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

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JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 1

### Two forces P and Q of magnitude 2F and 3F, respectively, are at an angle θ with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 1

4F2 + 9F2 + 12F2 cos θ = R2
4F2 + 36 F2 + 24 F2 cos θ = 4R2
4F2 + 36 F2 + 24 F2 cos θ
= 4(13F2 + 12F2cosθ) = 52 F2 + 48F2cosθ

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 2

### The actual value of resistance R, shown in the figure is 30Ω. This is measured in an experiment as shown using the standard formula R = V/1 , where V and I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is 5% less, then the internal resistance of the voltmeter is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 2

0.95 × 30 = 0.05 Rυ
Rυ = 19 × 30 = 570 Ω

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 3

### An unknown metal of mass 192 g heated to a temperature of 100ºC was immersed into a brass calorimeter of mass 128 g containing 240 g of water a temperature of 8.4ºC Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5ºC (Specific heat of brass is 394 J kg–1 K–1)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 3

192 × S × (100 – 21.5)
= 128 × 394 × (21.5 – 8.4)
+ 240 × 4200 × (21.5 – 8.4)
⇒ S = 916

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 4

A particle starts from the origin at time t = 0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5s?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 4

S = Area under graph
1/2× 2 × 2 + 2 × 2 + 3 × 1 = 9 m

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 5

The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 5

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 6

A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11V is connected across it is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 6

P = I2R
4.4 = 4 × 10–6 R
R = 1.1 × 106 Ω

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 7

The diameter and height of a cylinder are measured by a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 7

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 8

At some location on earth the horizontal component of earth's magnetic field is 18 x 10-6 T. At this location, magnetic neeedle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 8

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

The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you for license what broadcast frequency will you allot?

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

∴ Range of signal = 2250 Hz to 2750 Hz
Now check all options : for 2000 KHZ
fmod = 200 Hz
∴ Range = 1800 KHZ to 2200 KHZ

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

A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are Th and Tc respectively, then :

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

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

The electric field of a plane polarized electromagnetic wave in free space at time t= 0 is given by an expression

The magnetic field : (cis the velocity of light)

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

i.e. direction of 'c'.

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

Condiser the nuclear fission
Ne20 →  2He4 + C12
Given that the binding energy/nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement :

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

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 13

Two vectors have equal magnitudes. The magnitude of  is 'n' times the magnitude of  The angle between   is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 13

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 14

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 14

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 15

Consider a Young's double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength λ such that the first minima occurs directly in front of the slit (S1) ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 15

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 16

The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of cornea (7.8 mm). This surface separates two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus.

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 16

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 17

Half mole  of an ideal monoatomic gas is heated at constant pressure of 1atm from 20ºC to 90ºC. Work done by gas is close to :
(Gas constant R = 8.31 J /mol.K)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 17

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 18

A metal plate of area 1 × 10–4 m2 is illuminated by a radiation of intensity 16 mW/m2.The work function of the metal is 5eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be :
[1 eV = 1.6 × 10–19J]

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 18

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 19

Charges -q and +q located at A and B, respectively, constitute an electric dipole. Distance AB = 2a, O is the mid point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP = y and y >> 2a. The charge Q experiences and electrostatic force F. If O is now moved alons the equatorial line to P' such that OP = (y/3), the force on Q will be close to (y/3 >> 2a)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 19

Electric field of equitorial plane of dipole

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 20

Two stars of masses 3 × 1031 kg each, and at distance 2 × 1011m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star's rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is : (Take Gravitational constant G = 6.67 ×10–11 Nm2 kg–2)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 20

By energy convervation between 0 & ∞.

[M is mass of star m is mass of meteroite)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 21

A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be : (Assume that the highest frequency a person can hear is 20,000 Hz)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 21

For closed organ pipe, reson ate frequency is odd multiple of fundamental frequency.
∴ (2n + 1) f0 ≤ 20,000
(fo is fundamental frequency = 1.5 KHz)
∴ n = 6

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 22

A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 22

Applying torque equation about point P.

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 23

For the circuit shown below, the current through the Zener diode is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 23

Assuming zener diode doesnot undergo breakdown, current in circuit = 120/15000 = 8mA
∴ Voltage drop across diode = 80 V > 50 V.
The diode undergo breakdown.

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 24

Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, –2) and (0, –2). The work required to put a fifth charge Q at the origin of the coordinate system will be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 24

(Potential at ∞  = 0)

∴Work required to put a fifth charge Q at origin is equal to

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 25

A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency ω. If the radius of the bottle is 2.5 cm then ω close to : (density of water = 103 kg / m3)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 25

Extra Boyant force = δAxg
B0 + B × mg = ma
B = ma

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 26

A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates the work done by the capacitor on the slab is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 26

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 27

Two kg of a monoatomic gas is at a pressure of 4 × 104 N/m2 . The density of the gas is 8 kg /m3. What is the order of energy of the gas due to its thermal motion ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 27

Thermal energy of N molecule

order will 104

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 28

A particle which is experiencing a force, given by  undergoes a displacement of If the particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 28

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 29

The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor used as R1 has the colour code ( Orange, Red, Brown). The resistors R2 and R4 are 80Ω and 40Ω, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R3, would be :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 29

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 30

Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 30

For Ball
using parallel axis theorem.

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 31

An ide al ga s under goes isothe rma l compression from 5 m3 against a constant external pressure of 4 Nm–2. Heat released in this process is used to increase the temperature of 1 mole of Al. If molar heat capacity of Al is 24 J mol–1 K–1, the temperature of Al increases by :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 31

Work done on isothermal irreversible for ideal gas
= –Pext (V2 – V1)
= –4 N/m2 (1m3 – 5m3)
= 16 Nm
Isothermal process for ideal gas
ΔU = 0
q = –w
= –16 Nm
= – 16 J
Heat used to increase temperature of Aℓ
q = n Cm ΔT

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 32

The 71st electron of an element X with an atomic number of 71 enters into the orbital :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 32

Filling of electrons usually follow many rules, but what orbit the electron will enter is given the Aufbau rule.

This rule states that electrons will be filled in the atomic orbitals which are low in energy than the atomic orbitals which have high energy. The filling of atomic orbitals follow the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s and so on....

For the element X having atomic number 71 has 71 electrons.

The electronic configuration of this element is:

From the above configuration, it is visible that the 71st electron is entering the 5d orbital.

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 33

The number of 2-c entre-2-electron and 3-centre-2-electron bonds in B2H6, respectively, are :

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 34

The amount of sugar (C12H22O11) required to prepare 2 L of its 0.1 M aqueous solution is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 34

wt (C12H22O11) = 68.4 gram

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 35

Among the following reactions of hydrogen with halogens, the one that requires a catalyst is :

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 36

Sodium metal on dissolution in liquid ammonia gives a deep blue solution due to the formation of:

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 37

What will be the major product in the following mononitration reaction?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 37

amine is o-p directing

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 38

In the cell Pt(s)|H2(g, 1bar|HCl (aq)|Ag(s)|Pt(s) the cell potential is 0.92 when a 10–6 molal HCl solution is used. THe standard electrode potential of (AgCl/Ag,Cl) electrode is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 38

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 39

The major product of the following recation is:

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 40

The pair that contains two P–H bonds in each of the oxoacids is

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 41

The major product of the following reaction is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 41

SN2 reaction

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 42

The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 42

Co2+ -->d7
hs, n = 3,ls, n = 1

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 43

A compound of formula A2B3 has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 43

A2B3 has HCP lattice
If A form HCP, then 3th/4 of THV must occupied by B to form A2B2
If B form HCP, then 1th/3 of THV must occupied
by A to form A2B3

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 44

The reaction that is NOT involved in the ozone layer depletion mechanism in the stratosphere is:

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 45

The process with negative entropy change is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 45

N2(g) + 3H2(g) ⇔ 2NH3(g) ; Δng < 0

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 46

The major product of the following reaction is:

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 47

A reaction of cobalt(III ) chloride and ethylenediamine in a 1:2 mole ratio generates two isomeric products A (violet coloured) B (green coloured). A can show optical activity, B is optically inactive. What type of isomers does A and B represent ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 47

[Co(Cn)2 Cl2]Cl
cis --> Optically active
rans --> Optically in active

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 48

The major product obtained in the following reaction is :

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 49

Which of the following tests cannot be used for identifying amino acids ?

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 50

What is the IUPAC name of the following compound ?

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 51

Which is the most suitable reagent for the following transformation?

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 52

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

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 53

In the reaction of oxalate with permaganate in acidic medium, the number of electrons involved in producing one molecule of CO2 is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 53

10 e trans for 10 molecules of CO2 so per molecule of CO2 transfer of e is '1'

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 54

5.1g NH4SH is introduced in 3.0 L evacuated flask at 327°C. 30% of the solid NH4SH decomposed to NH3 and H2S as gases. The Kp of the reaction at 327°C is (R = 0.082 L atm mol–1K–1, Molar mass of S = 32 g mol/01, molar mass of N = 14g mol–1)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 54

NH4SH(s)⇔ NH3 (g) H2S(g)

α = 30% = .3
so number of moles at equilibrium

Now use PV = nRT at equilibrium
Ptotal × 3 lit = (.03 + .03) × .082 × 600
Ptotal = .984 atm
At equilibrium

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 55

The electrolytes usually used in the electroplating of gold and silver, respectively, are :

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 56

Elevation in the boiling point for 1 molal solution of glucose is 2 K. The depression in the freezing point of 2 molal solutions of glucose in the same solvent is 2 K. The relation between Kb and Kf is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 56

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 57

An aromatic compound 'A' having molecular formula C7H6O2 on treating with aqueous ammonia and heating forms compound 'B'. The compound 'B' on reaction with molecular bromine and potassium hydroxide provides compound 'C' having molecular formula C6H7N. The structure of 'A' is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 57

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 58

The ground state energy of hydrogen atom is –13.6 eV. The energy of second excited state He+ ion in eV is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 58

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 59

For an elementary chemical reaction,
the expression for is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 59

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 60

Haemoglobin and gold sol are examples of :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 60

Haemoglobin→ positive sol
Ag - sol → negative sol

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 61

Let If R(z) and I[z] respectively denote the real and imaginary parts of z, then :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 61

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 62

Let a1,a2,a3,....a10 be in G.P. with ai; > 0 for i = 1,2,...., 10 and S be the set of pairs (r,k), r k∈N (the set of natural numbers) for which

Then the number of elements in S, is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 62

Apply
C3 → C3 – C2
C2 → C2 – C1
We get   D = 0

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 63

The positive value of λ for whic h the co-efficient of x2 in the expression is 720, is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 63

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 64

The value of is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 64

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 65

The value of  where [t ] denotes the greatest integer less than or equal to t, is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 65

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 66

If the probability of hitting a target by a shooter, in any shot, is 1/3, then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than 5/6, is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 66

nmin = 5

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 67

If me an and standa rd devia tion of 5 observations x1, x2, x3,x4,x5 are 10 and 3, respectively, then the variance of 6 observations x1,x2, ....,x5 and –50 is equal to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 67

= 507.5

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 68

The length of the chord of the parabola x2 = 4y having equation x -√2y + 4√2 = 0 is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 68

Solving together we get

Similarly,

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 69

Let  where b > 0. Then the minimum value of is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 69

|A| = 2(2b2 + 2 – b2) – b(2b – b) + 1 (b2 – b2 – 1)
|A| = 2(b2 + 2) – b2 – 1
|A| = b2 + 3

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 70

The ta ngent to the cur ve, passing through the point (1,e) also passes through the point :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 70

T : y – e = 3e (x – 1)
y = 3ex – 3e + e
y = (3e)x – 2e

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 71

The number of values of θ∈(0,π) for which  the system of linear equations
x + 3y + 7z = 0
–x + 4y + 7z = 0
(sin 3θ)x + (cos 2θ) y + 2z = 0
has a non-trivial solution, is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 71

(8 – 7 cos 2θ) – 3(–2 – 7 sin 3θ) + 7 (– cos 2θ – 4 sin 3θ) = 0
14 – 7 cos 2θ + 21 sin 3θ – 7 cos 2θ – 28 sin 3θ = 0
14 – 7 sin 3θ – 14 cos 2θ = 0
14 – 7 (3 sin θ – 4 sin3 θ) – 14 (1 – 2 sin2 θ) = 0
–21 sin θ+ 28 sin3 θ + 28 sin2 θ = 0
7 sin θ [–3 + 4 sin2 θ + 4 sin θ] = 0
sin θ,
4 sin2 θ + 6 sin θ – 2 sin θ – 3 = 0
2 sin θ(2 sin θ + 3) – 1 (2 sin θ + 3) = 0

Hence, 2 solutions in (0, π) Option (4)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 72

If then f'(1/2) is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 72

Differentiate w.r.t. 'x'
f(x) = 2x + 0 – x2 f(x)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 73

Let f : (–1,1) → R be a function defined by If K be the set of all points at which f is not differentiable, then K has exactly :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 73

f : (–1, 1) → R

Non-derivable at 3 points in (–1, 1) Option (1)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 74

Let where r ≠ ±1. Then S represents :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 74

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 75

If then K is equal to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 75

∴  K = 225

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 76

Let N be the set of natural numbers and two functions f and g be defined as f,g : N→N such that :
and g(n) = n–(–1)n. The fog is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 76

∴  many one but onto Option (4)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 77

The values of λ such that sum of the squares of the roots of the quadratic equation, x2 + (3 – λ) x + 2 = λ has the least value is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 77

α + β = λ– 3
αβ = 2 – λ
α2 + β2 = (α + β)2 – 2αβ = (λ – 3)2 – 2(2 – λ)
= λ2 + 9 – 6λ – 4 + 2λ
= λ2 – 4λ + 5
= (λ – 2)2 + 1
∴ λ = 2

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 78

Two vertices of a triangle are (0,2) and (4,3). If its orthocentre is at the origin, then its third vertex lies in which quadrant?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 78

⇒  3b – 6 = –4a  ⇒  4a + 3b = 6   ……(ii)
From (i) and (ii)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 79

Two sides of a parallelogram are along the lines, x + y = 3 and x – y + 3 = 0. If its diagonals intersect at (2,4), then one of its vertex is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 79

C ⇒ (4, 5)
Now equation of BC is x – y = –1
and equation of CD is x + y = 9
Solving x + y = 9  and x – y = –3
Point D is (3, 6)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 80

Let  and be two given vectors where vectors  arenon-collinear. The value of λ for which vectors  are collinear, is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 80

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 81

The value of  is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 81

(Where tanA=20 , tanB=1)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 82

With the usua l notation, in Δ ABC, if ∠A + ∠B = 1200, a = √3 + 1 and √3 - 1, then the ratio ∠A :∠B, is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 82

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 83

The plane which bisects the line segment joining the points (–3,–3,4) and (3,7,6) at right angles, passes through which one of the following points?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 83

p : 3(x – 0) + 5 (y – 2) + 1 (z – 5) = 0
3x + 5y + z = 15
∴ Option (2)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 84

Consider the following three statements :
P : 5 is a prime number.
Q : 7 is a factor of 192.
R : L.C.M. of 5 and 7 is 35.
Then the truth value of which one of the following statements is true ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 84

It is obvious
∴ Option (4)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 85

On which of the following lines lies the point of intersection of the line, and the plane, x + y + z = 2 ?

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 85

General point on the given line is
x = 2λ + 4
y = 2λ + 5
z = λ + 3
Solving with plane,
2λ + 4 + 2λ + 5 + λ + 3 = 2
5λ + 12 = 2
5λ = –10

∴ Option (3)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 86

Let f be a differentiable function such that

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 86

∴ Option (1)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 87

A helicopter is flying along the curve given by y – x3/2 = 7, (x ≥ 0). A soldier positioned at the point (1/2,7) wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 87

Option (3)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 88

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

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 88

Put x3 = t
3x2 dx = dt

∴ f(x) = –1 – 4x3
Option (1)
(From the given options (1) is most suitable)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 89

The curve a mongst the family of curves, represented by the differential equation, (x2 – y2)dx + 2xy dy = 0 which passes through (1,1) is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 89

(x2 – y2) dx + 2xy dy = 0

Solving we get,

ln(v2 + 1) = –ln x + C
(y2 + x2) = Cx
1 + 1 = C ⇒ C = 2
y2 + x2= 2x
∴ Option (2)

JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 90

If the area of an equilateral triangle inscribed in the circle, x2 + y2 + 10x + 12y + c = 0 is 27√3 sq. units then c is equal to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Evening) - Question 90

C = 25

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