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# JEE Main 2020 Question Paper with Solution (9th January - Morning)

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 1

### Kinetic energy of the particle is E and it's De–Broglie wavelength is λ. On increasing it's KE by ΔE, it's new De–Broglie wavelength becomes λ/2 . Then ΔE is

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 1

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 2

### The dimensional formula of is

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 3

### Two immiscible liquids of refractive index √2 and 2√2 are filled with equal height h in a vessel. Then apparent depth of bottom surface of the container given that outside medium is air:

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 4

Three identical solid spheres each having mass 'm' and diameter 'd' are touching each other as shown in figure. Calculate ratio of moment of inertia about an axis (perpendicular to plane of paper) passing through point P and B as shown in figure. Given P is centroid of triangle ABC.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 4

Now ratio = 13 / 23

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 5

A solid sphere having radius R and Uniform charge density ρ has a cavity of radius R/2 as shown in figure. Find the ratio of magnitude of electric field at point A and B i.e.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 5

For a solid sphere

Electric field at point B = EB = E1A + E2A
E1A = Electric Field Due to solid sphere of radius R at point B =
E2A = Electric Field Due to solid sphere of radius R/2 (which having charge density –ρ)
E2A = R/2

EB =  E1A + E2A  =

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 6

Consider an infinitely long current carrying cylindrical straight wire having radius 'a'. Then the ratio of magnetic field at distance a/3 and 2a from axis of wire is.

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 7

Find current in the wire BC.

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 8

Two electromagnetic waves are moving in free space whose electric field vectors are given by  . A charge q is moving with velocity .Find the
net Lorentz force on this charge at t = 0 and when it is at origin.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 8

Magnetic field vectors associated with this electromagnetic wave are given by

by putting the value of
The net Lorentz force on the charged particle is
at t = 0 and at x = y = 0
t = 0 , x = y = 0

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 9

Two ideal di-atomic gases A and B. A is rigid, B has an extra degree of freedom due to vibration. Mass of A is m and mass of B is m/4. The ratio of molar specific heat of A to B at constant volume is :

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 9

Molar heat capacity of A at constant volume = 5R/2
Molar heat capacity of B at constant volume = 7R/2
Dividing both

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 10

An ideal liquid (water) flowing through a tube of non uniform cross section area at A and B are 40 cm2 and 20 cm2 respectively. If  pressure difference between A & B is 700 N/m2 then volume flow rate is :

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 10

using equation of continuity
40 VA = 20 VB
2VA =  VB
Using Bernoullies equation

Volume flow rate = 20 × 100 × VB = 2732 cm3/s

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 11

A screw gauge adv ances by 3mm in 6 rotations. There are 50 divisions on circular scale. Find least  count of screw gauge ?

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 11

Pitch = 3/6 = 0.5 mm
L.C. = =  = 0.01 mm = 0.001 cm

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 12

A telescope of aperture diameter 5m is used to observe the moon from the earth. Distance between the moon and earth is 4 × 105 km. Determine the minimum distance between two points on the moon's surface which can be resolved using this telescope. (Wave length of light is 5893 Å).

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 12

distance = O1O2 = dθ

distance = O1O2≈ 57.5 m
∴ answer from options = 60m
(minimum distance)

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 13

A particle of mass m is revolving around a planet in a circular orbit of radius R. At the instant the particle has velocity another particle of mass m/2 moving at velocity collides perfectly in-elastically with the first particle. The new path of the combined body will take is

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 13

Conserving momentum

vf < vorb (= v) thus the combined mass will go on to an elliptical path.

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 14

Two particles of same mass 'm' moving with velocities and  collide in-elastically. Find the loss in kinetic energy.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 14

Conserving momentum
on solving

Change in K.E

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 15

Three wav es of same intensity (I0) having initial phases rad respectively interfere at a point.Find the resultant Intensity

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 16

Particle moves from point A to point B along the line shown in figure under the action of force. . Determine the work done on the particle by in moving the particle from point A to point B

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 16

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 17

For the given P-V graph for an ideal gas, chose the correct V- T graph. Process BC is adiabatic.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 17

For process A – B Volume is constant
PV = nRT ; as P increases T increases
For process B – C
PVγ = Constant
TVγ-1 = Constant
For process C – A pressure is constant
V = kT

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 18

Given. Find vector parallel to electric field at position
[Note that ]

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 18

Since
must be antiparallel to
So ,
where λ is a arbitrary positive constant
Now

so

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 19

Which of the following statements are correct for moving charge as shown in figure.

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 20

Photons of wav elength 6556 Å falls on a metal surface. If ejected electrons with maximum K.E. moves in magnetic field of 3 × 10–4 T in circular orbit of radius 10–2m, then work function of metal surface is

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 20

= 1.1 eV

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 21

A rod of length 1 m is released from rest as shown in the figure below.

If ω of rod is √n at the moment it hits the ground, then find n.

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*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 22

If reversible voltage of 100 V is applied across an inductor, current in it reduces from 0.25A to 0A in 0.025ms. Find inductance of inductor (in mH).

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 22

∴ L = 100 × 10–4 H
= 10 mH

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 23

A wire of length l = 3m and area of cross section 10–2cm2 and breaking stress 48×107N/m2 is attached with block of mass 10kg. Find the maximum possible value of angular velocity with which block can be moved in circle with string fixed at one end.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 23

Solving

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 24

Position of a particle as a function of time is given as x2 = at2 + 2bt + c, where a, b, c are constants. Acceleration of particle varies with x–n then value of n is.

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 25

In the given circuit both diodes having zero forward resistance and built-in potential of 0.7 V. Find the potential of point E in volts.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 25

Let VB = 0
Right diode is reversed biased and left diode is forward biased
∴ VE = 12.7 – 0.7
= 12 Volt

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 26

Determine wavelength of electron in 4th Bohr's orbit ?

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 27

Which of the following species have one unpaired electron each?

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 28

For Br2(l) Enthalpy of atomisation = x kJ/mol Bond dissociation enthalpy of bromine = y kJ/mole
then

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 28

ΔHatomisation = ΔHvap + Bond energy
Hence x > y

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 29

Which of the following oxides are acidic, Basic Amphoteric Respectively.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 29

Non-metal oxides are acidic in nature alkali metal oxides are basic in nature Al2O3 is amphoteric.

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 30

Complex Cr(H2O)6Cln shows geometrical isomerism and also reacts with AgNO3 solution.
Given : Spin only magnetic moment = 3.8 B.M.
What is the IUPAC name of the complex.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 30

Cr(H2O)6Clncomplex)spin = 3.8 B.M.
From data of magnetic moment oxidation number of Cr should be +3
Hence complex is Cr(H2O)6Cl3.
Complex shows geometrical isomerism therefore formula of complex is [Cr(H2O)4Cl2]Cl . 2H2O.
It's IUPAC Name: Tetraaquadichloridochromium(III) chloride dihydrate.

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 31

The electronic configuration of bivalent Europium and trivalent cerium respectively is:  (Atomic Number : Xe = 54, Ce = 58, Eu = 63)

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 31

Eu2+ : [Xe]4f7
Ce3+ : [Xe]4f1

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 32

Ksp of PbCl2 = 1.6 × 10–5
On mixing 300 mL, 0.134M Pb(NO3)2(aq.) + 100 mL, 0.4 M NaCl(aq.)

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 32

= 0.105 × 10–2
=1.005 ×10–3
Q > Ksp

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 33

Which of the following can not act as both oxidising and reducing agent ?

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 33

As in H3PO4 Phosphorous is present it's maximum oxidation number state hence it cannot act as reducing agent.

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 34

​First Ionisation energy of Be is higher than that of Boron.
Select the correct statements regarding this
(i) It is easier to extract electron from 2p orbital than 2s orbital
(ii) Penetration power of 2s orbital is greater than 2p orbital
(iii) Shielding of 2p electron by 2s electron
(iv) Radius of Boron atom is larger than that of Be

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 34

Theory Based

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 35

[PdFClBrI]2– Number of Geometrical Isomers = n. For [Fe(CN)6]n–6, Determine the spin only magnetic moment and CFSE (Ignore the pairing energy)

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 35

Number of Geometrical Isomers in square planar [PdFClBrI]2– are = 3
Hence, n = 3

Fe3+ = 3d5,
According to CFT configuration is

CFSE = - 0 .4 Δ0 x nt2g + 0 .6 Δ0 x neg = -0.4 Δ0 × 5 = -2.0Δ0

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 36

A can reduce BO2 under which conditions.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 36

A + BO---> B + AO2
ΔG = -ve
Only above 1400°C

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 37

Rate of reaction in absence of catalyst at 700 K is same as in presence of catalyst at 500 K. If catalyst decreases activation energy barrier by 30 kJ/mole, determine activation energy in presence of catalyst. (Assume 'A' factor to be same in both cases)

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 37

Activation energy of the catalysed reaction = 105 – 30 = 75 kJ/mole

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 38

A substance 'X' having low melting point, does not conduct electricity in both solid and liquid state. 'X' can be :

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 38

CCl4 ---> Non-conductor in solid and liquid phase.

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 39

The major product for above sequence of reaction is :

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 40

Which of the following can give highest yield in Friedel craft reaction?

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Aniline form anilinium complex with lewis acid so phenol is most reactive among the given compounds for electrophilic substitution reaction.

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 41

What will be the major product ?

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 42

Which of the following is correct order for heat of combustion?

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 42

In isomers of hydrocarbon heat of combustion depends upon their stabilities. As the stability increases heat of combustion decreases.
Stability order

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 43

Write the correct order of basicity.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 43

Basicity is inversely proportional to electronegativity.

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 44

A, B, C, and D are four artificial sweetners.
(i) A & D give positive test with ninhydrin.
(ii) C  form precipitate with AgNO3 in the lassaigne extract of the sugar.
(iii) B & D give positive test with sodium nitroprusside.
Correct option is :`

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 44

A – Aspartame

B – Saccharine

C – Sucralose

D – Alitame

(i) A & D give positive test with ninhydrin because both have free carboxylic and amine groups.
(ii) C form precipitate with AgNO3 in the lassaigne extract of the sugar because it has chlorine atoms.
(iii) B & D give positive test with sodium nitroprusside because both have sulphur atoms.

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 45

Predict the compound (P) on the basis of above sequence of the reactions?
Where compound (P) gives positive Iodoform test.

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 46

Given a solution of HNO3 of density 1.4 g/mL and 63% w/w. Determine molarity of HNO3 solution.

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*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 47

Determine degree of hardness in term of ppm of CaCO3 of 10–3 molar MgSO4 (aq).

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10–3 molar MgSO4 = 10–3 moles of MgSO4 present in 1 L solutions.

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 48

Determine the amount of NaCl to be dissolved in 600g H2O to decrease the freezing point by 0.2°C
Given : kf of H2O = 2 k-m–1 density of H2O(l) = 1 g/ml

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 48

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 49

On passing a particular amount of electricity in AgNO3 solution, 108 g of Ag is deposited. What will be the volume of O2(g) in litre liberated at 1 bar, 273K by same quantity of electricity?

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 49

1F charge is required to deposit 1 mole of Ag

2F charge deposit → 1/2 mole
1F charge will deposit → 1/4 mole

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 50

Find percentage nitrogen by mass in Histamine ?

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 50

Structure of Histamine is
Molecular formula of Histamine is C5H9N3
Molecular mass of Histamine is 111
Percentage nitrogen by mass in Histamine =

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 51

Find the number of solution of log1/2 |sinx| = 2 – log1/2 |cosx| , x ∈ [0,2π]

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 51

log1/2 |sinx| = 2 – log1/2 |cosx|
log1/2 |sinx cosx| = 2
|sinx cosx| = 1/4
sin2x = ± 1/2

Number of solution = 8

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 52

If e1 and e2 are eccentricities of and , respectively and if the point (e1, e2) lies on ellipse 15x2 + 3y2 = k. Then find value of k

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 52

∴ k = 16

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 53

Find integration

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 53

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 54

If  = 1, |z| = 5/2 then value of |z + 3i| is

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JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 55

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 55

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 56

Value of is

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 56

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 57

Find the value of

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 57

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 58

If f(x) = a + bx + cx2 where a, b, c ∈ R then is

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 58

f(1) = a + b + c
f(0) = a

Now

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 59

If number of 5 digit numbers which can be formed without repeating any digit while tenth place of all of the numbers must be 2 is 336 k find value of k

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 59

Number of numbers
= 8 x 8 x 7 x 6 = 2688 = 336k => k = 8

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 60

A (3,–1), B(1,3), C(2,4) are vertices of ΔABC if  D is centroid of ΔABC and P is point of intersection of lines x + 3y - 1 = 0 and 3x - y + 1 = 0 then which of the following points lies on line joining D and P

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 60

D (2,2)
Point of intersection P
and equation of line DP
8x – 11y + 6 = 0

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 61

If f(x) is twice differentiable and continuous function in x ∈ [a,b] also f'(x) > 0 and f ''(x)  < 0 and c ∈ (a,b)
then is greater than

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 61

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 62

If plane
x + 4y – 2z = 1
x + 7y – 5z = β
x + 5y + αz = 5
intersects in a line (R × R × R) then α + β is equal to

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 62

(7α + 25) - (4α + 10) + (-20 + 14) = 0
3α + 9 = 0
α = - 3
Also Dz = 0  = 0
1(35 – 5β) - (15) + 1 (4β - 7) = 0
β = 13

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 63

For observations xi given and . If mean and variance of observations is λ & μ respectively then ordered pair (λ, μ) is

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 63

Mean
∴ λ = {Mean (xi – 5)} + 2 = 3
μ = var (xi – 5) =

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 64

In a bag there are 20 cards 10 names A and another 10 names B. Cards are drawn randomly one by one with replacement then find probability that second A comes before third B.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 64

AA + ABA + BAA + ABBA + BBAA + BABA

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 65

The negation of ‘ √5 is an integer or 5 is an irrational number’ is

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 65

√5 is not an integer and 5 is not an irrational Number ~ (p v q) = ~ p ∧ ~ q

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 66

If a circle touches y-axis at (0, 4) and passes through (2, 0) then which of the following can not be the tangent to the circle

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 66

equation of family of circle
(x – 0)2 + (y – 4)2 + λx = 0
passes (2, 0)
4 + 16 + 2λ = 0
λ= - 10
x2 + y2 – 10x – 8y + 16 = 0
centre (5, 4). R = = 5

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 67

If f'(x) = tan–1 (secx + tanx), x ∈ and f(0) = 0 then the value of f(1) is

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 67

f'(x) = tan–1 (secx + tanx) =
tan–1

= tan–1

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 68

A sphere of 10cm radius has a uniform thickness of ice around it. Ice is melting at rate 50cm3/min when thickness is 5cm then rate of change of thickness

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 68

Let thickness = x cm
Total volume v = 4/3 π(10 + x)3

Given dv/dt = 50cm3/min
At x = 5cm
50 = 4π (10 + 5)2dx/dt

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 69

Find number of real roots of equation e4x + e3x – 4e2x + ex + 1 = 0 is

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 69

Let ex = t ∈ (0, ∞)
Given equation t4 + t3 – 4t2 + t + 1 = 0

JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 70

If A = B = adj(A) and C = 3A then is

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 70

= 13 + 1 – 8 = 6

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 71

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 71

∴ at   x = 2,  y = 0
∴ 0 = 3 (2+1+c) => c = -3
at x = 3 , y = 3

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 72

Function is continuous at x = 0, find a + 2b.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 72

LHL = a + 3
f(0) = b
RHL =
∴ a = -2   b = 1 ∴ a + 2b = 0

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 73

Find the coefficient of x4 in (1 + x + x2)10

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 73

General term
for coefficient of  x4
=> β + 2γ = 4
γ = 0, β = 4 , α = 6

γ = 1, β = 2 , α = 7

γ = 2, β = 0 , α = 8

Total = 615

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 74

and are coplanar vectors and then value of λ is

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 74

= 0
a + 1 + a + a = 0 => a =

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 75

If points A (2, 4, 0), B(3, 1, 8), C(3, 1, –3), D(7, –3, 4) are four points then projection of line segment AB on line CD.

Detailed Solution for JEE Main 2020 Question Paper with Solution (9th January - Morning) - Question 75

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