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


74 Questions MCQ Test Additional Documents and Tests for JEE | JEE Main 2020 Question Paper with Solution (8th January - Evening)


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This mock test of JEE Main 2020 Question Paper with Solution (8th January - Evening) for JEE helps you for every JEE entrance exam. This contains 74 Multiple Choice Questions for JEE JEE Main 2020 Question Paper with Solution (8th January - Evening) (mcq) to study with solutions a complete question bank. The solved questions answers in this JEE Main 2020 Question Paper with Solution (8th January - Evening) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this JEE Main 2020 Question Paper with Solution (8th January - Evening) exercise for a better result in the exam. You can find other JEE Main 2020 Question Paper with Solution (8th January - Evening) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

Find magnetic field at O.

Solution:





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.

Solution:


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

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

Solution:

For Carnot engine using as refrigerator

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

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:
Solution:

QUESTION: 5

Output at terminal Y of given logic circuit. 
Solution:

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
Solution:





= 0.515 × 104 N

QUESTION: 7

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


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
Solution:




Alternative: 

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.
Solution:


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

K =  Radius of gyration

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.
Solution:


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)
Solution:


QUESTION: 12

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


QUESTION: 13

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





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 :
Solution:



Area of shaded trapezium
........(i)
......... (ii)



g = 8 m/s2

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 : 
Solution:


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:
Solution:

At focus, magnification is ∞ 

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.
Solution:

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 .
Solution:


time for collision
After t1

and 
at the time of collision 

and height from ground 
so time 

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 %)
Solution:

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.
Solution:

Initially m 
Velocity as a function of time = 
so wavelength 

*Answer can only contain numeric values
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.

Solution:


= 16.028 km/s

*Answer can only contain numeric values
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.

Solution:



= 4860 Å = 486 nm

*Answer can only contain numeric values
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.

Solution:

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

*Answer can only contain numeric values
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.

Solution:



*Answer can only contain numeric values
QUESTION: 25

An EMW is travelling along z-axis.

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

Solution:

E/B = c
E = B × c
= 15 N/c

QUESTION: 26

​Correct bond energy order of following is
Solution:

QUESTION: 27

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


For Li2+ 

QUESTION: 28

Given the following reaction sequence

A & B are respectively 
Solution:

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+
Solution:



QUESTION: 30

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

Number of neutrons

QUESTION: 31


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

log k = log A – 
slope

 
⇒ Eb > Ec > Ed > Ea 

QUESTION: 32

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

Only AgBr can exhibit both Schottky and Frenkel defect.

QUESTION: 33

​Given:
 
Basicity of B is: 
Solution:

Basicity = 1

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) 
Solution:

Theory based

QUESTION: 35

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

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]
Solution:

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

QUESTION: 37

​Assertion: pH of water increases on increasing temperature.
Reason: H2O → H+ + OH– is an exothermic process.
Solution:
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.
Solution:
QUESTION: 39

The major product of the following reactions is
Solution:

(Aromatic)

QUESTION: 40

Find the final major product of the following reactions
Solution:


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.

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?
Solution:

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

QUESTION: 42

Kjeldahl method cannot be used for :
Solution:

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.

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:
Solution:

QUESTION: 44

Formation of Bakelite follows:
Solution:

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


 

QUESTION: 45

​Products formed by hydrolysis of maltose are
Solution:

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

*Answer can only contain numeric values
QUESTION: 46

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

Solution:

*Answer can only contain numeric values
QUESTION: 47

Given : 
Determine at equilibrium

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

Solution:

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



= 101/3 = 2.1544

*Answer can only contain numeric values
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 )

Solution:

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

*Answer can only contain numeric values
QUESTION: 49

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

Solution:


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

∠90º = 8; ∠180º = 2 ⇒ Total = 10 

*Answer can only contain numeric values
QUESTION: 50

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

(Where A is the alkyne of lowest molecular mass)

Solution:



Number of atoms in one plane = 13

QUESTION: 51

Let  and is nonzero vector and  find 
Solution:


QUESTION: 52

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


α = – 96 and β = 36
∴ α - β = -132

QUESTION: 53

Differential equation of x2 = 4b(y + b), where b is a parameter, is
Solution:

2x = 4by' 
So. differential equation is 

QUESTION: 54

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

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 

QUESTION: 55

is equal to  
Solution:

Using L’Hospital

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 
Solution:

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

QUESTION: 57

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


∴ f(x) is a decreasing function
∴ 

QUESTION: 58

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

Characteristics equation of matrix ‘A’ is

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

QUESTION: 59

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

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

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.
Solution:

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

QUESTION: 61

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


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  

QUESTION: 62

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




QUESTION: 63

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

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 

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  
Solution:

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  

QUESTION: 65

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

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

QUESTION: 66

Let I = then
Solution:



1/3 < I < 1/√8

QUESTION: 67

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

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

QUESTION: 68

Which of the following is tautology-
Solution:

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

QUESTION: 69

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

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

⇒ b2 = 144
∴ hyperbola is  
∴ equation of normal is 
∴ putting we get 2x + 5y = 100 

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
Solution:

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
QUESTION: 71

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

Solution:

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

*Answer can only contain numeric values
QUESTION: 72

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

Solution:

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
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) 

Solution:


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

*Answer can only contain numeric values
QUESTION: 74

is equal to 

Solution:




1/4 [1568 + 420 + 28] = 504

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