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# JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April)

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## 90 Questions MCQ Test JEE Main & Advanced Mock Test Series | JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April)

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JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 1

### A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to dropto 1/1000 of the original amplitude is close to:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 1

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 2

### A cell of internal resistance r drives current through an external resistance R. Thepower delivered by the cell to the external resistance will be maximum when:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 2

Power maximum when R=r.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 3

### A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut–off, the coordinates of the centre of mass of the remaining portion will be

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 3

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 4

In a line of sight radio communication, a distance of about 50 km is kept betweenthe transmitting and receiving antennas. If the height of the receiving antenna is 70m, then the minimum height of the transmitting antenna should be:(Radius of the Earth = 6.4*106 m)

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 4

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 5

In a simple pendulum experiment for determination of acceleration due to gravity(g), time taken for 20 oscillations is measured by using a watch of 1 second leastcount. The mean value of time taken comes out to be 30 s. The length of pendulum ismeasured by using a meter scale of least count 1 mm and the value obtained is55.0 cm. The percentage error in the determination of g is close to:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 5

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 6

A parallel plate capacitor has 1μF capacitance. One of its two plates is given +2μCcharge and the other plate, +4μC charge. The potential difference developed acrossthe capacitor is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 6

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 7

Two magnetic dipoles X and Y are placed at a separation d, with their axes perpendicular to each ot.her. The dipole moment 0f Y is twice that of X. A particle of charge q is passing through their mid–point P, at angle θ = 45° with the
horizontal line, as shown in figure. What would be the magnitude of force on the particleat that instant ? (d is much larger than the dimensions of the dipole)

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 7

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 8

A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights hsph and hcyl on the incline. The ratio
is given by:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 8

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 9

A nucleus A, with a finite de–broglie wavelength λA, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of B. The de–Broglie wavelengths λB and λC of B and C are respectively :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 9

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 10

The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to:
[Boltzmann Constant kB =1.38 X 10–23 J/K Avogadro Number NA = 6.02 X 1026 /kg Radius of Earth : 6.4 X 106 m
Gravitational acceleration on Earth = 10 ms–2]

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 10

(But according to given data in the question no answer, avogadro number is given wrong )

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 11

A circuit connected to an ac source of emf e=e0 sin(100t) with t in seconds, gives aphase difference of π/4 between the emf e and current i. Which of the followingcircuits will exhibit this?

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 11

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 12

Young’s moduli of two wires A and B are in the ratio 7:4. Wire A is 2m long and hasradius R. Wire B is 1.5 m long and has radius 2 mm. If the two wires stretch by thesame length for a given load, then the value of R is close to :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 12

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 13

A body of mass m1 moving with an unknown velocity of , undergoes a collinearcollision with a body of mass m2 moving with a velocity . After collision, m1 andm2 move with velocities of and , respectively. If m2=0.5 m1 and v3 =0.5 v1​ ,then v1 is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 13

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 14

A particle starts from origin O from rest and moves with a uniform acceleration along the positive x– axis. Identify the figure that incorrectly represent the motion qualitatively. (a = acceleration, = velocity, x = displacement, t = time)

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 14

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 15

A positive point charge is released from rest at a distance ro from a positive line charge with uniform density. The speed of the point charge, as a function of instantaneous distance r from line charge, is proportional to:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 15

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 16

The electric field in a region is given by , where E is in NC–1 and x is in metres. The values of constants are A=20 SI unit and B=10 SI unit. If the potential at x=1 isV1 and that at x= –5 is V2, then V1 –V2is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 16

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 17

The ratio of mass densities of nuclei of 40Ca and 16O is close to:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 17

Radius of the nucleus is given by R= ROA1/3 So, Density constant

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 18

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 18

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 19

Two very long, straight, and insulated wires are kept at 90° angle from each other in xy–plane as shown in the figure.

These wires carry currents of equal magnitude I, whose directions are shown in the figure. The net magnetic field at point P will be:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 19

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 20

A common emitter amplifier circuit, built using an npn transistor, is shown in the figure. Its dc current gain is 250, RC  =1KΩ and VCC = 10V. What is the minimum base current for VCE to reach saturation?

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 20

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 21

If Surface tension (S), Moment of Inertia (I) and Planck's constant (h), were to betaken as the fundamental units, the dimensional formula for Linear momentumwould be:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 21

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 22

An electric dipole is formed by two equal and opposite charges q with separation d.The charges have same mass m. It is kept in a uniform electric field E. If it is slightly rotated from its equilibrium orientation, then its angular frequency ω is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 22

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 23

A rocket has to be launched from earth in such a way that it never returns. If E is the minimum energy delivered by the rocket launcher, what should be the mjnimum energy that the launcher should have if the same rocket is to be launched from the surface of the moon ? Assume that the density of the earth and the moon are equal and that the earth's volume is 64 times the volume of the moon.

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 23

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 24

A convex lens (of focal length 20 cm) and a concave mirror, having their principalaxes along the same lines, are kept 80 cm apart from each other. The concavemirror is to the right of the convex lens. When an object is kept at a distance of 30cm to the left of the convex lens, its image remains at the same position even if theconcave mirror is removed. The maximum distance of the object for which thisconcave mirror, by itself would produce a virtual image would be

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 24

Focal length of concave mirror is 10cm

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 25

In the circuit shown, a four–wire potentiometer is made of a 400 cm long wire,which extends between A and B. The resistance per unit length of the potentiometer wire is r= 0.01 Ω/cm If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A, the expected reading of the voltmeter will be:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 25

Total resistance of potentiometer wire is 4Ω

Total potential drop across wire is 4/6*3V =2V

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 26

A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5 m. When released, it slips off the table in a very short time τ =0.01 s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 26

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 27

Calculate the limit of resolution of a telescope objective having a diameter of 200cm,if it has to detect Light of wavelmgth 500 nm coming from a star.

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 27

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 28

In the figure shown, what is the current (in Ampere) drawn from the battery? You are given:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 28

Reff = 160/3

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 29

The magnetic field of an electromagnetic wave is given by:

The associated electric field will be:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 29

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 30

The given diagram shows four processes i.e., isochoric, isobaric, isothermal and adiabatic. The correct assignment of the processes, in the same order is given by:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 30

PVn = constant
If n = 0, isobaric
n = 1 isothermal
n = ∞ isochoric

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 31

The maximum prescribed concentration of copper in drinkmg water is:

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 32

The major product obtained in the following reaction is

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 32

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 33

Fructose and glucose can be distinguished by :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 33

Barfoed's test: Positive for Fructose and glucose.
Seliwanoff’s test: Positive for Fructose and negative for glucose.
Benedict’s test: Positive for Fructose and glucose.
Fehling’s test: Positive for Fructose and glucose.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 34

The structure of Nylon–6 is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 34

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 35

The covalent alkaline earth metal halide (X = Cl, Br, I) is :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 35

Beryllium halides are covalent due to more polarizing power of small Be2+ ion with more charge density.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 36

0.27 g of a long chain fatty acid was dissolved in 100 cm3 of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 cm. What is the height of the monolayer ? [Density of fatty acid = 0.9 g cm-3 ; p = 3 ]

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 36

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 37

For the following reactions, equilibrium constans are given:

The equilibrium constant for the reaction,

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 37

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 38

The correct statement about ICl5and ICl-4 is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 38

ICl5 have 4 B.P and 1 L.P. It is square pyramidal. ICl-4have 4 B.P. and 2 L.P. The two L.P occupy opposite corners of octahedral. So it square planar.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 39

The major product of the following reaction is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 39

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 40

The compound that inhibits the growth of tumors is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 40

cis–platini is used as anti–tumer agent.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 41

For a reaction scheme , if the rate of formation of B is set to be zero then the concentration of B is given by:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 41

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 42

Among the following molecules/ ions,

Which one is diamagnetic and has the shortest bond length?

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 42

Bond order

is diamagnetic since all electrons are paired and have shortest bond length due to triple bond.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 43

The major product of the following reaction is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 43

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 44

Calculate the standard cell potential (in V) of the cell in which following reaction takes place:

Fe2+ (aq) + Ag+ (aq) → Fe3+ (aq) + Ag (s)

Given that

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 44

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 45

The major product in the following reaction is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 45

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 46

The major product obtained in the following reaction is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 46

Intramolecular aldol condensation

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 47

Which one of the following alkenes when treated with HCl yields majority an anti Maxkovnikov product?

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 47

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 48

The calculated spin–only magnetic moments (BM) of the anionic and cationic species of [Fe(H2O)6]2 and [Fe(CN)6], respectively, are :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 48

The compound containing [Fe(H2O)6]2 cation and [Fe(CN)6] anion must be [Fe(H2O)6]2 [Fe(CN)6]. It contains 2[Fe(H2O)6]2+ cation and [Fe(CN)6]4– anion in which Iron is present as Fe2+ ion. In [Fe(H2O)6]2+ since H2O is weak ligand there will be 4 unpaired electrons with 4.9 BM while in [Fe(CN)6]4– all the electrons are paired as the CN is strong ligand with zero unpaired electrons and zero magnetic moment.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 49

The IUPAC symbol for the element with atomic number 119 would be:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 49

For 1 the abreviation is U (from unium) and for 9 is e (from enneium). So the IUPAC symbol is uue.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 50

The percentage composition of carbon by mole in methane is :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 50

% of C by mole = 1/5 x 100 = 20%

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 51

The Mond process is used for the:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 51

Nickel is purified by Mond’s process using the formation of unstable volatile compound. Ni(CO)4.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 52

The statement that is INCORRECT about the interstitial compounds is :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 52

They are relatively non reactive

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 53

Polysubstitution is a major drawback in

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 53

Polyalkylation – Products of Friedel–Crafts are even more reactive than starting material. Alkyl groups produced in Friedel–Crafts Alkylation are electron–donating substituents meaning that the products are more susceptible to electrophilic attack than reactant.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 54

For the solution of the gases w, x, y and z in water at 298 K, the Henrys law constants (KH) are 0.5, 2, 35 and 40 kbar, respectively. The correct plot for the given data is :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 54

KH for w, x, y, z are 0.5, 2, 35 & 40.
P = KH X
P = KH (1 - Xwater)
P = KH - KHXwater               (y = c - mx)
KH values increases in the order z > y > x > w.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 55

Consider the bcc unit cells of the solids 1 and 2 with the position of atoms as shown below. The radius of atom B is twice that of atom A. The unit cell edge length is 50% more in solid 2 than in 1. What is the approximate packing efficiency in solid 2?

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 55

√3a = 2r + 4r = 6r
a = 2√3r
Packing fraction =

i.e. 90% is the packed in the solid B

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 56

5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV = 28 JK-1 mol–1, calcuIate Δ U and Δ PV for this process. ( R = 8.0 JK -1 mol-1 )

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 56

ΔU = 28x5x100 = 14kJ
Δ(pV) = p2v2 - p1v2 = nR (T- T1)
= 5x8.314x100 ≈ 4KJ

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 57

The ion that has sp3d2 hybridization for the central atom, is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 57

ICl4- have 4 B.P. and 2 L.P (steric number 6). So central atom is involved in sp3d2 hybridization.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 58

If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ , then for 1.5 p momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function) :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 58

W0 is too small in comparision to K.E.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 59

Which of the following compounds will show the maximum ‘enol’ content ?

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 59

CH3COCH2COOC2H5    %enol = 0.00025
CH3COCH2CONH2         % enol = <7.5
CH3COCH2COCH3         %enol = 72
CH3COCH3                      %enol = 0.00025

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 60

The strength of 11.2 volume solution of H2O2 is:
[Given that molar mass of H=1 g mol–1 and O=16 g mol–1]

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 60

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 61

The tangent and the normal lines at the point ( √3,1) to the circle x2 + y2 = 4 and the x-axis form a triangle. The area of this triangle (in square units) is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 61

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 62

Let  and ,for some real x.Then  is possible if :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 62

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 63

If three distinct numbers a, b, c are in G.P. and the equations ax2 + 2bx +c =0 and dx2 + 2ex + f = 0 have a common root, then which one of the following statements is correct ?

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 63

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 64

The sum  is equal to

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 64

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 65

If the system of linear equations
x – 2y + kz =1
2x + y + z = 2
3x – y – kz =3
has a solution (x, y, z), z ≠ 0, then (x, y) lies on the straight line whose equation is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 65

From the given eqs, (1) + (3) → 4x – 3y – 4 = 0

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 66

The tangent to the parabola y2=4x at the point where it intersects the circle x2 + y2 = 5 in the first quadrant, passes through the point :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 66

y2 = 4x,x2 + y2 = 5 ⇒ x2 + 4x -5= 0
x =1 (x ≠-5) ⇒ y =2 Q  (y>0)
Eq of tangent is 2y = 2(x+1) ⇒ x – y + 1 = 0

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 67

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

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 67

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 68

Suppose that the points (h, k), (1, 2) and (–3, 4) lie on the line L1. If a line L2 passing through the points (h, k) and (4, 3) is perpendicular to L1, then k/h equals:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 68

eq of L1 is x + 2y = 5 ⇒

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 69

Let the numbers 2, b, c be in an A.P. and  If det(A) ∈ [2,16],then c lies in the interval:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 69

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 70

The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 70

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 71

Two vertical poles of heights, 20 m and 80 m stand apart on a horizontal plane. The height (in meters) of the point of intersection of the Lines joining the top of each pole to the foot of the other, from this horizontal plane is :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 71

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 72

Let , where g is a non–zero even function. If f(x+5)=g(x), then  equals:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 72

g(x) = f '(x) & g (- x) = g (x) = f (x + 5) = f (5 - x) and also f(x) is odd.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 73

Let  be a diiferentiable function Satisfying f '(3) + f '(2) = 0 . Then  is equal to

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 73

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 74

Let  be deﬁned as

where [t] denotes the greatest integer less than or equal to t. Then, f is discontinuous at:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 74

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 75

let  and A(α) is area of the region S (α) . If for a λ, 0 < λ < 4, A(λ) : A(4) = 2 : 5 , then λ equals:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 75

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 76

The number of integral values of m for which the equation (1 + m2) x2 - 2 (1 + 3m) x + (1 + 8m) = 0 has no real root is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 76

D < 0 ⇒ 8m3 -8m2 + 2m > 0 ⇒ 2m (2m-1)2 > 0 ⇒ m > 0

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 77

If the eccmtricity of the standard hyperbola passing through the point (4, 6) is 2, then the equation of the tangent to the hyperbola at (4, 6) is :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 77

Eq. of hyperbola is    it passes through (4,6) ⇒ a2=4.
Eq. of tangent is 2x–y=2.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 78

If , then  is equal to:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 78

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 79

The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 79

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 80

The number of four–digit numbers strictly greater than 4321 that can be formed using the digits 0, 1, 2, 3, 4, 5 (repetition of digits is allowed) is :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 80

Total no. = 310.

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 81

Let  be written as f(x) =f1(x) +f2(x), where f1(x) is an even function and f2(x) is an odd function. Then f1(x + y) +f1(x – y) equals:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 81

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 82

The vector equation of the plane through the line of intersection of the planes x+y+ z= 1 and 2x+3y+4z=5 which is perpendicular to the plane x – y+ z =0 is :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 82

Line of intersection of the planes x+y+z=1, 2x+3y+4z=5 is

Eq. of required plane is

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 83

Which one of the following statements is not a tautology ?

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 83

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 84

Given that the slope of the tangent to a curve y = y ( x) at any point (x,y) is 2y/xIf the curve passes through the centre of the circle x2 + y2 – 2x – 2y = 0, then its equation is :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 84

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 85

If a point R(4, y, z) lies on the line segment joining the points P (2,-3, 4) and Q(8, 0, 10), then the distance of R from the origin is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 85

Here, P, Q, R are collinear ⇒

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 86

In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at (0, 5 √3) , then the length of its latus rectum is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 86

Here, given ellipse is vertical ellipse & b – a = 5, b2 – a2 = 75

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 87

If the fourth term in the binomial expansion of  is equal to 200,and x >1, then the value of x is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 87

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 88

If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is:

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 88

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 89

If f(1) = 1, f '(1) = 3, then the derivative of f(f(f ( x))) + (f ( x))2 of x=1 is

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 89

JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 90

A student scores the following marks in ﬁve tests : 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is :

Detailed Solution for JEE Main 2019 Paper - 2 with Solution (After Noon, 8 April) - Question 90

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