JEE Main Question Paper 2020 With Solutions (7th January - Morning)


75 Questions MCQ Test JEE Main Mock Test Series 2020 & Previous Year Papers | JEE Main Question Paper 2020 With Solutions (7th January - Morning)


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This mock test of JEE Main Question Paper 2020 With Solutions (7th January - Morning) for JEE helps you for every JEE entrance exam. This contains 75 Multiple Choice Questions for JEE JEE Main Question Paper 2020 With Solutions (7th January - Morning) (mcq) to study with solutions a complete question bank. The solved questions answers in this JEE Main Question Paper 2020 With Solutions (7th January - Morning) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this JEE Main Question Paper 2020 With Solutions (7th January - Morning) exercise for a better result in the exam. You can find other JEE Main Question Paper 2020 With Solutions (7th January - Morning) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

A block of mass m is suspended from a pulley in form of a circular disc of mass m & radius R. The system is released from rest, find the angular velocity of disc when block has dropped by height h. (there is no slipping between string & pulley)

Solution:


v = ωR (no slipping)



 

QUESTION: 2

Three point masses 1kg, 1.5 kg, 2.5 kg are placed at the vertices of a triangle with sides 3cm, 4cm and 5cm as shown in the figure. The location of centre of mass with respect to 1kg mass is :

Solution:

Take 1kg mass at origin


QUESTION: 3

In a single slit diffraction set up, second minima is observed at an angle of 60°. The expected position of first minima is  

Solution:

For 2nd minima
d sinθ = 2λ
sinθ = (given)
......(1)
So for 1st minima is
d sinθ = λ
sinθ = (from equation (i))
θ = 25.65° (from sin table)
θ ≈ 25°

QUESTION: 4

There are two infinite plane sheets each having uniform surface charge density +σ C/m2. They are inclined to each other at an angle 30° as shown in the figure. Electric field at any arbitrary point P is:
   

Solution:


QUESTION: 5

A parallel plate capacitor with plate area A & plate separation d is filled with a dielectric material of dielectric constant given by k = k0(1 + αx). Calculate capacitance of system: (given αd << 1)

Solution:

Capacitance of element 

Capacitance of element 

Given αd << 1

QUESTION: 6

A long solenoid of radius R carries a time dependent current I = I0 t(1 – t). A ring of radius 2R is placed coaxially near its centre. During the time interval 0 ≤ t ≤ 1, the induced current IR and the induced emf VR in the ring vary as:

Solution:


and 

 

QUESTION: 7

If 10% of intensity is passed from analyser, then, the angle by which analyser should be rotated such that transmitted intensity becomes zero. (Assume no absorption by analyser and polarizer).

Solution:

So θ > 45° and 90 – θ < 45º so only one option is correct i.e. 18.4º
angle rotated should be = 90° – 71.6° = 18.4°

QUESTION: 8

Three moles of ideal gas A with  is mixed with two moles of another ideal gas B with . The of mixture is (Assuming temperature is constant)

Solution:


on rearranging we get




γmix = 1.42

QUESTION: 9

Given magnetic field equation is B = 3 × 10–8 sin(ωt + kx + φ)then appropriate equation for electric field (E) will be :

Solution:

 (speed of light in vacuum)
E0 = B0C = 3 × 10–8 × 3 × 108
= 9 N/C
So E = 9 sin (ωt + kx + φ)

QUESTION: 10

There is a LCR circuit, If it is compared with a damped oscillation of mass m oscillating with force constant k and damping coefficient 'b'. Compare the terms of damped oscillation with the devices in LCR circuit.

Solution:

In damped oscillation 
ma + bv + kx = 0


In the circuit

Comparing equation (i) and (ii)
m = L, b = R, k = 1/c

QUESTION: 11

A lift can hold 2000kg, friction is 4000N and power provided is 60HP. (1 HP = 746W) Find the maximum speed with which lift can move up.

Solution:

QUESTION: 12

A H–atom in ground state has time period T = 1.6 × 10–16 sec. find the frequency of electron in first excited state 

Solution:


QUESTION: 13

Magnification of compound microscope is 375. Length of tube is 150mm. Given that focal length of objective lens is 5mm, then value of focal length of eyepiece is:

Solution:

Case-I
If final image is at least distance of clear vision


Case-II
If final image is at infinity

fe = 22 mm 

QUESTION: 14

1 litre of a gas at STP is expanded adiabatically to 3 litre. Find work done by the gas. Given γ = 1.40 and 31.4= 4.65

Solution:


now work done 
Closest ans is 90.5 J

QUESTION: 15

A string of length 60 cm, mass 6gm and area of cross section 1mm2 and velocity of wave 90m/s. Given young's modulus is Y = 1.6 × 1011 N/m2. Find extension in string. 

Solution:


after substituting value of μ,v,l,A and Y we get
Δl = 0.3 mm

QUESTION: 16

Which of the following gate is reversible 

Solution:

A logic gate is reversible if we can recover input data from the output eg. NOT gate

QUESTION: 17

A thin uniform rod is of mass M and length L. Find the radius of gyration for rotation about an axis passing through a point at a distance of L/4 from centre and perpendicular to rod.

Solution:


QUESTION: 18

A satellite of mass 'M' is projected radially from surface of earth with speed 'u'. When it reaches a height equal to radius of earth, it ejects a rocket of mass M/10 and itself starts orbiting the earth in circular path 
of radius 2R, find the kinetic energy of rocket.

Solution:




Kinetic energy 

QUESTION: 19

The current 'i' in the given circuit is

Solution:


QUESTION: 20

A current carrying circular loop is placed in an infinite plane if φ1 is the magnetic flux through the inner region and φ0 is magnitude of magnetic flux through the outer region, then  

Solution:

As magnetic field lines always form a closed loop, hence every magnetic field line creating magnetic flux in the inner region must be passing through the outer region. Since flux in two regions are in opposite direction,
∴ φi = - φ0

*Answer can only contain numeric values
QUESTION: 21

Consider a loop ABCDEFA with coordinates A (0, 0, 0), B(5, 0, 0), C(5, 5, 0), D(0, 5, 0) E(0, 5, 5) and F(0, 0, 5). Find magnetic flux through loop due to magnetic field 


Solution:



φ = (3 × 25) + (4 × 25)  = 175 weber

*Answer can only contain numeric values
QUESTION: 22

A Carnot's engine operates between two reservoirs of temperature 900K and 300K. The engine performs 1200 J of work per cycle. The heat energy delivered by the engine to the low temperature reservoir in a cycle is:


Solution:

*Answer can only contain numeric values
QUESTION: 23

A non-isotropic solid metal cube has coefficients of linear expansion as 5 × 10–5/°C along the x-axis and 5 × 10–6/°C along y-axis and z-axis. If coefficient of volume expansion of the solid is C × 10–6/°C then the value of C is 


Solution:

*Answer can only contain numeric values
QUESTION: 24

A particle is released at point A. Find its kinetic energy at point P. (Given m = 1 kg and all surfaces are frictionless)


Solution:

As only consevative internal force acts upon the mass and earth system, thus we can say mechanical energy is conserved. Thus we get that net loss in PE = net gain in KE
Loss in PE = mg Δh = 1 x 10 x 1 = 10J
Thus gain in KE = 10J

*Answer can only contain numeric values
QUESTION: 25

On a photosensitive metal of area 1 cm2 and work function 2eV, light of intensity 6.4 × 10–5 W/cm2 and wavelength 310 nm is incident normally. If 1 out of every 103 photons are successful, then number of photoelectrons emitted in one second is 10x. Find x  


Solution:

Energy of photon. E == 4eV  >  2eV (so photoelectric effect will take place)
= 4 × 1.6 × 10–19 = 6.4 × 10–19 Joule
No. of photons falling per second

No. of photoelectron emitted per second

QUESTION: 26

Solution:


2 × 0.34 = + 1 x 0.522
Eº1 = 0.68 – 0.522
Eº1 = 0.158

QUESTION: 27

Correct order of electron gain enthalpy (kJ/mole) of F, Cl, Br, I

Solution:
QUESTION: 28

Arrange the following in order of their pKb value
  (B)  CH3–NH–CH3    (C) CH3–CH=NH  

Solution:

Option "A" represent Guanadine, the conjugate acid of which is resonance stabilised. The option 'B' is aliphatic amine, here the 'N' is sp3 whereas in option 'C' the 'N' is sp2, hence B is more basic than C.

QUESTION: 29

1-Methylethylene oxide Product 'X' will be –

Solution:

--excess HBr---> 


 

QUESTION: 30

Correct order of Intermolecular forces

Solution:
QUESTION: 31

Hex-3-ynal  (X), formed product X will be: 

Solution:


QUESTION: 32

  , formed product 'X' is used as:

Solution:



Methyl orange is used as an indicator in acid base titration.

QUESTION: 33

In which of the following Saytzeff product will not be formed as major product ?
(A)                         (B)   
(C)                  (D) 

Solution:





 

QUESTION: 34

Match the column
Column-I                                        Column-II
(A) Thiamine                                  (P) Scurvy
(B) Riboflavin                                 (Q) Beri Beri
(C) Pyridoxine                                (R) Cheilosis
(D) Ascorbic acid                            (S) Convulsions 

Solution:

QUESTION: 35

​Atomic radius of Ag is similar to

Solution:

Atomic radius of Ag is closest to Au.

QUESTION: 36

Correct IUPAC name of [Pt(NH3)2Cl(CH3NH2 )]Cl is:

Solution:

The IUPAC name of [Pt(NH3)2Cl(CH3NH2 )]Cl is diamminechlorido(methylamine)platinum(II)chloride.
The ligands present are ammine, chloro and methyl amine. The ligands are named according to the alphabetical order.
The prefix di indicates two.
The oxidation state of platinum is +2. The oxidation state is written in roman numerals inside parenthesis.

QUESTION: 37

Vapour pressure of pure CS2 and CH3COCH3 are 512 mm of Hg and 312 mm of Hg respectively. Total vapour pressure of mixture is 600 mm of Hg then find incorrect statement:

Solution:

Above mixture of liquids show positive deviation from Raoult's Law

QUESTION: 38

​Purest form of commercial iron is:

Solution:

Purest form is wrought iron.

 

QUESTION: 39


Mixture of above three organic compound was subjected to aq NaHCO3 and followed by dil NaOH. compounds which will be soluble in given solvent will be : 

Solution:


QUESTION: 40

Which theory can explain bonding of Ni(CO)4:

Solution:

QUESTION: 41

​n = 5, ms = + 1/2 How many orbitals are possible:

Solution:
QUESTION: 42

In zeolites & synthetic resin method which will be more efficient in removing permanent hardness of water :

Solution:
QUESTION: 43

Oxidation state of potassium in K2O, K2O2 & KO2 are respectively –

Solution:

QUESTION: 44

Decreasing order of dipole moment in CHCl3, CCl4 & CH4 is –

Solution:

QUESTION: 45

Amongst the following which is not a postulate of Dalton's atomic theory

Solution:

1) Elements are composed of extremely small particles called atoms that are indivisible and indestructible
2)All atoms of a given element are identical; they have the same size, mass, and chemical properties
3) Atoms of 1 element are different from the atoms of all other elements
4)Compounds are composed of atoms of more than 1 element. The relative number of atoms of each element in a given compound is always the same.
5)Chemical reactions only involve the rearrangement of atoms. Atoms are not created or destroyed during chemical reactions.

*Answer can only contain numeric values
QUESTION: 46

Half life of 90Sr is 6.93 years. In a child body 1 μg of 90Sr dopped in place of calcium, how many years will it take to reduce its concentration by 90% (Assume no involvement of Sr in metabolism).


Solution:

*Answer can only contain numeric values
QUESTION: 47

Each of solution A and B of 100 L containing 4 g NaOH and 9.8 g H2SO4. Find pH of solution which is obtain by mixing 40 L solution of A and 10 L solution of B.


Solution:


*Answer can only contain numeric values
QUESTION: 48

A(l) → 2B(g)
ΔU = 2.1 kcal, ΔS = 20 cal/k, T = 300 K.
Find ΔG (in kcal)

ΔU = 2.1 kcal, ΔS = 20 cal/k, T = 300 K.


Solution:

ΔH = ΔU + ΔngRT
= 2.1 × 103 + 2(2) (300)
= 2100 + 1200
= 3300 cal
ΔG = ΔH - TΔS
= 3300 – (300) (20)
= 3300 – 6000
= –2700 cals = –2.7 kcal

*Answer can only contain numeric values
QUESTION: 49

Cl2 on reaction with hot & conc. NaOH gives two chlorine having products X and Y. On treatment with AgNO3, X gives precipitate. Determine average bond order of Cl and O bond in 'Y' ?


Solution:

*Answer can only contain numeric values
QUESTION: 50

Number of chiral centers in chloramphenicol is : 


Solution:

QUESTION: 51

If f(x) is continuous and differentiable in x ∈ [ -7, 0] and f'(x) ≤ 2∈ [-7, 0], also f(-7) = -3 then range of f(-1) + f(0) 

Solution:

Lets use LMVT for x ∈ [-7, -1]

Also use LMVT for x ∈ [-7, 0]

∴ f(0) + f(-1) ≤ 20

QUESTION: 52

If y = mx + 4 is common tangent to parabolas y2 = 4x and x2 = 2by. Then value of b is 

Solution:

y = mx + 4  ……(i)
y2 = 4x tangent
 y = mx + a/m
⇒ y = mx + 1/m ………(ii)
from (i) and (ii)
4 = 1/m ⇒ m =1/4
So line y = + 4 is also tangent to parabola x2 = 2by, so solve

⇒ 2x2 – bx – 16b = 0
⇒ D = 0
⇒ b2 - 4 × 2 × (-16b) = 0
⇒ b2 + 32 × 4b = 0 
b = –128, b = 0 (not possible)

QUESTION: 53

If α and β are the roots of equation (k+1) tan2x - √2λ tanx = 1 - k and tan2 (α+β) = 50. Find value of λ

Solution:



λ = 10

QUESTION: 54

Find image of point (2, 1, 6) in the plane containing points (2, 1, 0), (6, 3, 3) and (5, 2, 2) 

Solution:

Plane is x + y – 2z = 3

⇒ (x, y, z) = (6, 5, -2)

QUESTION: 55

Let (x)k + (y)k = (a)where a, k > 0 and then find k

Solution:


k – 1 =
 k == 2/3

QUESTION: 56

If  g(x) = x2 + x – 1 and g(f(x)) = 4x2 – 10x + 5, then find f.

Solution:


QUESTION: 57

If z = x + iy and  real part  = 1 then locus of z is  

Solution:

QUESTION: 58

Let y = f(x) is a solution of differential equation and f(0) = 0 then f(1) is equal to :  

Solution:

ey = t


Put x = 0, y = 0 then c = 1
ey–x = x + 1
y = x + ln(x + 1)
at x = 1 , y = 1 + ln(2) 

QUESTION: 59

​If α is a roots of equation x2 + x + 1 = 0 and A = then A31 equal to : 

Solution:


=> A4 = I 
=> A30 = A28 × A3 = A3 

*Answer can only contain numeric values
QUESTION: 60

The six digit numbers that can be formed using digits 1, 3, 5, 7, 9 such that each digit is used at least once.


Solution:

1, 3, 5, 7, 9
For digit to repeat we have 5C1 choice And six digits can be arrange in ways.
Hence total such numbers = 

QUESTION: 61

The area that is enclosed in the circle x2 + y2 = 2 which is not common area enclosed by y = x & y2 = x is

Solution:

Total area – enclosed area 



QUESTION: 62

If sum of all the coefficient of even powers in (1– x + x2 – x3 ……..x2n) (1 + x + x2 +x3 ……….+ x2n) is 61 then n is equal to

Solution:

Let (1– x + x2 …..) (1 + x + x2 ……) = a0 + a1 x + a2x2 +……
put  x = 1
1(2n+1) = a0 + a1 + a2 +...,.a2n ...(i)
put  x = –1
(2n+1) x 1 = a0 - a1 + a2 +....a2n.....(ii)
Form (i) + (ii)
4n + 2= 2(a0 + a2 +….)
= 2 x 16
⇒  2n+1 = 61
⇒ n = 30

*Answer can only contain numeric values
QUESTION: 63

If variance of first n natural numbers is 10 and variance of first m even natural numbers is 16 then the value of m + n is


Solution:


⇒ n2 - 1 = 120
⇒ n = 11
Var (2, 4, 6,.....,2m) = 16
⇒ var (1, 2,....,m) = 4
⇒ m2 - 1 = 48
⇒ m = 7
⇒ m + n = 18 

*Answer can only contain numeric values
QUESTION: 64

Evaluate 


Solution:

Put 3x/2 = t



 

QUESTION: 65

​If A(1, 1), B(6, 5), C are vertices of ΔABC. A point P is such that area of ΔPAB, ΔPAC, ΔPBC are equal, also , then length of PQ is 

Solution:

P will be centroid of ΔABC
 

QUESTION: 66

​(p → q) ∧ (q → ~p) is equivalent to

Solution:


Clearly (p → q) ∧ (q → ~p) is equivalent to ~p

QUESTION: 67

​Find greatest value of k for which 49k + 1 is factor of 1 + 49 + 492 …..(49)125

Solution:


 

*Answer can only contain numeric values
QUESTION: 68

If f(x) = |2 – |x – 3|| is non differentiable in x ∈ S.Then value of  is   


Solution:

∵ f(x) is non differentiable at x = 1,3,5
∑ f(f(x)) = f(f(1) + f(f(3)) + f (f(5))
= 1 + 1 + 1
= 3 

QUESTION: 69

If  system of equations
2x + 2ay + az = 0
2x + 3by + bz = 0
2x + 4cy + cz = 0
have non-trivial solution
then

Solution:

For non-trivial solution

(3bc – 4bc) – (2ac – 4ac) + (2ab – 3ab) = 0
–bc + 2ac – ab = 0
ab + bc = 2ac 
a, b, c in H.P.
 in A.P

QUESTION: 70

If sum of 5 consecutive terms of 'an A.P is 25 & product of these terms is 2520. If one of the terms is – 1/2 then the value of greatest term is

Solution:

Let terms be a – 2d, a –d, a, a + d , a + 2d .
sum = 25 

⇒  5a = 25
⇒ a = 5
Product = 2520
⇒ (5–2d) (5 – d) 5(5+d)  (5+2d) =2520
⇒ (25 - 4d2) (25 - d2) = 504
⇒ 625 -100d- 25d2 + 4d4 = 504
⇒ 4d4 - 125d2 + 625 -504 = 0
⇒ 4d4 - 125d2 + 121 = 0
⇒ 4d4 - 121d2 - 4d2 +121 = 0
⇒ (d2-1) (4d2 - 121) = 0
d = ±1, d = ± 11/2
d = ±1,  does not give -1/2 as a term
∴ d = 11/2
∴ Largest term = 5 + 2d = 5 + 11 = 16    

QUESTION: 71

Let
lies in plane of

of a bisectors angle between& then  

Solution:

angle bisector can be or 



Compare with 


Not in option so now consider



Compare with 

QUESTION: 72

Given f(a + b + 1 – x) = f(x) R then the value of is equal to 

Solution:


x → a + b - x
[∵ put x → x + 1 in given equation]
(1) + (2)
2I = 


QUESTION: 73

If distance between the foci of an ellipse is 6 and distance between its directrices is 12, then length of its latus rectum is 

Solution:

2ae = 6 and 
⇒ ae = 3 and a/e = 6
⇒ a2 = 18
⇒ b2 = a2 - a2e2 = 18 - 9 = 9

QUESTION: 74

An unbiased coin is thrown 5 times. Let X be a random variable and k be the value assigned to X for  k = 3, 4, 5 times Head occurs consecutively and otherwise the value of X is assigned –1. What is value of expectation.

Solution:


k = no. of times head occur consecutively
Now expectation

QUESTION: 75

If  when then find dy/dα at α = 5π/6

Solution:



= –1 – cota 
 
⇒  will be = 4