complex that has highest crystal field splitting energy (), is :
1) K3 [Co (CN)6] 2) [Co(NH3)5(H2O)]Cl3
3) K2 [CoCl4] 4) [Co (NH3)5Cl]Cl2
Ans. 4
Sol: The field strength of ligand “CN” is high.
33. The metal that forms nitride by reacting directly with N2 of air, is:
1) K 2) Cs 3) Li 4) Rb
Ans. 3
Sol: Among Alkali metals “Li” reacts with air forming both oxide and nitride.
34. In which of the following processes, the bond order has increased and paramagnetic
character has changed to diamagnetic?
1) 2 2N N 2)NO NO 3)
22 2O O 4) 2 2O O
Ans. 2
Sol: Bond order Magnetic behaviour
No 2.5 paramagnetic
NO 3 diamagnetic
35. The major product of the following reaction is:
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
1) 2) 3) 4)
OH
3CH
O
Ans: 4
Sol:
OH
OH 3CH
3CH
C
O2CH
2CH
2CH OH
O
OH
3CH
3AlCl
O
O
36. The transition element that has lowest enthalpy of atomisation, is :
1) Zn 2) Cu 3) V 4) Fe
Ans: 2
Sol:
TiVCr
Mn
Fe Co Ni
Cu
AtomisationH
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
37. Which of the following combination of statements is true regarding the
interpretation of the atomic orbitals ?
(a) An electron in an orbital of high angular momentum stays away from the
nucleus than an electron in the orbital of lower angular momentum.
(b) For a given value of the principal quantum number, the size of the orbit is
inversely proportional to the azimuthal quantum number.
(c) According to wave mechanics, the ground state angular momentum is equal to2
h
.
(d) The plot of Vs r for various azimuthal quantum numbers, shows peak
shifting towards higher r value.
1) (b), (c) 2) (a), (d) 3) (a), (b) 4) (a), (c)
Ans: 4
Sol: 2
hmvr n
-constant
If mvr is high n is high so electron stays away from the nucleus.
For ground state 1n .
38. The tests performed on compound X and their inferences are:
Test Inference
(a) 2,4- DNP test Coloured precipitate
(b) Iodoform test Yellow precipitate
(c) Azo-dye test No dye formation
Compound 'X' is:
1) 2) 3) 4)
Ans: 2
Sol: 2,4DNP test is given by aldehydes and ketones
Iodo form test given by C
O
3CH
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
Azo dye test not given by
3CH3CH
N
AR
39. The major product formed in the
following reaction is:
1) 3CH
O
H
OH
2)
3) 4)
Ans: (no option)
Sol:
CHO3 2OHCH CHO CH CHO
2CHOH CH CHO
40. For the reaction, 2A B products, when the concentrations of A and B both
wrere doubled, the rate of the reaction increased from –1 –10.3 mol L s to
– –12.4 lmolL s . When the concentration of A alone is doubled, the rate increased
from 1 –10.3 molL s to –1 –10.6 mol L s
Which one of the following statements is correct ?
1) Order of the reaction with respect to Bis2
2) Order of the reaction with respect to Ais2
3) Total order of the reaction is 4
4) Order of the reaction with respect to B is 1
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
Ans: 4
Sol: 1x y
R K A B
0.3 (1)
2.4 2 2 (2)
0.6 2 (3)
x y
x y
x Y
K A B
K A B
K A B
Order w.r.t B is (2)
From eq (1)&(3)
1 1
2 2
1
x
x
From eq (2)&(3)
4 (2)
2
y
y
41. The correct sequence of amino acids present in the tripeptide given below is :
1) Leu - Ser – Thr 2) Thr - Ser- Leu
3) Thr - Ser - Val 4) Val - Ser – Thr
Ans: 4
Sol:
3CH CH CH CHO
2NH
3CH 2NH
CH C OH
O
2CH OH2NH
OOH
3CH CH CH C OH
Valine serine Threonine
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
42. The correct statement regarding the given Ellingham diagram is:
1) At 800°C, Cu can be used for the extraction of Zn from ZnO
2) At 500 C, coke can be used for the extraction of Zn from ZnO
3) Coke cannot be used for the extraction of Cu from Ca2O.
4) At 1400°C, Al can be used for the extraction of Zn from ZnO
Ans: 4
Sol: At 01400 G for the formation of 2 3Al O is more negative compound to that of “zero”.
43. For the following reaction, the mass of water produced from 445 g of C57H110O6 is :
2C57H110O6(s) + 163O2(g)114CO2(g) + 110 H2OP(1)
1) 495 g 2) 490 g 3) 890 g 4) 445 g
Ans: 1
Sol: 57 110 6 2 2 2
16357 55
2C H O O CO H O
445
0.5890compoundn moles
Due to equation 1 mole of compound gives 55 moles of 2H O so 0.5 moles will give
0.5 55moles
2 0.5 55 18weight of H O formed g 495grms
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
44. The correct match between Item I and Item II is :
Item I Item II
(A) Benzaldehyde (P) Mobile phase
(B) Alumina (Q) Adsorbent
(C)Acetonitrile (R) Adsorbate
1) (A) (Q);(B) (R);(C) (P)
2) (A) (P); (B) (R); (C) (Q)
3) (A) (Q); (B) (P); (C) (R)
4) (A) (R); (B) (Q); (C) (P)
Ans: 4
Sol: Conceptual
45. The increasing basicity order of the following compounds is :
(A) 3 2 2CH CH NH (B) (C) (D)
1) (D)<(C)<(A)<(B) 2) (A)<(B)<(D)<(C)
3) (A)<(B)<(C)< (D) 4) (D)<(C)<(B)<(A)
Ans: 1
Sol: Donating groups will increase the basicity of amines, Ph N
3CH
H is least basic due to
delocalization of electron pair.
46. For coagulation of arscnious sulphide sol, which one of the following salt solution
will be most effective?
1) AlCl3 2) NaCl 3) BaCl2 4) Na3PO4
Ans: 1
Sol: To coagulate a negative colloid cation with higher charge is more effective.
47. At100°C, copper (Cu) has FCC unit cell structure with cell edge length of x
Å. What is the approximate density of Cu (in g cm–3) at this temperature?
[Atomic Mass of Cu = 63.55u]
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
1) 3
105
X 2)
3
211
X 3)
3
205
X 4)
3
422
X
Ans: 4
Sol: 3 23 8 3
4 63.5
(6 10 )( 10 )a
ZMd
N a X
= 3
3 3
4 63.5 421.71610 /
6gm cm
X X
48. The major product obtained in the following reaction is :
1) 2)
3) 4)
Ans: 3
Sol:
3 3CH C O C CH
2NH
O
OH OH
3NH C CH
OO
NH H
3CH C O
3O C CH
OOH
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
49. Which of the following conditions in drinking water causes
methemoglobinemia?
1) > 50ppm of lead 2) > 100 ppm of sulphate
3) > 50 ppm of chloride 4) > 50 ppm of nitrate
Ans: 4
Sol: Drinking water containing > 50 ppm of nitrate causes methemoglobinemia.
50. Homoleptic octahedral complexes of a metal ion 'M3+' with three monodentate ligands
L1, L2 and L3 absorb wavelengths in the region of green, blue and red respectively. The
increasing order of the ligand strength is:
1) L2 < L1 < L3 2) L3 < L2 < L1
3) L3 < L1 <L2 4) L1 < L2 < L3
Ans: 2
Sol: If ligand strength is high crystal field splitting is high so energy absorbed is high so wave
length is low So L3 < L2 < L1
51. The product formed in the reaction of cumene with O2 followed by treatment with dil. HCl
are:
1) 2)
3) 4)
Ans: 3
Sol:
3CH C 3CH3CH C 3CH
OO H
OH
3CH C 3CH
O
H
2O 3H O
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
52. The temporary hardness of water is due to :-
` 1) Ca (HCO3)2 2) NaCl 3) Na2SO4 4) CaCl2
Ans: 1
Sol: Hydrogen carbonates of calcium and magnesium causes temporary hardness
53. The entropy change associated with the conversion of 1 kg of ice at 273 K to water
vapours at 383 K is:
(Specific heat of water liquid and water vapour are 4.2 kJ K–1 kg–1 and 2.0 kJ K–1 kg–1 ;
heat of liquid fusion and vapourisation of water are 344 kJ kg–1 and 2491 kJ kg–1,
respectively). (log 273 = 2.436, log 373 = 2.572, log 383 = 2.583)
1) 7.90 kJ kg–1 K–1 2) 2.64 kJ kg–1 K–1
3) 8.49 kJ kg–1 K–1 4) 9.26 kJ kg–1 K–1
Ans: 3
Sol: 2 ( ) 2 ( )
(273 ) (273 )
S lH O H O
K K
0 33 1
1
344 101.26 10
273f
f
HS KJK
T
2 ( ) 2 ( )
(273 ) (383 )
l lH O H O
K K
33
2
10 383344 18 10 2.303log 0.677
18 273S KJ
2 ( ) 2 ( )
(383 ) (383 )
l gH O H O
K K
1
3
24916.5
383V
b
HS KJK
T
1.26 0.677 6.5 8.49S
54. The pH of rain water, is approximately :
1) 6.5 2) 7.5 3) 5.6 4) 7.0
Ans: 3
Sol: Rain water is acidic so-5.6
55. If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium
constant (K) for the reaction 2 2( ) ( ) ( ) ( )Zn s cu aq Zn aq Cu s at 300 K
is approximately.
(R = 8 JK–1 mol–1, F = 96000 C mol–1)
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
1) e 160 2) e320 3) e–160 4) e–80
Ans: 1
Sol: 0 0 lnG nFE RT k 0 lnRT
E knf
1608 300 965002 ln ln
2 96500 600k k k e
56. A solution containing 62 g ethylene glycol in 250 g water is cooled to –10°C. If Kf for
water is 1.86 K kg mol–1, the amount of water (in g) separated as ice is:
1) 32 2) 48 3) 16 4) 64
Ans: 4 Sol: 01
1.86 7.440.25f f fT K m T C
New freezing point
0 110 1.86
water
molC
m
1.860.186 186
10250 186 64
waterm kg g
gm gm gm
57. When the first electron gain enthalpy (egH) of oxygen is –141 kJ/mol, its gain
enthalpy is:
1) almost the same as that of the first
2) negative, but less negative than the first
3) a positive value
4) a more negative value than the first
Ans: 3
Sol: Conceptual
58. The major product of the following reaction is :
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
1) 2) 3) 4)
Ans: 3
Sol:
2C NH
O
2 3CH CH
2 /i Br hv
ii KOH N-H
O
3CH
2C NH
O
3CH CH2
OH
H O
Br
O
3CH CH
Br
NH
59. Which of the following compounds is not aromatic?
1) 2) 3) 4) Ans: 3
Sol: not satisfying (4n+2) It is anti aromatic. 60. Consider the following reversible chemical reactions:
1
2
2 2
2 2
( ) ( ) 2 ( )......(1)
6 ( ) 3 ( ) 3 ( ).....(2)
K
K
A g B g AB g
AB g A g B g
The relation between K1 and K2 is:
1) 3
2 1K K 2) 3
2 1K K 3) K1K2 = 3 4) 1 2
1
3K K
Ans: 2
Sol: 3
2 1 31
1( )K K
K
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
MATHEMATICS 61. The number of all possible positive integral values of for which the roots of the
quadratic equation, 26x 11x 0 are rational numbers is:
1) 4 2) 5 3) 3 4) 2
Ans. 3
Sol. 11 121 24
x12
121-24 is a perfect square
= 3,4,5
62. Let a i j 2k,
1 2b b i b j 2k, and c 5i j 2k,
be three vectors such that
the projection vector of b
on a
is a
. If a b
is perpendicular to c
, then b
is equal to:
1) 6 2) 32 3)4 4) 22
Ans. 1
Sol. a b is to c a b .c 0 a.c b.c 0
1 2
1 2
5 1 2 5b b 2 0
5b b 10........ 1
Projection b on a
is 2
b.aa a
a
1 21 2
b b 21 b b 2...... 2
4
Solve eq. 1& 2
63. If
8 6
22 7
5x 7xf x dx, x 0 ,
x 1 2x
and f 0 0, then the value of f 1 is:
1)1
2 2)
1
2 3)
1
4 4)
1
4
Ans. 4
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
Sol.
8 6
214 5 7
5x 7xf x dx
x x x 2
6 8
25 7
5x 7xdx
x x 2
5 7x x 2 t 6 85x 7x dx dt
7
2 2 7
dt 1 xf x c c
t t x 1 2x
1 1f 0 0 c 0, f 1
1 1 2 4
64. If both the roots of the quadratic equation 2x mx 4 0 are real distinct and they lie in
the interval 1,5 then m lies in the interval:
1) 4,5 2) 5, 4 3) 3,4 4) 5,6
Ans. 1
Sol. 51
f 1 .f 5 0
1 m 4 25 5m 4 0
29m 5 m 0
5
5
++
29
5 29m ,5 ,
5
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
And 20 m 16 0
++
44
m , 4 4,
44 5 29
5
29m , 4 4,5 ,
5
65. Let a,b and c be the th th7 ,11 and th13 terms respectively of a non- constant A.P. If these
are also the three consecutive terms of a G.P., then a
c is equal to:
1)2 2)4 3)7
13 4)
1
2
Ans. 2
Sol. A 10d A 12d
rA 6d A 10d
where a,b,c are A.P.also G.P.
A+10d=r (A+6d)……….(1)
A+12d=r (A+10d)……..(2)
Eq (2) - eq (1) 2d = r ( 4d)
1
r2
2 2
a a 14
c ar r
66. Let 0z be a root of the quadratic equation, 2 81 930 0x x 1 0, If z 3 6iz 3iz ,
Then arg z is equal to:
1)4
2)0 3)
6
4)
3
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
Ans. 1
Sol. Roots of 2x x 1 0 is 2,
0
81 93
z
z 3 6i 3i 3 6i 3i 3 3i
Arg 1 1z Tan
1 4
67. If 0 x2
, then the number of values of x for which sin x sin 2x sin3x 0, is:
1)1 2)4 3)2 4)3
Ans. 3
Sol. z 2sin 2x cos x sin 2x 0
1
sin 2x 0 cosx2
2x 0, ,2 ........ x3
x 0, , ......2
x 0,3
68. For each x R, let x be the greatest integer less than or equal to x. Then
x 0
x x x sin xlim
x
is equal to:
1)sin 1 2)-sin 1 3)1 4)0
Ans. 2
Sol. x 0 x 0 h,h 0 and x 1
x 0
h 1 h sin 1lim sin1
h
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
69. If
t t t
t t t t t
t t t
e e cos t e sin t
A e e cos t e sin t e sin t e cos t
e 2e sin t 2e cos t
, then A is:
1)not invertible for any t R 2) invertible only if t .
3)invertible for all t R 4) invertible only if t .2
Ans. 3
Sol. t t t
1 cos t sin t
A e e e 1 cos t sin t sin t cos t
1 2sin t 2cos t
t
1 cos t sin t
e 0 2cos t sin t 2sin t cos t
0 2sin t cos t 2cos t sin t
te 0
70. If /3
0
tan 1d 1 k 0 ,
2ksec 2
then the value of k is:
1)4 2)2 3)1
2 4)1
Ans. 4
Sol. /3
0
tan secd
2ksec .ssec
2sec t
2sec tan 2tdt
22
211
t dt 1 1
t2k.t.t 2k
1 1 1
1 12k 2 2
71. If x=3 tan t and y = 3sec t, then the value of 2
2
d y
dx at t
4
, is:
1)1
6 2)
1
6 2 3)
3
2 2 4)
1
3 2
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
Ans. 2
Sol. 2dx dy3sec t 3sec t tan t
dt dt
dy tan t
sin tdx sec t
2
2 2
d y dt 1cos t cos t.
dx dx 3sec t
72. The coefficient of 4t in the expansion of
361 t
1 t
is:
1)14 2)15 3)10 4)12
Ans. 2
Sol. 361 t 1 t
r
6 12 18 rc1 3t 3t t 3 r 1 t
2
6 12 18 rc1 3t 3t t r 2 t
2
4ct coefficient 6
73. Let S be the set of all triangles in the xy-plane, each having one vertex at the origin and
the other two vertices lie on coordinate axes with integral coordinates. If each triangle in
S has area 50sq. units, then the number of elements in the set S is:
1)18 2)32 3)9 4)36
Ans. 3
Sol. 1
ab 50 ab 1002
Possible integer of a,b are
a:1 2 4 5 10
b:100 50 25 20 10
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
(0,0) (a,0)
(0,b)
No of points 9
74. The equation of the plane containing the straight line x y z
2 3 4 and perpendicular to the
plane containing the straight lines x y z
3 4 2 and
x y z
4 2 3 is:
1)3x 2y 3z 0 2) x 2y z 0 3)5x 2y 4z 0 4) x 2y 2z 0
Ans. 2
Sol.
x y z
2 3 4 0
8 1 10
x 30 4 y 20 32 z 2 24 0
x 2y z 0
75. An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the
drawn ball green, then a red ball is added to the urn and if the drawn ball is red, then a
green ball is added to the urn; the original ball is not returned to the urn. Now a second
ball is drawn at random from it. The probability that the second ball is red, is
1)32
49 2)
27
49 3)
21
49 4)
26
49
Ans. 1
Sol. A box contain 5R, 2G balls
1E a green ball in drawn no of balls in box 6R,1G
2E a red ball in drawn no of balls in box 4R,3G
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
A= Second ball is red
2 6 5 4p A
7 7 7 7
76. Let f : 0,1 R be such that f xy f x .f y , for all x,y 0,1 , and f 0 0. if
y y x satisfies the differential equation, dyf x
dx with y 0 1, then
1 3y y
4 4
is equal to:
1)2 2)4 3)5 4)3
Ans. 4
Sol. x 0 y f 0 f 0 f 0 1
Put y = 0 f 0 f x f 0 f x 1
dy1 y x c y 0 1
dx
1 0 c c 1 y x 1
1 3 1 3
y y 1 1 34 4 4 4
77. Let A 4, 4 and B 9,6 be points on the parabola, 2y 4x . Let C be chosen on the arc
AOB of the parabola, where O is the origin, such that the area of ACB is maximum.
Then, the area (in sq. units) of ACB , is:
1)1
302
2)1
314
3)32 4)3
314
Ans. 2
Sol. 2A 4, 11 B 9,6 C t ,2t
2
2
4 4 11
ACB 9 6 1 30 5t 5t 52
t 2t 1
d 1 1 1
0 t c ,1 area 31dt 2 4 4
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
78. If 1x sin sin10 and 1y cos cos10 , then y x is equal to:
1)10 2)0 3) 4) 7
Ans. 3
Sol. 1x sin sin10 3 10
1y cos cos10 4 10 y x
79. The area of the region A x, y : 0 y x x 1 and 1 x 1 in sq. units, is:
1)2 2)1
3 3)
4
3 4)
2
3
Ans. 3
Sol. 2
2
y x x 1
x 1, x 0
x 1 x 0
0 1
2 2
1 0
area x 1dx x 1dx
-1 1
80. If the system of linear equations
x 4y 7z g
3y 5z h
2x 5y 9z k
Is consistent, then:
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
1) g h k 0 2) 2g h k 0 3) g 2h k 0 4) g h 2k 0
Ans. 2
Sol. 3 3 1
1 4 7 g
0 3 5 h R R 2R
2 5 9 k
3 3 2
1 4 7 g
0 3 5 h R R R
0 3 5 k 2g
1 4 7 g
0 3 5 h
0 0 0 2g h k
system of eqs consistent 2g+h+k = 0
81. The sum of the following series
2 2 2 2 2 2 2 2 2 29 1 2 3 12 1 2 3 4 15 1 2 ....51 6 .....
7 9 11
up to 15 terms, is:
1)7830 2)7820 3)7510 4)7520
Ans. 3
Sol. 2 2 2
thn
3n 1 2 .....nn term t
2n 1
23n n n 1 n n 1
6 2
Sum of n terms 3 2n n
1S t n n
2
Put n = 15
82. The logical statement ~ ~ p q p r ~ q r is equivalent to:
1) ~ p r 2) ~ p ~ q r 3) p ~ q r 4) p r ~ q
Ans. 4
Sol. p ~ q p r ~ q r
p ~ q r ~ q r
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
p r ~ q
83. Let f be a differentiable function from R to R such that 3/2f x f y 2 x y , for all
x, y R, If f 0 1 then 1
2
0
f x dx is equal to:
1)1 2)0 3)2 4)1
2
Ans. 1
Sol. 1/2
x y x y
f x f ylim lim2 x y
x y
1 1f y 0 f y 0 f y k
Given f 0 1 f x 1
1 1
12
00 0
f x dx 1dx x 1
84. A hyperbola has its centre at the origin, passes through the point (4,2) and has transverse
axis of length 4 along the x- axis. Then the eccentricity of the hyperbola is:
1)2 2)2
3 3)
3
2 4) 3
Ans. 2
Sol. Given 2a = 4 a 2
Eq. of hyperbola 2 2
2 2
x y1
a b
Passing through (4,2)
2
2
16 4 41 b
4 b 3
Eccentricity 2 2
2
a be
a
85. The number of natural numbers less than 7,000 which can be formed by using the digits
0,1,3,7,9 (repetition of digits allowed) is equal to:
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
1)250 2)372 3)374 4)375
Ans. 3
Sol. Single digit no.s = =4
2 digit no.s = =4 5=20
3 digit no.s = =425=100
4 digit no.s which are less than 7000 is =2555 = 250
Total no 4+20+100+250=374
86. A data consists of n observation:
n
2
1 2 n ii 1
x ,x ,.....,x . x 1 9n
and n
2
ii 1
x 1 5n
, then the standard deviation of this
data is:
1) 5 2) 7 3)2 4)5
Ans. 2
Sol. 2 2
i ix 1 x 1 9n 5n
2i i2 x 1 14n x 6n
2 2
i ix 1 x 1 9n 5n
i4 x 4n xi n
S.D22
i ix xvar
n n
87. If the circles 2 2 2x y 16x 20y 164 r and 2 2x 4 y 7 36 intersect at two
distinct points, then:
1) r 11 2) r 11 3) 0 r 1 4)1 r 11
Ans. 4
Sol. 21 1c 8,10 r 64 100 164 r r
2 2c 4,7 r 6
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
Two circles intersect 1 2 1 2 1 2r r c c r r
88. Let A x R : xisnot positive int eger . Define a function 2xf : A R as f x
x 1
then f is:
1)not injective 2)neither injective nor surjective
3)injective but not surjective 4) surjective but not injective
Ans. 4
Sol. One – one onto
1 2f x f x f x y
1 2
1 2
2x 2x
x 1 x 1
2xy 2x xy y
x 1
1 2x x y
xy 2
one – one fn Range = R-{2}
Codamain =R
not onto function
89. Let the equations of two sides of a triangle be 3x 2y 6 0 and 4x 5y 20 0 . If the
orthocenter of this triangle is at (1,1), then the equation of its third side is:
1) 26x 61y 1675 0 2)122y 26x 1675 0
3) 26x 122y 1675 0 4)122y 26x 1675 0
Ans. 3
Sol. Line to 4x 5y 20 and
A
B
3x-2y+6=0 4x+5y-20=0
(1,1)
Passing through (1,1) is
2 0 1 9 _ Jee-Main Question Paper_ Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081
www. srichaitanya.net, iconcohyd@ srichaitanyacollege.net
5x 4y 3....... 1
And line to 3x 2y 6 0 and passes (1,1) in 2x 3y 5........... 2
Solve AC and eq. (2)35
, 102
Solve AB and eq. (1) 33
13,2
Then eq. of BC 26x 122y 1675 0
90. If the lines x ay b,z cy d and x a 'z b ', y c 'z d ' are perpendicular, then:
1) cc' a a ' 0 2) bb' cc' 1 0 3) aa ' c c' 0 4) ab' bc 1 0
Ans. 3
Sol. Eq. of lines x b y q d
a 1 c
1 1
1 1
x b y q d
a 1 c
are ler
aa ' 1 cc' 0
Top Related