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

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JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 1

### A paramagnetic substance in the form of a cube with sides 1 cm has a magnetic dipole moment of 20 × 10–6 J/T when a magnetic intensity of 60 × 103 A/m is applied. Its magnetic susceptibility is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 1

= 0.33 × 10–3 = 3.3 × 10–4

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 2

### A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 2

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 3

### Seven capacitors, each of capacitance 2μF, are to be connected in a configuration to obtain an effective capacitance of (6/13) μF. Which of the combinations, shown in figures below, will achieve the desired value ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 3

Therefore three capacitors most be in parallel to get 6 in

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 4

An electric field of 1000 V/m is applied to an electric dipole at angle of 45°. The value of electric dipole moment is 10–29 C.m. What is the potential energy of the electric dipole?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 4

= –PE cos θ
= –(10–29) (103) cos 45º
= – 0.707 × 10–26 J
= –7 × 10–27 J.

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 5

A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10–2 m. The relative change in the angular frequency of the pendulum is best given by :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 5

Angular frequency of pendulum

s = angular frequency of support]

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 6

Two rods A and B of identical dimensions are at temperature 30°C. If A is heated upto 180°C and B upto T°C, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 6

T = 230º C

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 7

In a double-slit experiment, green light (5303 Å) falls on a double slit having a separation of 19.44 μm and a width of 4.05 μm. The number of bright fringes between the first and the second diffraction minima is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 7

For diffraction
location of 1st minime

Now for interference Path difference at P.

path difference at Q

So orders of maxima in between P & Q is
5, 6, 7, 8, 9
So 5 bright fringes all present between P & Q.

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 8

An amplitude modulated signal is plotted below :-

Which one of the following best describes the above signal?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 8

Analysis of graph says
(1) Amplitude varies as 8 – 10 V or 9 ± 1
(2) Two time period as 100 μs (signal wave) & 8 μs (carrier wave)

= 9 ± 1sin (2π × 104t) sin 2.5π × 105 t

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

In the circuit, the potential difference between A and B is :-

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

Potential difference across AB will be equal to battery equivalent across CD

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 10

A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by [Given permittivity of space ∈0 = 9 × 10-12 SI units, Speed of light c = 3 × 108 m/s]:-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 10

Intensity of EM wave is given by

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

A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2 Then :-

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

Maximum kinetic energy at lowest point B is given by
K = mgl (1 – cos θ)
where θ= angular amp.

K1 = mgℓ (1 - cos θ)
K2 = mg(2ℓ) (1 - cos θ)
K2 = 2K1.

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 12

In a hydrogen like atom, when an electron jumps from the M - shell to the L - shell, the wavelength of emitted radiation is λ. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 12

For M → L steel

for N → L

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 13

If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's modulus will be :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 13

Now from dimension

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 14

A particle moves from the point  m, at t = 0, with an initial velocity  ms–1.It is acted upon by a constant force which produces a constant acceleration What is the distance of the particle from the origin at time 2s?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 14

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 15

A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is √3, then the angle of incidence is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 15

i = e

by Snell's law

i = 60

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 16

A galvanometer having a resistance of 20 Ω and 30 divisions on both sides has figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt, is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 16

Rg = 20Ω
NL = NR = N = 30

Igmax = 0.005 × 30
= 15 × 10–2 = 0.15
15 = 0.15 [20 + R]
100 = 20 + R
R = 80

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 17

The circuit shown below contains two ideal diodes, each with a forward resistance of 50Ω. If the battery voltage is 6 V, the current through the 100 Ω resistance (in Amperes) is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 17

I = 6/300 = 0.002 (D2 is in reverse bias)

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 18

When 100 g of a liquid A at 100°C is added to 50 g of a liquid B at temperature 75°C, the temperature of the mixture becomes 90°C. The temperature of the mixture, if 100 g of liquid A at 100°C is added to 50 g of liquid B at 50°C, will be :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 18

100 × SA × [100 – 90] = 50 × SB × (90 – 75)
2SA = 1.5 SB

Now, 100 × SA × [100 – T] = 50 × SB (T – 50)

300 – 3T = 2T – 100
400 = 5T
T = 80

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 19

The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2s. The period of oscillation of the same pendulum on the planet would be :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 19

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 20

The region between y = 0 and y = d contains a magnetic field A particle of mass m and charge q enters the region with a velocity  the acceleration of the charged particle at the point of its emergence at the other side is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 20

In equation, entry point of particle is no given Assuming particle center from (0, d)

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 21

A thermometer graduated according to a linear scale reads a value x0 when in contact with boiling water, and x0/3 when in contact with ice. What is the temperature of an object in 0 °C, if this thermometer in the contact with the object reads x0/2 ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 21

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 22

A string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string) :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 22

40 + f = m(Rα) .....(i)
40 × R – f × R = mR2α
40 – f = mRα   ...... (ii)
From (i) and (ii)

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 23

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is incresed by ΔT. The amount of heat absorbed by gas is (R is gas constant) :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 23

VT = K

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 24

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 24

......(i)
......(ii)
(i) – (ii)

= 1V

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 25

A metal ball of mass 0.1 kg is heated upto 500°C and dropped into a vessel of heat capacity 800 JK-1 and containing 0.5 kg water. The initial temperature of water and vessel is 30°C. What is the approximate percentage increment in the temperature of the water ?
[Specific Heat Capacities of water and metal are, respectively, 4200 Jkg–1K–1 and 400 JKg–1K–1]

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 25

0.1 × 400 × (500 – T) = 0.5 × 4200 × (T – 30) + 800 (T – 30)
⇒ 40(500 - T) = (T - 30) (2100 + 800)
⇒ 20000 - 40T = 2900 T - 30 × 2900
⇒ 20000 + 30 × 2900 = T(2940)
T = 30.4°C

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 26

The magnitude of torque on a particle of mass 1kg is 2.5 Nm about the origin. If the force acting on it is 1 N, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians) :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 26

2.5 = 1 × 5 sin θ
sin θ = 0.5 = 1/2
θ = π/6

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 27

In the experimental set up of metre bridge shown in the figure, the null point is obtained at a distance of 40 cm from A. If a 10Ω resistor is connected in series with R1, the null point shifts by 10 cm. The resistance that should be connected in parallel with (R1 + 10) Ω such that the null point shifts back to its initial position is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 27

.......(1)

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 28

A circular disc D1 of mass M and radius R has two identical discs D2 and D3 of the same mass M and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO', passing through the centre of D1, as shown in the figure, will be:-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 28

= 3 MR2

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 29

A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 29

Total length L will remain constant
L = (3a) N (N = total turns) and length of winding = (d) N (d = diameter of wire)

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 30

A particle of mass m and charge q is in an electric and magnetic field given by

The charged particle is shifted from the origin to the point P(x = 1 ; y = 1) along a straight path. The magnitude of the total work done is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 30

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 31

The correct option with respect to the Pauling electronegativity values of the elements is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 31

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 32

Along the period electronegativity increases The homopolymer formed from 4-hydroxy-butanoic acid is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 32

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 33

The correct match between Item I and Item II is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 33

(A) Ester test (Q) Aspartic acid (Acidic amino  acid)
(B) Carbylamine (S) Lysine [NH2 group present]
(C) Phthalein dye  (P) Tyrosine {Phenolic group present)

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 34

Taj Mahal is being slowly disfigured and discoloured. This is primarily due to

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 34

Taj mahal is slowely disfigured and discoloured due to acid rain.

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 35

The major product obtained in the following conversion is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 35

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 36

The number of bridging CO ligand (s) and Co-Co bond (s) in CO2(CO)8, respectively are :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 36

Bridging CO are 2 and Co – Co bond is 1.

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 37

In the following compound,

the favourable site/s for protonation is/are :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 37

Localised lone pair e.

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 38

The higher concentration of which gas in air can cause stiffness of flower buds?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 38

Due to acid rain in plants high concentration of SO2 makes the flower buds stiff and makes them fall.

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 39

The correct match between item I and item II is :-

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 40

The radius of the largest sphere which fits properly at the centre of the edge of body centred cubic unit cell is: (Edge length is represented bv 'a')

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 40

r = 0.067 a

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 41

Among the colloids cheese (C), milk (M) and smoke (S), the correct combination of the dispersed phase and dispersion medium, respectively is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 41

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 42

The reaction that does NOT define calcination is:-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 42

Calcination in carried out for carbonates and oxide ores in absence of oxygen. Roasting is carried out mainly for sulphide ores in presence of excess of oxygen.

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 43

MgO(s) + C(s) → Mg(S) + CO(g), for which ΔrHº = + 491.1 kJ mol–1 and ΔrSº = 198.0 JK–1 mol–1 , is not feasible at 298 K. Temperature above which reaction will be feasible is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 43

= 2480.3 K

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 44

Given the equilibrium constant :
KC of the reaction :
Cu (s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s ) is 10 × 1015, calculate the E0cell of this reaction at 298 K

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 44

At equilibrium

= 0.059×8
= 0.472 V

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 45

The hydride that is NOT electron deficient is:-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 45

(1) B2H6 : Electron deficient
(2) A1H3 : Electron deficient
(3) SiH4 : Electron precise
(4) GaH3 : Electron deficient

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 46

The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by
ΔrGº = A – BT
Where A and B are non-zero constants. Which of the following is TRUE about this reaction ?

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 47

K2HgI4 is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 47

For  K2[HgI4]
i = 1+ 0.4 (3–1)
= 1.8

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 48

The de Broglie wavelength (λ) associated with a photoelectron varies with the frequency (v) of the incident radiation as, [v0 is thre shold frequency] :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 48

For electron
(de broglie wavelength)
By photoelectric effect
hν = hν0 + KE
KE = hν –hν0

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 49

The reaction 2X → B is a zeroth order reaction. If the initial concentration of X is 0.2 M, the half-life is 6h. When the initial concentration of X is 0.5 M, the time required to reach its final concentration of 0.2 M will be:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 49

For zero order
[A0]–[At] = kt
0.2 – 0.1 = k×6

t = 18 hrs.

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 50

A compound 'X' on treatment with Br2/NaOH, provided C3H9N, which gives positive carbylamine test. Compound 'X' is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 50

Thus [X] must be amide with one carbon more than in amine.
Thus [X] is CH3CH2CH2CONH2

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 51

Which of the following compounds will produce a precipitate with AgN03?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 51

as it can produce aromatic cation so will produce precipitate with AgNO3.

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 52

The relative stability of +1 oxidation state of group 13 elements follows the order :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 52

Due to inert pair effect as we move down the group in 13th group lower oxidation state becomes more stable.
Al < Ga < In < Tℓ

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 53

Which of the following compounds reacts with ethylmagnesium bromide and also decolourizes bromine water solution

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 53

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 54

Match the following items in column I with the corresponding items in column II.

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 54

Na2CO3.10H2O → Solvay process
Mg(HCO3)2 → Temporary hardness
NaOH → Castner-kellner cell
Ca3Al2O6 → Portland cement

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 55

25 ml of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2 M aqueous NaOH solution?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 55

HCl with Na2CO3
Eq. of HCl = Eq. of Na2CO3

Eq of HCl = Eq. of NaOH

V = 25 ml

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 56

In the above sequence of reactions,  respectively, are :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 56

A → MnO2
D → KIO3

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 57

The coordination number of Th in K4[Th(C2O4)4(OH2)2] is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 57

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 58

The major product obtained in the following reaction is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 58

LiAlH4 will not affect C=C in this compound.

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 59

The major product of the following reaction is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 59

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 60

For the equilibrium, 2H2O ⇔H3O+ + OH,  the value of ΔGº at 298 K is approximately :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 60

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 61

If the point (2, α, β) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x – 5y = 15, then 2α – 3β is equal to :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 61

Normal vector of plane

equation of plane is 5(x–7)+ 2y–3(z– 6) = 0 5x + 2y – 3z = 17

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 62

Let α and β be the roots of the quadratic equation x2 sin θ – x (sin θ cos θ + 1) + cos θ = 0 (0 < θ < 45º), and α < β. Then  is equal to :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 62

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 63

Let K be the set of all real values of x where the function f(x) = sin |x| – |x| + 2(x – π) cos |x| is not differentiable. Then the set K is equal to :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 63

ƒ(x) = sin|x|–|x| + 2(x – π) cosx
∵ sin|x| – |x| is differentiable function at x=0
∴ k = f

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 64

Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 64

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 65

If the area of the triangle whose one vertex is at the vertex of the parabola, y2 + 4(x – a2)= 0 and the other two vertices are the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of 'a' is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 65

Vertex is (a2,0)
y2 = –(x – a2) and x = 0 ⇒ (0,±2a)
Area of triangle is = 1/2 4a. ( a2) 250
⇒ a3 = 125 or a= 5

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 66

The integral  equals :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 66

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 67

​​Let (x + 10)50 + (x – 10)50 = a0 + a1x + a2x2 + ..... + a50 x50, for all x∈R, then a2/ais equal to:-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 67

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 68

Let a function f : (0, ∞) → (0, ∞) be defined by Then f is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 68

⇒ ƒ(x) is not injective
but range of function is [0,∞)
Remark : If co-domain is [0,∞), then ƒ(x) will be surjective

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 69

Let S = {1, 2, ...... , 20}. A subset B of S is said to be "nice", if the sum of the elements of B is 203. Then the probability that a randomly chosen subset of S is "nice" is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 69

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 70

Two lines   intersect at the point R. The reflection of R in the xy-plane has coordinates :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 70

Point on L1 (λ+ 3, 3λ – 1, –λ+ 6)
Point on L2 (7μ – 5, –6μ + 2, 4μ + 3)
⇒ λ + 3 = 7μ – 5 ...(i)
3λ – 1 = –6μ + 2 ...(ii) ⇒ λ = –1, μ=1
point R(2,–4,7)
Reflection is (2,–4,–7)

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 71

The number of functions f from {1, 2, 3, ..., 20} onto {1, 2, 3, ....., 20} such that f(k) is a multiple of 3, whenever k is a multiple of 4, is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 71

ƒ(k) = 3m (3,6,9,12,15,18)
for k = 4,8,12,16,20 6.5.4.3.2 ways
For rest numbers 15! ways
Total ways = 6!(15!)

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 72

Contrapositive of the statement "If two numbers are not equal, then their squares are not equal." is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 72

Contrapositive of p → q is ~q → ~p

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 73

The solution of the differential equation, dy/dx = (x-y)2 , when y(1) = 1, is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 73

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 74

Let A and B be two invertible matrices of order 3 × 3. If det(ABAT) = 8 and det(AB–1) = 8, then det (BA–1 BT) is equal to :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 74

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 75

If  where C is a constant of integration, then f(x) is equal to :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 75

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 76

A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, the  is equal to :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 76

p (probability of getting white ball) = 30/40

and standard diviation

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 77

If in a parallelogram ABDC, the coordinates of A, B and C are respectively (1, 2), (3, 4) and (2, 5), then the equation of the diagonal AD is:-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 77

co-ordinates of point D are (4,7)
⇒ line AD is 5x – 3y + 1 = 0

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 78

If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 78

2b = 5 and 2ae = 13

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 79

The area (in sq. units) in the first quadrant bounded by the parabola, y = x2 + 1, the tangent to it at the point (2, 5) and the coordinate axes is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 79

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 80

Let  and  respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is 3√2 , then the sum of all possible values of β is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 80

Angle bisector is x – y = 0

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 81

If  = (a + b + c) (x + a + b + c)2, x ≠ 0 and a + b + c ≠ 0, then x is equal to :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 81

= (a + b + c)(a + b + c)2
⇒ x = –2(a + b + c)

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 82

Let Sn = 1 + q + q2 + ....... + qn and  where q is a real number and q ≠ 1. If 101C1 + 101C2.S1 + ...... + 101C101.S100 = αT100, then α is equal to :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 82

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 83

A circle cuts a chord of length 4a on the x-axis and passes through a point on the y-axis, distant 2b from the origin. Then the locus of the centre of this circle, is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 83

Let equation of circle is
x2 + y2 + 2ƒx + 2ƒy + e = 0, it passes through (0, 2b)
⇒ 0 + 4b2 + 2g × 0 + 4ƒ + c = 0
⇒ 4b2 + 4ƒ + c = 0 ...(i)

g2 – c = 4a2 ⇒ c = ( g2- 4a2 )
Putting in equation (1)
⇒ 4b2 + 4ƒ + g2 – 4a2 = 0
⇒ x2 + 4y + 4(b2 – a2) = 0, it represent a parabola.

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 84

If 19th term of a non-zero A.P. is zero, then its (49th term) : (29th term) is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 84

a + 18d = 0 ...(1)

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 85

Let  x∈R, where a, b and d are non-zero real constants. Then :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 85

ƒ(x) is an increasing function.

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 86

Let z be a complex number such that |z| + z = 3 + i (where i = √-1). Then |z| is equal to :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 86

|z| + z = 3 + i
z  = 3 – |z| + i
Let 3 – |z| = a ⇒ |z| = (3 – a)

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 87

All x satisfying the inequality (cot–1 x)2 – 7 (cot–1 x) + 10 > 0, lie in the interval:-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 87

cot–1x > 5, cot–1x < 2
⇒ x < cot5, x > cot2

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 88

Given  for a ΔABC with usual notation. If then the ordered triad (α, β, γ) has a value :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 88

b + c = 11λ, c + a = 12λ, a + b = 13λ
⇒ a = 7λ, b = 6λ, c = 5λ
(using cosine formula)

α : β : γ ⇒ 7 : 19 : 25

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 89

Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 89

using AM ≥ GM

JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 90

Detailed Solution for JEE Main 2019 Question Paper with Solutions (11th January - Evening) - Question 90

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