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# JEE Main 2020 Question Paper with Solution (8th January - Evening)

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## 74 Questions MCQ Test Additional Study Material for JEE | JEE Main 2020 Question Paper with Solution (8th January - Evening)

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JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 1

### Find magnetic field at O.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 1

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 2

### Position of particle as a function of time is given as . Choose correct statement about and whereare velocity and acceleration of particle at time t.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 2

= ω (–sinωt cosωt + cosωt sinωt) = 0
so

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 3

### ​A Carnot engine, having an efficiency of η = 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 3

For Carnot engine using as refrigerator

It is given η = 1/10
⇒
⇒
So, Q2 = 90 J    (as W = 10 J)

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 4

Two uniformly charged solid spheres are such that E1 is electric field at surface of 1st sphere due to itself. E2 is electric field at surface of 2nd sphere due to itself. r1, r2 are radius of 1st and 2nd sphere respectively. If then ratio of potential at the surface of spheres 1st and 2nd due to their self charges is:

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JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 5

Output at terminal Y of given logic circuit.

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JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 6

Velocity of a wave in a wire is v when tension in it is 2.06 × 104 N. Find value of tension in wire when velocity of wave become v/2

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 6

= 0.515 × 104 N

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 7

n mole of He and 2n mole of O2 is mixed in a container. Then  will be

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

​A uniform solid sphere of radius R has a cavity of radius 1m cut from it if centre of mass of the system lies at the periphery of the cavity then

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

Alternative:

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

A solid sphere of mass m= 500gm is rolling without slipping on a horizontal surface. Find kinetic energy of a sphere if velocity of centre of mass is 5 cm/sec.

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

K.E. of the sphere = Translational K.E + Rotational K.E.

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 10

Two liquid columns of same height 5m and densities ρ and 2ρ are filled in a container of uniform cross sectional area. Then ratio of force exerted by the liquid on upper half of the wall to lower half of the wall is.

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JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 11

Two square plates of side 'a' are arranged as shown in the figure. The minimum separation between plates is 'd' and one of the plate is inclined at small angle α with plane parallel to another plate. The capacitance of capacitor is (given α is very small)

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 11

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 12

​In YDSE path difference at a point on screen is λ/8. Find ratio of intensity at this point with maximum
intensity.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 12

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 13

In the given circuit switch is closed at t = 0. The charge flown in time t = TC (where TC is time constant).

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JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 14

A particle is dropped from height h = 100 m, from surface of a planet. If in last 1/2 sec of its journey it covers 19 m. Then value of acceleration due to gravity that planet is :

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 14

........(i)
......... (ii)

g = 8 m/s2

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 15

A charge particle of mass m and charge q is released from rest in uniform electric field. Its graph between velocity (v) and distance travelled (x) will be :

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 15

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 16

An object is moving away from concave mirror of focal length f starting from focus. The distance of an object from pole of mirror is x. The correct graph of magnitude of magnification(m) verses distance x is:

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 16

At focus, magnification is ∞

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 17

​In full scale deflection current in galvanometer of 100 Ω resistance is 1 mA. Resistance required in series to convert it into voltmeter of range 10 V.

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JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 18

There are two identical particles A and B. One is projected vertically upward with speed √2gh from ground and other is dropped from height h along the same vertical line. Collision between them is perfectly inelastic. Find time taken by them to reach the ground after collision in terms of .

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 18

time for collision
After t1

and
at the time of collision

and height from ground
so time

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 19

​Length of a simple pendulum is 25.0 cm and time of 40 oscillation is 50 sec. If resolution of stop watch is 1 sec then accuracy is g is (in %)

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JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 20

An electron is moving initially with velocityin uniform electric field . If initial wavelength of electron is λ0 and mass of electron is m. Find wavelength of electron as a function of time.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 20

Initially m
Velocity as a function of time =
so wavelength

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 21

An asteroid of mass m (m << mE) is approaching with a velocity 12 km/s when it is at distance of 10 R from the centre of earth (where R is radius of earth). When it reaches at the surface of Earth, its velocity is (Nearest Integer) in km/s.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 21

= 16.028 km/s

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 22

In H–spectrum wavelength of 1st line of Balmer series is λ = 6561Å. Find out wavelength of 2nd line of same series in nm.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 22

= 4860 Å = 486 nm

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 23

There are three containers C1, C2 and C3 filled with same material at different constant temperature. When we mix then for different volume then we get some final temperature as shown in the below table. So find value of final temperature θ as shown in the table.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 23

............(i)
.......(ii)
.........(iii)
and θ1 + θ2 + θ3 = (1 + 1 + 1)θ....(iv)
from (1) + (2) + (3)
1 + 3θ2 + 3θ3 = 450
⇒ θ1 + θ2 + θ3 = 150
from (4) equation 150 = 3θ
θ = 50ºC

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 24

Two batteries (connected in series) of same emf 10 V of internal resistances 20Ω and 5Ω are connected to a load resistance of 30Ω. Now an unknown resistance x is connected in parallel to the load resistance. Find value of x so that potential drop of battery having internal resistance 20Ω becomes zero.

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*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 25

An EMW is travelling along z-axis.

& Frequency of wave is 25 Hz, then electric field in volt/m.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 25

E/B = c
E = B × c
= 15 N/c

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 26

​Correct bond energy order of following is

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JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 27

Determine Bohr's radius of Li2+ ion for n = 2. Given (Bohr's radius of H-atom = a0)

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For Li2+

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 28

Given the following reaction sequence

A & B are respectively

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JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 29

​Correct order of magnetic moment (spin only) for the following complexes
(a) [Pd(PPh3)2Cl2]
(b) [Ni(CO)4
(c) [Ni(CN)4]2–
(d)[Ni(H2O)6]2+

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 29

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 30

​Determine total number of neutrons in three isotopes of hydrogen.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 30

Number of neutrons

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 31

Compare Ea (activation energy) for a, b, c and d.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 31

log k = log A –
slope

⇒ Eb > Ec > Ed > E

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 32

​Which of the following exhibit both Frenkel & Schottky defect?

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 32

Only AgBr can exhibit both Schottky and Frenkel defect.

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 33

​Given:

Basicity of B is:

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 33

Basicity = 1

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 34

​Which reaction does not occurs in the blast furnace in the metallurgy of Fe
(A) CaO + SiO2 → CaSiO3
(B) Fe2O3 + CO → Fe3O4 + CO2
(C) FeO + SiO2 → FeSiO3
(D)

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 34

Theory based

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 35

Correct order of radius of elements is:
C, O, F, Cl, Br

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JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 36

​Amongs the following which will show geometrical isomerism.
(a) [Ni(NH3)5Cl]+
(b) [Ni(NH3)4ClBr]
(c) [Ni(NH3)3Cl]+
(d) [Ni(NH3)2(NO2)2]

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 36

Ma4bc can show 2 G.I.
Ma2b2 can show 2 G.I.
(Square planar)

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 37

​Assertion: pH of water increases on increasing temperature.
Reason: H2O → H+ + OH is an exothermic process.

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 38

Assertion: It has been found that for hydrogenation reaction the catalytic activity increases from group-5 to group-11 metals with maximum activity being shown by groups 7-9 elements of the periodic table.
Reason: For 7-9 group elements adsorption rate is maximum.

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 39

The major product of the following reactions is

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 39

(Aromatic)

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 40

Find the final major product of the following reactions

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 40

On prototaoon, OH dispatches as H2O making a carbocation. The H+ will shift to 3° carbon for a more stable carbocation and then elimination will occur to form.

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 41

There are two compounds A and B of molecular formula C9H18O3. A has higher boiling point than B. What are the possible structures of A and B?

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 41

In (A), extensive inter-molecular H-bonding is possible while in (B) there is no Inter-molecular H-bonding.

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 42

Kjeldahl method cannot be used for :

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 42

Kjeldahl method is not applicable to nitro or diazo groups present in the ring, as nitrogen atom can't be converted to ammonium sulfate under the reaction conditions.

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 43

A compound X that adds 2 hydrogen molecules on hydrogenation. The compound X also gives 3 - oxohexanedioic acid on oxidative ozonolysis. The compound 'X' is:

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 43

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 44

Formation of Bakelite follows:

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 44

Formation of Bakelite follows electrophilic substitution reaction of phenol with formaldehyde followed by condensation.

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 45

​Products formed by hydrolysis of maltose are

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 45

Maltose on hydrolysis gives 2 moles of α-D-glucose.

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 46

Temperature of 4 moles of gas increases from 300 K to 500 K find 'Cv' if ΔU = 5000 J.

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*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 47

Given :
Determine at equilibrium

For cell reaction Sn | Sn 2+ || Pb 2+ | Pb
take = 0.06 V

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 47

At Equilibrium state.
Ecell = 0 ; Eºcell = 0.01 V

= 101/3 = 2.1544

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 48

Given following reaction,
NaClO3 + Fe → O2 + FeO + NaCl
In the above reaction 492 L of O2 is obtained at 1 atm & 300 K temperature.
Determine mass of NaClO3 required (in kg).
(R = 0.082 L atm mol–1 K–1 )

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 48

mol of NaClO3 = mol of O2
mol of O
= 20 mL
mass of NaClO3= 20 × 106.5 = 2130 g

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 49

Complex [ML5] can exhibit trigonal bipyramidal and square pyramidal geometry. Determine total number of 180º, 90º & 120º L-M-L bond angles.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 49

∠120º = 3; ∠90º = 6; ∠180º = 1
→Total = 10

∠90º = 8; ∠180º = 2 ⇒ Total = 10

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 50

How many atoms lie in the same plane in the major product (C)?

(Where A is the alkyne of lowest molecular mass)

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 50

Number of atoms in one plane = 13

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 51

Let  and is nonzero vector and  find

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JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 52

​Let coefficient of x4 and x2 in the expansion of is α and β then α -β is equal to

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 52

α = – 96 and β = 36
∴ α - β = -132

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 53

Differential equation of x= 4b(y + b), where b is a parameter, is

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 53

2x = 4by'
So. differential equation is

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 54

​Image of (1, 2, 3) w.r.t a plane is then which of the following points lie on the plane

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 54

d.r of normal to the plane
10/3, 10/3, 10/3
1, 1, 1
midpoint of P and Q is
equation of plane x + y + z = 1

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 55

is equal to

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 55

Using L’Hospital

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 56

Let P be the set of points (x, y) such that x2 ≤ y ≤ – 2x + 3. Then area of region bounded by points in set P is

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 56

Point of intersection of y = x2 & y = – 2x + 3 is
obtained by x2 + 2x – 3 = 0
⇒ x = - 3, 1
So, Area =
= 12 + 8 – 28/3 = 32/3

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 57

Let f(x) = → R then range of f(x) is (where [ . ] denotes greatest integer function)

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 57

∴ f(x) is a decreasing function
∴

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 58

​Let A = and I = then value of 10 A–1 is –

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 58

Characteristics equation of matrix ‘A’ is

x2 – 6x – 10 = 0
∴ A2 - 6A - 10I = 0
⇒ 10A–1 = A – 6I

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 59

Solution set of 3x(3x –1) + 2 = |3x –1| + |3x – 2| contains

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 59

Let 3x = t
t(t –1) + 2 = |t –1| + |t –2|
t2 – t + 2 = |t –1| + |t –2|

are positive solution
t = a
3x = a
x = log3 a so singleton set

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 60

​Mean and variance of 20 observation are 10 and 4. It was found, that in place of 11, 9 was taken by mistake find correct variance.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 60

................... (1)
................... (2)
= 104 × 20 = 2080
Actual mean =
Variance =
= 106 – 102.01 = 3.99

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 61

λx + 2y + 2z = 5
2λx + 3y + 5z = 8
4x + λy + 6z = 10
for the system of equation check the correct option.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 61

D = (λ + 8) ( 2 - λ)
for λ = 2

= 5[18 – 10] – 2 [48 – 50] + 2 (16 – 30]
= 40 + 4 – 28 ≠ 0
No solutions for λ = 2

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 62

For an A.P. T10 = 1/20; T20 = 1/10 Find sum of first 200 term.

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 62

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 63

Let α = and a = . If a and b are roots of quadratic equation then quadratic equation is

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 63

α = ω, b = 1 + ω3 + ω6 + .......... = 101
a = (1 + ω) (1 + ω2 + ω4 + ......... ω198 + ω200)

Equation: x2 – (101 +1)x + (101) × 1 = 0
⇒ x2 - 102x + 101 = 0

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 64

Let f(x) is a three degree polynomial for which f '(–1) = 0, f ''(1) = 0, f(–1) = 10, f(1) = 6 then local minima of f(x) exist at

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 64

Let f(x) = ax3 + bx2 + cx + d
a = 1/4     d = 35/4
b = -3/4    c = -9/4
⇒ f(x) = a(x3 – 3x2 – 9x) + d
f ' (x) = 3/4 (x2 – 2x – 3)
⇒ f ' (x) = 0 ⇒ x = 3, –1

local minima exist at x = 3

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 65

Let A and B are two events such that P(exactly one) = 2/5, P(A ∪ B) = then P(A ∩ B) =

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 65

P(exactly one) = 2/5
⇒ P(A) + P(B) - 2P(A ∩ B) = 2/5
P (A ∪ B) = 1/2
⇒ P(A) + P(B) – P(A ∩ B) = 1/2
∴ P(A ∩ B) = 1/2 - 2/5 = (5-4)/10 = 1/10

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 66

Let I = then

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 66

1/3 < I < 1/√8

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 67

Normal at (2, 2) to curve x2 + 2xy – 3y2 = 0 is L. Then perpendicular distance from origin to line L is

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 67

x2 + 2xy – 3y2 = 0
x2 + 3xy – xy – 3y2 = 0
(x – y) (x + 3y) = 0
x – y = 0 x + 3y = 0
(2, 2) satisfy x – y = 0
Normal : x + y = λ
λ = 4
Hence x + y =4
perpendicular distance from origin = = 2√2

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 68

Which of the following is tautology-

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 68

(~p ∧ q) → (p ν q)
~{(~p ∧ q) ∧ (~p ∧ ~q)}
~{~p ∧ f}

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 69

​If a hyperbola has vertices (±6, 0) and P(10, 16) lies on it, then the equation of normal at P is

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 69

Vertex is at (±6, 0)
∴ a = 6
Let the hyperbola is
Putting point P(10, 16) on the hyperbola

⇒ b= 144
∴ hyperbola is
∴ equation of normal is
∴ putting we get 2x + 5y = 100

JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 70

If y = mx + c is a tangent to the circle (x – 3)2 + y2 = 1 and also the perpendicular to the tangent to the circle x2 + y2 = 1 at , then

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 70

Slope of tangent to x2 + y2 = 1 at
x2 + y2 = 1
2x + 2yy' = 0

y = mx + c is  tangent of x2 + y2 = 1
so m = 1
y = x + c
now distance of (3, 0) from y = x + c is
= 1
c2 + 6c + 9 = 2
c2 + 6c + 7 = 0

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 71

Let  = 1/7 and = 1/√10 where α, β ∈ . Then tan (α + 2β) is equal to

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 71

and = 1/√10
tan α = 1/7
sin β = 1/√10
tanβ = 1/3
tan2β =
tan (α + 2β) =

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 72

The number of four letter words that can be made from the letters of word "EXAMINATION" is

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 72

EXAMINATION
2N, 2A, 2I, E, X, M, T, O
Case I  All are different so = 8.7.6.5 = 1680
Case II 2 same and 2 different so
= 3.21.12 = 756
Case III 2 same and 2 same so = 3.6 = 18
Total = 1680 + 756 + 18 = 2454

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 73

Let the line y = mx intersects the curve y2 = x at P and tangent to y2 = x at P intersects x-axis at Q. If area (ΔOPQ) = 4, find m (m > 0)

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 73

2ty = x + t2
Q(–t2, 0)
= 4
|t|3 = 8
t = ± 2 (t > 0)
m = 2

*Answer can only contain numeric values
JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 74

is equal to

Detailed Solution for JEE Main 2020 Question Paper with Solution (8th January - Evening) - Question 74

1/4 [1568 + 420 + 28] = 504

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