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# JEE Main 2019 Question Paper with Solutions (12th January - Morning)

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JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 1

### Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length l and mass m. The rod is pivoted at its centre 'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 1

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 2

### A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K1 and that of the outer cylinder is K2. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 2

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 3

### A travelling harmonic wave is represented by the equation y (x, t) = 10–3 sin (50 t + 2x), where x and y are in meter and t is in seconds. Which of the following is a correct statement about the wave? The wave is propagating along the

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 3

y= a sin(ωt + kx)

⇒ wave is moving along –ve x-axis with speed

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 4

A straight rod of length L extends from x = a to x=L + a. The gravitational force is exerted on a point mass 'm' at x = 0, if the mass per unit length of the rod is A + Bx2, is given by:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 4

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 5

A light wave is incident normally on a glass slab of refractive index 1.5. If 4% of light gets reflected and the amplitude of the electric field of the incident light is 30V/m, then the amplitude of the electric field for the wave propogating in the glass medium will be:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 5

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 6

The output of the given logic circuit is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 6

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 7

In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the energy dissipated in the circuit in terms of capacitance 'C' and total charge 'Q' is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 7

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 8

A particle of mass m moves in a circular orbit in a central potential field   If Bohr's
quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number n as:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 8

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

Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P1 and P2 respectively, then:

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

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 10

A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision.
The subsequent motion of the combined body will be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 10

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

Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is:

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

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

A passeng er train o f leng th 60m travels a t a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/hr.
The ratio of times taken by the passenger train to completely cross the freight train when : (i) they are moving in the same direction, and (ii) in the opposite directions is :

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

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 13

An ide al gas occupies a volu me of 2m3 at a pressure of 3 × 106 Pa. The energy of the gas is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 13

Energy

Considering gas is monoatomic i.e. f = 3 E. = 9 × 106 J
Option-(4)

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 14

A 100 V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index?

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JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 15

The galvanometer deflection, when key K1 is closed but K2 is open, equals θ0 (see figure).
On closing K2 also and adjusting R2 to 5Ω, the deflection in galvanometer becomes

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 15

Case I

Case II

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 16

A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle 60° with ground level. But he finds the aeroplane right vertically above his position.
If is the speed of sound, speed of the plane is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 16

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 17

A proton and an a-particle (with their masses in the ratio of 1:4 and charges in the ratio of 1:2) are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii rp : rα of the circular paths described by them will be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 17

KE = qΔV

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 18

A point source of light, S is pla ce d at a distance L in front of the centre of plane mirror of width d which is hanging vertically on a wall.
A man walks in front of the mirror along a line parallel to the mirror, at a distance 2L as shown below. The distance over which the man can see the image of the light source in the mirror is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 18

3d

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 19

“The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 micrometer diameter of a wire is ”

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 19

Least count of main scale of a screw gauge = 1mm = 10-3 m
Pitch of screw gauge = least count of main scale of screw gauge = 10-3 m
Diameter of wire = 5 μm
It is clear that; 5μm is much smaller than least count of main scale. For minimum number of division of circular scale,
Least count of screw gauge = diameter of wire = 5 μm = 5 × 10-6 m
Least count of screw gauge = pitch/number of division of circular scale
⇒5 × 10-3 = 10-3/N
⇒N = 1000/5 = 200

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 20

A simple pendulum, made of a string of length l and a bob of mass m, is released from a small angle θ0. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle θ1. Then M is given by :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 20

By momentum conservation

By componendo divided

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 21

What is the position and nature of image formed by lens combination shown in figure? (f1, f2 are focal lengths)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 21

For first lens

For  second lens

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 22

In the figure shown, a circuit contains two identical resistors with resistance R = 5W and an inductance with L = 2mH. An ideal battery of 15 V is connected in the circuit. What will be the current through the battery long after the switch is closed?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 22

Ideal inductor will behave like zero resistance long time after switch is closed

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 23

Determine the electric di pole mo ment of the system of three charges, placed on the vertices of an equilateral triangle, as shown in the figure:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 23

|P1| = q(d)
|P2| = qd
|Resultant| = 2 P cos30º

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 24

The position vector of the centre of mass of a symmetric uniform bar of negligible area of cross-section as shown in figure is :

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JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 25

As sh own i n t he fi gure, tw o i nfi ni tel y lon g, identical wires are bent by 90° and placed in such a way that the segments LP and QM are along the x-axis, while segments PS and QN are parallel to the y-axis. If OP = OQ = 4cm, and the magnitude of the magnetic field at O is 10–4 T, and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at O will be (μ0 = 4π × 10–7 NA– 2) :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 25

Magnetic field at ‘O’ will be done to ‘PS’ and ‘QN’ only
i.e. B0 = BPS + BQN →  Both inwards Let current in each wire = i

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 26

In a meter bridge, the wire of length 1 m has a non-uniform cross-section such that, the variation  of its resistance R with length ℓ is  Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 26

For the given wire :
where  C = constant.
Let resistance of part AP is R1 and PB is R2

∴  By balanced WSB concept.

Putting R1 = R2

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 27

For the given cyclic process CAB as shown for a gas, the work done is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 27

Since P–V indicator diagram is given, so work done by gas is area under the cyclic diagram.

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 28

An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5W.The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire, is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 28

Let current flowing in the wire is i.

If resistance of 10 m length of wire is x

then

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 29

A part icle A of mass 'm' and charg e 'q ' is accelerated by a potential difference of 50 V.Another particle B of mass '4 m' and charge 'q' is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths  is close to:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 29

K.E. acquired by charge = K = qV

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 30

There is a uniform spher ically sym metric surface charge density at a distance R0 from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V(R(t)) of the distribution as a function of its instantaneous radius R (t) is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 30

At any instant 't'
Total energy of charge distribution is constant

Also the slope of v-s curve will go on decreasing

∴ Graph is correctly shown by option(1)

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 31

Water samples with BOD values of 4 ppm and 18 ppm, respectively, are

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 31

Clean water have BOD value of less than 5 ppm whereas highly polluted water could have BOD value of 17 ppm or more.

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 32

Given

Temperature/K

On the basis of data given above, predict which of the following gases shows least adsorption on a definite amount of charcoal?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 32

More easily liquefiable a gas is (i.e. having higher critical temperature), the more readily it will be adsorbed.

∴ Least adsorption is shown by H2 (least critical temperature)

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 33

The metal d-orbitals that are directly facing the ligands in K3[Co(CN)6] are

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 33

During splitting in octahederal co-ordination entities, dx2 – y2  and orbitals point towards the direction of ligands (i.e. they experience more repulsion and their energy is raised)

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 34

A metal on combustion in excess air forms X. X upon hydrolysis with water yields H2O2 and O2 along with another product. The metal is

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JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 35

The correct order for acid strength of compounds  is as follows :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 35

Order of acidic strength is

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 36

The hardness of a water sample (in terms of equivalents of CaCO3) containing 10–3 M CaSO4 is (molar mass of CaSO4 = 136 g mol–1)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 36

10–3 M CaSO4 ≌ 10-3 M CaCO3

10–3 M CaCO3 means 10–3 moles of CaCO3 are present in 1L

ie 100 mg of CaCO3 is present in 1L solution.
Hardness of water = Number of milligram of CaCO3  per litre of water.

∴ Hardness of water = 100 ppm

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 37

In the following reaction

The best combination is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 37

∴ Best combination is HCHO and MeOH

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 38

Poly-β-hydroxybutyrate-co-β-hydroxyvalerate (PHBV) is a copolymer of ___.

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 38

∴ Monomers of PHBV are 3-Hydroxybutanoic acid and 3-Hydroxypentanoic acid.

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 39

The molecule that has minimum/no role in the formation of photochemical smog, is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 39

NO, O3 and HCHO are involved in the formation photochemical smog.
N2  has no role in photochemical smog

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 40

The increasing order of reactivity of the following compounds towards reaction with alkyl halides directly is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 40

Reactivity of compounds (nucleophiles) with alkyl halides will depend upon the availability of lone pair of electrons on nitrogen (amines or acid amides)

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 41

cannot be prepared by

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 41

Reaction (3) gives primary alcohol which is different from tertiary alcohol given by the remaining reactions.

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 42

Two solids dissociate as follows

The total pressure when both the solids dissociate simultaneously is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 42

∴  P1(P1 + P2) + P2(P1 + P2) = x + y

⇒  (P1 + P2)2 = x + y

∴ Total pressure  atm at equilibrium

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 43

The standard electrode potential Eo and its temperature coefficient  for a cell are 2 V and – 5×10–4 VK–1 at 300 K respectively. The cell reaction is

Zn(s) + Cu2+(aq) →  Zn2+ (aq) + Cu(s)

The standard reaction enthalpy (ΔfHo ) at 300 K in kJ mol-1 is,
[Use R = 8 JK-1 mol-1 and F = 96,000 C mol-11]

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 43

so,

Cell reaction :

Zn(s) + Cu2+(aq) →  Zn2+ (aq) + Cu(s)

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 44

Decomposition of X exhibits a rate constant of 0.05 mg/year. How many years are required for the decomposition of 5 mg of X into 2.5 mg?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 44

Rate constant of decomposition of  X = 0.05 μg/year From unit of rate constant, it is clear that the decomposition follows zero order kinetics.
For zero order kinetics,

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 45

In the Hall-Heroult process, aluminium is formed at the cathode. The cathode is made out of

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 45

In Hall-Heroult process, steel vessel with carbon lining acts as cathode.

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 46

What is the work function of the metal if the light of wavelength 4000 Å generates photoelectrons of velocity 6 × 105 ms–1 from it?

(Mass of electron = 9 × 10–31 kg
Velocity of light = 3 × 108 ms–1
Planck’s constant = 6.626 × 10–34 Js
Charge of electron = 1.6 × 10–19 JeV–1)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 46

∴ Work function = 3.1 - 1 = 2.1 eV

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 47

Among the following four aromatic compounds, which one will have the lowest melting point?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 47

In general, polarity increases the intermolecular force of attraction and as a result increases the melting point.

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 48

In the following reactions, products A and B are

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 48

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 49

The pair of metal ions that can give a spin only magnetic moment of 3.9 BM for the complex [M(H2O)6]Cl2, is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 49

μ = 3.9 BM

So, the central metal ion has 3 unpaired electrons.
∴ Configuration is either d3 or d7 as H2O is a weak field ligand.

V2+ has d3 configuration.
Co2+ has d7 configuration

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 50

In a chemical reaction, the initial concentration of B was 1.5 times of the concentration of A, but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant (K) for the aforesaid chemical reaction is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 50

given

3 – 2x = 2 – x
⇒ x = 1

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 51

The major product of the following reaction

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 51

DIBAL-H followed by hydrolysis converts nitrile to aldehyde and ester to aldehyde and alcohol.

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 52

For a diatomic ideal gas in a closed system, which of the following plots does not correctly describe the relation between various thermodynamic quantities?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 52

Cp and Cv for ideal gases are dependant on temperature only. So, Cp will not change with pressure.

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 53

The volume of gas A is twice than that of gas B. The compressibility factor of gas A is thrice than that of gas B at same temperature. The pressure of the gases for equal number of moles are

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 53

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 54

Among the following compounds most basic amino acid is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 54

Lysine is the most basic among the given amino acids

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 55

Mn2(CO)10 is an  organometallic compound due to the presence of

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 55

It is organometallic compound due to presence of Mn – C bond.

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 56

The major product of the following reaction is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 56

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 57

Iodine reacts with concentrated HNO3 to yield Y along with other products. The oxidation state of iodine in Y, is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 57

Conc. HNO3 oxidises I2 to iodic acid (HIO3).

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 58

The element with Z = 120 (not yet discovered) will be an/a

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 58

Element with Z = 120 will belong to alkaline earth metals.

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 59

Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y.
If molecular weight of X is A, then molecular weight of Y is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 59

(Since density of solutions are not given therefore assuming molality to be equal to molarity and given % as % W/V)

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 60

50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 60

2 × 50 × 0.5 = 25 × M

⇒ M = 2

∴ Moles of NaOH in 50 mL

∴ Weight = 4 grams

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 61

For x >1, i f (2x)2y = 4e2x–2y, then   is equal to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 61

(2x)2y = 4e2x–2y
2yℓn2x = ℓn4 + 2x – 2y

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 62

The sum of the distinct real values of m, f or which the vec tors,    are co-planer, is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 62

sum of distinct solutions = –1

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 63

Let S be the set of all points in (–π,π) at which the function, f(x) = min {sinx, cosx} is not differentiable. Then S is a subset of which of the following?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 63

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 64

The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now from an A.P. Then the sum of the original three terms of the given G.P. is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 64

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 65

The integral  is equal to : (where C is a constant of integration)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 65

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 66

Let  If  then A is equal to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 66

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 67

Let S = {1,2,3, ... ., 100}. The number of nonempty subsets A of S such that the product of elements in A is even is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 67

S = {1,2, 3------100}
= Total non empty subsets-subsets with product of element is odd

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 68

If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observation is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 68

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 69

If a variable line, 3x+4y–λ=0 is such that the two circles x2 + y2 – 2x – 2y + 1 = 0 and x2+y2–18x–2y+78 = 0 are on its opposite sides, then the set of all values of l is the interval :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 69

Centre of circles are opposite side of line

(3 + 4 – λ) (27 + 4 – λ) < 0
(λ – 7) (λ – 31) < 0
λ ∈ (7, 31)

distance from S1

distance from S2

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 70

A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of   is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 70

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 71

let C1 and C2 be the centres of the circles x2+y2–2x–2y–2 = 0 and x2+y2–6x–6y+14 = 0 respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral PC1QC2 is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 71

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 72

In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 72

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 73

If the stra ight line, 2x–3y+17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, b), then β equals :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 73

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 74

Let f and g be continuous functions on [0, a] such that f(x) = f(a–x) and g(x)+g(a–x)=4, then   is equal to :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 74

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 75

The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12–x2 such that the rectangle lies inside the parabola, is :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 75

f(a) = 2a(12 – a)2

f'(a) = 2(12 – 3a2)
maximum at a = 2
maximum area = f(2) = 32

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 76

The Boolean expression   is equivalent to:

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 77

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 77

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 78

Considering only the principal values of inverse functions, the set

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 78

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 79

An ordered pair(α,β) for which the system of linear equations (1+α)x + βy+z = 2
αx+(1+β)y+z = 3
αx+βy+2z = 2   has a unique solution is

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 79

For unique solution

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 80

The area (in sq. units) of the region bounded by the parabola, y = x2 + 2 and the lines, y = x + 1, x = 0 and x = 3, is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 80

Req. area

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 81

If λ be the ratio of the roots of the quadratic equation in x, 3m2x2+m(m–4)x+2 = 0, then the least value of m for which   , is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 81

3m2x2 + m(m – 4) x + 2 = 0

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 82

If the vertices of a hyperbola be at (–2, 0) and (2, 0) and one of its foci be at (–3, 0), then which one of the following points does not lie on this hyperbola?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 82

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 83

If  is a purely imaginary numberand |z| = 2, then a value of α is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 83

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 84

Let P(4, –4) and Q(9, 6) be two points on the parabola, y2= 4x and let X be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of ΔPXQ is maximum. Then this maximum area (in sq. units) is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 84

y2 = 4x
2yy' = 4

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 85

the perpendicular distance from the origin to the plane containing the two lines,  and  is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 85

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 86

The maximum value of  for any real value of θ is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 86

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 87

A tetrahe dron has ver tices P(1, 2, 1), Q(2, 1, 3), R(–1,1,2) and O(0, 0, 0). The angle between the faces OPQ and PQR is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 87

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 88

Lety = y(x) be the solution of the differential equation,  If 2y(2) = loge4–1, then y(e) is equal to :-

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 88

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 89

Let  and Q = [qij] be two 3×3 matrices such that Q–P5 = I3. Then   is equal to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 89

Aliter

JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 90

Consider three boxes, each containing 10 balls labelled 1,2,....,10. Suppose one ball is randomly drawn from each of the boxes.
Denote by ni, the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (12th January - Morning) - Question 90

No. of ways = 10C3 = 120

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