JEE  >  JEE Main & Advanced Mock Test Series  >  JEE Main 2019 Question Paper with Solutions (9th January - Evening) Download as PDF

JEE Main 2019 Question Paper with Solutions (9th January - Evening)


Test Description

90 Questions MCQ Test JEE Main & Advanced Mock Test Series | JEE Main 2019 Question Paper with Solutions (9th January - Evening)

JEE Main 2019 Question Paper with Solutions (9th January - Evening) for JEE 2022 is part of JEE Main & Advanced Mock Test Series preparation. The JEE Main 2019 Question Paper with Solutions (9th January - Evening) questions and answers have been prepared according to the JEE exam syllabus.The JEE Main 2019 Question Paper with Solutions (9th January - Evening) MCQs are made for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main 2019 Question Paper with Solutions (9th January - Evening) below.
Solutions of JEE Main 2019 Question Paper with Solutions (9th January - Evening) questions in English are available as part of our JEE Main & Advanced Mock Test Series for JEE & JEE Main 2019 Question Paper with Solutions (9th January - Evening) solutions in Hindi for JEE Main & Advanced Mock Test Series course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt JEE Main 2019 Question Paper with Solutions (9th January - Evening) | 90 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study JEE Main & Advanced Mock Test Series for JEE Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you?
JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 1

Two plane mirrors arc inclined to each other such that a ray of light incident on the first mirror (M1) and parallel to the second mirror (M2) is finally reflected from the second mirror (M2) parallel to the first mirror (M1). The angle between the two mirrors will be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 1


Assuming angles between two mirrors be θ as per geometry,
sum of anlges of Δ
3θ = 180°
θ = 60°

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 2

In a Young's double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength λ = 500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range –30°≤θ≤30°is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 2


Pam difference dsinθ = nλ
where d = seperation of slits
λ = wave length
n = no. of maximas
0.32 × 10–3 sin 30 = n × 500 × 10–9
n = 320
Hence total no. of maximas observed in angular range –30°≤ θ ≤ 30° is
maximas = 320 + 1 + 320 = 641

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 3

At a given instant, say t = 0, two radioactive substances A and B have equal activities. The ratio RB/RA  of their activities after time t itself decays with time t as e-3t [f the half-life of A is m2, the half-life of B is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 3

Half life of A = ℓn2

at t = 0  RA = RB

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 4

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 Vo changes by : (assume that the Ge diode has large breakdown voltage)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 4

Initially Ge & Si are both forward biased so current will effectivily pass through Ge diode with a drop of 0.3 V if "Ge" is revesed then current will flow through "Si" diode hence an effective drop of (0.7 – 0.3) = 0.4 V is observed.

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 5

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 

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 5

Frequency of torsonal oscillations is given by


f2 = 0.8 f1

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 6

A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature of 27°C. Amount of heat transferred to the gas, so that RMS velocity of molecules is doubled, is about :
[Take R = 8.3 J/ K mole]

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 6

Q = nCvΔT as gas in closed vessel

Q = 10000 J = 10 kJ

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 7

A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 7



According to the question, U = k


∴ Correct answer is (3)

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 8

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 :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 8

Frequency of the sound produced by flute,

Velocity of observer, 

∴ frequency detected by observer, f' =

= 335.56 × 2 = 671.12
∴ closest answer is (4)

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 9

In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accomodated for transmitting TV signals of band width 6 MHz are (Take velocity of light c = 3 × 108m/s,h = 6.6 × 10–34 J-s)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 9


= 3.75 × 1014 Hz
1% of f = 0.0375 × 1014 Hz
= 3.75 × 1012 Hz = 3.75 × 106 MHz
number of channels = 
∴ correct answer is (4)

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 10

Two point charges qand  are placed on the x-axis at x = l m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 10


Let are the vaues of electric field dueto q1 & q2 respectively magnitude of


E2 = 9 × 103 V/m










∴ correct answer is (3)

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 11

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K1, K2, K3, K4 arranged as shown in the figure. The effective dielectric constant K will be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 11





in the same way we get,



on comparing equation (i) to equation (ii), we get

This does not match with any of the options so probably they have assumed the wrong combination




However this is one of the four options.
It must be a "Bonus" logically but of the given options probably they might go with (4)

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 12

A rod of length 50cm is pivoted at one end. It is raised such that if makes an angle of 30° from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s–1) will be (g = 10ms–2)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 12


Work done by gravity from initial to final position is,


According to work energy theorem




∴ correct answer is (1)

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 13

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 (BL) to that at the centre of the coil (BC), i.e. R   will be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 13


L = 2πR L = N × 2πr
R = Nr


JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 14

The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 x 103 km) is E1 and kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal, is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 14

Usurface + E1 = Uh
KE of satelite is zero at earth surface & at height h



Gravitational attraction




JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 15

The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 15

Average energy density of magnetic field,

is maximum value of magnetic field.
Average energy density of electric field,

now,



uE = uB
since energy density of electric & magnetic field is same, energy associated with equal volume will be equal.
uE = uB

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 16

A series AC circuit containing an inductor (20 mH), a capacitor (120 μF) and a resistor (60Ω) is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in 60 s is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 16

R = 60Ω f = 50Hz, ω = 2πf = 100 π




Q = P.t = 8.64 × 60 = 5.18 × 102

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 17

Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of light) is proportional to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 17




on comparing the powers of M, L, T
– x + y = 0 ⇒  x = y
3x + 2y + z = 0 ⇒  5x + z = 0 .. ..(i)
–2x – y – z = 1 ⇒ 3x + z = –1 .. .(ii)
on solving (i) & (ii) 

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 18

The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 × 107)ct + sin(6.28 × 107)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 ?
(c = 3 × 108ms–1, h = 6.6 × 10–34 J-s)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 18

B = B0sin (π × 107C)t + B0sin (2π × 107C)t
since there are two EM waves with different frequency, to get maximum kinetic energy we take the photon with higher frequency
B1 = B0sin(π × 107C)t

B2 = B0sin(2π × 107C)t v2 = 107C
where C is speed of light  C = 3 × 108 m/s v2 > v1
so KE of photoelectron will be maximum for photon of higher energy.
v2 = 107C Hz
hv  = φ + KEmax
energy of photon
Eph = hn = 6.6 × 10-34 × 107 × 3 × 109
Eph = 6.6 × 3 × 10–19J

KEmax = Eph – φ
= 12.375 – 4.7 =  7.675 eV ≈ 7.7 eV

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 19

Charge is distributed within a sphere of radius R with a volume charge density  where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 19






JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 20

Two Carnot engines A and B are operated in series. The first one, A, receives heat at T1(= 600 K) and rejects to a reservoir at temperature T2. The second engine B receives heat rejected by the first engine and, in turn, rejects to a heat reservoir at T3(= 400 K). Calculate the temperature T2 if the work outputs of the two engines are equal :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 20


w1 = w2
Δu1 = Δu2
T3 – T2 = T2 – T1
2T2 = T1 + T3
T2 = 500 K

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 21

A carbon resistance has a following colour code. What is the value of the resistance ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 21


R = 53 × 104 ± 5% = 530 kΩ ± 5%

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 22

A force acts on a 2 kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 22

x = 3t2 + 5

v = 6t + 0
at t = 0 v = 0
t = 5 sec v = 30 m/s
W.D. = ΔKE

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 23

The position co-ordinates of a particle moving in a 3-D coordinate system is given by x = a cosωt y = a sinωt and z = aωt The speed of the particle is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 23

vx = –aωsinωt   ⇒  vy = aωcosωt
vz = aω  

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 24

In the given circuit the internal resistance of the 18 V cell is negligible. If R1 = 400 Ω, R3 = 100 Ω and R4 = 500 Ω and the reading of an ideal voltmeter across R4 is 5V, then the value R2 will be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 24


V4 = 5V

V3 = i1R3 = 1V
V3 + V4 = 6V = V2
V1 + V3 + V4 = 18V
V1 = 12 V

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 25

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 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms –2)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 25


at equation

F = 100 N

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 26

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 a1 and a2 respectively. Then 'v' is equal to

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 26

For A & B let time taken by A is t0
from ques.
vA – vB = v = (a1 – a2)t0 – a2t .. ..(i)


 .. ..(ii)
putting t0 in equation


JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 27

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 :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 27

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 28

The top of a water tank is open to air and its water level is maintained. It is giving out 0.74 m3 water per minute through a circular opening of 2 cm radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 28

In flow volume = outflow volume





JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 29

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 :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 29


LC = 0.5 × 10–2 mm
+ve error = 3 × 0.5 × 10–2 mm
= 1.5 × 10–2 mm = 0.015 mm
Reading = MSR + CSR – (+ve error)
= 5.5 mm + (48 × 0.5 × 10–2) – 0.015
= 5.5 + 0.24 – 0.015 = 5.725 mm

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 30

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 =1.6 × 10–19C)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 30


mv = qBR .. .. (i)
Path is straight line
it qE = qvB
E = vB .. ..(ii)
From equation (i) & (ii)

m = 2.0 × 10–24 kg

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 31

lood reducing nature of H3PO2 ttributed to the presence of:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 31

H3PO2 is good reducing agent due to presence
of two P–H bonds. 

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 32

The complex that has the highest crystal field splitting energy (Δ), is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 32

As complex K3[Co(CN)6] hav e CNligand which is strongfield ligand amongst the given ligands in other complexes.

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 33

The metal that forms nitride by reacting directly with N2 of air, is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 33

Only Li react directly with N2 out of alkali metals
6Li + N2 → 2Li3N

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 34

In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 34

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 35

The major product of the following reaction is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 35

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 36

The transition element that has lowest enthalpy of atomisation, is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 36

Since Zn is not a transition element so transition element having lowest atomisation energy out of Cu, V, Fe is Cu.

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 37

Which of the following combina tion 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 h equal to h/2π
(d) The plot of ψ Vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value.

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 38

The tests performed on compound X and their inferences are:

Compound 'X' is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 38

→ 2,4 – DNP test is given by aldehyde on ketone
→ Iodoform test is given by compound having

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 39

The major product formed in the following reaction is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 39

Aldehyde reacts at a faster rate than keton during aldol and stericall less hindered anion will be a better nucleophile so sefl aldol at  will be the major product.

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 40

For the reaction, 2A + B → products, when the concentrations of A and B both wrere doubled, the rate of the reaction increased from 0.3 mol L–1s–1 to 2.4 mol L–1 s–1. When the concentration of A alone is doubled, the rate increased from 0.3 mol L–1s–1 to 0.6 mol L–1s–1
Which one of the following statements is correct ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 40

r  = K[A]x[B]y
⇒ 8 = 23 = 2x+y
⇒ x + y = 3 ...(1)
⇒ 2 = 2x 
⇒ x = 1, y = 2
Order w.r.t. A = 1
Order w.r.t. B = 2

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 41

The correct sequence of amino acids present in the tripeptide given below is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 41



JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 42

The correct statement regarding the given Ellingham diagram is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 42

According to the given diagram Al can reduce ZnO.
3ZnO+2Al → 3Zn+Al2O3

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 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)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 43

moles of C57H110O6(s) = 445/890 = 0.5 moles
2C57H110O6(s) + 163 O2(g) → 114 CO2(g) + 110 H2O(l)


= 495gm

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 44

The correct match between Item I and Item II is :

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 45

The increasing basicity order of the following compounds is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 45

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 46

For coagulation of arscnious sulphide sol, which one of the following salt solution will be most effective ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 46

Sulphide is –ve charged colloid so cation with maximum charge will be most effective for coagulation.
Al3+ > Ba2+ > Na+  coagulating power.

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 47

At 100°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]

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 47

FCC unit cell Z = 4


JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 48

The major product obtained in th e following reaction is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 48

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 49

Which of the following conditions in drinking water causes methemoglobinemia?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 49

Concentration of nitrate >50 ppm in drinking water causes methemoglobinemia

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 50

Homoleptic octahedral complexes of a metal ion 'M3+' with three monodentate ligands and L1, L2, L3 absorb wavelengths in the region of green, blue and red respectively. The increasing order of the ligand strength is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 50

Order of λabs - L3>L1>L2
So ΔO order will be L2 > L1 > L3

So order of ligand strength will be L2>L1>L3

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 51

The product formed in the reaction of cumene with O2 followed by treatment with dil. HCl are:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 51

Cummene hydroperoxide reaction

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 52

The temporary hardness of water is due to:-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 52

Ca (HCO3)2 is re ponsible for temporary hardness of water

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 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).

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 53




ΔStotal =  9.26 kJ kg–1 K–1

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 54

The pH of rain water, is approximately :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 54

pH of rain water is approximate 5.6

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 55

If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction Zn(s) + Cu2+(aq) ⇔ Zn2+(aq) + Cu(s) at 300 K is approximately.
(R = 8 JK–1 mol–1 , F = 96000 C mol–1)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 55

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 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 :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 56

ΔTf = Kf . m

W = 0.186 kg
ΔW = (250 – 186) = 64 gm

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 57

When the first electron gain enthalpy (ΔegH) of oxygen is –141 kJ/mol, its second electron gain enthalpy is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 57

Second electron gain enthalpy is always positive for every element.
O(g)+ e →O¯2(g)   ;  ΔH = positive

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 58

The major product of the following reaction is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 58


JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 59

Which of the following compounds is not aromatic?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 59


Do not have (4n + 2) π electron It has 4n π electrons
So it is Anti aromatic.

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 60

Consider the following reversible chemical reactions :

The relation between K1 and K2 is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 60


JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 61

Let f be a differentiable function from R to R such that |f (x) - f(y)|≤2|x-y|3/2, for all x, y ε R. If f(0) = 1 then is  equal to

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 61


JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 62

If then the value of k is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 62



JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 63

The coefficient of t4 in the expansion of is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 63

(1 – t6)3 (1 – t)–3
(1 – t18 – 3t6 + 3t12) (1 – t)–3
⇒ cofficient of t4 in (1 – t)–3 is
3+4–1C4 = 6C2 = 15

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 64

For each xεR, let [x] be the greatest integer less than or equal to x. Then
 is equal to

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 64


JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 65

If both the roots of the quadratic equation x2 - mx + 4 = 0 are real and distinct and they lie in the interval [1 5] then m lies in the interval:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 65



No option correct : Bonus
* If we consider αβ∈ (1,5) then option (1) is correct.

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 66

If

Then A is -

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 66


JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 67

The area of the region A = [(x,y) : 0 ≤ y ≤ x|x| +1 and -1 ≤ x ≤ 1] in sq. units, is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 67

The graph is a follows

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 68

Let z0 be a root of the quadratic equation, x2 + x + 1 = 0. If z = 3 + 6iz081 - 3iz093, then arg z is equal to:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 68

z0 = ω or ω2 (where ω is a non-real cube root of unity)
z = 3 + 6i(ω)81 – 3i(ω)93
z = 3 + 3i

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 69

Let  and be three vectors such that the projection vector of  on  is perpendicular to  is equal to:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 69


⇒ 5b1 + b2 = – 10 ...(2)
from (1) and (2) ⇒ b1 = –3 and b2 = 5

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 70

Let A(4,–4) and B(9,6) be points on the parabola, y2 + 4x. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of ΔACB is maximum. Then, the area (in sq. units) of ΔACB, is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 70



JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 71

The logical statementis equivalent to:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 71

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 72

An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is 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 :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 72

E1 : Event of drawing a Red ball and placing a green ball in the bag
E2 : Event of drawing a green ball and placing a red ball in the bag
E : Event of drawing a red ball in second draw

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 73

If 0 ≤ x < π/2, then the number of values of x for which sin x-sin2x+sin3x = 0, is 

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 73

sinx – sin2x + sin3x = 0
⇒ (sinx + sin3x) – sin2x = 0
⇒ 2sinx. cosx – sin2x = 0
⇒ sin2x(2 cosx – 1) = 0
⇒ sin2x = 0 or cosx = 1/2
⇒ x = 0, π/3

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 74

The equation of the plane containing the straight line x/2 = y/3 = z/4 and perpendicular to the plane containing the straight lines x/3 = y/4 = z/2 and x/4 = y/2 = z/3 is: 

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 74

Vector along the normal to the plane containing the lines

vector perpendicular to the vectors  and 
so, required plane is 26x – 52y + 26z = 0
x – 2y + z = 0

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 75

Let the equations of two sides of a triangle be 3x – 2y + 6 = 0 and 4x + 5y – 20 = 0. If the orthocentre of this triangle is at (1,1), then the equation of its third side is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 75


Equation of AB is 3x – 2y + 6 = 0
equation of AC is 4x + 5y – 20 = 0
Equation of BE is 2x + 3y – 5 = 0
Equation of CF is 5x – 4y – 1 = 0
⇒ Equation of BC is 26x – 122y = 1675

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 76

If x = 3 tan t and y = 3 sec t, then the value of 

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 76



JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 77

If x = sin–1(sin10) and y= cos –1(cos 1 0), then y–x is equal to:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 77


x = sin–1(sin 10) = 3π – 10

y = cos–1(cos10) = 4π – 10
y – x = π

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 78

If the lines x = ay+b, z = cy + d and x=a'z + b', y = c'z + d' are perpendicular, then:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 78

Line x = ay + b, z = cy + d 
⇒ 
Line x = a'z + b', y = c'z + d'

Given both the lines are perpendicular
⇒ aa' + c' + c = 0

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 79

The number of all possible positive integral values of α for which the roots of the quadratic equation, 6x2–11x+α = 0 are rational numbers is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 79

6x2 – 11x + α = 0
given roots are rational
⇒ D must be perfect square
⇒121 – 24α = λ2
⇒ maximum value of α is 5
α = 1 ⇒ λ ∉ I
α = 2 ⇒ λ ∉ I
α = 3 ⇒ λ ∈ I   ⇒ 3 integral values
α = 4 ⇒ λ ∈ I
α = 5 ⇒ λ ∈ I

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 80

A hyperbola has its centre at the or igin, passes through the point (4,2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 80



2a = 4   a = 2

Passes through (4,2)

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 81

Let A={xεR:x is not a positive integer}
Define a function f :A→R as f(x) = 2x/x-1 then f is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 81



⇒ ƒ is one-one but not onto

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 82

If and f(0) = 0, then the value of f(1) is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 82




JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 83

If the circles x2 + y2 – 16x–20y + 164 = r2 and (x–4)2 + (y – 7)2 = 36 intersect at two distinct points, then:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 83

x2 + y2 – 16x – 20y + 164 = r2
A(8,10), R1 = r
(x – 4)2 + (y – 7)2 = 36
B(4,7), R2 = 6
|R1 – R2| < AB < R1 + R2
⇒ 1 < r < 11

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 84

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:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 84

Let A(α,0) and B(0,β) be the vectors of the given triangle AOB
⇒ |αβ| = 100
⇒ Number of triangles
= 4 × (number of divisors of 100)
= 4 × 9 = 36

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 85

The sum of the follwing series 
+...up to 15 terms, is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 85




= 7820

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 86

Let a, b and c be the 7th, 11th and 13th 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:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 86

a = A + 6d
b = A + 10d
c = A + 12d
a,b,c are in G.P.
⇒ (A + 10d)2 = (A + 6d) (a + 12d)


 

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 87

If the system of linear equations
x–4y+7z = g    
3y – 5z = h
–2x + 5y – 9z = k
is consistent, then 

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 87

P1 ≡ x – 4y + 7z – g = 0
P2 ≡ 3x – 5y – h = 0
P3 ≡ –2x + 5y – 9z – k = 0
Here Δ = 0
2P1 + P2 + P3 = 0 when 2g + h + k = 0

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 88

 Let f:[0,1]→R be such that f(xy) = f(x).f(y) for all x,y,ε[0,1], and f(0)≠0. If y = y(x) satisfies  the differential equation, dy/dx = f(x) with y(0) = 1, then  is equal to

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 88

ƒ(xy) = ƒ(x). ƒ(y)
ƒ(0) = 1 as ƒ(0) ≠ 0
⇒ ƒ(x) = 1

⇒ y = x+c
At, x = 0, y = 1 ⇒ c = 1
y = x + 1

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 89

A data consists of n observations:
x1, x2, ....., xn. If  and , then the standard deviation of this data is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 89





⇒ variance = 6 – 1 = 5
⇒ Standard diviation = √5

JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 90

The number of natural numbers less than 7,000 which can be formed by using the digits 0,1,3,7,9 (repitition of digits allowed) is equal to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (9th January - Evening) - Question 90


Number of numbers = 53 – 1

2 ways for a4
Number of numbers = 2 × 53
Required number = 53 + 2 × 53 – 1
= 374

Use Code STAYHOME200 and get INR 200 additional OFF
Use Coupon Code
Information about JEE Main 2019 Question Paper with Solutions (9th January - Evening) Page
In this test you can find the Exam questions for JEE Main 2019 Question Paper with Solutions (9th January - Evening) solved & explained in the simplest way possible. Besides giving Questions and answers for JEE Main 2019 Question Paper with Solutions (9th January - Evening), EduRev gives you an ample number of Online tests for practice