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

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

### A uniform metallic wire has a resistance of 18 Ω and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is :

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

Req between any two vertex will be

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

### A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass 'm' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is :

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

At height r from center of earth. orbital velocity

∴ By energy conservation

(At infinity, PE = KE = 0)

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

### A solid metal cube of edge length 2 cm is moving in the positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis is :

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

Potential difference between two faces perpendicular to x-axis will be

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

A parallel plate capacitor is of area 6 cm2 and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants K1, = 10, K2 = 12 and K3 = 14. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be :

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

Let dielectric constant of material used be K.

⇒ K = 12

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

A 2 W carbon resistor is color coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is :

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

P = i2R.
∴ for imax, R must be minimum
from color coding R = 50×102Ω
∴ imax = 20mA

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

In a Young's double slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle 1/40 rad by using light of wavelength λ1. When the light of wavelength λ2 is used a bright fringe is seen at the same angle in the same set up. Given that λand λare in visible range (380 nm to 740 nm), their values are :

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

Path difference = d sinθ ≈ dθ
= 0.1 x 1/40 mm = 2500nm
or bright fringe, path difference must be integral multiple of λ.
∴ 2500 = nλ1 = mλ2
∴ λ1 = 625, λ2 = 500 (from m=5) (for n = 4)

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

A magnet of total magnetic moment 10-2 î A-m2 is placed in a time varying magnetic field, B î (costωt) where B = l Tesla and ω = 0.125 rad/ s. The work done for reversing the direction of the magnetic moment at t = 1 second, is :

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

Work done,

= 2 × 10–2 × 1 cos(0.125)
= 0.02 J

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

To mop-clean a floor, a cleaning machine presses a circular mop of radius R vertically down with a total force F and rotates it with a constant angular speed about its axis. If the force F is distributed uniformly over the mop and if coefficient of friction between the mop and the floor is μ the torque, applied by the machine on the mop is :

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

Consider a strip of radius x & thickness dx, Torque due to friction on this strip.

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

Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:

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

at t = 0, A0 = dN/dt = 1600 C/s
at t = 8s, A = 100 C/s

Therefor half life is t1/2 = 2 sec
∴ Activity at t = 6 will be 1600 (1/2)= 200 C/s

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

If the magnetic field of a plane electromagnetic wave is given by (The speed of light = 3 × 108/m/s)
then the maximum electric field associated with it is :

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

E0 = B0 × C
= 100 × 10–6 × 3 × 108
= 3 × 104 N/C

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

A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c ) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be :

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

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

Water flows into a large tank with flat bottom at the rate of 10–4 m3s–1. Water is also leaking out of a hole of area 1 cm2 at its bottom. If the height of the water in the tank remains steady, then this height is:

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

Since height of water column is constant therefore, water inflow rate (Qin)
= water outflow rate
Qin = 10–4 m3s–1

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

A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : (g =10ms–2)

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

Time taken for the particles to collide,

Speed of wood just before collision = gt = 10 m/s & speed of bullet just before collision v-gt = 100 – 10 = 90 m/s
Now, conservation of linear momentum just before and after the collision -
–(0.02) (1v) + (0.02) (9v) = (0.05)v
⇒ 150 = 5v
⇒ v = 30 m/s
Max. height reached by body h = v2/2g

∴  Height above tower = 40 m

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

The density of a material in SI units is 128 kg m-3. In certain units in which the unit of length is 25 cm and the unit of mass is 50 g, the numerical value of density of the material is :

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

= 40 units

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

To get output '1' at R, for the given logic gate circuit the input values must be :

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

To make O/P P + Q must be 'O' SO, y = 0 x = 1

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

A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in fig. Work done by normal reaction on block in time t is :

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

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

A heat source at T= 103 K is connected to another heat reservoir at T=102 K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is 0.1 WK-1 m-1, the energy flux through it in the steady state is :

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

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

A TV transmission tower has a height of 140 m and the height of the receiving antenna is 40 m. What is the maximum distance upto which signals can be broadcasted from this tower in LOS(Line of Sight) mode ? (Given : radius of earth = 6.4 x 106m).

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

Maximum distance upto which signal can be broadcasted is

where hT and hR are heights of transmiter tower and height of receiver respectively.
Putting all values -

on solving, dmax = 65 km

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

A potentiometer wire AB having length L and resistance 12 r is joined to a cell D of emf ε and internal resistance r. A cell C having emf ε/2 and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in fig. shows no deflection is:

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

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

An insulating thin rod of length ℓ has a x linear charge density λ (x) =  on it. The rod is rotated about an axis passing through the origin (x = 0) and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magnetic moment of the rod is :

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

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

Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is :

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

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

A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to :

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

Velocity of wave on string

Now, wavelength of wave

Separation b/w successive nodes,

= 20 cm

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

A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is f1. If the speed of the train is reduced to 17 m/s, the frequency registered is f2. If speed of sound is 340 m/s, then the ratio f1/f2 is :

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

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

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10–12m, the minimum electron energy required is close to :

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

{λ = 7.5 × 10–12}

KE = 25 Kev

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

A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is:

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

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

A plano convex lens of refractive index µ1 and focal length f1 is kept in contact with another plano concave lens of refractive index µ2 and focal length f2. If the radius of curvature of their spherical faces is R each and f1 = 2f2, then µ1 and µ2 are related as

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

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

Two electric dipoles, A, B with respective dipole moments and placed on the x-axis with a separation R, as shown in the figure

The distance from A at which both of them produce the same potential is :

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

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

In the given circuit the cells have zero internal resistance. The currents (in Amperes) passing through resistance R1, and R2 respectively, are:

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

i1 = 10/20 = 0.5A
i2 = 0

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

In the cube of side 'a' shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be:

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

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

Three Carnot engines operate in series between a heat source at a temperature T1 and a heat sink at temperature T4 (see figure). There are two other reservoirs at temperature T2, and T3, as shown, with T2 > T2 > T3 > T4 . The three engines are equally efficient if:

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

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

Two pi and half sigma bonds are present in:

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

⇒ BO = 2.5 ⇒ [π - Bond = 2 & σ - Bond = 1/2]
N2 ⇒ B.O. = 3.0 ⇒ [π - Bond = 2 & σ - Bond = 1]
= B.O. ⇒ 2.5 ⇒ [π - Bond = 1.5 & σ - Bond = 1]
O2 ⇒ B.O. ⇒ 2 ⇒ [π - Bond ⇒ 1 & σ - Bond = 1]

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

The chemical nature of hydrogen preoxide is :-

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

H2O2 act as oxidising agent and reducing agent in acidic medium as well as basic medium.
H2O2 Act as oxidant :-
(In acidic medium)
(In basic medium)
H2O2 Act as reductant :-
(In acidic medium)
(In basic medium)

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

Which dicarboxylic acid in presence of a dehydrating agent is least reactive to give an anhydride :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Morning) - Question 33

7 membered cyclic anhydride (Very unstable)

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

Which premitive unit cell has unequal edge lenghs (a ≠ b ≠ c) and all axial angles different from 90° ?

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

In Triclinic unit cell
a ≠ b ≠ c & α ≠ β ≠ γ ≠ 90°

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

Wilkinson catalyst is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Morning) - Question 35

Wilkinsion catalyst is [(ph3P)3RhCl]

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

The total number of isotopes of hydrogen and number of radioactive isotopes among them, respectively, are :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Morning) - Question 36

Total number of isotopes of hydrogen is 3

and only 31H or 3T is an Radioactive element.

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

The major product of the following reaction is

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

Example of E2 elimination and conjugated diene is formed with phenyl ring in conjugation which makes it very stable.

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

The total number of isomers for a square planar complex [M(F)(Cl)(SCN)(NO2)] is :

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

The total number of isomers for a square planar complex [M(F)(Cl)(SCN)(NO2)] is 12.

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

Hall-Heroult's process is given by "

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Morning) - Question 39

In Hall-Heroult's process is given by
2Al2O3 + 3C → 4Al + 3CO2

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

The value of Kp/KC for the following reactions at 300K are, respectively :
(At 300K, RT = 24.62 dm3atm mol–1)
N2(g) + O2(g) ⇔ 2NO(g)
N2O4(g) ⇔ 2NO2(g)
N2(g) + 3H2(g) ⇔ 2NH3(g)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Morning) - Question 40

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

If dichloromethane (DCM) and water (H2O) are used for differential  extraction, which one of the following statements is correct?

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

The type of hybridisation and number of lone pair(s) of electrons of Xe in XeOF4, respectively, are :

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

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

The metal used for making X-ray tube window is :

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

"Be" Metal is used in x-ray window is due to transparent to x-rays.

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

Consider the given plots for a reaction obeying Arrhenius equation (0°C < T < 300°C) : (k and Ea are rate constant and activation energy, respectively)

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Morning) - Question 44

On increasing Ea, K dec reases

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

Water filled in two glasses A and B have BOD values of 10 and 20, respectively. The correct statement regarding them, is :

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

Two glasses "A" and "B" have BOD values 10 and "20", respectively.
Hence glasses "B" is more polluted than glasses "A".

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

The increasing order of the pKa values of the following compounds is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Morning) - Question 46

Acidic strength is inversely proportional to pka.

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

Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapor pressures of pure A and pure B are 7 × 103 Pa and 12 × 103 Pa, respectively. The composition of the vapor in equilibrium with a solution containing 40 mole percent of A at this temperature is :

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

yB = 0.72

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

Consider the following reduction processes :
Zn2+ + 2e → Zn(s); E° = – 0.76 V
Ca2+ + 2e → Ca(s); E° = – 2.87 V
Mg2+ + 2e → Mg(s); E° = – 2.36 V
Ni2+ + 2e → Ni(s); E° = – 0.25 V
The reducing power of the metals increases in the order :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Morning) - Question 48

Higher the o xidation potentia l bet ter will be reducing power.

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

The major product of the following reaction is:

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Morning) - Question 49

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

The electronegativity of aluminium is similar to :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Morning) - Question 50

E.N. of Al = (1.5) ≌ Be (1.5)

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

The decreasing order of ease of alkaline hydrolysis for the following esters is :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Morning) - Question 51

More is the electrophilic character of carbonyl group of ester faster is the alkaline hydrolysis.

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

A process has ΔH = 200 Jmol–1 and ΔS = 40 JK–1mol–1. Out of the values given below, choose the minimum temperature above which the process will be spontaneous :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Morning) - Question 52

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

Which of the graphs shown below does not represent the relationship between incident light and the electron ejected form metal surface ?

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

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

Which of the following is not an example of heterogeneous catalytic reaction?

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

Then is no catalyst is required for combustion of coal.

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

The effect of lanthano id contraction in the lanthanoid series of elements by and large means :

Detailed Solution for JEE Main 2019 Question Paper with Solutions (10th January - Morning) - Question 55

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

The major product formed in the reaction given below will be :

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

The correct structure of product 'P' in the following reaction is :

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

Asn–Ser is dipeptide having following structure

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

Which hydrogen in compound (E) is easily replaceable during bromination reaction in presence of light :

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

The major product 'X' formed in the following reaction is :

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

A mixture of 100 m mol of Ca(OH)2 and 2g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. The mass of calcium sulphate formed and the concentration of OH in resulting solution, respectively, are: (Molar mass of Ca(OH)2, Na2SO4 and CaSO4 are 74, 143 and 136 g mol–1, respectively; Ksp of Ca(OH)2 is 5.5 × 10–6)

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

Ca(OH)2  + Na2SO4 → CaSO4 + 2NaOH 100 m mol 14 m mol         —    — —     — 14 m mol 28 m mol

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

Consider a triangular plot ABC with sides AB=7m, BC=5m and CA=6m. A vertical lamp-post at the mid point D of AC subtends an angle 30° at B. The height (in m) of the lamp-post is:

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

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

Let f : R → R be a function such that f(x) = x3+x2f'(1) + xf''(2)+f'''(3), x∈R. Then f(2) equal :

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

f(x) = x3 + x2 f'(1) + x f''(2) + f'''(3)
⇒f'(x) = 3x2 + 2xf'(1) + f''(x) .. ...(1)
⇒ f''(x) = 6x + 2f'(1) .....(2)
⇒ f'''(x) = 6 .....(3)
put x = 1 in equation (1) :
f'(1) = 3 + 2f'(1) + f''(2) .....(4)
put x = 2 in equation (2) :
f''(2) = 12 + 2f'(1) .....(5)
from equation (4) & (5) :
–3 – f'(1) = 12 + 2f'(1)
⇒ 3f'(1) = –15
⇒ f'(1) = –5  ⇒  f''(2) = 2  .. ..(2)
put x = 3 in equation (3) :
f''(3) = 6
∴ f(x) = x3 – 5x2 + 2x + 6
f(2) = 8 – 20 + 4 + 6 = –2

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

If a circle C passing th rough the point (4, 0) touches the circle x2 + y2 + 4x – 6y = 12 externally at the point (1, –1), then the radius of C is :

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

x2 + y2 + 4x – 6y – 12 = 0
Equation of tangent at (1, –1)
x – y + 2(x + 1) – 3(y – 1) – 12 = 0
3x – 4y – 7 = 0
∴ Equation of circle is
(x2 + y2 + 4x – 6y – 12) + λ(3x – 4y – 7) = 0
It passes through (4, 0) :
(16 + 16 – 12) + λ(12 – 7) = 0
⇒ 20 + λ(5) = 0
⇒ λ = –4
∴ (x2 + y2 + 4x – 6y – 12) – 4(3x – 4y – 7) = 0
or x2 + y2 – 8x + 10y + 16 = 0

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

In a class of 140 students numbered 1 to 140, all even numbered students opted mathematics course, those whose number is divisible by 3 opted Physics course and  theose whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is :

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

Let n(A ) = number of students opted Mathematics = 70,
n(B) = number of students opted Physics = 46,
n(C) = number of students opted Chemistry = 28,
n(A ∩ B) = 23,
n(B ∩ C) = 9,
n(A ∩ C) = 14,
n(A ∩ B ∩ C) = 4,
Now n(A ∪ B ∪ C)
= n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C)
– n(A ∩ C) + n(A ∩ B ∩ C)
= 70 + 46 + 28 – 23 – 9 – 14 + 4 = 102
So number of students not opted for any course
= Total – n(A ∪ B ∪ C)
= 140 – 102 = 38

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

The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is :

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

= 7 × 90 + 24 = 654

Total = 654 + 702 = 1356

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

Let  and  be three vectors such thatis perpendicular to Then a possible value of (λ123) is :-

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

Now check the options, option (2) is correct

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

The equation of a tangent to the hyperbola 4x2–5y2 = 20 parallel to the line x–y = 2 is :

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

slope of tangent = 1
equation of tangent

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

If the area enclosed between the curves y=kx2 and x=ky2, (k>0), is 1 square unit. Then k is:

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

Area bounded by y2 = 4ax & x2 = 4by, a, b ≠ 0 is
by using formula :

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

Let

Let S be the set of points in the interval (–4,4) at which f is not differentiable. Then S:

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

f(x) is not differentiable at x = {–2,–1,0,1,2}
⇒ S = {–2, –1, 0, 1, 2}

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

If the parabolas y2=4b(x–c) and y2=8ax have a common normal, then which one of the following is a valid choice for the ordered triad (a,b,c)

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

Normal to these two curves are y = m(x - c) - 2bm - bm3, y = mx - 4am - 2am3
If they have a common normal (c + 2b) m + bm3 = 4am + 2am3 Now (4a - c - 2b) m = (b - 2a)m3
We get all options are correct for m = 0 (common normal x-axis)
Ans. (1), (2), (3), (4)
Remark :
If we consider question as If the parabolas y2 = 4b(x - c) and y2 = 8ax
have a common normal other than x-axis, then
which one of the following is a valid choice for the ordered triad (a, b, c) ?
When m≠ 0 : (4a - c - 2b) = (b - 2a)m2

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

The sum of all values of  satisfying sin2 2θ + cos4 2θ = 3/4 is:

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

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

Let z1 and z2 be any two non-zero complex numbers such that 3|z1| = 4 |z2|. If then :

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

Now all options are incorrect
Remark :
There is a misprint in the problem actual problem should be :
" Let zand z2 be any non-zero complex number such that 3|z1| = 2|z2|.

∴ Im(z) = 0
Now option (4) is correct.

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

If the system of equations
x+y+z = 5
x+2y+3z = 9
x+3y+αz = β
has infinitely many solutions, then β–α equals:

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

for infinite solutions D = 0 ⇒ α = 5

⇒ 2 +β -15 = 0 ⇒ β-13= 0
on β = 13 we get Dy = Dz = 0
α = 5, α = 13

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

The shortest distance between the point  and the curve y = √x, (x> 0) is :

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

Let points

So minimum distance is

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

Consider the quad ra tic equa tion (c–5)x2–2cx + (c–4) = 0, c≠5. Let S be the set of all integral values of c for which one root of the equation lies in the interval (0,2) and its other root lies in the interval (2,3). Then the number of elements in S is :

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

Let f(x) = (c – 5)x2 – 2cx + c – 4
∴ f(0)f(2) < 0 .....(1)
& f(2)f(3) < 0 .....(2)
from (1) & (2)
(c – 4)(c – 24) < 0
& (c – 24)(4c – 49) < 0

∴ s = {13, 14, 15, ..... 23}
Number of elements in set S = 11

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

then k equals :

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

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

Let d∈R, and θ∈[0,2π]. If the minimum value of det(A) is 8, then a value of d is :

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

= (2 + sinθ)(2 + 2d - sinθ) - d(2 sin θ - d)
=4 + 4d – 2sinθ + 2sinθ+2dsinθ – sin2θ–2dsinθ+d2
=d2 + 4d + 4 – sin2θ
=(d + 2)2 – sin2θ
For a given d, minimum value of det(A) =  (d + 2)2 – 1 = 8
⇒ d = 1 or –5

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

If the third term in the binomial expansion of  equals 2560, then a possible value of x is :

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

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

If the line 3x + 4y – 24 = 0 intersects the x-axis at the point A and the y-axis at the point B, then the incentre of the triangle OAB, where O is the origin, is

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

7r – 24 = ±5r
2r = 24  or 12r + 24
r = 14, r = 2
then incentre is (2, 2)

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

The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is :

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

Let two observations are x1 & x2
mean =
⇒ 1 + 3 + 8 + x1 + x2 = 25
⇒ x1 + x2 = 13 ....(1)
variance

by (1) & (2)
(x1 + x2)2 – 2x1x2 = 97
or x1x2 = 36
∴ x1 : x2 = 4 : 9

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

A point P moves on the line 2x – 3y + 4 = 0. If Q (1,4) and R (3,–2) are fixed points, then the locus of the centroid of ΔPQR is a line :

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

Let the centroid of ΔPQR is (h, k) & P is (α, β), then
and
α = (3h – 4) β = (3k – 4)
Point P(α, β) lies on line 2x – 3y + 4 = 0
∴ 2(3h – 4) – 3(3k – 2) + 4 = 0
⇒ locus is 6x – 9y + 2 = 0

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

If  and then equals :

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

Given

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

The plane passing through the point (4, –1, 2) and parallel to the lines and also passes through the point :

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

Let be the normal vector to the plane passing through (4, –1, 2) and parallel to the lines L1 & L2

∴ Equation of plane is
–1(x – 4) – 1(y + 1) + 1(z – 2) = 0
∴ x + y – z – 1 = 0
Now check options

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

Let If I is minimum then the ordered pair (a, b) is :

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

Let f(x) = x2(x2 – 2)

As long as f(x) lie below the x-axis, definite integral will remain negative,
so correct value of (a, b) is (-√2,√2) for minimum of I

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

If 5, 5r, 5r2 are the l engths of the sides of a triangle, then r cannot be equal to :

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

r = 1 is obviously true.
Let 0 < r < 1
⇒ r + r2 > 1
⇒ r2 + r - 1 > 0

When r > 1

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

Consider the statement : "P(n): n2 – n + 41 is prime." Then which one of the following is true?

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

P(n) : n2 – n + 41 is prime
P(5) = 61 which is prime
P(3) = 47 which is also prime

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

Let A be a point on the line and B(3, 2, 6) be a point in the space. Then the value of µ for which the vector is parallel to the plane x -4y +3z=1 is :

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

Let point A is

and point B is (3, 2, 6)
then
which is parallel to the plane x – 4y + 3z = 1
∴ 2 + 3µ – 12 + 4µ + 12 – 15µ = 0
8µ = 2
µ = 1/4

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

For each t∈R, let [t] be the greatest integer less than or equal to t. Then,

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

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

An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered 1,2,3,...,9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is:

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

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

Let n≥2 be a natural number and 0<θ<π/2.
Then is equal to :
(Where C is a constant of integration)

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

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